Voltage Divider Circuit vs. Current Limiting Resistor
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Kamal Advani
Oct 25, 2014, 11:15:57 PM10/25/14
to connected-commu...@googlegroups.com
Hi Folks,
Been experimenting with batteries and LEDs and learning basics.
When you have a battery that has a voltage larger than a LED you want
to turn on, often we put in a current limiting resistor so it doesn't
bust the LED. I mistakenly used to think that the resistor would bring
down the voltage post that point as well (at least.. using a
multimeter across the resistor contacts show the same voltage as the
battery... unless I am doing something silly...). Based on my
observations:
1. Voltage drop across a resistor does not occur with a single
resistor in a closed circuit (no other component besides battery).
2. If I have a battery, a resistor, a LED -- voltage across resistor +
voltage across LED = battery voltage, thus the drops occur.
My questions then:
1. Conceptually, why is there no drop in case 1? Is it because there
is no load as such, and the current passing through is the maximum the
battery produces? I'm having trouble understanding this idea of
current draw by different components I guess. A LED has a current draw
of say 20mA, what happens in a closed circuit with only a resistor?
2. When do I need a voltage divider vs. just a current limiting resistor.
As a real example, I'm playing with a solid state relay (if you
remember my post from yonks ago, I am back on my thermostat project!),
which is LED triggered (input), and so I have not needed a voltage
divider circuit. The datasheet specifies that the relay input forward
voltage must not exceed 1.65 V. And with the current limiting
resistor, I get the same voltage drop behaviour as with Case 2 with
the regular/plain LED. Why don't we need a voltage divider circuit for
LEDs?
Thanks for your help, as usual.
Regards,
Kamal
--
Out beyond ideas of wrongdoing and rightdoing, there is a field. I
will meet you there.
-- Rumi
Been experimenting with batteries and LEDs and learning basics.
When you have a battery that has a voltage larger than a LED you want
to turn on, often we put in a current limiting resistor so it doesn't
bust the LED. I mistakenly used to think that the resistor would bring
down the voltage post that point as well (at least.. using a
multimeter across the resistor contacts show the same voltage as the
battery... unless I am doing something silly...). Based on my
observations:
1. Voltage drop across a resistor does not occur with a single
resistor in a closed circuit (no other component besides battery).
2. If I have a battery, a resistor, a LED -- voltage across resistor +
voltage across LED = battery voltage, thus the drops occur.
My questions then:
1. Conceptually, why is there no drop in case 1? Is it because there
is no load as such, and the current passing through is the maximum the
battery produces? I'm having trouble understanding this idea of
current draw by different components I guess. A LED has a current draw
of say 20mA, what happens in a closed circuit with only a resistor?
2. When do I need a voltage divider vs. just a current limiting resistor.
As a real example, I'm playing with a solid state relay (if you
remember my post from yonks ago, I am back on my thermostat project!),
which is LED triggered (input), and so I have not needed a voltage
divider circuit. The datasheet specifies that the relay input forward
voltage must not exceed 1.65 V. And with the current limiting
resistor, I get the same voltage drop behaviour as with Case 2 with
the regular/plain LED. Why don't we need a voltage divider circuit for
LEDs?
Thanks for your help, as usual.
Regards,
Kamal
--
Out beyond ideas of wrongdoing and rightdoing, there is a field. I
will meet you there.
-- Rumi
Luke Weston
Oct 26, 2014, 12:04:47 AM10/26/14
to connected-commu...@googlegroups.com
Hi Kamal,
Basically, if you have a battery and a resistor, there are no other
components in the circuit, so all the battery voltage is measured
across the resistor. I will try and explain as best I can, quickly.
Hope this is helpful.
Well, actually, if you have a battery, a piece of wire, a resistor,
and other piece of wire, then effectively you have three resistors in
series, so you can measure the voltage across one, and the other, and
the other, and all three voltages should add up to the voltage across
the battery.
However, in practice, the resistance of the wire is so low that you'll
probably measure negligible voltage drop across it with an average
multimeter.
If you have two resistors in series, then you can measure two
different voltages across them, which add up to the battery voltage.
Or you can try a couple of resistors in series and a battery and a LED
and measure the voltage across each of them.
This is called Kirchoff's voltage law, and it is really just a
statement of potential energy conservation - but usually the textbooks
(and Wikipedia) get bogged down explaining the concepts underneath a
lot of fancy language and maths which isn't helpful to explain the
basic concept or required in all contexts.
http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws#Kirchhoff.27s_voltage_law_.28KVL.29
The current through a resistor connected to a battery can be
calculated just using Ohm's law from the resistance and the voltage
across the resistor (which in the case of just a single resistor is
the battery voltage.)
A diode like a LED is a bit more complicated. When the diode is biased
with some current flowing through it in the forward direction, it has
a voltage drop across it, and that voltage drop remains consistent
over a range of current, even if the current changes a bit. The
voltage-current relationship is not completely linear like it is for a
resistor. It can be assumed that, within a certain limit, the voltage
drop is always a constant. It depends on the semiconductor properties,
the bandgap etc, but let's say for the sake of argument we have a blue
LED and we say the forward voltage is 3V.
If the battery voltage is 5V, say, and we know the LED's forward
voltage drop is 3V, the voltage across the resistor, in a simple
series circuit with only one loop with a LED, resistor and battery,
must just be the difference which is 2V. And knowing that, and knowing
Ohm's law, we can choose to set the resistance so that the bias
current through the LED is at some value we choose... R = V / I, where
V is the voltage across the resistor. So if we want, say, 2 mA, we can
choose a 1k resistor. (Which is a good default value for general
indicator LEDs in 5V or 3.3V circuits.)
A LED doesn't have a "fixed" current draw, but it will have a maximum
safe operating current (typically say 20mA) - it is that forward
voltage which is a property of the LED material/construction, and it
is that choice of resistor that sets the current in the circuit.
But using a single resistor is only a good approach in circuits where
the voltage drop is known to be fixed - in series with a single diode
to bias it, for example lighting up a LED, or in a zener diode
regulator circuit, or a similar voltage reference diode circuit, or a
temperature sensor like the LM335, where you know what the voltage
drop across the diode is reasonably well.
If you look at a voltage divider with two resistors, it works based on
the assumption that the "load" circuit has zero current flowing into
it... if that other circuit is tapping off any noticable amount of
current then the current through the two series resistors is not equal
so the voltage division does not behave the way you think it does. So
a voltage divider is not a good general purpose voltage regulator...
even if they could actively regulate voltage it would still be really
inefficient for any substantial load current, and burn a lot of power
dissipated in the resistors.
But voltage dividers are very commonly used to adjust voltages in
circuits that do have very high input impedance, where there is no
significant current flowing into the load circuit - for example
biasing transistors, setting up feedback networks that control the
gain in opamp amplifier circuits, on the inputs to opamps to provide a
"reference" voltage cheaply with relatively low precision, or to
provide feedback into a comparator at a certain reference voltage
(which is a divided fraction of the output voltage) to set the output
voltage of a voltage regulator or power supply circuit.
Cheers,
Luke
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Basically, if you have a battery and a resistor, there are no other
components in the circuit, so all the battery voltage is measured
across the resistor. I will try and explain as best I can, quickly.
Hope this is helpful.
Well, actually, if you have a battery, a piece of wire, a resistor,
and other piece of wire, then effectively you have three resistors in
series, so you can measure the voltage across one, and the other, and
the other, and all three voltages should add up to the voltage across
the battery.
However, in practice, the resistance of the wire is so low that you'll
probably measure negligible voltage drop across it with an average
multimeter.
If you have two resistors in series, then you can measure two
different voltages across them, which add up to the battery voltage.
Or you can try a couple of resistors in series and a battery and a LED
and measure the voltage across each of them.
This is called Kirchoff's voltage law, and it is really just a
statement of potential energy conservation - but usually the textbooks
(and Wikipedia) get bogged down explaining the concepts underneath a
lot of fancy language and maths which isn't helpful to explain the
basic concept or required in all contexts.
http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws#Kirchhoff.27s_voltage_law_.28KVL.29
The current through a resistor connected to a battery can be
calculated just using Ohm's law from the resistance and the voltage
across the resistor (which in the case of just a single resistor is
the battery voltage.)
A diode like a LED is a bit more complicated. When the diode is biased
with some current flowing through it in the forward direction, it has
a voltage drop across it, and that voltage drop remains consistent
over a range of current, even if the current changes a bit. The
voltage-current relationship is not completely linear like it is for a
resistor. It can be assumed that, within a certain limit, the voltage
drop is always a constant. It depends on the semiconductor properties,
the bandgap etc, but let's say for the sake of argument we have a blue
LED and we say the forward voltage is 3V.
If the battery voltage is 5V, say, and we know the LED's forward
voltage drop is 3V, the voltage across the resistor, in a simple
series circuit with only one loop with a LED, resistor and battery,
must just be the difference which is 2V. And knowing that, and knowing
Ohm's law, we can choose to set the resistance so that the bias
current through the LED is at some value we choose... R = V / I, where
V is the voltage across the resistor. So if we want, say, 2 mA, we can
choose a 1k resistor. (Which is a good default value for general
indicator LEDs in 5V or 3.3V circuits.)
A LED doesn't have a "fixed" current draw, but it will have a maximum
safe operating current (typically say 20mA) - it is that forward
voltage which is a property of the LED material/construction, and it
is that choice of resistor that sets the current in the circuit.
But using a single resistor is only a good approach in circuits where
the voltage drop is known to be fixed - in series with a single diode
to bias it, for example lighting up a LED, or in a zener diode
regulator circuit, or a similar voltage reference diode circuit, or a
temperature sensor like the LM335, where you know what the voltage
drop across the diode is reasonably well.
If you look at a voltage divider with two resistors, it works based on
the assumption that the "load" circuit has zero current flowing into
it... if that other circuit is tapping off any noticable amount of
current then the current through the two series resistors is not equal
so the voltage division does not behave the way you think it does. So
a voltage divider is not a good general purpose voltage regulator...
even if they could actively regulate voltage it would still be really
inefficient for any substantial load current, and burn a lot of power
dissipated in the resistors.
But voltage dividers are very commonly used to adjust voltages in
circuits that do have very high input impedance, where there is no
significant current flowing into the load circuit - for example
biasing transistors, setting up feedback networks that control the
gain in opamp amplifier circuits, on the inputs to opamps to provide a
"reference" voltage cheaply with relatively low precision, or to
provide feedback into a comparator at a certain reference voltage
(which is a divided fraction of the output voltage) to set the output
voltage of a voltage regulator or power supply circuit.
Cheers,
Luke
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Andy Gelme
Oct 26, 2014, 12:11:15 AM10/26/14
to connected-commu...@googlegroups.com
hi All,
On 2014-10-26 14:15 , Kamal Advani wrote:
> When you have a battery that has a voltage larger than a LED you want
> to turn on, often we put in a current limiting resistor so it doesn't
> bust the LED. I mistakenly used to think that the resistor would bring
> down the voltage post that point as well (at least.. using a
> multimeter across the resistor contacts show the same voltage as the
> battery... unless I am doing something silly...).
If you have a resistor and (correctly oriented) LED in series connected
to a battery, there will be a voltage drops across both the resistor and
the LED (not just the resistor), which can be measured by your multimeter.
> Based on my observations:
>
> 1. Voltage drop across a resistor does not occur with a single
> resistor in a closed circuit (no other component besides battery).
>
> 2. If I have a battery, a resistor, a LED -- voltage across resistor +
> voltage across LED = battery voltage, thus the drops occur.
>
> My questions then:
> 1. Conceptually, why is there no drop in case 1? Is it because there
> is no load as such, and the current passing through is the maximum the
> battery produces?
If you put a multimeter across the resistor, you will measure a voltage
drop ... which is equivalent to the battery voltage, because (apart from
the battery) there is only the resistor in your circuit.
The resistor *is* the load.
See Ohm's law V = IR. The battery provides V. The resistor provides
R. And, those two values will determine the current I ... flowing
through the resistor.
> I'm having trouble understanding this idea of current draw by different components I guess.
Different component types, e.g resistor or capacitor or inductor or
various different semiconductors (diode, transistor) ... will have
different behavior in a circuit. Searching for a good on-line
electronics introduction on-line will provide you with that information.
> A LED has a current draw of say 20mA, what happens in a closed circuit with only a resistor ?
A (correctly oriented) LED will have a constant (forward) voltage drop
over a range of current flowing through the LED, specific to that LED
type (colour). Until the current is too high and then the LED dies.
See
http://dangerousprototypes.com/docs/Basic_Light_Emitting_Diode_guide#Voltage_drop_or_forward_voltage
Typically, 20 mA is used in discussion, because it is a safe amount of
current to flow through a typical LED ... and have a long life and good
brightness.
Let's assume an LED voltage drop of 2 V. This means that if you have a
3 V power source, then the LED will drop 2 V ... forcing there to be a 1
V drop across your resistor ...
3 V (power supply) - 2 V (LED) = 1 V (resistor)
Now, you choose a resistor value to ensure that the current flowing
through the LED doesn't exceed the maximum allowed current ... and also
is an amount of current that provides the desired brightness, i.e it
does not have to be running at the maximum allowed current.
Using Ohm's law ... and the voltage drop across the resistor and the
desired current, you get ...
V = I R ... R = V / I ... R = 1 V / 0.020 A ... R = 50 Ohms.
In this circuit, you'd choose a 50 Ohm resistor to limit the current
flowing through the entire circuit (including the LED) to be 20 milliAmps.
> 2. When do I need a voltage divider vs. just a current limiting resistor.
For simple LED applications such that you are talking about here ...
usually not. Because it is an exercise in "limiting current", not
"limiting voltage".
> As a real example, I'm playing with a solid state relay (if you
> remember my post from yonks ago, I am back on my thermostat project!),
> which is LED triggered (input)
What do you mean by "LED triggered" ?
For all of this email, you've been using the LED as an output (as people
commonly do).
Are you now saying that you really wanted to use it as an input (which
is possible, LEDs also detect light) ? This is different case.
Note: You may need to do some more reading of electronics tutorials and
searching for similar applications. Too hard to provide a complete
electronics course via email :)
--
-O- cheers = /\ /\/ /) `/ =
--O -- http://www.geekscape.org --
OOO -- an...@geekscape.org -- http://twitter.com/geekscape --
On 2014-10-26 14:15 , Kamal Advani wrote:
> When you have a battery that has a voltage larger than a LED you want
> to turn on, often we put in a current limiting resistor so it doesn't
> bust the LED. I mistakenly used to think that the resistor would bring
> down the voltage post that point as well (at least.. using a
> multimeter across the resistor contacts show the same voltage as the
> battery... unless I am doing something silly...).
to a battery, there will be a voltage drops across both the resistor and
the LED (not just the resistor), which can be measured by your multimeter.
> Based on my observations:
>
> 1. Voltage drop across a resistor does not occur with a single
> resistor in a closed circuit (no other component besides battery).
>
> 2. If I have a battery, a resistor, a LED -- voltage across resistor +
> voltage across LED = battery voltage, thus the drops occur.
>
> My questions then:
> 1. Conceptually, why is there no drop in case 1? Is it because there
> is no load as such, and the current passing through is the maximum the
> battery produces?
drop ... which is equivalent to the battery voltage, because (apart from
the battery) there is only the resistor in your circuit.
The resistor *is* the load.
See Ohm's law V = IR. The battery provides V. The resistor provides
R. And, those two values will determine the current I ... flowing
through the resistor.
> I'm having trouble understanding this idea of current draw by different components I guess.
various different semiconductors (diode, transistor) ... will have
different behavior in a circuit. Searching for a good on-line
electronics introduction on-line will provide you with that information.
> A LED has a current draw of say 20mA, what happens in a closed circuit with only a resistor ?
over a range of current flowing through the LED, specific to that LED
type (colour). Until the current is too high and then the LED dies.
See
http://dangerousprototypes.com/docs/Basic_Light_Emitting_Diode_guide#Voltage_drop_or_forward_voltage
Typically, 20 mA is used in discussion, because it is a safe amount of
current to flow through a typical LED ... and have a long life and good
brightness.
Let's assume an LED voltage drop of 2 V. This means that if you have a
3 V power source, then the LED will drop 2 V ... forcing there to be a 1
V drop across your resistor ...
3 V (power supply) - 2 V (LED) = 1 V (resistor)
Now, you choose a resistor value to ensure that the current flowing
through the LED doesn't exceed the maximum allowed current ... and also
is an amount of current that provides the desired brightness, i.e it
does not have to be running at the maximum allowed current.
Using Ohm's law ... and the voltage drop across the resistor and the
desired current, you get ...
V = I R ... R = V / I ... R = 1 V / 0.020 A ... R = 50 Ohms.
In this circuit, you'd choose a 50 Ohm resistor to limit the current
flowing through the entire circuit (including the LED) to be 20 milliAmps.
> 2. When do I need a voltage divider vs. just a current limiting resistor.
usually not. Because it is an exercise in "limiting current", not
"limiting voltage".
> As a real example, I'm playing with a solid state relay (if you
> remember my post from yonks ago, I am back on my thermostat project!),
> which is LED triggered (input)
For all of this email, you've been using the LED as an output (as people
commonly do).
Are you now saying that you really wanted to use it as an input (which
is possible, LEDs also detect light) ? This is different case.
Note: You may need to do some more reading of electronics tutorials and
searching for similar applications. Too hard to provide a complete
electronics course via email :)
--
-O- cheers = /\ /\/ /) `/ =
--O -- http://www.geekscape.org --
OOO -- an...@geekscape.org -- http://twitter.com/geekscape --
Kamal Advani
Oct 26, 2014, 3:52:27 AM10/26/14
to connected-commu...@googlegroups.com
On Sun, Oct 26, 2014 at 3:04 PM, Luke Weston <reindeer...@gmail.com> wrote:
> Basically, if you have a battery and a resistor, there are no other
> components in the circuit, so all the battery voltage is measured
> across the resistor. I will try and explain as best I can, quickly.
> Hope this is helpful.
>
Luke, thanks a heap for taking the time to answer. I think I
> Basically, if you have a battery and a resistor, there are no other
> components in the circuit, so all the battery voltage is measured
> across the resistor. I will try and explain as best I can, quickly.
> Hope this is helpful.
>
understand the voltage drop concepts better now, and I had been
learning about KVL (and using that to test out the voltage points).
With the single resistor circuit, I was so localised on the component
that I forgot that across the resistor in that case is basically
across the circuit! Speak about missing the forest for the trees.
> A LED doesn't have a "fixed" current draw, but it will have a maximum
> safe operating current (typically say 20mA) - it is that forward
> voltage which is a property of the LED material/construction, and it
> is that choice of resistor that sets the current in the circuit.
that there is a range, so it adjusts its resistance based on V and
I...
> If you look at a voltage divider with two resistors, it works based on
> the assumption that the "load" circuit has zero current flowing into
> it... if that other circuit is tapping off any noticable amount of
> current then the current through the two series resistors is not equal
> so the voltage division does not behave the way you think it does. So
> a voltage divider is not a good general purpose voltage regulator...
> even if they could actively regulate voltage it would still be really
> inefficient for any substantial load current, and burn a lot of power
> dissipated in the resistors.
it can't regulate voltage... is it more because the current is not
regulated? As in, over time, the current is not steady when using
resistor-based voltage dividers making it unfeasible for many loads?
Thanks for the overview of how they are used. Your next paragraph (not
quoted) mentions op amps and similar, I have not touched analogue
circuits, and I think this is where a lot of interesting theory is
applied.
Thanks again, this has cleared up a lot of things, especially diodes,
which I have read about on and off (no pun intended), but I think the
pieces are coming together in my head, especially Ohm's law in various
contexts.
Geoff
Oct 26, 2014, 4:02:34 AM10/26/14
to connected-commu...@googlegroups.com
Hi Kamal
You may find trying out your circuit using some circuit simulation software helpful, and then attend a night (or Saturday) at CCHS to build it up. We have a range of resistors and LED's. You will need a breadboard (solder-less) and some hookup wire.
Circuit simulation software:
Breadboard, wire, etc:
Regards
Geoff
Kamal Advani
Oct 26, 2014, 4:08:39 AM10/26/14
to connected-commu...@googlegroups.com
Hi Andy,
On Sun, Oct 26, 2014 at 3:11 PM, Andy Gelme <an...@geekscape.org> wrote:
>> My questions then:
>> 1. Conceptually, why is there no drop in case 1? Is it because there
>> is no load as such, and the current passing through is the maximum the
>> battery produces?
>
> If you put a multimeter across the resistor, you will measure a voltage
> drop ... which is equivalent to the battery voltage, because (apart from
> the battery) there is only the resistor in your circuit.
>
> The resistor *is* the load.
ACK. I see that now. Thanks. :-)
>
> See Ohm's law V = IR. The battery provides V. The resistor provides
> R. And, those two values will determine the current I ... flowing
> through the resistor.
>
>> I'm having trouble understanding this idea of current draw by different components I guess.
>
> Different component types, e.g resistor or capacitor or inductor or
> various different semiconductors (diode, transistor) ... will have
> different behavior in a circuit. Searching for a good on-line
> electronics introduction on-line will provide you with that information.
Hmm, I do try my best to look things up... but sometimes some answers
raise more questions, especially when some basics are not yet really
that well internalised, I'm just a tad dense :-).
With my comment on current draw, specifically, I'm just beginning to
get a feel for the dynamic nature of certain components, and how they
interact within the context of Ohm's Law. I acknowledge it was most
vague, and I should have focussed on specifics.
> See
> http://dangerousprototypes.com/docs/Basic_Light_Emitting_Diode_guide#Voltage_drop_or_forward_voltage
Neat, thanks!
>
> Typically, 20 mA is used in discussion, because it is a safe amount of
> current to flow through a typical LED ... and have a long life and good
> brightness.
Thanks for the worked out example, in my circuit, I have gotten my
calculations correct thankfully, so this was a good confirmation. It
was really the 'why' behind certain behaviour that I was struggling
with.
>> 2. When do I need a voltage divider vs. just a current limiting resistor.
>
> For simple LED applications such that you are talking about here ...
> usually not. Because it is an exercise in "limiting current", not
> "limiting voltage".
I see.
>> As a real example, I'm playing with a solid state relay (if you
>> remember my post from yonks ago, I am back on my thermostat project!),
>> which is LED triggered (input)
>
> What do you mean by "LED triggered" ?
>
> For all of this email, you've been using the LED as an output (as people
> commonly do).
>
> Are you now saying that you really wanted to use it as an input (which
> is possible, LEDs also detect light) ? This is different case.
Bad explanation on my part, sorry. Case 1 was a simple LED scenario.
With case 2, I have a solid-state relay whose input side is a LED
(opto-isolated). Given that, I was wondering if I needed a voltage
divider, or I could treat it like any other LED and use a simple,
single current limiting resistor.
> Note: You may need to do some more reading of electronics tutorials and
> searching for similar applications. Too hard to provide a complete
> electronics course via email :)
I understand, and I have been, I'm possibly just too vague at points.
Once more, thanks for taking the time to reply.
On Sun, Oct 26, 2014 at 3:11 PM, Andy Gelme <an...@geekscape.org> wrote:
>> My questions then:
>> 1. Conceptually, why is there no drop in case 1? Is it because there
>> is no load as such, and the current passing through is the maximum the
>> battery produces?
>
> If you put a multimeter across the resistor, you will measure a voltage
> drop ... which is equivalent to the battery voltage, because (apart from
> the battery) there is only the resistor in your circuit.
>
> The resistor *is* the load.
>
> See Ohm's law V = IR. The battery provides V. The resistor provides
> R. And, those two values will determine the current I ... flowing
> through the resistor.
>
>> I'm having trouble understanding this idea of current draw by different components I guess.
>
> Different component types, e.g resistor or capacitor or inductor or
> various different semiconductors (diode, transistor) ... will have
> different behavior in a circuit. Searching for a good on-line
> electronics introduction on-line will provide you with that information.
raise more questions, especially when some basics are not yet really
that well internalised, I'm just a tad dense :-).
With my comment on current draw, specifically, I'm just beginning to
get a feel for the dynamic nature of certain components, and how they
interact within the context of Ohm's Law. I acknowledge it was most
vague, and I should have focussed on specifics.
> See
> http://dangerousprototypes.com/docs/Basic_Light_Emitting_Diode_guide#Voltage_drop_or_forward_voltage
Neat, thanks!
>
> Typically, 20 mA is used in discussion, because it is a safe amount of
> current to flow through a typical LED ... and have a long life and good
> brightness.
calculations correct thankfully, so this was a good confirmation. It
was really the 'why' behind certain behaviour that I was struggling
with.
>> 2. When do I need a voltage divider vs. just a current limiting resistor.
>
> For simple LED applications such that you are talking about here ...
> usually not. Because it is an exercise in "limiting current", not
> "limiting voltage".
>> As a real example, I'm playing with a solid state relay (if you
>> remember my post from yonks ago, I am back on my thermostat project!),
>> which is LED triggered (input)
>
> What do you mean by "LED triggered" ?
>
> For all of this email, you've been using the LED as an output (as people
> commonly do).
>
> Are you now saying that you really wanted to use it as an input (which
> is possible, LEDs also detect light) ? This is different case.
With case 2, I have a solid-state relay whose input side is a LED
(opto-isolated). Given that, I was wondering if I needed a voltage
divider, or I could treat it like any other LED and use a simple,
single current limiting resistor.
> Note: You may need to do some more reading of electronics tutorials and
> searching for similar applications. Too hard to provide a complete
> electronics course via email :)
Once more, thanks for taking the time to reply.
Robert Powers
Oct 26, 2014, 4:10:29 AM10/26/14
to connected-commu...@googlegroups.com
As in, over time, the current is not steady when using
resistor-based voltage dividers making it unfeasible for many loads?
Vin-|
|
R1
|
-----.----- - Vout
| |
R2 RL
| |
-----.-----
|
GND
Not that. Here's one simple way to evaluate things:
Basically if you were to add a load (RL) to the 'lower' leg of a resistor divider (R2), you've put something in parallel with R2 which has some given resistance. Now do your circuit analysis and assume that you want significant current through the leg RL. Ohm's Law will tell you that if you want this current to be significant, either the voltage must be high or the resistance low. Throw out high voltage because you want to set a point, usually low. So you're stuck with low resistance. But it's in parallel with R2, so if you have a low resistance, what is the equivalent resistance below R1 (at Vout). Now what happens to your resistor divider?
Vout = (R2 || RL) * Vin/(R1+(R2||RL))
Hint - R2||RL (equivalent resistance of R2 in parallel with RL) ~ RL for RL<< (much smaller than) R2.
Run through setting values for each and find why you believe the statement - "Resistor dividers make very poor power supplies."
Oh, and you don't get to chose the resistance of your load in real life (indeed, it will dynamically change in this setup for anything not a simple resistor).
Cheers,
Bob
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Kamal Advani
Oct 26, 2014, 8:38:08 AM10/26/14
to connected-commu...@googlegroups.com
On Sun, Oct 26, 2014 at 7:02 PM, Geoff <geoff.le...@gmail.com> wrote:
> Hi Kamal
>
> You may find trying out your circuit using some circuit simulation software
> helpful, and then attend a night (or Saturday) at CCHS to build it up. We
> have a range of resistors and LED's. You will need a breadboard
> (solder-less) and some hookup wire.
>
> Circuit simulation software:
> https://www.circuitlab.com/
> http://en.wikipedia.org/wiki/List_of_free_electronics_circuit_simulators
Thanks for the idea and links, Geoff, I'll have a look through, sounds
> Hi Kamal
>
> You may find trying out your circuit using some circuit simulation software
> helpful, and then attend a night (or Saturday) at CCHS to build it up. We
> have a range of resistors and LED's. You will need a breadboard
> (solder-less) and some hookup wire.
>
> Circuit simulation software:
> https://www.circuitlab.com/
> http://en.wikipedia.org/wiki/List_of_free_electronics_circuit_simulators
like an interesting learning tool.
I do have a breadboard and some components which I am experimenting on.
Luke Weston
Oct 26, 2014, 8:52:52 AM10/26/14
to connected-commu...@googlegroups.com
OK, so exactly what sort of solid-state relay module have you got?
Usually you don't need to provide an external resistor or any
components, and they will specify an easy to use, broad input voltage
range such as 3-30V DC.
You need to check the data for the particular device you're using, though.
They usually have some sort of active current bias circuit inside on
the input side that drives the infrared LED, so you're not connected
into that LED directly, and the current remains at a good value across
that wide range of acceptable voltages. The internal circuit will be
something roughly like that shown in Figure 6:
https://www.fairchildsemi.com/application-notes/AN/AN-3003.pdf
Usually you don't need to provide an external resistor or any
components, and they will specify an easy to use, broad input voltage
range such as 3-30V DC.
You need to check the data for the particular device you're using, though.
They usually have some sort of active current bias circuit inside on
the input side that drives the infrared LED, so you're not connected
into that LED directly, and the current remains at a good value across
that wide range of acceptable voltages. The internal circuit will be
something roughly like that shown in Figure 6:
https://www.fairchildsemi.com/application-notes/AN/AN-3003.pdf
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Kamal Advani
Oct 26, 2014, 9:03:29 AM10/26/14
to connected-commu...@googlegroups.com
On Sun, Oct 26, 2014 at 11:52 PM, Luke Weston
<reindeer...@gmail.com> wrote:
> OK, so exactly what sort of solid-state relay module have you got?
http://www.avagotech.com/pages/en/optocouplers_plastic/solid_state_relay_photo_mosfet/assr-1510-003e/
<reindeer...@gmail.com> wrote:
> OK, so exactly what sort of solid-state relay module have you got?
Datasheet (PDF): http://www.avagotech.com/docs/AV02-0036EN
> Usually you don't need to provide an external resistor or any
> components, and they will specify an easy to use, broad input voltage
> range such as 3-30V DC.
resistor (well, a series to get about 600 ohms), using a 9V battery.
The plan is to hook it up to a RPi at 3.3V with an appropriate
resistor.
> You need to check the data for the particular device you're using, though.
>
> They usually have some sort of active current bias circuit inside on
> the input side that drives the infrared LED, so you're not connected
> into that LED directly, and the current remains at a good value across
> that wide range of acceptable voltages. The internal circuit will be
> something roughly like that shown in Figure 6:
> https://www.fairchildsemi.com/application-notes/AN/AN-3003.pdf
circuits. But it looks like mine, based on the diagrams, has input
points direct to the internal LED?
Kamal Advani
Oct 26, 2014, 9:19:47 AM10/26/14
to connected-commu...@googlegroups.com
On Sun, Oct 26, 2014 at 7:10 PM, Robert Powers <rdpo...@gmail.com> wrote:
>> As in, over time, the current is not steady when using
>> resistor-based voltage dividers making it unfeasible for many loads?
>
>
> Vin-|
> |
> R1
> |
> -----.----- - Vout
> | |
> R2 RL
> | |
> -----.-----
> |
> GND
>
> Not that. Here's one simple way to evaluate things:
Awesome, thanks for the exercise and hints, I'll work it out. :-)
>> As in, over time, the current is not steady when using
>> resistor-based voltage dividers making it unfeasible for many loads?
>
>
> Vin-|
> |
> R1
> |
> -----.----- - Vout
> | |
> R2 RL
> | |
> -----.-----
> |
> GND
>
> Not that. Here's one simple way to evaluate things:
Rob Gannon
Oct 26, 2014, 9:50:13 PM10/26/14
to connected-commu...@googlegroups.com
Hey Luke, Andy, Geoff, Bob,
For those of us that are like me, and would love to understand all of this stuff but have a bit of trouble getting my head around it all when reading wiki pages, web tutorials and blog posts; would any of you guys want to run some training sessions or workshops on some of the basics. I'm sure there must be plenty of others that would be interested in that sort of thing. Something aimed at the beginner that can lead into actually designing and building something would be awesome.
Rob.
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damie...@gmail.com
Oct 26, 2014, 10:12:48 PM10/26/14
to CCHS
I second Rob's request.
Voltage dividers are a useful "building-block" of circuit design.
Thanks to Luke, Andy, Geoff, and Bob for the detailed explanations and links. I'd find a training session on this area useful, and appreciate some sort of workshop / hands-on demo that puts the theory into practice.
Cheers,
Damien
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Nick Meinhold
Oct 26, 2014, 10:34:24 PM10/26/14
to connected-commu...@googlegroups.com
+1 to Rob's workshop for beginners request. I would be very happy to help with any of the preparation type work required to make such a thing happen.
Nick
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Andy Gelme
Oct 26, 2014, 11:06:48 PM10/26/14
to connected-commu...@googlegroups.com
hi All,
On 2014-10-27 12:50 , Rob Gannon wrote:
> For those of us that are like me, and would love to understand all of
> this stuff but have a bit of trouble getting my head around it all
> when reading wiki pages, web tutorials and blog posts; would any of
> you guys want to run some training sessions or workshops on some of
> the basics. I'm sure there must be plenty of others that would be
> interested in that sort of thing. Something aimed at the beginner that
> can lead into actually designing and building something would be awesome.
Clearly there is enough interest ... that this should happen.
The first two things to determine would be when and desired content ...
1) I'd suggest that the first time we run such a workshop should be a
Saturday afternoon (more time and less busy than the week nights).
2) Rather than have a massive email thread, let's use the CCHS Wiki ...
http://www.hackmelbourne.org/wiki/workshop:electronics_introduction
... please add your suggestions.
Having a "hands-on" exercises would be good, so that it doesn't become
only presentation and talk.
On 2014-10-27 12:50 , Rob Gannon wrote:
> For those of us that are like me, and would love to understand all of
> this stuff but have a bit of trouble getting my head around it all
> when reading wiki pages, web tutorials and blog posts; would any of
> you guys want to run some training sessions or workshops on some of
> the basics. I'm sure there must be plenty of others that would be
> interested in that sort of thing. Something aimed at the beginner that
> can lead into actually designing and building something would be awesome.
The first two things to determine would be when and desired content ...
1) I'd suggest that the first time we run such a workshop should be a
Saturday afternoon (more time and less busy than the week nights).
2) Rather than have a massive email thread, let's use the CCHS Wiki ...
http://www.hackmelbourne.org/wiki/workshop:electronics_introduction
... please add your suggestions.
Having a "hands-on" exercises would be good, so that it doesn't become
only presentation and talk.
Kamal Advani
Oct 31, 2014, 10:46:08 PM10/31/14
to connected-commu...@googlegroups.com
Hi Robert,
Pardon the delayed response, amid work and stuff. Reply-inline:
circuits together, thanks for the starting point. I deliberately did
not look at the formula you provided, since I wanted to derive it from
basics on my own. But please do verify my calculation below, any tips,
further points appreciated. FYI, my online reference has been
http://www.allaboutcircuits.com/
Example calculation:
Let:
Vin = Vtotal = 40V
R1 = 200 ohms,
R2 = 60 ohms
RL = 20 ohms
Then:
Rp = R2||RL = 1 / (1/60 + 1/20) = 15 ohms
Rtotal = R1 + Rp = 200 + 15 = 215 ohms
Max I = Vin / Rtotal = 40 / 215 = 0.186A
VdropR1 = I * R1 = 0.186 * 200 = 37.2V
Alternative, 'serialising' the circuit as R1 and Rp, current is the
same across both resistors in a series circuit.
I = Vtotal/Rtotal
VdropR1 = Vtotal/Rtotal * R1 => Vtotal * R1/Rtotal = 40 * 200/215 = 37.209
VdropRp = Vin - VdropR1 = 40 - 37.2 = 2.8V // based on KVL
Alternative:
Vtotal * Rp/Rtotal = 40 * 15/215 = ~2.8V
R2 and RL thus have 2.8V across each due to electrically common points.
I2 = 2.8 / 60 = 0.0467A
IL = 2.8 / 20 = 0.14A
Iptotal = I2 + IL = ~0.186A (equivalent to I across R1) // I believe
this splitting across junctions is KCL?
i.e. if RL was not in parallel, it would have gotten 0.186 rather just 0.14.
>
> Hint - R2||RL (equivalent resistance of R2 in parallel with RL) ~ RL for
> RL<< (much smaller than) R2.
Hmm... interesting... thanks. Thus RL must be really low to get most
of the current coming from Vout.
Digression: I have yet to figure out how the total resistance formula
for parallel resistance is derived... not important for usage, but
curious.
>
> Run through setting values for each and find why you believe the statement -
> "Resistor dividers make very poor power supplies."
So, essentially, because the load needing the reduced voltage will
effectively always be in parallel to the second/subsequent in-series
resistor, the current to it will always be branched, and relative to
the in-parallel resistances of the load and dividing resistor, you can
never get the same current as you would from a direct voltage source
(i.e. not using the divider to do the drop, making it in series) of
the same voltage.
>
> Oh, and you don't get to chose the resistance of your load in real life
> (indeed, it will dynamically change in this setup for anything not a simple
> resistor).
>
Understood, thanks again for helping me understand this. :-)
Pardon the delayed response, amid work and stuff. Reply-inline:
>> As in, over time, the current is not steady when using
>> resistor-based voltage dividers making it unfeasible for many loads?
>
>
> Vin-|
> |
> R1
> |
> -----.----- - Vout
> | |
> R2 RL
> | |
> -----.-----
> |
> GND
>
I did the numbers, and thus learnt about evaluating series + parallel
>> resistor-based voltage dividers making it unfeasible for many loads?
>
>
> Vin-|
> |
> R1
> |
> -----.----- - Vout
> | |
> R2 RL
> | |
> -----.-----
> |
> GND
>
circuits together, thanks for the starting point. I deliberately did
not look at the formula you provided, since I wanted to derive it from
basics on my own. But please do verify my calculation below, any tips,
further points appreciated. FYI, my online reference has been
http://www.allaboutcircuits.com/
Example calculation:
Let:
Vin = Vtotal = 40V
R1 = 200 ohms,
R2 = 60 ohms
RL = 20 ohms
Then:
Rp = R2||RL = 1 / (1/60 + 1/20) = 15 ohms
Rtotal = R1 + Rp = 200 + 15 = 215 ohms
Max I = Vin / Rtotal = 40 / 215 = 0.186A
VdropR1 = I * R1 = 0.186 * 200 = 37.2V
Alternative, 'serialising' the circuit as R1 and Rp, current is the
same across both resistors in a series circuit.
I = Vtotal/Rtotal
VdropR1 = Vtotal/Rtotal * R1 => Vtotal * R1/Rtotal = 40 * 200/215 = 37.209
VdropRp = Vin - VdropR1 = 40 - 37.2 = 2.8V // based on KVL
Alternative:
Vtotal * Rp/Rtotal = 40 * 15/215 = ~2.8V
R2 and RL thus have 2.8V across each due to electrically common points.
I2 = 2.8 / 60 = 0.0467A
IL = 2.8 / 20 = 0.14A
Iptotal = I2 + IL = ~0.186A (equivalent to I across R1) // I believe
this splitting across junctions is KCL?
i.e. if RL was not in parallel, it would have gotten 0.186 rather just 0.14.
>
> Hint - R2||RL (equivalent resistance of R2 in parallel with RL) ~ RL for
> RL<< (much smaller than) R2.
of the current coming from Vout.
Digression: I have yet to figure out how the total resistance formula
for parallel resistance is derived... not important for usage, but
curious.
>
> Run through setting values for each and find why you believe the statement -
> "Resistor dividers make very poor power supplies."
effectively always be in parallel to the second/subsequent in-series
resistor, the current to it will always be branched, and relative to
the in-parallel resistances of the load and dividing resistor, you can
never get the same current as you would from a direct voltage source
(i.e. not using the divider to do the drop, making it in series) of
the same voltage.
>
> Oh, and you don't get to chose the resistance of your load in real life
> (indeed, it will dynamically change in this setup for anything not a simple
> resistor).
>
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