# Led and Resistor in parallel, can you explain it?

#### JosXD

Joined Mar 16, 2022
63
Hello,

Today I was watching a gif image of a transistor NOT Gate

They used a npn transistor, but it seems that here the output is not actually the emitter pin, it is the Led.

I was thinking the only way to turn the Led off was to equal the voltage of both led's pins, but then I figure it out that the npn transistor might have a small resistance from colletor-emitter that is in series with a 1k ohms resistor which causes a big drop voltage when transistor in fully on.

So I tried to do a similar exercise with the stuff I have at home, I don't have that kind of transistor so I just used 2 resistors, one led and 3NiMh batteries in a battery holder.

I calculated first the voltage at the resistor divider point by removing the led of the circuit: (with Dv I mean drop voltage not sure if is correct)
Rt = 680+2.2k = 2880 ohms
I = 3.6V / 2.88k Ohms = 0.00125A
R1Dv = 0.00125A * 680 omhs = 0.85v
R2Dv = 0.00125A * 2.2k omhs = 2.75v
R1Dv + R2Dv = 3.6v

So in the resistor divider point the voltage is 2.75v without the led attached to the circuit, was 2.75-2.76v measured with the multimeter, then I connected the RGB 2 pin led, sometimes voltage in the divider point was 2.63v and sometimes 2.39v and 2.08v due the forward voltage of each color tho.

What I would like to know it is how the current flows, I mean at the startup does the current first flows throught the led or R2? how it is stablished the voltage in the resistor divider point when there is a led in parallel to R2 out of formulas, is there a kind of visual flow of electricity for this example like those of water?

It is a bit confusing I can think first it is defined by the resistor divider and if there is enough voltage to turn on the led, then the led redefines the dropout voltages with its forward voltage.

If I change the R1 value to 970 omhs and R2 to 680 omhs, the voltage in the divider point is 1.482v so in this case the led doesn't turn on.

I'm not an electrical or circuit engineer.

Last edited:

#### Ya’akov

Joined Jan 27, 2019
8,973
Welcome to AAC.

You might be overthinking.

If you remove the transistor and is base resistor from the circuit what do you get?
If you replace the transistor and resistor with a wire from the 1kΩ resistor to the low side of the battery, what do you get?

#### LowQCab

Joined Nov 6, 2012
3,937
You can not replace a Transistor with a Resistor and get the same results.
When the Transistor is fully turned-on by adequate Base-Current,
the Voltage across the LED will be somewhere around ~0.6-Volts.
.
.
.

#### crutschow

Joined Mar 14, 2008
34,047
I'm not an electrical or circuit engineer
I figure it out that the npn transistor might have a small resistance from colletor-emitter that is in series with a 1k ohms resistor which causes a big drop voltage when transistor in fully on.
I assume you pulled the value of 2.2kΩ for R2 that replaces the transistor from thin air?
No discrete bipolar transistor you can buy has anywhere near that high a resistance when fully on (saturated).
I suggest you look at a typical transistor data sheet, such as the common 2N3904, and see the collector-emitter saturation voltage.

#### Tonyr1084

Joined Sep 24, 2015
7,777
The reason why the LED shuts off when the transistor is on is because the transistor provides a pathway back to the battery negative terminal. Current will follow the path of least resistance. When the transistor is on the LED is off because all the current is being directed past the LED, not through it. When the transistor shuts off then the only path back to the battery negative terminal is through the LED, and the LED illuminates.

The transistor is merely a switch. Granted, it's not a perfect switch and you WILL see some voltage across the LED when the transistor is on but that's because of the PN Junctions. There's always a voltage drop of some measurable value.

#### WBahn

Joined Mar 31, 2012
29,864
If you want to, very roughly, approximate the transistor with a parallel resistor, then you need to use reasonable choices -- one value for when the input voltage is 0 V and a different one when it is 9 V. The exact values aren't critical, but they need to be in a proper range.

For the case when Vin = 0 V, the only current flowing in the transistor collector is leakage current, which is typically well under 100 nA. If your LED has a forward voltage of 2.6 V, then this would be equivalent to 26 MΩ. So just use the largest resistor you have -- even a 100 kΩ resistor should work fine. Or just remove it entirely.

For the case when Vin = 9 V, the transistor will be in hard saturation and the Vce of the transistor will be in the 200 mV range (more like 50 mV in practice), which can be approximated using a 0.03 Ω resistor. So just use the smallest resistor you have -- anything less than 100 Ω should work. Or just use a wire.

These values are for the original circuit (the one with the 9 V battery). For your circuit with the 3.6 V battery, the resistor values would be different, but the end result is the same -- use a very high value resistor when the transistor would be off, and a very low value resistor when the transistor would be on. As a rule of thumb, make the first at least 100x the collector resistor (so >68 kΩ) and the second no more than 0.01x that value (so <6.8 Ω, though 10 Ω should work okay).

#### dl324

Joined Mar 30, 2015
16,684
I was thinking the only way to turn the Led off was to equal the voltage of both led's pins,
To turn off, the voltage across the LED doesn't need to be 0V. It just needs to be low enough for the number of photons emitted to be zero/undetectable.
but then I figure it out that the npn transistor might have a small resistance from colletor-emitter that is in series with a 1k ohms resistor which causes a big drop voltage when transistor in fully on.
I wouldn't recommend thinking of a transistor this way (even though transistor is a contraction of transresistance). When the transistor is off, ignoring leakage current, current doesn't flow through the transistor and all of it goes through the LED. When the transistor is on, it conducts virtually all of the current through the 1k resistor. Depending on the transistor, it will have a collector-emitter saturation voltage of a tenth of a volt, or less, at a collector current of 9mA. That should be sufficient to turn the LED off.
So I tried to do a similar exercise with the stuff I have at home, I don't have that kind of transistor so I just used 2 resistors, one led and 3NiMh batteries in a battery holder.
You can't replace a transistor with passive devices. That circuit isn't special and any general purpose NPN transistor can be substituted.
They used a npn transistor, but it seems that here the output is not actually the emitter pin, it is the Led.
The collector is the output. It wouldn't make sense for the emitter to be the output because it's grounded and can't change.

#### JosXD

Joined Mar 16, 2022
63

I think I explained it wrong,

I'm not going to build any circuit so I won't replace the transistor with a resistor, I just wanted to understand the concept of that part where you bypass the current from the led.

And yes it depends if the voltage at certain resistor divider point it is enough to turn on the led, and back to the transistor I remember mosfets have a low resistance between drain and source, so I thought the same is for the npn transistor between collector and emitter, that path should have a low resistance, so there is a dropout voltage enough to not turn the led on ( Image 1 when transistor is hard on ).

What I would like to know, out of formulas, in reality where the current will flow first between R2 or the Led? ( in the second image ).

#### dl324

Joined Mar 30, 2015
16,684
I'm not going to build any circuit so I won't replace the transistor with a resistor, I just wanted to understand the concept of that part where you bypass the current from the led.
It still doesn't make sense to use resistors or think of the transistor being a resistor.

When the transistor is off, essentially all of the current flows in the LED. I say essentially because a non-ideal transistor has leakage current, but we ignore it for first pass analysis.

When the transistor is on, essentially all of the current flows in the transistor.
What I would like to know, out of formulas, in reality where the current will flow first between R2 or the Led? ( in the second image).
Analysis is iterative. You make some assumptions. If they don't bear out, you adjust them.

If you assume that the LED is on and that its forward voltage is 2V. That give
$$I_{R1} = \frac{V_{R1}}{R_1} = \frac{3.6V-2V}{680\Omega} = 2.3mA$$

At that current, the forward voltage of the LED would be more like 1.8V:

Since the assumption for LED voltage drop is wrong, the calculations need to be redone with 1.8V across R1. That will give a different forward voltage for the LED, and a different current in R2. That iteration, or the next, should be close enough and I leave the exercise to you.

Normally LED color affects the forward voltage range, but this Siemens LED gives the same range regardless of color:

Last edited:

#### WBahn

Joined Mar 31, 2012
29,864

I think I explained it wrong,

I'm not going to build any circuit so I won't replace the transistor with a resistor, I just wanted to understand the concept of that part where you bypass the current from the led.

And yes it depends if the voltage at certain resistor divider point it is enough to turn on the led, and back to the transistor I remember mosfets have a low resistance between drain and source, so I thought the same is for the npn transistor between collector and emitter, that path should have a low resistance, so there is a dropout voltage enough to not turn the led on ( Image 1 when transistor is hard on ).

What I would like to know, out of formulas, in reality where the current will flow first between R2 or the Led? ( in the second image ).
Since the two devices are in parallel, they will always have the same voltage across them and the total current will split between them.

In theory, there will always be some current in each of them. But if the voltage is below the Vf of the LED, the current in the LED will be negligibly small. For a normal diode, a common rule of thumb is that the current will change by an order of magnitude for each 120 mV of change in forward voltage. I don't know how well that applies to typical LEDs, let along the RBG one that you have (do you have a manufacturer and part number for it). It's probably in the ballpark at low currents, but likely deviates considerably at higher currents. But let's go with it.

Say that the LED has 10 mA at a Vf of 2.24 V. If that voltage is lowered to 2 V, the current will be decreased to 100 µA, At 1.76 V, it will be down to 1 µA. Lower it to 1 V and you are in the vicinity of 1 pA.

From a practical perspective, you can probably call it zero current by the time you get to 2 V.

#### MrChips

Joined Oct 2, 2009
30,486
Think of the transistor as an ideal SPST switch.
When the switch is open, all the current flows through the LED.
When the switch is closed, all the current flows through the switch.

#### Tonyr1084

Joined Sep 24, 2015
7,777
A few years ago I made a video where I powered several parallel LED's through a single resistor. In this video I show what happens when you redirect current through other devices. In the video's case the "Other" devices are also LED's with lower forward voltages. When you have two components (LED's in the video) in parallel and each depends on certain parameters then one will hog all the current and effectively shunt the other one off.

I start with red LED's in parallel then add a green LED. The green LED doesn't light because all the current is being shunted through the red LED's. When the red LED's are removed from the circuit then and only then does the green LED light.

For those who have seen the video before and those who see it for the first time - I do not advocate the use of parallel LED's and a single control resistor. It's just not the right way to do it. Yeah, I did it and it worked, but it's still not the right approach. Each LED should have its own dedicated resistor in series with its associated LED. That's the right way. Nevertheless, assume the red LED (with the lowest forward voltage) is doing the same thing the transistor is doing in your schematic. The transistor has a forward voltage drop of about (likely) 600mV (0.6Vf) whereas the green LED has a 2.92 Vf average. The current is going to take the path of least resistance.

#### BobTPH

Joined Jun 5, 2013
8,664
What I would like to know, out of formulas, in reality where the current will flow first between R2 or the Led? ( in the second image ).
It does not work that way. Current flows immediately in both devices. The current divides such that both have the same voltage across them.

Why do you think it would flow first in one or the other?

#### JosXD

Joined Mar 16, 2022
63
Okay guys so the forward voltage of the led will vary depending on the current throught the led?

I was doing something similar like Tony's video, but a 1kohms as R1 and then 2 parellel componentes a warm white led and a 680ohms as R2, when everything was plugged into the breadboard the led was off, but inmmediately I removed R2 the led went on, then I plug R2 and led went off.

And BobTPH that's a wonderful question, it is because lack of undestanding, as I read for example an specific led a "Cree XHP50.3 3V" it says tipical forward voltage 2.8V and I thought you need strictly a minimum of 2.8V to turn it on or to start drive a current thought it.

To that sum it I was thinking the series resistor should solved first to set a voltage divider, also was thinking as the 2.8V Forward Voltage as a barrier, but yes weird thoughts myself knew it was wrong, 'cuz current won't do a thing first and also the electricity initially comes with a pressure of 3.6v enough to go trought the led, but know I know higher or lower than a certain led FV it will cross the led just with different currents and if it too low the led won't light.

This was the experiment I did today:

With my multimeter, the meter measured 1.47V in the resistor divider, led was off, I remove R2 and ofc Led went on, plugged R2 again Led went off, If the led and R2 were not part of the circuit then R1 attached to gnd, the circuit will draw ~3.6mA.

Voltage at resistor divider remains the same 1.47V doesn't matter if the led is plugged or not.

I think okay across R1 we can supply 3.6mA but then oh! another resistor, now forget about the led, so I sum R1+R2 then 3.6V/1.68Kohms = ~2.14mA, without the led there will be a current 2.14mA in the circuit.

Now here comes the tricky part for me, let's add the Led again, now there is another path where current can flow throught, so I think 3.6mA can be supplied after R1 and would mean the led will take 3.6mA-2.14mA = 1.46mA, but that's wrong because the led is off.

Can I calculate that circuit If I don't know the forward voltage of the led?

Last edited:

#### MrChips

Joined Oct 2, 2009
30,486
No. You cannot use simple math and Ohm's Law to calculate the current or voltage in that circuit.
That is because the LED is not an ohmic device. It does not follow Ohm's Law.

The voltage across R2 is 1.46V with or without the LED because the LED is not conducting any appreciable current.

#### dl324

Joined Mar 30, 2015
16,684
so the forward voltage of the led will vary depending on the current throught the led?
Yes.

Here's a graph from an old HP databook. I prefer this graph because it shows the exponential nature of the forward voltage, in a traditional way, and that it's similar to other, non-light emitting diodes.

This is what Siemens prefers:

#### MrChips

Joined Oct 2, 2009
30,486
Your math and logic is flawed.

Yes, a 1kΩ resistor can pass 3.6mA from a 3.6V supply, only when the total resistance is 1kΩ.

Read this blog and gain some insight into the meaning and usefulness of load line.

#### MrChips

Joined Oct 2, 2009
30,486
As an experiment, leave R1 = 1kΩ
Try increasing the value of R2 in steps until the LED turns on.
At each step, measure the voltage across R2 (which is the same as the voltage across the LED).
With this information, you will be able to plot the LED current vs LED voltage.

#### WBahn

Joined Mar 31, 2012
29,864
Okay guys so the forward voltage of the led will vary depending on the current throught the led?

I was doing something similar like Tony's video, but a 1kohms as R1 and then 2 parellel componentes a warm white led and a 680ohms as R2, when everything was plugged into the breadboard the led was off, but inmmediately I removed R2 the led went on, then I plug R2 and led went off.

And BobTPH that's a wonderful question, it is because lack of undestanding, as I read for example an specific led a "Cree XHP50.3 3V" it says tipical forward voltage 2.8V and I thought you need strictly a minimum of 2.8V to turn it on or to start drive a current thought it.

To that sum it I was thinking the series resistor should solved first to set a voltage divider, also was thinking as the 2.8V Forward Voltage as a barrier, but yes weird thoughts myself knew it was wrong, 'cuz current won't do a thing first and also the electricity initially comes with a pressure of 3.6v enough to go trought the led, but know I know higher or lower than a certain led FV it will cross the led just with different currents and if it too low the led won't light.

This was the experiment I did today:

View attachment 308957

With my multimeter, the meter measured 1.47V in the resistor divider, led was off, I remove R2 and ofc Led went on, plugged R2 again Led went off, If the led and R2 were not part of the circuit then R1 attached to gnd, the circuit will draw ~3.6mA.

Voltage at resistor divider remains the same 1.47V doesn't matter if the led is plugged or not.

I think okay across R1 we can supply 3.6mA but then oh! another resistor, now forget about the led, so I sum R1+R2 then 3.6V/1.68Kohms = ~2.14mA, without the led there will be a current 2.14mA in the circuit.

Now here comes the tricky part for me, let's add the Led again, now there is another path where current can flow throught, so I think 3.6mA can be supplied after R1 and would mean the led will take 3.6mA-2.14mA = 1.46mA, but that's wrong because the led is off.

Can I calculate that circuit If I don't know the forward voltage of the led?
Forget the 3.6 mA. The only way to get 3.6 mA in the circuit is if the ONLY component between the supply and ground is the 1 kΩ resistor.

The forward voltage of the LED isn't going to change by much over a significant span of currents. You will usually see more variation from one LED to another.

Is the LED you are using that CREE that says the Vf is typically 2.8 V? Let's assume so.

First, figure out the voltage across where the LED will go but without the LED. You've already done that and got 1.46 V. That means that when you put the LED across R2 that the voltage is too low and no appreciable current will flow in the LED.

Now let's choose something higher, like a 4.7 kΩ resistor.

Without the LED, the voltage across R2 would be about 2.97 V. That means that when you put the LED across it, that there is enough voltage for current to flow through it. But it will clamp the voltage across it to something close to 2.8 V. Let's assume that it does. This means that there is (3.6 V - 2.8 V) = 0.8 V across R1, which means that about 0.8 mA is flowing through it. There is also 2.8 V across R2, which means that 0.6 mA flowing through it, which leaves the remaining current of 0.2 mA flowing through the LED.

If you remove R2 completely and just have R1 and the LED, then all 0.8 mA will flow in the LED, which may or may not be enough to see any light coming out of it.

Your 3.6 V source voltage is really too low to get good current regulation in an LED that is going to have 2.8 V across it. Raising that to 5 V or more would be much better (though it will consume more power at the same current).

#### JosXD

Joined Mar 16, 2022
63
Okay guys I did again the circuit with different resistors values, now R1 = 680omhs and R2 = 2.2Komhs, led remains the same warm white, without the Led multimeter measured 2.76V in the resistor divider, when plugged the Led into bread board the voltage sag to 2.53V, this is confuse tbh but yeah that's how the Vfw Led works.

When you say an ideal Led you mean a perfect Led on paper with a exact Vfw at what it will turn on?

I was reading the Thevenin's theorem, but haven't made a calculation yet.

dl324
The HP's grapichs looks way easier to understand.

MrChips
I don't have so many Resistors values, I should buy some potentiometers to try, but yes I have a larger value R to turn it on.

WBanh
I understand the calculations you did there now, so according to the measures my meter did when led is connected, 3.6V-2.53V = 1.07V, so 1.07V across R1, this will allow a current throught R1 of 1.07V/680ohms = 1.573mA, since V across Led is 2.53v for R2 is the same 2.53V/2.2Kohms = 1.15mA, 1.573mA-1.15mA = 0.423mA left and that goes to the led.

I think it's starting to make sense although consfusing, and let's asume I didn't have a circuit, led datasheet, and a meter, just a exam paper with a schematic, a pencil and a calculator, there we will need an ideal led to solve it right?

I don't know If I'm an extreme case or if it's normal for a newbie.

Last edited: