Linear Voltage Regulators, Batteries, and PWM motor control
P. Douglas Reeder
I'm running my robot electronics from six alkaline cells, for a
voltage of 9 volts (fresh). A linear voltage regulator supplies 5V
power for the electronics, while the motors are run directly from
batteries, using an H-bridge (SK10001 Push-Pull Four Channel Driver)
and PWM. There is a 1000uF capacitor across the batteries, and
another from 5V to ground.
For testing, can I supply the 5V from a power supply? What happens if
I supply nothing to the motor power supply? What happens if I supply 5V
to on both sides of the voltage regulator?
Also, can I hook two identical LED's in parallel (and thus in series
with a single current limiting resistor), or must they have separate
current limiting resistors?
Doug Reeder Internet: ree...@reed.edu
Div, Grad & Curl USENET: ...!tektronix!reed!reeder
programming & derivative work
I am actively seeking scientific programming contracts.
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Div, Grad & Curl USENET: ...!tektronix!reed!reeder
programming & derivative work
I am actively seeking scientific programming contracts.
Tom Kite
>ree...@reed.edu (P. Douglas Reeder) writes:
>>Also, can I hook two identical LED's in parallel (and thus in series
>>with a single current limiting resistor), or must they have separate
>>current limiting resistors?
>
>You may want to adjust the normal resistor value to take into account the
>half-volt drop across the other LED. This will guarantee that the exact
>same *current* flows through both LEDs... which means each will shine just
>as bright as the other.
>
> Vs Each LED drops about 0.6 Volts,
> ^ so if the LED's rated current
> | Rc LED LED is "Id" then "Rc" comes out to:
> +---/\/\/\/------>|------>|---+
> | Vs - 0.6 - 0.6
> _ Rc = ----------------
> = Gnd Id
Red LEDs drop about 2V, not 0.6V! Regular diodes drop about 0.6V. We can
understand this by using the energy relation:
Energy of light at frequency f = hf (h=Planck's constant)
Energy dropped by electrons across V = qV (q=electron charge)
Since all the energy appears as light we have:
hf = qV
Hence
V = hf/q
Or, using f=c/(wavelength):
V = hc/q(wavelength)
For red LEDs, the wavelength is around 650nm; h is 6.6e-34, c is 3e8, q is
1.6e-19. This gives
V = 1.9 V
Similarly, for green (550nm) we have V=2.3V and for blue (470nm) we have
V=2.7V. Note that the forward voltage may be even larger than this at
high currents because of Ohmic resistance.
Hope this helps out.
Thomas Kite
Teaching Assistant, EE Senior Design Lab
UT Austin
Ted Dunning
In article <26n677...@uwm.edu> ri...@ee.uwm.edu (Rick Miller) writes:
...
Putting 5 Volts on both sides of the linear regulator shouldn't cause any
problems though.
most regulators require some amount of head room. if you put in 5V,
you can't get 5V out. some have less than a volt of headroom
required, but most require nearly 2V.
>Also, can I hook two identical LED's in parallel (and thus in series
>with a single current limiting resistor), or must they have separate
>current limiting resistors?
They wouldn't work very well, because they wouldn't be *electrically*
identical. One would be just a little more conductive than the other,
and would shine brighter. The other would be dimmer, and may not be
bright enough to be seen.
this is correct.
You may want to adjust the normal resistor value to take into account the
half-volt drop across the other LED.
the voltage drop across an led is more like 2V! this varies with
color and construction.
Vs Each LED drops about 0.6 Volts,
^ so if the LED's rated current
| Rc LED LED is "Id" then "Rc" comes out to:
+---/\/\/\/------>|------>|---+
| Vs - 0.6 - 0.6
_ Rc = ----------------
= Gnd Id
Tre simple, chu ne? Check your LEDs' packaging for more accurate V-drop.
Just add more " - 0.6 " terms to the numerator for more LEDs in series.
again... replace 0.6 with 2 here.
john r strohm
>You may want to adjust the normal resistor value to take into account the
>half-volt drop across the other LED.
> Vs Each LED drops about 0.6 Volts,
> ^ so if the LED's rated current
> | Rc LED LED is "Id" then "Rc" comes out to:
> +---/\/\/\/------>|------>|---+
> | Vs - 0.6 - 0.6
> _ Rc = ----------------
> = Gnd Id
>
>Tre simple, chu ne? Check your LEDs' packaging for more accurate V-drop.
>Just add more " - 0.6 " terms to the numerator for more LEDs in series.
One PICKY detail. LEDs typically drop between 1.7 and 2.4 VDC, depending
on the LED. The generic red LEDs usually drop 1.7 V. The generic lab for
this is to take a known current-limiting resistor and a handful of LEDs,
and MEASURE the voltage drop with a halfway-decent DVM (Radio Shack works
fine for this). (In fact, Radio Shack used to sell grab bags of various
color LEDs, so you could pick up all the supplies on one mall trip.)
The voltage drop is a function of the LED material, not the brightness.
(Incidentally, I once used this series LED trick to make some flashing
tuxedo studs, using a flasher LED and two conventional LEDs in series.)
Rick Miller
>voltage of 9 volts (fresh). A linear voltage regulator supplies 5V
>power for the electronics, while the motors are run directly from
>batteries, using an H-bridge (SK10001 Push-Pull Four Channel Driver)
>and PWM. There is a 1000uF capacitor across the batteries, and
>another from 5V to ground.
9V all the way down to 5V?!? You're wasting your batteries, Doug.
That linear regulator will be almost *doubling* your control circuit's
energy consumption due to the large voltage drop ...for nothing but heat.
>For testing, can I supply the 5V from a power supply? What happens if
>I supply nothing to the motor power supply? What happens if I supply 5V
>to on both sides of the voltage regulator?
Hmmm... Depends on how you've got stuff wired.
How big are those power-supply capacitors?
What kind of current do you expect your motors to draw (stall current)?
I'd advise you to use *different* *batteries* for motors and for controls.
High-current PWM can do nasty things when it's drawing from your controls'
supply. Then you can use only, oh, four AA-size batteries (or some sort
of small, 6 Volt cell) for your control circuitry.
Putting 5 Volts on both sides of the linear regulator shouldn't cause any
problems though.
>Also, can I hook two identical LED's in parallel (and thus in series
>with a single current limiting resistor), or must they have separate
>current limiting resistors?
They wouldn't work very well, because they wouldn't be *electrically*
identical. One would be just a little more conductive than the other,
and would shine brighter. The other would be dimmer, and may not be
bright enough to be seen.
Better yet, put them in *SERIES* with each other and with a single resistor.
You may want to adjust the normal resistor value to take into account the
half-volt drop across the other LED. This will guarantee that the exact
same *current* flows through both LEDs... which means each will shine just
as bright as the other.
Vs Each LED drops about 0.6 Volts,
^ so if the LED's rated current
| Rc LED LED is "Id" then "Rc" comes out to:
+---/\/\/\/------>|------>|---+
| Vs - 0.6 - 0.6
_ Rc = ----------------
= Gnd Id
Tre simple, chu ne? Check your LEDs' packaging for more accurate V-drop.
Just add more " - 0.6 " terms to the numerator for more LEDs in series.
Using a somewhat higher resistance may decrease the brightness a bit, but
it may also save you some power. So, do you want your robot to roam for a
good long time, or do you just want der blinkenlights?
RICK MILLER <ri...@ee.uwm.edu> Voice: +1 414 221-3403
P.O. BOX 1759 FAX: +1 414 221-4744
MILWAUKEE, WI Send a postcard and I'll send one back.
53150-1759 USA Sendu bildkarton kaj mi retrosendos unu.
Nathan Stratton
wast so much power. If you wanted a crude supply you could use a LM741
comparator and use zener diode to give you a ref voltage.
I have just built a H bridge, but used NPN and PNP instead of all NPN. Well
howdo I absorb the spike from the motor. I would like to use diodes, but
how do you connect them?
Nathan Stratton
ns...@quercus.gsfc.nasa.gov
Jim Siwek
>>
>>I'm running my robot electronics from six alkaline cells, for a
>>voltage of 9 volts (fresh). A linear voltage regulator supplies 5V
>>power for the electronics, while the motors are run directly from
>>batteries, using an H-bridge (SK10001 Push-Pull Four Channel Driver)
>>and PWM. There is a 1000uF capacitor across the batteries, and
>>another from 5V to ground.
>
>9V all the way down to 5V?!? You're wasting your batteries, Doug.
>That linear regulator will be almost *doubling* your control circuit's
>energy consumption due to the large voltage drop ...for nothing but heat.
Linear voltage regulators are integrated circuits that take an input voltage
and transform it to a regulated output voltage through a series of transistor
circuits. The voltage drop of 4 volts is not lost as heat, and the only
power that is lost is the power used to supply the internal circuitry in order
to self bias the circuits for a steady 5V. In fact, maximum input ratings
for a typical IC regulator are 28V, so 9V is a typical voltage to apply
to a 5V regulator IC.
>I'd advise you to use *different* *batteries* for motors and for controls.
>High-current PWM can do nasty things when it's drawing from your controls'
>supply. Then you can use only, oh, four AA-size batteries (or some sort
>of small, 6 Volt cell) for your control circuitry.
The 100uF capacitors across the batteries would eliminate the problems
of the PWM drawing current. This is a battery application, so I'm sure
the current is not all *that* high, anyways...
>Putting 5 Volts on both sides of the linear regulator shouldn't cause any
>problems though.
What if the minimum input voltage of the regulator IC is 7.5 V? The output
may clamp to 0V, shorting out your power supply.
Jim
MILLS,JOHN M.
In article <1993Sep9.1...@lmpsbbs.comm.mot.com> you write:
>In article <26n677...@uwm.edu> ri...@ee.uwm.edu (Rick Miller) writes:
>>ree...@reed.edu (P. Douglas Reeder) writes:
>>>
>>>I'm running my robot electronics from six alkaline cells, for a
>>>voltage of 9 volts (fresh). A linear voltage regulator supplies 5V
>>9V all the way down to 5V?!? You're wasting your batteries, Doug.
>>That linear regulator will be almost *doubling* your control circuit's
>>energy consumption due to the large voltage drop ...for nothing but heat.
That was right.
>
>Linear voltage regulators are integrated circuits that take an input voltage
>and transform it to a regulated output voltage through a series of transistor
>circuits. The voltage drop of 4 volts is not lost as heat, and the only
>power that is lost is the power used to supply the internal circuitry in order
>to self bias the circuits for a steady 5V. In fact, maximum input ratings
>for a typical IC regulator are 28V, so 9V is a typical voltage to apply
>to a 5V regulator IC.
BZZZT -- You can easily figure the power loss from series regulation: it's
at least I * (Delta E), where I is the current to the regulated load and
Delta E is the difference between supply and regulated output voltages. The
additional loss will be bias and inefficiency, plus any stray current paths.
You may elect to dissipate this power (as heat) in another component, passive
or active, but you _are_ giving away the power. You're right about the
usual voltage drop. There are two reasons:
(1) forward voltage across a few junctions is usually a volt or two, and
(2) if you're running on filtered, but unregulated, rectified AC, you may
have to give up some headroom to limit ripple in your regulated output.
That translates into more rms power loss, and therefore _heat_ (but
that isn't a problem where the primary source is batteries).
Next time you look at a linear regulator data sheet, check the power
dissipation rating, usually given with and without heat sinking. If you
have one burning a few watts, _carefully_ touch it to verify it's dissipating
power.
Switchers can have efficiencies of 80%, perhaps higher, (fairly) independent
of regulation drop and current drain. SCR power amps can approach +/- 1.0,
which means you pump power back into the net when (for example) you actively
brake a DC motor. That can come as a surprize to the rest of your system!
You will probably draw some flames for this, but ignoring Kirschoff's and
Ohm's laws in the same post probably deserves some flames. ;->)
Regards --jmm--
--
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uucp: ...!{decvax,hplabs,ncar,purdue,rutgers}!gatech!prism!jm59
Internet: john.m...@gtri.gatech.edu
Say "Goodnight," Gracie.
Somesh Rao
>howdo I absorb the spike from the motor. I would like to use diodes, but
>how do you connect them?
Why don't you use a Power FET in a H bridge? The one's that I used about 5 years back
could be driven by a CMOS gate (well, a couple of inverters really).
I just cannot remember the name of the manufacturer (I have not done much
hardware in the last 5 years), but if you wish, I could call up my
old company and check with a couple of colleagues.
--
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Somesh Rao som...@syl.nj.nec.com
NEC Systems Lab, 609-734-6115 (Voice)
4 Independence Way, 609-734-6002 (FAX)
Princeton, NJ 08540
#include <std-disclaimer.h>
I speak for myself. I do not speak for NEC and NEC does not speak for me.
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Nathan Stratton
sitting around. The problem I have is not with the H, but with the spike
caused when the motor stops and the electromagnetic field collapses. How
do I solve this problem??????
Nathan Stratton
ns...@quercus.gsfc.nasa.gov
KEN E WILLMOTT
the power drivers. The polarity is easy to remember - it's just the
connection that has the diode reverse biased. If it wasn't, the
power supplies would short out. That's just a mnemonic attitude.
(N channel enhancement MOSFET shown)
V+
|
*--
| |
| |- |
| _ <---- protection diode
_____| |- ^
| |
*--
|
|------------> load
|
*--
| |
| |- |
| _ <---- protection diode
_____| |- ^
| |
*--
|
|
V-
Now, the neat part is that a lot of MOSFETs have these diodes
built in, so you needn't bother adding them externally.
-Ken Willmott
Tom Kite
Jim Siwek <jim...@ecs.comm.mot.com> wrote:
>
>Linear voltage regulators are integrated circuits that take an input voltage
>and transform it to a regulated output voltage through a series of transistor
>circuits. The voltage drop of 4 volts is not lost as heat, and the only
>power that is lost is the power used to supply the internal circuitry in order
>to self bias the circuits for a steady 5V. In fact, maximum input ratings
>for a typical IC regulator are 28V, so 9V is a typical voltage to apply
>to a 5V regulator IC.
This is plain wrong. Of course the 4V (multiplied by the load current) is
lost as heat! The higher the input voltage to the regulator, the bigger
the drop across the regulator, and the more power is wasted. By
regulating 9V down to 5V, 4/9 of the power is wasted in the regulator.
Perhaps you could consider using a low-dropout regulator running off 7.5V
or maybe even 6V.
>
>>I'd advise you to use *different* *batteries* for motors and for controls.
>>High-current PWM can do nasty things when it's drawing from your controls'
>>supply. Then you can use only, oh, four AA-size batteries (or some sort
>>of small, 6 Volt cell) for your control circuitry.
>
>The 100uF capacitors across the batteries would eliminate the problems
>of the PWM drawing current. This is a battery application, so I'm sure
>the current is not all *that* high, anyways...
The 100uF caps may reduce the problems, but if there is any reasonable
current in the PWM side they will not eliminate them. However, if the
regulator is reasonable there shouldn't be any trouble.
>>Putting 5 Volts on both sides of the linear regulator shouldn't cause any
>>problems though.
>
>What if the minimum input voltage of the regulator IC is 7.5 V? The output
>may clamp to 0V, shorting out your power supply.
Also wrong. If the input to the regulator drops below the minimum
(usually 2.5 or 3V above the output voltage) the output will not drop to
0V, but will just dip below the required output voltage, 5V in this case.
That is, ripple will appear at the output.
Nathan Stratton
using a 100 uf cap it will not work vary well. The only way to solve the
problem is to use a separate supply. I also believe he is correct in
saying that the rest of the input voltage is dissipated as heat.
Nathan Stratton
ns...@quercus.gsfc.nasa.gov
P. Douglas Reeder
The linear voltage regulator in my circuit does waste some power, but
most of the power goes to the motors without regulation, so the total
inefficiency is not too bad. The other important point is that six
alkaline cells do give 9V when fresh, but only 6V at exhaustion. The
regulator insures that the circuits all work properly whatever the
battery state.
P. Douglas Reeder
:In article <26n677...@uwm.edu>, Rick Miller <ri...@ee.uwm.edu> wrote:
:>
:> Each LED drops about 0.6 Volts,
:Red LEDs drop about 2V, not 0.6V!
:
: V = 1.9 V
:
:Similarly, for green (550nm) we have V=2.3V and for blue (470nm) we have
:V=2.7V. Note that the forward voltage may be even larger than this at
:high currents because of Ohmic resistance.
Some actual measurments of a random assortment of LED's I got cheap:
IR 1.25V
red 2V
orange 1.9V
yellow 2V
green 2V
Thanks for the suggestions, people! The original LED's are now in
series, and thus drawing a lot less current from the batteries.
I also put a power on indicator LED in series with them, costing no
additional current.
MILLS,JOHN M.
>The last H bridge I built I used MOSFET's, I had the transistor I used
>sitting around. The problem I have is not with the H, but with the spike
>caused when the motor stops and the electromagnetic field collapses. How
>do I solve this problem??????
I know this is a real problem, and don't know a pat solution. Could you put
a Zener diode across the gate to protect it? This would only work if the
that gate was always biased in the same direction. A bit of parallel resistance
would also limit the dynamic spike, at the cost of power efficiency. A pair of
back-to-back Zeners across the motor coils would limit the spike, but might
limit your forward drive voltage as well. They could go across the gate, too.
Any of these components needs adequate heat-sinking if you're absorbing a lot
of energy. Post any good solution, please.
The problem is avoided in on-line SCR designs, but there the line commutation
frequency limits your performance, and in some applications that's unacceptable.
"Generic" PWM amps are easier to come by than SCR amps, I think, and they
run on DC. You may also need line isolation, which burdens the SCR design with
a heavy (60 Hz) transformer, where you can drive the PWM with an isolated
switching power supply using light (> 50KHz) transformers.
Gary Richard Cane
>>I have just built a H bridge, but used NPN and PNP instead of all NPN. Well
>>howdo I absorb the spike from the motor. I would like to use diodes, but
>>how do you connect them?
>
>Why don't you use a Power FET in a H bridge? The one's that I used about 5 years back
>could be driven by a CMOS gate (well, a couple of inverters really).
>I just cannot remember the name of the manufacturer (I have not done much
>hardware in the last 5 years), but if you wish, I could call up my
>old company and check with a couple of colleagues.
>
Harris Semiconductor makes power logic level FETs. These devices can
attach directly to your TTL/CMOS logic without any special interface.
They are very easy to use and have protection diodes inherently
between the drain and source. The RFP50N05L FET can handle 50amps
continuous and 130amps pulsed. This makes these devices very useful
for motor control.
Gary.
Dave Platt
Actually, it's a function of both... as a previous poster noted, there's
a significant ohmic effect in the LEDs, and their measured forward
voltage _does_ vary at different current levels.
From the '89 edition of the HP optoelectronics book, I note that the HP
high-efficiency red LED (HLMP-L250) has a forward voltage as low as 1.6
volts when it's passing less than a milliamp of current, rises to 2.0
volts somewhere in the neighborhood of 16 mA, and gets up as high as 3.0
volts at currents of 60 mA or so (which is within the peak forward
current range but beyond the average forward current range, and so can
be reached only during low-duty-cycle pulses).
The HLMP-L350 (yellow) and HLMP-L550 (green) LEDs have similar
forward-voltage-vs.current curves. If you parallel a green and a red
LED in this family, and use a single current-limiting resistor, the two
LEDs will not draw equal currents... at a forward voltage of 2 volts,
the red will be drawing 16 mA and the green will draw only about 4 mA.
If, however, you send pulses of about 120 mA through the pair, they'll
both draw about 60 mA each... their forward-voltage-vs.-current curves
intersect at about that point.
In practice, I've found that you _can_ parallel two LEDs of the same
family with a single series current-limiting resistor, and get fairly
close current-matching. Brightness variations due to differing forward
currents seem to be less than variations inherent in the devices.
--
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Domain: dpl...@ntg.com UUCP: ...netcomsv!ntg!dplatt
USNAIL: New Technologies Group Inc. 2470 Embarcardero Way, Palo Alto CA 94303
John Nagle
>>ree...@reed.edu (P. Douglas Reeder) writes:
>>>
>>>I'm running my robot electronics from six alkaline cells, for a
>>>voltage of 9 volts (fresh). A linear voltage regulator supplies 5V
>>>power for the electronics, while the motors are run directly from
>>>batteries, using an H-bridge (SK10001 Push-Pull Four Channel Driver)
>>>and PWM. There is a 1000uF capacitor across the batteries, and
>>>another from 5V to ground.
>>
>>9V all the way down to 5V?!? You're wasting your batteries, Doug.
>>That linear regulator will be almost *doubling* your control circuit's
>>energy consumption due to the large voltage drop ...for nothing but heat.
There must be some cheap DC-DC converters for automotive applications
that accept low-quality power in the 6-12V range and put out clean +5.
This is becoming a common application. I've seen such units advertised,
but in the $50-100 range. But I suspect there are cheaper units available,
what with all the stuff going into cars. Any part numbers?
John Nagle
Bruce Rowen
>The last H bridge I built I used MOSFET's, I had the transistor I used
>sitting around. The problem I have is not with the H, but with the spike
>caused when the motor stops and the electromagnetic field collapses. How
>do I solve this problem??????
Power mosfets are easy to damage with overvoltage. Zener diodes are
the best snubbers to use since they have the fast reverse recovery
times mandated by the fast switching time of the mosfet. Most common
diodes will not start to conduct until the spike has already exceeded
the devices rating, thereby damaging the device.
What sucks about snubbers is that you are just wasteing energy. I have
seen many motor driver subroutines that the author was so proud of the
way they could control a set of (stepper) motors, but were confused as
to why the motors performed so badly. They gave no consideration to
the dynamics of the power flow through the devices. Their routine
would energize all motors at the same time (major shock on the power
supply), and update all motor phase states on the same time slice.
This is fine from a code point of view, but horrible from an
electronic point of view. One can make use of the energy spike from a
collapsing field in one motor (phase) to help turn on the next motor
phase. Remember that it takes CURRENT to produce torque. The fastest
way to maximize the dI/dt in a motor coil with fixed resistance and
inductance is to dynamically control the voltage. This is why you want
a motor power supply to be current regulated not voltage regulated.
When the motor phase first switches on, you want the voltage to be as
high as practical to get a fast dI/dt. As the current reaches the
desired level, back off on the voltage. With a PWM controlled motor
running 0-100% duty cycles, you are limiting the power supply voltage
to provide max rated current into the motor when full on. A better
approach is to run the supply voltage at say 10x the motor rating and
have a PWM duty cycle of (approx) 0-10%.
A little understanding of the energy dynamics in your motor
system and some judicious software timing can turn a vibrating mass of
stepper or PWM motors into a smooth running system. The energy spike
from a collapsing field can be directed into the next motor or phase
to give it a boost. You may even be able to eliminate snubbers
alltogether.
$0.02
--
-------------------------------------------------------------------------------
Bruce Rowen National Radio Astronomy Observatory
Scientific Programmer Array Operations Center
bro...@aoc.nrao.edu P.O. Box O Socorro, NM 87801
(505)385-7329 (505)385-7000
-------------------------------------------------------------------------------
Steve Harmon
There are several single chip solutions to this problem that are cheap
(actually, a single chip + a small inductor and diode). An example is the
MAX730 +5V Step-Down Current-Mode PWM regulator. The chip is a effectively
a step-down switching power supply which only requires a hand full of extra
components (a 100uH inductor, a 1N5817 diode, a resistor and a few caps).
The MAX730 will provide a 5 volt output over a 5.2 to 11.0 volt input.
Under the right conditions (i.e. with the right component selection) this
type of supply can provide up to 1Amp of 5V power.
Linear Technology is another good source for step-down switching regulator
chips. An example is the LTC1174-5 which will provide 5 volts @ 400ma from
a 9 volt battery @ 90% efficiency. This particular device will even provide
a low battery output that could be of some use.
--
| Steve Harmon @ AT&T Tridom Marietta, Georgia.
| INET: s...@eng.tridom.com
| UUCP: ..gatech!emory!tridom!srh
| VOICE: (404) 426-4261
|
Chuck McManis
>I'm running my robot electronics from six alkaline cells, for a
>voltage of 9 volts (fresh). A linear voltage regulator supplies 5V
>power for the electronics, while the motors are run directly from
#1) Linear regulators are real energy hogs when you drop this much voltage
across them. Contact either National or Maxim for their switching
regulators. Both are 8 pin minidips that with a couple of additional
components will create a 5V switching regulator. .5A is pretty cheap.
(ask Nat'l if they will send you their "Switching Regulator Design Kit"
which has a .5A 5V regulator sample on a PC board ready to use.)
> using an H-bridge (SK10001 Push-Pull Four Channel Driver)
>and PWM. There is a 1000uF capacitor across the batteries, and
^^^^
>another from 5V to ground.
You need to reconsider the 1000uF capacitor. The motors act like inductors
(or coils) with a moving core. Using a capacitor in parallel with them
basically builds a LC bridge that given the proper signal (ie a PWM signal
at the resonant frequency of the resulting tank circuit) will cause a
current of several tens of amps to build up bouncing between the capacitor
and the motor. Smoke-a-roo as they say. Basically the voltage drop across
the motor windings is L*di/dt (first derivative of the change in current
times the inductance) clearly when you turn the PWM voltage full on you
get a current rush and to satisfy this relationship the voltage across
the motor windings goes way up (on a square wave, di/dt on the leading
and trailing edges is "infinite"). Anyway on capacitors the relationship
is the opposite, the current is C * dv/dt. So when you put a capacitor
across the terminals what you are trying to accomplish is that the
capacitor will take up some of the current to mitigate the voltage spike
produced when you slammed the motor on. You don't want it so big that
when the motor inductor is saturated (and the voltage across it effectively
becomes zero) and you remove the current that the capacitor supplies
the current for a nice big spike. To cut a non-interesting story short
(math is just math after all) use something like a .002 uF capacitor
(non-polarized because this thing will produce posititve and negative
voltages (the trailing edge of a square wave produced a negative di/dt)
>For testing, can I supply the 5V from a power supply?
Yes, as long as there is a common ground reference.
> What happens if I supply nothing to the motor power supply?
Depends on the PWM circuit. If its a MOSFET and you leave the source
"open" probably not a whole lot. Don't tie it to ground or you'll
reverse bias the gate source junction and toast it. Its safer to
connect Vmotor to the PWM circuit and just not hook the motors up
the the output. Use light bulbs if you want to see what motors
would be running.
> What happens if I supply 5V to on both sides of the voltage regulator?
Then the voltage drop across the regulator becomes 0 and it just sits
there.
>Also, can I hook two identical LED's in parallel (and thus in series
>with a single current limiting resistor), or must they have separate
>current limiting resistors?
LEDs "drop" a certain voltage, typically .2 to .7 volts, lets
say yours are .5V. After that voltage drop they will suck as much
current as you can feed through them until they melt. They typically
have a typical current rating of 20 - 40mA (let's say its 30mA).
Now if you put 'n' LEDs in series, that is a voltage drop of n * .5v/LED.
Now subtract that from the source voltage (9V in your example? or
5V ?) Let's say its 9v and you're running 2 LEDs. 2 * .5V = 1V
9V - 1V = 8V. Now that means there will be an 8volt drop over the
resistor. Since ohm's law is E=IR and you know E (voltage = 8v) and you
know I (30mA) solve for R = 8/.03 = 266.66 ohms. You can probably
use a 270 ohm resistor. If you were driving them from the 5V supply
(not recommended since they would just use up the current capacity of
your linear regulator) it would be 4v (5v - 1v) / 0.03A = 133.33 ohms
pick something close and use that.
Simple no ?
--
--Chuck McManis Mr. NIS+ Sunsoft
uucp: {anywhere}!sun!cmcmanis BIX: <none> Internet: cmcm...@Eng.Sun.COM
These opinions are my own and no one elses, but you knew that didn't you.
Chuck McManis
> There must be some cheap DC-DC converters for automotive applications
>that accept low-quality power in the 6-12V range and put out clean +5.
National LM2575 et al. about $1.50 each, require one inductor ($1.00) and
one capacitor ($.50). You can get 5v @500mA for about 5$ including the
PC board if you build in sufficient scale.
--Chuck
Nathan Stratton
10 amps and +12 V at 5 amps, any suggestions?
Nathan Stratton
ns...@quercus.gsfc.nasa.gov
Ted Dunning
In article <m979p2...@appserv.Eng.Sun.COM> cmcm...@yikes.Eng.Sun.COM (Chuck McManis) writes:
> using an H-bridge (SK10001 Push-Pull Four Channel Driver)
>and PWM. There is a 1000uF capacitor across the batteries, and
^^^^
>another from 5V to ground.
You need to reconsider the 1000uF capacitor. [colloquial
description of resonance deleted]
this is correct.
Basically the voltage drop across
the motor windings is L*di/dt (first derivative of the change in current
times the inductance)
right.
clearly when you turn the PWM voltage full on you
get a current rush and to satisfy this relationship the voltage across
the motor windings goes way up (on a square wave, di/dt on the leading
and trailing edges is "infinite").
this is wrong. and contradicts the previous sentence. when you apply
a voltage step to an inductor whose initial current is zero, the
current increases from zero at a rate which is initially v/L. as the
current reaches the value limited by whatever series resistance is
present, this rate of current increase decreases.
Anyway on capacitors the relationship
is the opposite, the current is C * dv/dt.
this is also correct.
the current for a nice big spike. To cut a non-interesting story short
(math is just math after all) use something like a .002 uF capacitor
(non-polarized because this thing will produce posititve and negative
voltages (the trailing edge of a square wave produced a negative di/dt)
your math doesn't seem to be the same math the rest of us use.
anyway, the rationale for the small cap is to get rid of brush noise
and has little to do with the inductance of the motor. putting the
cap directly on the motor and using some ferrite beads right there is
also a good idea.
> What happens if I supply nothing to the motor power supply?
Depends on the PWM circuit. If its a MOSFET and you leave the source
"open" probably not a whole lot. Don't tie it to ground or you'll
reverse bias the gate source junction and toast it.
?? this is an interesting claim. on a mosfet, the gate is insulated
from both the drain and source. you may have been thinking of a JFET
(present day would *never* be used to control a motor larger than a
mosquito).
>Also, can I hook two identical LED's in parallel (and thus in series
>with a single current limiting resistor), or must they have separate
>current limiting resistors?
LEDs "drop" a certain voltage, typically .2 to .7 volts, lets
say yours are .5V. After that voltage drop they will suck as much
current as you can feed through them until they melt. They typically
have a typical current rating of 20 - 40mA (let's say its 30mA).
oh noo... *please* don't start that again.
LEDs drop nearly 2 volts (as any number of posters have said).
James Bach
> >9V all the way down to 5V?!? You're wasting your batteries, Doug.
> >That linear regulator will be almost *doubling* your control circuit's
> >energy consumption due to the large voltage drop ...for nothing but heat.
>
> Linear voltage regulators are integrated circuits that take an input voltage
> and transform it to a regulated output voltage through a series of transistor
> circuits. The voltage drop of 4 volts is not lost as heat, and the only
> power that is lost is the power used to supply the internal circuitry in order
> to self bias the circuits for a steady 5V. In fact, maximum input ratings
> for a typical IC regulator are 28V, so 9V is a typical voltage to apply
> to a 5V regulator IC.
>
Pffft, where do you think the other 4V goes to?
The way a "linear" regulator works is by changing the equivalent
resistance of the "pass transistor" between the "input" and "output"
of the regulator; the delta in voltage (9-5=4) times the current thru
the regulator (i.e. that used by the "load") gives you the power
dissipated by the regulator. If the only "power lost" in the
regulator was that caused by the "self bias" circuitry inside the
part, then why do some regulators come in crappy little plastic TO-92
packages, some in meaty TO-220, and others in monsterous TO-3 cans?
I'm sure it isn't because the bias circuitry inside the TO-3 cans is
very inefficient compared to the TO-92 parts!
--
James C. Bach Ph: (317)-451-0455 The views & opinions expressed
Advanced Project Engr. GM-NET: 8-322-0455 herein are mine alone, and are
Powertrain Strategy Grp Amateur Radio: WY9F NOT endorsed, sponsored, nor
Delco Electronics Corp. Just say NO to UNIX! encouraged by DE or GM.
James Bach
In article <26n677...@uwm.edu>, ri...@ee.uwm.edu (Rick Miller) writes:
> Better yet, put them in *SERIES* with each other and with a single resistor.
> You may want to adjust the normal resistor value to take into account the
> half-volt drop across the other LED. This will guarantee that the exact
> same *current* flows through both LEDs... which means each will shine just
> as bright as the other.
>
> Vs Each LED drops about 0.6 Volts,
> ^ so if the LED's rated current
> | Rc LED LED is "Id" then "Rc" comes out to:
> +---/\/\/\/------>|------>|---+
> | Vs - 0.6 - 0.6
> _ Rc = ----------------
> = Gnd Id
>
> Tre simple, chu ne? Check your LEDs' packaging for more accurate V-drop.
> Just add more " - 0.6 " terms to the numerator for more LEDs in series.
Most LEDs I've ever run across have a Vd=1.7V to 2.4V, not 0.6V!
Gary Richard Cane
>I am using 2 130 ah battries in my robot.
Wow. That is a big battery. I think the first thing to do
is put a fuse on that thing.
>I need to get regulated +5 V at
>10 amps and +12 V at 5 amps, any suggestions?
>
>Nathan Stratton
>ns...@quercus.gsfc.nasa.gov
I have put DC/DC converters in parallel before to increase the
total output current. The problem with this is that if the converters
don't all have exactly the same output voltage, some will sink the current
instead of all sourcing it. It may be a good idea to put a diode
on each converter so that a neighboring converter does not sink
current. Also, if a diode is used for protection, be careful
about using inductive loads -- you may get a voltage increase
like those found in step-up converters. Another idea I have
used before is to put a very low value resistor on the output
of each DC/DC converter, and then connect all the resistors together.
These resistors will drop any voltage difference between converters.
Gary.
John Nagle
>I have put DC/DC converters in parallel before to increase the
>total output current. The problem with this is that if the converters
>don't all have exactly the same output voltage, some will sink the current
>instead of all sourcing it. It may be a good idea to put a diode
>on each converter so that a neighboring converter does not sink
>current. Also, if a diode is used for protection, be careful
>about using inductive loads -- you may get a voltage increase
>like those found in step-up converters. Another idea I have
>used before is to put a very low value resistor on the output
>of each DC/DC converter, and then connect all the resistors together.
>These resistors will drop any voltage difference between converters.
There are DC/DC converters that are designed to be connected in
parallel. Lambda makes some. But it's not just a matter of hooking
the outputs together; the control circuitry has to be are interconnected
so the switchers cooperate instead of oscillating or acting as sinks for
each other.
Can you divide the load?
Putting a resistor after each output should work, but the voltage
after the resistor will no longer be regulated.
John Nagle
Nathan Stratton
problem I have is when I disconnect power to the motor, It takes the motor
a second or so to stop. How do I solve the problem without getting a new
motor? I need a geared motor because the robot weighs about 300 pounds, but
I also need acceracy. I can't predict how long the motor will take to stop
because is reacts differently to different surfaces.
Thanks for any help.
Nathan Stratton
ns...@quercus.gsfc.nasa.gov
Mato Hattori
you that it is absolutely natural phenomennon that things won't stop
suddenly. You should control revolution of the motor with a certain
feedback circuit. Don't let motors move suddenly.
If you just want to shorten the duration of the motor stopping, let
the two terminal of the motor electrically be short as soon as you cut
off the drivr voltage.
Hope this helps,
--
Mato Hattori Computer System Research Department,
email : hat...@csrd.nkk.co.jp Electronics Research Center, NKK Corporation
voice : 81+44-322-6483 (office) 1-1 Minamiwatarida Kawasaki-ku
: 81+44-246-9665 (home) Kawasaki-city, 210 JAPAN
Rick Miller
One simple way to stop a motor quickly is to short its leads together.
So, if you're using a relay/solenoid to turn your motors on and off, you
should arrange things such that the motor's leads are shorted together
when it's not being driven. That may be tricky if you're driving the
motor bi-directionally. Try putting just one of the N.C. contacts from
each relay (FWD & REV) in series with each other from one motor terminal
to the other.
RICK MILLER <ri...@ee.uwm.edu> Voice: +1 414 221-3403
P.O. BOX 1759 FAX: +1 414 221-4744
MILWAUKEE, WI Send a postcard and I'll send one back.
53150-1759 USA Sendu bildkarton kaj mi retrosendos unu.
Nathan Stratton
Ok, I would like to brake the motor by shorting both leads of the motor to
ground. Is this a good way to do the job? Also how much current do you get
when you short out a motor? I a worried that the transistors might fry if
I do this.
Thank you for your help.
Nathan Stratton
ns...@quercus.gsfc.nasa.gov
Bruce Rowen
Nathan Stratton
ns...@quercus.gsfc.nasa.gov
Did you consider (or are you using) dynamic brakeing?
Dynamic brakeing is where you dump the power from the motor when it is
acting as a generator into a load.
The load can be a dead short across the motor power leads if the motor
is small and the transistor is robust, or you can use some scheme to
put the power back into the batteries.
NATHAN PHILLIP STEWART
>Ok, I would like to brake the motor by shorting both leads of the motor to
>ground. Is this a good way to do the job? Also how much current do you get
>when you short out a motor? I a worried that the transistors might fry if
>I do this.
Is this what was meant? I figured he meant shorting the motor across itself,
disconnected to anything else - using back emf as a brake. Is this what was
intended? Would back emf be strong enough to do it?
NAte
KEN E WILLMOTT
back into the batteries. We have streetcars and subways here that
do that (only they don't have batteries :-> ).
If you're driving the motor through an H bridge, that seems a little
tricky, however, turning either the top or bottom pair of drivers
should short the motor out nicely. No additional components, at
least major ones, are needed.
The braking effect will be limited by the efficiency of the motor.
-Ken
francis deck
>I like Bruce Rowan's idea about dynamic braking by dumping power
>back into the batteries. We have streetcars and subways here that
>do that (only they don't have batteries :-> ).
>
shuts down its superconducting bending magnets by dumping the
power back onto the grid, and that this can be detected 100's of
miles away by generating plants that are actively synchronized to
the grid!
f.deck
physics
notre dame
Does the Tevatron qualify as the world's biggest motor?
John Nagle
>back into the batteries. We have streetcars and subways here that
>do that (only they don't have batteries :-> ).
>If you're driving the motor through an H bridge, that seems a little
>tricky, however, turning either the top or bottom pair of drivers
>should short the motor out nicely. No additional components, at
>least major ones, are needed.
Where is the energy going to go? Shorting the motor is not
an acceptable approach to dynamic braking. Something is going to give;
probably either the FETs in the H-bridge or the commutator.
Better speed controls for R/C racing cars do use dynamic braking,
and it really does recharge the battery.
John Nagle
KEN E WILLMOTT
>>I like Bruce Rowan's idea about dynamic braking by dumping power
>>back into the batteries. We have streetcars and subways here that
>>do that (only they don't have batteries :-> ).
>>If you're driving the motor through an H bridge, that seems a little
>>tricky, however, turning either the top or bottom pair of drivers
>>should short the motor out nicely. No additional components, at
>>least major ones, are needed.
> Where is the energy going to go? Shorting the motor is not
>.......
John Nagle
Into the motor windings.
-Ken
Rick Miller
More accurately, the energy goes to HEAT. That heat will be dissipated
mostly from whatever component in the "shorted" circuit which has the
highest resistance. ...that could very well be the "short" itself,
depending upon how you short it. If you short it through your H-bridge,
you may burn it out!
KEN E WILLMOTT
>>>cs91...@ariel.yorku.ca (KEN E WILLMOTT) writes:
>>
>>> Where is the energy going to go? Shorting the motor is not
>>>.......
>> John Nagle
>>
>>Into the motor windings.
>>
>> -Ken
>More accurately, the energy goes to HEAT. That heat will be dissipated
>mostly from whatever component in the "shorted" circuit which has the
>highest resistance. ...that could very well be the "short" itself,
>depending upon how you short it. If you short it through your H-bridge,
>you may burn it out!
>RICK MILLER <ri...@ee.uwm.edu> Voice: +1 414 221-3403
Gee, I thought it stayed in the windings, until nobody was looking,
then made everything blow up. :-)
Of course some energy will be dissipated in the FETs, but they will
(should?) be matched to the current requirements of the motor. The
same goes for the heatsink they're on.
The back EMF dissipation should (must?) be after all, less than
the dissipation involved in powering the motor. Also, it is
transient.
The stiffening of the motor due to loading will also cause a lot
of heat loss in the gear train, especially if the gear ratio is
high. Those RC cars have low ratios for high speeds, whereas most
robots have high ratios for low speeds. This makes regenerative
braking less efficient.
Also, think about what kind of circuit is required to accomplish
regen. braking. It would be *much* more complicated than my
suggestion.
Perhaps the original poster could describe the mass, gear retio,
motor specs, and driving circuit, so that we could comment more
specifically.
-Ken Willmott
Nathan Stratton
Well, Rick you won. They fried, They were 10 amp darlington power transistors.
They were controlling a 6 amp PM motor. I think I will just slow the motor
down before it stops. That will eliminate some of the problem. I am 17 and
do not have a major job and can't keep experimenting with this H-Bridge.
Nathan Stratton
ns...@quercus.gsfc.nasa.gov
John Little - Nihon Sun Repair Depot
% >depending upon how you short it. If you short it through your H-bridge,
% >you may burn it out!
% >
Standard procudure with the spindle motor on (physically) big disk
drives used to be to use a dual pole relay to switch the bridge out of
circuit and short the motor windings through some "dumping" circuit.
In older drives the dumping circuit was simply a very high wattage, low
value resistor.
I'm sure someone will correct me if the grey cells are playing tricks,
but I seem to remember the CDC 300MB removable-pack (washing-machine)
drive used this arrangement. Maybe someone could give us an exact
reference from the manuals?
--
------------------------------------------------------------------------
| John Little - gai...@Japan.Sun.COM - Sun Microsystems. Atsugi, Japan |
------------------------------------------------------------------------
Fong Choon Chiak
If you short the motor directly, the main resistance is the armature
resistance, the shorting current will be large, usually larger than the rating
of the H bridge. I think you can chop the transistor to control the current
and the braking torque.
Another simple method is just to use a dump resistor in dynamic braking.
James Bach
When you want the robot to STOP, instead of just removing power to the
motor, try applying a "dead short" (or a very low resistance) across
the motor. By just removing power (leaving the terminals "float") the
motor becomes a generator (driven by the inertial of the robot), with
virtually NO load on it; thus you will develop a large voltage across
the motor's terminals. By placing a short on the terminals, the
energy being generated from the inertia is converted into heat (in the
windings of the motor, and/or the low resistance placed across the
terminals).
If you want to prove this to yourself on a smaller scale, take a small
DC motor (like from a toy) with a shaft you can spin freely with your
fingers; preferably with some sort of weighted disk/gear on it (so you
can see it spin for a few seconds). Spin the motor with the wires
"open"; note that it takes a few seconds to "spin down". Now short
the wires together; notice that the shaft "feels" tighter/harder to
spin, and it spins down MUCH faster (nearly instantly?).
I don't know what the rest of your driver circuit looks like, so I
can`t recommend a good way of doing this "short out" trick (called
"braking" in the literature). If you are doing a simple "on/off"
low-side driver (motor connected to V+ and a FET switch to ground)
then the solution might be as easy as adding a diode across the motor
so that when the low-side FET turns off the current that was flowing
in the motor recirculates thru the diode. Not as good as a short
circuit, but much better than open. The ideal solution would be to
put a big FET across the motor, which is on when the other is off. If
you PWM'd this FET then you could control the rate at which the robot
slowed down/stopped.
James Bach
In article <27crii$j...@skates.gsfc.nasa.gov>, ns...@quercus.gsfc.nasa.gov (Nathan Stratton) writes:
>
> Ok, I would like to brake the motor by shorting both leads of the motor to
> ground. Is this a good way to do the job? Also how much current do you get
> when you short out a motor? I a worried that the transistors might fry if
> I do this.
>
>
You can only get out what you put in . . . if you were supplying 10A
to the motor when you turned-off the driver, then (unless the robot is
going down-hill) the current coming from the motor should be 10A or
less. Any transistor type that you used to DRIVE the motor should be
capable of "STOPPING" the motor too.
Jeroen Belleman
c2x...@kocrsv01.delcoelect.com (James Bach) writes:
>
>You can only get out what you put in . . . if you were supplying 10A
>to the motor when you turned-off the driver, then (unless the robot is
>going down-hill) the current coming from the motor should be 10A or
>less. Any transistor type that you used to DRIVE the motor should be
>capable of "STOPPING" the motor too.
>--
>James C. Bach Ph: (317)-451-0455 The views & opinions expressed
Halt there, that is not true. Permanent magnet DC motors have an
approximately linear torque vs. current characterisitic. Stopping a
motor quickly requires a large torque and hence a large current.
This current can easily exceed the steady running current by a large
factor.
In fact, shorting a motor to stop it is really quite brutal. You don't
want to do that with an expensive servo motor, for example.
Jeroen Belleman
jer...@dxcern.cern.ch
Greg Campbell
>Please forgive me if you feel this offensive. I would like to tell
>you that it is absolutely natural phenomennon that things won't stop
>suddenly. You should control revolution of the motor with a certain
>feedback circuit. Don't let motors move suddenly.
>
>If you just want to shorten the duration of the motor stopping, let
>the two terminal of the motor electrically be short as soon as you cut
>off the drivr voltage.
Another idea is to stop the motor by ramping down a PWM signal to it.
This way you have control over the movement (also a good way of starting
up the motor - little or no jerking).
Greg Campell
>
>Hope this helps,
>--
>Mato Hattori Computer System Research Department,
>email : hat...@csrd.nkk.co.jp Electronics Research Center, NKK Corporation
>voice : 81+44-322-6483 (office) 1-1 Minamiwatarida Kawasaki-ku
> : 81+44-246-9665 (home) Kawasaki-city, 210 JAPAN
--
---------------------------------------------------------------------------
Greg Campbell Idacom - an operation of Hewlett-Packard
email: gr...@idacom.hp.com Voice: (403) 462-4545 Fax: (403) 462-4869
---------------------------------------------------------------------------
John Nagle
>In article <27crii$j...@skates.gsfc.nasa.gov>, ns...@quercus.gsfc.nasa.gov (Nathan Stratton) writes:
>> Ok, I would like to brake the motor by shorting both leads of the motor to
>> ground. Is this a good way to do the job? Also how much current do you get
>> when you short out a motor? I a worried that the transistors might fry if
>> I do this.
>You can only get out what you put in . . . if you were supplying 10A
>to the motor when you turned-off the driver, then (unless the robot is
>going down-hill) the current coming from the motor should be 10A or
>less. Any transistor type that you used to DRIVE the motor should be
>capable of "STOPPING" the motor too.
Er, no. You can get far more current out than you put in; just
run the motor up slowly, and then try to brake it fast, which is what
a dead short will do. All that kinetic energy turns into heat in the
motor windings and semiconductors. Energy really is conserved, guys.
You might be able to get away with this for a
very small motor, but if you try it with anything with serious inertia,
you're going to overheat something, either the semiconductors or
motor windings. With a large motor, you might well get a fire.
Incidentally, this is a good time to mention that PC board material
comes in fire-resistant and non-fire-resistant forms. High-power
stuff should always be on fire-resistant board. This can be the
difference between a scorched board and a serious fire.
Put in a resistor sized to limit the current to the maximum the
semiconductors can handle, and with a wattage rating appropriate to the
job. You probably need a power ceramic type for even a small motor.
Remember to provide ventilation for the resistor; it's going to
get warm.
>James C. Bach
>Advanced Project Engr.
>Powertrain Strategy Grp
>Delco Electronics Corp.
Powertrain Strategy Group? Delco Division of General Motors?
I'd expect more respect for power dissipation from that end of the industry.
John Nagle
Topi Maurola
switching converters and a big capasitor:
When braking the motor, connect it to "boost regulator" which's duty cycle
is controlled by input voltage (motor's output voltage) which is kept constant
(e.g. 4 V), and connect its output to a big capasitor (e.g. 1500 uF/350 V)
which will be loaded by motors inertia (and robot's also).
This capasitor would also be connected to power supply system so that when
it's voltage is high enough (e.g. >30 V), the power is taken form it by using
a step-down converter until it's energy is used.
This system would have better efficiency than with re-chargeable batteries,
which are not needed. The step-up switcher's efficiency increases with output
voltage, and if capasitors minimum voltage is kept, braking would start from
that with high efficiency. Drawback is that efficiency drops with input voltage
which is tried to keep low for high braking power. But if the brakinbg power
needed isn't very high and a motor isn't lo-voltage, a few volts (3 ~ 4) could
do it. Then the voltage drop (and power lost) in the motor wouldn't dominate.
I'd use Linear's LT1070/LT1074 switchers and low-resistance, high-current
fets. For further reading, try to get Linear's "Linear Applications Handbook"
which has _many_ different designs with theory and math.
The following could be boost switcher used in system. Inductor and fet
values depends on motor and wanted brake-level. Brake power can be changed
by changing 3V +-input to op.amp. 3V sets brake voltage to ~3.7V (Brake voltage
~=2*op.amp.+.input-2.35V). Be sure that inductor is ment for switching purposes
and diode is fast enough (e.g. schotty). Possible FET could be IRF340 which
is 400V/25A. Also an overcurrent shut-down should be added!
__________________
motor | |
| || |
GND--[M]-----||--GND |
| || |
8 inductor |
8 |
GND fet | || |
| |-----|>|--||--GND |
1kohm-------| diode || |
+12V | |-| + - |
| | | big cap. |
|--------------| --- |
|VIN VSW | - |
| | |
| LT1074 | /| |
| | /+|--3V |
|GND VC FB |___/ | |
|______________| | \ | |
| | | \-|--10kohm------------
| | | \| |
--- --- | |
- --- 10kohm-|
|
---
-
Topi Maurola
Finland, Europe
Yi-Hsiu Wang
Can anyone tell me what kind of supporting tools does 68HC11 have? What
I currently know is EVB, EVBu, and EVM. Is there any emulator or software
package? By the way, has anyone tried to operate EVB, EVBu or EVM in 3.3V?
Thanks for any information.
Yi-Hsiu
StarLab/Stanford Univ
Kevin White
all be done with the same circuitry. I have not built a reversible circuit
but I have done one that worked successfully for a R/C car for forward
motion only. Peak currents were in the 50 amp range at 7.2 volts. Power
MOSFETs these days are almost ideal devices and make excellent power
switches.
power --------
| |
|- |
FET 1 -| - +
|- ^
| |
|------Inductor--
| | |
|- | |
FET 2 -| - + Motor
|- ^ |
| | |
Common ---------------------
This is a releatively conventional half bridge output stage to drive the motor.
The drive to FET1 and FET2 should be in antiphase.
(Be careful about gate to source voltage rating of FETs, especially the
upper one as it is difficult to drive). The frequency of the drive should
be about 50KHz for small motors. Bigger motors should probably have the
inductor, small ones do not need it as the inductance of the rotor is adequate
and not too lossy.
When the upper FET is on, current flows into the motor and increases linearly,
when it turns off the current then continues flowing through the lower FET.
Note that no power is coming from the power supply! The duty cycle of the
drive determines the average voltage presented to the motor. This is varied
for normal speed control.
To achieve regenerative braking the duty cycle is reduced until this
average voltage is less than the Back-EMF of the motor. Under this condition
when the FET2 is conducting current will flow from the motor and build up in
the inductor, when FET2 stops conducting, the voltage at the junction of
FET1 and FET2 will increase, and current will flow through FET1 and into the
power supply. This energy is taken from the motor and thus slows it down.
As the motor slows down, the duty cycle has to be reduced, it is not practical
to go too low, probably about 10% is the limit. Below that limit the amount
of current from the motor is probably low enough that you could turn FET2
full on to stop the motor.
This same technique can be extended to a full bridge to get bidirectional
control and braking.
The diodes shown are desirable to give more latitude in drive generation.
The substrate diodes in the MOSFETs may be adequate although they are not
very good diodes. I prefer schottkey diodes here.
I have found it is easy to destroy MOSFETs with overvoltage with circuits
of this kind, even a few inches of wire can have enough inductance to wreak
havoc when you have 50 AMPS flowing with rates of change of many Amps per
nanosecond. Voltage pulses of hundreds of volts were common until I
put a few good tantalum electrolytics across the supply lines, close to the
power transistors.
Kevin White
James Bach
In article <1993Sep21....@dxcern.cern.ch>, jer...@dxcern.cern.ch (Jeroen Belleman) writes:
> Halt there, that is not true. Permanent magnet DC motors have an
> approximately linear torque vs. current characterisitic. Stopping a
> motor quickly requires a large torque and hence a large current.
> This current can easily exceed the steady running current by a large
> factor.
> In fact, shorting a motor to stop it is really quite brutal. You don't
> want to do that with an expensive servo motor, for example.
>
> Jeroen Belleman
> jer...@dxcern.cern.ch
I hang my head in shame . . . . you and most of the other replys to my
reply are most surely correct! I failed to take into account the
inertia of the "system"; my quick reply made assumptions (which were
incorrect) based on MY experience with very small motors with
virtually NO inertial loading. Obviously a robot with a lot of mass
has a lot of inertia, and therefore a lot of energy (1/2 MV^2) that
needs to be dissipated during the braking process.
To take one of the respondants "hammer and nail" rebutals, and to
convert it into an electrical equivalent . . . your swinging of the
hammer is like the application of power to the motor to make the robot
start moving; the energy is applied as a small force applied over a
large period of time. The hitting of the hammer on the nailhead is
like your trying to brake the robot; you need to dissipate the energy
in a very brief period of time, thus applying an impulsive "shock" to
your system (in this analogy the block of wood seems to be equivalent
to your driver IC, which seemed to get "nailed" when you tried to use
it to "short out" the motor for braking).
I found my copy of the SGS-Thomson "Smart Power Application Manual",
and they have a nice section detailing driving DC motors with
H-switched (bridges), including braking. They recommend a "chopper"
braking scheme which essentially applies the "short" to the motor in a
PWM'd sort-of way which controls the energy dissipated in the driver
to what the driver can handle. I also dug-out my International
Rectifier "Power MOSFET HEXFET DatBook", and it also has a section
describing DC motor drive; it is unidirectional, and uses a FET to V+
to "drive" the motor, and a FET across the motor terminals to "Regen"
(brake) the motor by dumping it's current (brute force) like I
suggested initially. Obviously this device needs to handle the
current caused by the inertia of the motor on spin-down; perhaps a
good way to perform a "current limit" on this FET is to use a "logic
level" FET with a sense resistor in the Source leg; pick the value of
the sense resistor so that when the current thru it nears the spec
limit of the FET the IR drop of the resistor reduces the Vgs and
turns-off the FET. For instance, I have the data-sheet for the IRL530
logic-level power FET (TO-220 pkg, Rds(on)=.16 Ohm, Id=14A cont. (10A
at 100C) with Vgs=5V, 100V break-down voltage, Vgs(th)=1V,
Vgs(max)=10V. Looking at the curves the drain current will fall from
14A to .1A as Vgs falls from about 3.75 to 1.5V. Thus if you wanted
to limit the current thru the FET to, say 5A during braking, then a
sense resistor of about .4 Ohms would make Vs rise to about 5Ax.4=2V,
and thus Vgs would fall to 3.0V (assumes 5V "logic" drive), which is
the point in the curve that shows Id=5A. 10A limiting would require
about .15Ohm sense resistor. Obviously the higher you make this
"braking current" the faster the robot will stop; your limiting factor
will be the type of FET you use as your "short".
I'm not sure what sort of driver the
original poster was using, and whether he tried to just "short" the
motor for the entire braking period (perhaps his hardware allows him
to PWM the driver), nor how well he had it heat-sinked. Was it a
discrete circuit or an integrated driver (i.e. power-package
multi-legged SIP)? Perhaps a review of the data/handbooks I've
mentioned will help.
In any event, I appologize for my hasty (and incorrect) initial reply.
Sorry you fried a driver, too. As the saying goes, "Good advice costs
money, but dumb looks are still free!".
--
James C. Bach Ph: (317)-451-0455 The views & opinions expressed
James Bach
> >James C. Bach
> >Advanced Project Engr.
> >Powertrain Strategy Grp
> >Delco Electronics Corp.
>
> Powertrain Strategy Group? Delco Division of General Motors?
> I'd expect more respect for power dissipation from that end of the industry.
>
> John Nagle
Sorry, we deal with "gasoline" motors! :-)
In general, all the inductors we energize here have no mechanical
inertia (at least none like a heavy robot); generally we turn
"solenoids" on and off, and although they have a pintle which moves,
the "back EMF" from them tend to be negligible compared to the energy
stored in the inductance of the coil (which we DO take very seriously).
KEN E WILLMOTT
>> >James C. Bach
>> >Advanced Project Engr.
>> >Powertrain Strategy Grp
>> >Delco Electronics Corp.
>>
>> Powertrain Strategy Group? Delco Division of General Motors?
>> I'd expect more respect for power dissipation from that end of the industry.
>>
>> John Nagle
>Sorry, we deal with "gasoline" motors! :-)
>In general, all the inductors we energize here have no mechanical
>inertia (at least none like a heavy robot); generally we turn
>"solenoids" on and off, and although they have a pintle which moves,
>the "back EMF" from them tend to be negligible compared to the energy
>stored in the inductance of the coil (which we DO take very seriously).
>--
>James C. Bach Ph: (317)-451-0455 The views & opinions..
Of course, the original poster never said anything about having
a ramped startup. That's what prompted me to suggest shorting directly.
I figured, if he could get the thing moving with circuit X, then he
could probably make it stop with circuit X (in a nutshell).
-Ken
Dave Medin
|> % >
|> % >depending upon how you short it. If you short it through your H-bridge,
|> % >you may burn it out!
|> % >
|>
|> Standard procudure with the spindle motor on (physically) big disk
|> drives used to be to use a dual pole relay to switch the bridge out of
|> circuit and short the motor windings through some "dumping" circuit.
|> In older drives the dumping circuit was simply a very high wattage, low
|> value resistor.
For old CDC disk packs, they actually had a separate dynamic brake
which when DC was applied to it, generated eddy currents internally thus
stopping the drive pretty quickly. An amp at 12 volts made this brake heard to
turn by hand at any appreciable speed. The device was about doughnut-sized,
and slipped on the drive shaft. It was not part of the motor, which may
have been AC synchronous anyway (no braking when shorted). I may still
have it in my junk collection somewhere...
--
--------------------------------------------------------------------
Dave Medin--SSD Networking Phone: (205) 730-3169 (w)
Intergraph Corp. (205) 837-1174 (h)
M/S GD3004 Internet: dtm...@ingr.com
Huntsville, AL 35894 UUCP: ...uunet!ingr!b30!catbyte!dtmedin
* The opinions expressed here are mine (or those of my machine)
Mark Bayern
In article <1993Sep20.1...@kocrsv01.delcoelect.com>, James Bach (c2x...@kocrsv01.delcoelect.com) writes:
>
>In article <27crii$j...@skates.gsfc.nasa.gov>, ns...@quercus.gsfc.nasa.gov (Nathan Stratton) writes:
>>
>> Ok, I would like to brake the motor by shorting both leads of the motor to
>> ground. Is this a good way to do the job? Also how much current do you get
>> when you short out a motor? I a worried that the transistors might fry if
>> I do this.
>>
>>
>
>You can only get out what you put in . . . if you were supplying 10A
>to the motor when you turned-off the driver, then (unless the robot is
>going down-hill) the current coming from the motor should be 10A or
>less. Any transistor type that you used to DRIVE the motor should be
>capable of "STOPPING" the motor too.
>--
Wrong! 'You can only get out what you put in' is correct, but you
must think of it in terms of energy (stored in the magnetic field,
and in the kinetic energy of the system). Putting a short across
the motor can result in very high currents! Of course, the high
current will only exist only for a short period of time. <grin>
Mark
Nathan Stratton
I had my 2 rear wheels connected to my rear axal. I the problem I had was
to turn the robot left or right. One wheel needs to spin slower then the
other axal. When this dose not happen you get big problems with a 300 pound
robot. I fixed the problem by letting one wheel spin freely.
The problem I have now is in programing the robot. I would like to program
the robot in C or FORTH. Dose anybody know of any books or other info on
robot programming.
Nathan Stratton
ns...@quercus.gsfc.nasa.gov
Michael Covington
>I have this big problem with my robot. I am using PWM to control my motors,
>but I can not make the robot to go in a strate line. Dose anyone know of
>a cheep compass I can interface with my robot to tell the robot when it is
>drifting?? I whould like to take the drift and compensate for it.
It would then go in circles around any magnetic object, wouldn't it? :)
--
< Michael A. Covington, Assc Rsch Scientist, Artificial Intelligence Programs >
< The University of Georgia, Athens, GA 30606-7415 USA mcov...@ai.uga.edu >
<>< ----------------------------------------------------------------------- ><>
< For info about U.Ga. degree programs, email GRA...@UGA.CC.UGA.EDU (not me) >
Nathan Stratton
but I can not make the robot to go in a strate line. Dose anyone know of
a cheep compass I can interface with my robot to tell the robot when it is
drifting?? I whould like to take the drift and compensate for it.
Nathan Stratton
ns...@quercus.gsfc.nasa.gov
Chuck McManis
>but I can not make the robot to go in a strate line. Dose anyone know of
>a cheep compass I can interface with my robot to tell the robot when it is
>drifting?? I whould like to take the drift and compensate for it.
The easiest way to do this is to provide some sort of feedback on the motors.
You would do this by providing some mechanism that would let the computer
driving the motors know how many turns the wheels have gone through. There
are several examples in the literature (the mobile robots: inspiration to
implementation book is good for this, as are the Sams books by Mark Robillard)
You can use optical sensing or magnetic (see this months issue of Computer
Craft magazine for a base that uses magnets and a hall effect switch.
Many of your questions are answerable by any good text on robotics, allow
me to reccomend the following, you can get them at your local library or
through direct mail. If you take this list to the library they may be able
to order copies for their shelves.
Title : Mobile Robots: Inspiration to Implementation
Authors : Joesph L. Jones, Anita M. Flynn
Publisher: AK Peters Ltd
ISBN # : 1-56881-011-3
Notes : This is the most current book and an overall good reference source.
Title : The Robot Builder's Bonanza, 99 inexpensive robotics projects
Author : Gordon Mccomb
Publisher: TAB Books Inc
ISBN # : 0-8306-2800-2
Notes : Published in '87 it is showing its age (uses older parts) but
contains a wealth of good ideas and they are pretty cheap to
build. You can get this book for 99 cents by joingin the
electronic's book club. (look for ads in most electronics
hobbiest magazines)
Title : Microprocessor Based Robotics
Author : Mark J. Robillard
Publisher: Howard W. Sams & Co Inc
ISBN # : 0-672-22050-4
Notes : Published in '83 it is even more dated than the previous but
covers interesting subjects like hands and such. Chapter 5
is dedicated to position feedback systems.
Title : Android Design
Author: Martin ?
Notes: Its out of print and my copy is on loan but it has an excellent
discussion of the _physics_ of robotic platforms. Calculating
torques, power requirements, and drive configurations.
--
--Chuck McManis Mr. NIS+ Sunsoft
uucp: {anywhere}!sun!cmcmanis BIX: <none> Internet: cmcm...@Eng.Sun.COM
These opinions are my own and no one elses, but you knew that didn't you.