Help with understanding astable multvibrator circuit.
John Wilson
I've breadboarded the following circuit and it works correctly. However, I
am unsure as to _why_ it works correctly. :)
The schematic is located at:
http://webpages.charter.net/jmw74/circuit.JPG
I don't understand how the values of R1 and R4 are being calculated. The
forward voltage of the LED's is 2 volts with a forward current of 20 mA, so
according to Ohm's Law, the value of the resistor would be (9-2)/.020, which
would yield a 350 ohm resistor. Why does the schematic call for a 470 ohm
resistor?
R2 and R3, as I understand, act as a current limiter to the caps, and when
the caps are fully charged, it provides current to the base of the
transistors. A high ohm value is used to charge up the caps as a very slow
rate, until the caps stored charge equals 9V, at which point, current will
be diverted to the base of the transistors since the caps will be "full".
As far as the transistors are concerned, when current is applied to the
base, it will allow current to flow from the transistors collector to its
emitter, thus completing the circuit and allowing the LED's to light.
This is where my understanding begins to falter. In regard to the cap and
the current flowing from R2 or R3, it would make sense to me that while the
cap is charging, there is no flow to the base of the transistor. Once the
cap is fully charged, the flow to the cap will essentially be "blocked", and
current will then flow to the base of the transistor. Is this corrent?
When would the cap actually discharge and where does it discharge _to_?
Thank you for your time,
John
John Popelish
>
> Hello.
>
> I've breadboarded the following circuit and it works correctly. However, I
> am unsure as to _why_ it works correctly. :)
>
> The schematic is located at:
> http://webpages.charter.net/jmw74/circuit.JPG
>
> I don't understand how the values of R1 and R4 are being calculated. The
> forward voltage of the LED's is 2 volts with a forward current of 20 mA, so
> according to Ohm's Law, the value of the resistor would be (9-2)/.020, which
> would yield a 350 ohm resistor. Why does the schematic call for a 470 ohm
> resistor?
To allow a current that is most of the rated 20 milliamps while
certainly being below it. If you drop the current to 10 milliamps,
there is a barely perceptible lowering of the light output. Try it.
> R2 and R3, as I understand, act as a current limiter to the caps, and when
> the caps are fully charged, it provides current to the base of the
> transistors. A high ohm value is used to charge up the caps as a very slow
> rate, until the caps stored charge equals 9V, at which point, current will
> be diverted to the base of the transistors since the caps will be "full".
Keep in mind that the function of R2 and 3 involve the diode
properties of the base emitter junctions. For example, when TR1 first
switches on, the ~9 volt negative change in voltage on the left side
of C1 is communicated to the right side of C1, lowering whatever
voltage was there by about 9 volts. Since, right before TR1 turned
on, TR2 had to have been on, that means that the voltage on the right
side of C1 had to have been about .6 volts positive (compared to the
bottom rail). So right after TR1 switches on, the voltage on the
right side of C1 gets shoved down to about -8.4 volts (assuming the
base emitter junction of TR2 can stand that much voltage without
breaking down and clamping that peak voltage to something less than
this).
That big negative voltage is what must be charged back up to +.6 volts
or so, before TR2 again begins to turn on. Of course, the two sides
of the circuit are symmetrical.
> As far as the transistors are concerned, when current is applied to the
> base, it will allow current to flow from the transistors collector to its
> emitter, thus completing the circuit and allowing the LED's to light.
Yes. The collector current can be expected to be many times (50 to
500, say, depending on the current gain of that particular transistor)
the base current, as long as the collector voltage is more positive
than the base. Once the collector current is pulled lower than the
base, it begins to steal the base drive current (because the collector
to base junction gets forward biased), and the effective gain goes way
down.
> This is where my understanding begins to falter. In regard to the cap and
> the current flowing from R2 or R3, it would make sense to me that while the
> cap is charging, there is no flow to the base of the transistor.
As long as the base voltage is below +.6 volts, yes, all the current
through the resistor is being used to raise the voltage on the
capacitor.
> Once the
> cap is fully charged, the flow to the cap will essentially be "blocked", and
> current will then flow to the base of the transistor. Is this corrent?
Actually, a better way to say that is that once the capacitor has been
charged up to +.6 volts or so, the base emitter junction detours all
the resistor current and none is left to charge the capacitor
further. At that point, the transistor turns on, driving its
collector capacitor voltage negative, which switches the base resistor
current for the other transistor back into its capacitor, leaving none
for the other base, so the other transistor is switched off. This
action causes the other capacitor to see a positive voltage change
(because the LED current that had been going into the collector is now
free to dump into the capacitor). This drives a hugh pulse of base
current into the transistor that was just beginning ot turn on, and
charging base capacitor very quickly to about a little less than +9
volts on one end (don't forget the LED drop) and +.6 volts on the
other end. This is the setup for the turn off process when the off
transistor turns back on after its base voltage reaches the forward
bias threshold.
> When would the cap actually discharge and where does it discharge _to_?
In one direction its charge current passes through LED, its series
resistor, and the forward biased base emitter junction. It charges in
the other direction through the collector to emitter path and the
recharge resistor, R2 or 3.
> Thank you for your time,
> John
You are welcome.
--
John Popelish
Martin Pickering {UK}
John Popelish <jpop...@rica.net> wrote:
>bottom rail). So right after TR1 switches on, the voltage on the
>right side of C1 gets shoved down to about -8.4 volts (assuming the
>base emitter junction of TR2 can stand that much voltage without
>breaking down and clamping that peak voltage to something less than
>this).
It probably won't. -8.4 volts is really over the limit for a
general-purpose silicon transistor. I think the "designer" kept
things simple by omitting the reverse Vbe protection diodes that
should be interposed between each emitter and negative rail. In fact,
if the diodes were fitted, it would probably slow down the frequency,
allowing a reduction in capacitor value to get the same frequency as
without the diodes (= lower cost and better reliability since, I
believe, continually stressing the be junction can reduce the gain
ove a period of time).
These considerations don't matter for a one-off hobby circuit but
they could be of interest for a mass-produced unit if reliability and
repeatability were important.
Just my 2 cents worth.
Martin
Martin T. Pickering B.Eng.
http://www.satcure-focus.com
________________________
John Popelish
>
> In article <3D5EC1FE...@rica.net>,
> John Popelish <jpop...@rica.net> wrote:
>
> >bottom rail). So right after TR1 switches on, the voltage on the
> >right side of C1 gets shoved down to about -8.4 volts (assuming the
> >base emitter junction of TR2 can stand that much voltage without
> >breaking down and clamping that peak voltage to something less than
> >this).
>
> It probably won't. -8.4 volts is really over the limit for a
> general-purpose silicon transistor. I think the "designer" kept
> things simple by omitting the reverse Vbe protection diodes that
> should be interposed between each emitter and negative rail. In fact,
> if the diodes were fitted, it would probably slow down the frequency,
> allowing a reduction in capacitor value to get the same frequency as
> without the diodes (= lower cost and better reliability since, I
> believe, continually stressing the be junction can reduce the gain
> ove a period of time).
Since the half period is based on charging the cap up from the large
negative bias, if you clamp the bias t oa smaller negative value, then
the charging time is shorter.
> These considerations don't matter for a one-off hobby circuit but
> they could be of interest for a mass-produced unit if reliability and
> repeatability were important.
Yes, I would worry about eventually degrading the current gain of the
transistors by repeatedly driving the base emitter junction into zener
breakdown.
> Just my 2 cents worth.
--
John Popelish
John Wilson
I've been tring to figure out this circuit as best I can. Below
is what I believe is occuring at any one instance. I'm really
trying to understand this and believe that I am very close
to doing so.
The circuit schematic is:
http://webpages.charter.net/jmw74/CIRCUIT.GIF
Any comments on my analysis of this circuit would be
greatly appreciated. :)
=======================
A 9 volt source is applied. As D1, R2, R3, and D2 are wired in
parallel, they each receive a full 9 volts, assuming we discard
any wire resistance.
R1 and R4 act as current and voltage limiters so we don't
overload the requirements of the LED's. Assuming that the
forward voltage of a LED is 2 volts and a forward current
maximum of 20mA, the resistors are calculated by Ohm's law
as follows:
R=V/I = (9V - 2V) / 20ma = (9V - 2V) / .020 = 350 Ohms.
A 350 Ohm resistor would provide the maximum allowable voltage
and current to the LED, however, to build in a "margin of error",
a higher rated resistor should be used.
Using a 470 Ohm resistor as in the circuit diagram, the current
and voltage delivered to the LED is:
I = V/R = 9 / 470 = 0.015 A = 15mA.
V = 9 - (IR) = 9 - (0.015 * 470) = 9 - 7.05 = 1.95 V.
R2 and R3 provide the following:
I = V/R = 9 / 4700 = 0.0019 A = 1.9mA.
V = 9 - IR = 9 - (.0019 * 4700) = 9 - 8.93 = 0.07 V.
At this point, we need to assume that C1 is discharging and C2 is
charging. C2 must already have a minimum stored capacity of at least
~0.06 volts at this point.
C1
--
* Discharging
* A negative voltage is placed along C1 and Q2's base due to the
discharge of C1, effectively shutting off Q2 due to the voltage
being less than the required ~+0.06 volts to Q2's base.
C2
--
* Charging
* R3 is proving ~ +0.07 volts between C2 and Q1's base. Once the
voltage is equalized at about ~ +0.06 and +0.07 volts, it meets
the required voltage for Q1 to "turn on", thus providing Q1's
collector to emitter flow and turning the LED D1 on.
Now, Q1 is "on", Q2 is "off", C1 is discharging, and C2 is charging.
While all this is happening, R2 is delivering voltage between C1
and Q2's base. Eventually, the negative voltage on this line will
rise to ~0.06 volts, which will trigger the following events:
* Q2's base will be properly biased, triggering voltage flow between
its collector and base. C2 will discharge through Q2, placing a
negative voltage across C2 and Q1, turning "off" Q1.
* D2 will turn on.
* D1 will turn off.
R3 will eventually raise the voltage between C2 and Q1's base high
enough ( ~0.6V ) to trigger Q1 to "open" again, discharging C1 and
placing a negative voltage across C1 and Q2's base, turning Q2 off.
The process will repeat.
John Popelish
>
> Hello again.
Hi. Did you receive the reply I made to your first post?
This is a bit circular. The first equation is wrong. 9/470 = .019,
but the resistor does not have a 9 volt drop, but only the part of 9
volts that is left over after the LED and transistor have used up
whatever they will. If we assume that the transistors are saturated,
they can be expected to drop less than .5 volt, and the LED will drop
about 1.5 volts (red about 1.5, green closer to 2 volts), so just
assume those drops, and subtract them from 9 and use that remainder to
calculate the resistor current. It them comes out very close to .015
amps. I think you did this correctly in your first post.
> R2 and R3 provide the following:
>
> I = V/R = 9 / 4700 = 0.0019 A = 1.9mA.
> V = 9 - IR = 9 - (.0019 * 4700) = 9 - 8.93 = 0.07 V.
Again, you are assuming all the voltage is dropped across the
resistors, and this would be the case, only if something was forcing
the voltage at the bottom of those resistors to be zero. so it is no
wonder that you calculate that voltage to be zero within the round off
error of your math. The actual voltage across these resistors will
depend on what else is in series with that current and there are two
cases. In one case, the voltage will be positive enough for the base
emitter junction to forward bias, and this is a low resistance state
for that junction, so it will hog all the current through the resistor
and clamp the voltage to about .6 volts. The other case would be if
the capacitor were holding the voltage below .6 volts, so that the
transistor base emitter junction were reverse biased, and so allowed
no current to pass through, so all the resistor current would have to
flow into the capacitor. And this does not produce a stable voltage,
but a positive ramp as the capacitor charges.
> At this point, we need to assume that C1 is discharging and C2 is
> charging. C2 must already have a minimum stored capacity of at least
> ~0.06 volts at this point.
I guess you have to start somewhere, but a better way would be to
assume that both capacitors start with zero voltage across them, and
see what that implies. Remember that a capacitor passes a current
that is proportional to how fast the voltage is changing across it.
This implies that at the instant the battery is connected, R1 and R4
supply current to the capacitors, because the other end of both are
clamped to no more than .6 volts by base emitter junctions, and both
transistors are turned on by these currents. But to the extent that
each transistor is turned on, the currents from R1 and R4 are detoured
away from the capacitors into the transistors. Whichever transistor
has the highest current gain (assuming all other components are
perfectly matched pairs) will have the lowest collector voltage, and
the higher voltage on the other one will give the high gain one more
capacitor current to turn it on more, and the lower gain one will see
no voltage rise through its capacitor and be turned off more, so the
situation will quickly turn into one saturated on transistor and one
cut off transistor.
> C1
> --
> * Discharging
> * A negative voltage is placed along C1 and Q2's base due to the
> discharge of C1, effectively shutting off Q2 due to the voltage
> being less than the required ~+0.06 volts to Q2's base.
Actually, the only way this negative voltage can appear is if the
collector it is attached to suddenly pulls down to zero after that
capacitor was pre charged positive on the collector end, and more
negative on the base end, so that both sides of the cap go negative by
the same amount at the same time. The collector end goes from about
+9 to zero, and the base end goes from about +.6 to -8.4, but the
original +8.4 volts across the cap is unchanged for a brief instant.
> C2
> --
> * Charging
> * R3 is proving ~ +0.07 volts between C2 and Q1's base. Once the
> voltage is equalized at about ~ +0.06 and +0.07 volts, it meets
> the required voltage for Q1 to "turn on", thus providing Q1's
> collector to emitter flow and turning the LED D1 on.
This turn on event would have been the cause of the sudden voltage
change applied to the cap we just talked about.
> Now, Q1 is "on", Q2 is "off", C1 is discharging, and C2 is charging.
>
I agree that C1 is discharging, since it has a large voltage across it
that is being slowly eliminated by current from R2. But C2 is pretty
static, since it has no source of current to change its voltage. All
the current from R3 is being sucked up by the base of Q1.
> While all this is happening, R2 is delivering voltage between C1
> and Q2's base.
The current through R2 is raising the voltage on the base end of C1
from the starting -8.4 toward the turn on voltage around +.6.
Eventually, the negative voltage on this line will
> rise to ~0.06 volts, which will trigger the following events:
Make that +.6
> * Q2's base will be properly biased, triggering voltage flow between
> its collector and base.
Voltages are applied, then current flows. The base voltage causes
current to flow from base to emitter. The current gain of the
transistor causes a much larger current to flow through the collector
to the emitter.
> C2 will discharge through Q2, placing a
> negative voltage across C2 and Q1, turning "off" Q1.
The negative voltage swing at the collector end of C2 will be passed
to the base end so that the total drop is unchanged. Since C2 started
out with almost +9 volts on its collector end, and +.6 volts on its
base end (for a voltage difference of about 8.4 volts) this same
voltage difference must still be there after the collector pulls one
end down to near zero volts. That means that the base end must now be
at about -8.4 volts. That certainly turns off Q1.
> * D2 will turn on.
> * D1 will turn off.
Right.
> R3 will eventually raise the voltage between C2 and Q1's base high
> enough ( ~0.6V )
+.6 volts.
> to trigger Q1 to "open" again,
to turn on again (if a switch is the analogy, you have in mind, we
would say that that switch closed at that point). Open usually refers
to an open circuit, i.e. no current through that path.
> discharging C1
Just relocating the voltage that was already charged across C1.
> and
> placing a negative voltage across C1 and Q2's base, turning Q2 off.
>
> The process will repeat.
You are getting very close.
--
John Popelish
John Wilson
"John Popelish" <jpop...@rica.net> wrote in message
news:3D5FDBA5...@rica.net...
> Hi. Did you receive the reply I made to your first post?
No, unfortunately, all I received were fragments of your message when a
Martin Pickering replied to your post.
[...]
>> The process will repeat.
>You are getting very close.
I'm also rather impatient. :) The circuit in general is still quite
confusing to me, and based on all the corrections that you've made to what I
originally wrote, some of which I don't understand at this point (but I've
saved your reply and hope that it will make more sense soon), I think I need
to refocus on more basic circuits and start smaller.
I've only been at this for about a week, with no backgound in electronics at
all, but I hope that at some point, things will "click".
On a side note, do you think that those small kits that are made make good
learning tools? All I have here are a bunch of components (resistors, caps,
diodes, transistors, etc.), but no IC's. Most of the schematics that I see
floating around the web make use of one IC or another, which I do not
possess. I was just curious as to the learning potential of these kits, as
they are somewhat inexpensive (cost is an issue), and I was wondering if
they would make good learning tools.
Thank you very much for your time. You have helped quite a lot.
- John
John Popelish
>
> "John Popelish" <jpop...@rica.net> wrote in message
> news:3D5FDBA5...@rica.net...
> > Hi. Did you receive the reply I made to your first post?
>
> No, unfortunately, all I received were fragments of your message when a
> Martin Pickering replied to your post.
I have attempted to email a copy of it to you.
>
> [...]
> >> The process will repeat.
> >You are getting very close.
>
> I'm also rather impatient. :) The circuit in general is still quite
> confusing to me, and based on all the corrections that you've made to what I
> originally wrote, some of which I don't understand at this point (but I've
> saved your reply and hope that it will make more sense soon), I think I need
> to refocus on more basic circuits and start smaller.
Since this circuit is just two copies of a 5 component function, it is
pretty simple, already. Feel free to repost my comments, with any
questions that arise right where I lose you. This will help me figure
out exactly what concepts you are having trouble with.
> I've only been at this for about a week, with no backgound in electronics at
> all, but I hope that at some point, things will "click".
I guessed as much from your sloppy use of voltage and current
descriptions. Voltage is always a difference between two points in a
circuit (or between one point and an implied reference point). All
components connected in parallel must share the shame voltage.
Current passes through a path (from one point to another). All
components in series must share the same current.
> On a side note, do you think that those small kits that are made make good
> learning tools? All I have here are a bunch of components (resistors, caps,
> diodes, transistors, etc.), but no IC's. Most of the schematics that I see
> floating around the web make use of one IC or another, which I do not
> possess. I was just curious as to the learning potential of these kits, as
> they are somewhat inexpensive (cost is an issue), and I was wondering if
> they would make good learning tools.
>
> Thank you very much for your time. You have helped quite a lot.
>
Op amps and comparators are good functional blocks for beginners to
work with, as are simple logic functions, like gates and flip flops
(especially the almost indestructible 4000 CMOS series). If you see
circuits based on these components, they should be very educational.
Highly integrated chips are so obscure that you learn less by playing
with them than by working with the more primitive functions.
--
John Popelish
Rich Mechaber
link to the post and replies that set me straight:
Rich Mechaber