Alternating potentiometers in Parallel to the base of an NPN Transistor

It’s that time of the week again where Jonny pretends he understands transistors. :partying_face:

See below a schematic.

Screen Shot 2024-03-27 at 2.20.44 pm

I have an JK Flip Flop driven by a clock such that when Q and Q-compliment oscillate HIGH then LOW.

Q and Qc then drive two variable resistors, such that if one is receiving volts, the other is not.
These then drive the bass of a tranistor (with a 1k in series to set a minimum).

The transistor drives an LED.

Let’s set up a scenario.

Assume that VR1 is all the way up, i.e. 10k and VR2 is all the way down, i.e. 0k

  • When Q is HIGH the base of the transistor is receiving 1.06ma (5v/4700Ω)
  • When Qc is HIGH the base of the transistor is receiving 0.34ma (5v/14700Ω)

I predict that the LED is oscillating in brightness because …
Ie = Ib + Ic thus when Ib increases Ic must decrease to maintain the equation.


  • Do I need diodes between VR2-R1 & VR1-R1?
  • Is my prediction true?
  • Given components only draw the amps they require how can I calculate the expected current at the emitter? Is that the kind of thing I would find in a data sheet? What is the keyword I’d hunt for?

Thanks in advance.
Pix :heavy_heart_exclamation:

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Hi Pix.
You are dead set on complicating things.

If you are going to do this the answer is YES. you have the potential to get a short between Q and Q- which would not do the IC much good.
One solution would be to use ganged pots and connect so there is no possibility of a short.
Another is to use 1 pot and connect the ends to Q and Q- and the slider to R1. This would achieve your purpose (whatever that is) but the transistor would not change much if at all.

What is R2 doing except raise the emitter voltage when collector current increases. That has to be added to R3 for calculating So the LED current limiting resistor now becomes 400Ω instead of 300Ω.

That is true up to a point. But the Ib CONTROLS the Ic. An increase in Ib will result in an increase in Ic by a factor equal to hfe (or DC current gain) until the transistor saturates and turns hard on. So if you turn one side up and the other side down the transistor will not change much at all.

0.34mA could be enough to turn the transistor full on. So no change.

Doubt it. The voltage drop across R2 will always try to be about 0.6V less than base voltage so I just can’t think at the moment what is likely to happen. You have in effect an emitter follower circuit with a load in the collector circuit. Not sure of what actually happens here a bit of chicken and egg situation I think.

Do you want to switch the LED on and off. If so just remove R2 and the 2 pots and connect a suitable value resistor to Q or Q- and switch the transistor on and off.

If you want the LED to change brightness just do the above and connect a suitable resistor across the transistor from emitter to collector. The value of this resistor chosen to allow the LED to glow dimly with the transistor OFF.
Cheers Bob
PS. Read up a bit more on transistor operation.

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Haha! I am my own worst enemy.

Some Context.

Below is the example I’m working off.
This is from The Art of Electronics.
I feel comfortable with this example.

What I want to work with is an collection of D-latches where the output of each latch is in series with a potentiometer. All those pots lead to a transistor and I’m wanting to amplify the current. I can change the brightness of my LED by changing which latch I turn on (where I’ll need to ensure only one can be on at a time).

I believe this is possible and my toy example above is my attempt at a simplified version of my final idea (start small).
All I feel I have done is swap out the 10k resistor in AoE example with two variable resistors.
I’ve then put a Latch in the middle to ensure that only one is on at a time :slight_smile:

I’m not disagreeing with you I’m just playing devils advocate to help me understand.

  • Ie = Ib + Ic
  • BUT *Ib controlls Ic

Wouldn’t Ie grow arbitrarily large as Ib receives more and more current.

If Ie is variable; and I just want to amplify current, wouldn’t this diagram, where the LED is being controlled at the Emitter make much more sense? (since if Ib grows then Ic grown then Ie grows).

Screen Shot 2024-03-27 at 5.32.35 pm

Again, not saying I’m right, rather “I know I’m wrong, what assumption have I made that means I’m wrong?”

(thanks again Bob).

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Ah ok, maybe I should be looking into much higher value for R1.
I’ll check the data sheet for my transistor and pick a more conservative value.
Hopefully I can find a transistor where 20-30kohms gives me a good range of values.

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Hi Pix

Yes until the transistor reaches saturation.

You could try it like in that circuit but you might have to transfer R2 and the LED to the collector side and raise R2 to about 330Ω

Be careful you don’t ever get VR1 and VR2 at minimum or you will get a short between Q and Q- and probably destroy the IC. Or put a 1N4148 diode in series with each leg.
Cheers Bob

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I just understood saturation!

Why would I choose the collector side over the emitter side? Aren’t they proportional to one another?

Yep that. makes sense to me. Why that specific diode?

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Hi Pix

You will have to read up on different transistor configurations. There are 3 basic, common emitter which is the most common, common collector which you have here (also known as emitter follower) and common base which is the least used. There have been volumes written about this and is too long winded to conduct a course here. To be truthful these days I would have to look up most of the details.

In a nutshell you have 3 voltage drops, across the transistor (varies), across the LED (fairly constant value depends on colour) and across R2. As the current Ic rises so does the voltage across R2 and the LED thus trying to turn OFF the transistor. Think of it as nearly 100% negative feedback. This makes the input resistance at the base seem very high (the advantage of this circuit) with unity voltage gain or actually a small loss but you can have a power gain.

All very confusing and I am not sure if you will light the LED as you want. If id doesn’t move R2 and the LED as I suggested.

Usually because they are a good fast small signal diode and very suited to this sort of application. Used to be known as 1N914 and so common they are sold in packs of 100 for about $9. If you haven’t got some of these in your parts bin you should have. “Normal” diodes such as the 1N400X series are not much good for fast switching and when the frequency gets up a bit they look just like a piece of wire.
Cheers Bob


Ah! Those are excellent google search terms.
I have seen a section on emitter follower in the AoE. Maybe that’s a chapter I’ll read up on before proceeding.

Anyway, I think I get the ideas here.
My mistake was that I assumed the current at the emitter would be a fixed value but that’s not the case.

Thanks Bob.
Always so helpful.

Pix :heavy_heart_exclamation:

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This is what I think I’ll attempt to build.

Screen Shot 2024-03-27 at 8.26.06 pm
I’ll let everyone know if I’m successful.
Not sure what Transistor I’ll use yet… maybe a 2N2222…

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Good as any, pretty robust.
Cheers Bob

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Lets see if I understand.
A particular D-latch will be activated to change the brightness of the LED or the current to some other device. The potentiometers are to set the amount of current for each D-Latch. And the test circuit is to simulate two D-latches, one on the other off.

In my opinion feeding this to one transistor will not work very well, each potentiometer will affect the others. Using a separate transistor for each D-latch would be better. The common point would then be the LED connected to the load resistors of each transistor.



Rob and I spoke about using 1N4148 Diodes to isolate the variable resistors from the base of the transistor. You’re idea would work as well since the transistor will “act like a diode” when the output of the latch is low. Maybe when I come to bread boarding this thing I’ll experiment with both… it’s not like transistors are expensive…

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Ran out of time with previous post (RL commitments). Wanted to say the following.

The diagram posted earlier from a book shows the transistor being used as a simple switch (On and Off). Varying the current drive of a transistor to vary the intensity level of a LED is more tricky.

The following is the BD135 graph of base/emitter voltage to collector current.

0.6V generates about 2mA, 0.65V about 20mA, 0.71V about 100mA.
My point is the voltage range is quite small and would require careful selection of components.
Some biasing might also be required to ensure the voltage level on the base is within the range.

Another consideration is the varying hfe of transistors, even within the same batch. For the BD135 the data sheet says it can vary from 25 to 250. Therefore the circuit should be designed so varying hfe values do not affect circuit operation. Changing a transistor should not change the way the circuit works.

It is beyond my electronics engineering skills to analyse the circuit and how it would work. There is software you can put the circuit into and it will show voltage and current levels and signals.



Hi Pix
As James just pointed out the hfe of transistors varies over quite a large range. So if you find a resistor combination that works with one transistor it may not work with another one.

The solution to this is to provide negative feedback to take control of this gain…This is provided with a resistor from collector back to base. To do this the load (LED and R2) must be transferred to the collector circuit. Emitter Grounded.

There is one bit of a problem. This provides control over the VOLTAGE gain with the collector as the output point. Now the LED is a current operated device that will have a reasonably steady forward voltage drop with changing current so the only bit that will change is the drop across the resistor. The system is normally used with a resistor as the load but you might be complicated by the addition of a virtual static drop across the LED.

The gain would be feedback resistor divided by source resistor, similar to OpAmp behaviour. The source resistor in this case being the series base resistor.

Say you want a current of 10mA.
R2 would be (Ω) (5V minus (LED forward volt drop + Transistor ON voltage drop)) / 10mA (0.01)
The LED drop will depend on colour.
Transistor drop saturated could be about 0.2V
Which would work out to be somewhere about 180Ω to 270Ω.

If we work on a gain of 10 then for 100kΩ feedback resistor use a base resistor of 10kΩ and increase the base resistor to reduce gain (and hopefully collector current).

You will have to experiment here as this is not a normal use. A LED is usually ON or OFF. You will find the current quite critical when trying to use this method to control brightness as along with any LED non linearity you have the human eye to contend with. That is why LED brightness is usually controlled with a PWM signal.

The feedback resistor / source resistor combination is used to control the gain of say an audio pre-amp but the transistor would have a resistor load, no LED.

As James just pointed out this whole thing is pretty iffy. A simulator might show something and if I get time I might try that. But I think an experimental build would be the best bet.
Cheers Bob


Hmmm. That’s great info.
We all know my love of 𝖀𝖓𝖓𝖊𝖈e̷̡͖̩͉̅̀ͅ𝖘𝖘𝖆𝖗y̶̭̑͌ 𝕺𝖛𝖊𝖗-𝕮ǫ̷̢̙͒̈̎𝖒𝖕𝖑𝖎𝖈𝖆𝖙𝖎𝖔𝖓 but now I think my design isn’t just over designed it isn’t robust and has too many conditionals.

Ok this might work. I don’t mind my load going back to the collector side. :slight_smile:

Sure. There is nothing special about this LED.
How about instead of using the transistor amplifier to dim an LED I drive the frequency of a 555 timer in astable configuration? Is that a more appropriate indicator? I now have a multi-meter that can measure frequency :slight_smile:

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Pix haven’t you read the previous posts. There is something special about all LEDs. Each different colour has a different forward voltage drop. When using 24V or 48V or some of the larger values this difference can be swallowed up. But when the voltage is only 5V this is a significant portion as LED voltage drops can vary between 1.8V and 2.8V or roughly half what is available. Thus calculating for this is essential.

The upshot is don’t go swanning off brushing this aside as if it makes no difference. No resistor recommendations have been made for a current limiting resistor up to date as you have not specified what the colour is and what current you would like to use.

The use of diodes poses the same problem as it doesn’t take many of these coupled with LEDs etc to swallow up the whole 5V.

The problem gets very much more critical when using the now popular 3.3V supply.

I have a very good reason for sticking to 5V for any experimental work as it proves to be less troublesome when you start introducing some of these bits. Can convert to 3.3V when everything is sorted if required.
Cheers Bob

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I’m sorry Bob I wasn’t clear.
I was trying to communicate that an my use of an LED was just as an indicator for success. That is, i just used an LED to test whether the circuit was working or not. I could use anything as a measure. :slight_smile:
I’m not implying that all LEDs are the same in practice. I’m aware that different colors and styles behave differently. :slight_smile:

Here I’m listing an alternate indicator to the LED.

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Hi Pix
Just had a look at the data sheet for the flip flop 74LS73. You have 'Clear" pin 6 grounded. Nothing will work.
Cheers Bob
Clock to pin 5.
Pins 6, 7, 10 connect to VCC (5V)
Outputs will toggle on the falling edge of the clock signal.

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I think @Pixmusix made the assumption that when an output of 74LS73 is HIGH, it delivers current but when LOW it does nothing. In fact, a LOW will sink current if there is a source to sink from. Hence @Robert93820 warning not to wind both VR1 and VR2 to minimum. Current will flow from whichever Q is high to the one that is low, effectively a short circuit. That’s the reason for the recommendation to add diodes to prevent current flowing into the low Q.

The whole circuit is messy to analyze because the transistor has an unknown current multiplier (Hfe) effect. A solution is to do away with the transistor altogether and connect the potentiometers directly to the LED. A 74LS73 can source less than 1mA, maybe not enough for satisfactory brightness (although I use scavenged christmas light LEDs with 0.5mA and they are quite bright). It can sink 8mA which should be enough to light most recent LEDs. So connecting the LED to the +5V via 270 ohm (to limit the maximum possible current), and having a diode and potentiometer to each Q from the LED should work fine. The maximum current can be worked out as 5V - drop through LED (at least 1.8) - drop through the protection diode (0.6) - internal drop through Q (nominal 0.6) = about 2V across 270 ohm = 7.4mA. The current through the LED now goes to whichever Q is low.

I’m also not sure if transistor saturation is explained. A transistor will sink through the collector connection whatever the base current is multiplied by Hfe. So if 20uA (microamps) go into the base, and the Hfe is 100, the transistor will sink 2mA through the collector regardless of the voltage applied to the collector with the caveat that the collector has to have a voltage a little higher than the emitter. The transistor acts like a variable resistor adjusting to whatever resistance limits the current to 2mA. And as a resistor, it dissipates heat. If the base current is increased enough, and there’s enough voltage on the collector, the transistor will destroy itself.

But transistors usually have some sort of resistive load in the collector circuit. As the base current increases, the collector current increases until the voltage drop across the load is great enough that the collector voltage is only slightly higher than the emitter voltage. At that point, increasing the base current has no effect, the transistor is almost a short circuit. That’s called saturation.

Hope this is helpful.


Hiya @Alan73922
Thanks for taking the time to respond.

Diodes and 74LSxx chips

The way I understood it: HIGH means the 74LSxx chip is routing the output to power (i.e. 5v) and LOW means the 74LSxx chip is routing me to ground. Is that the same thing as what you’re saying here or is that not quite right?

This was the main reason for starting this post. Just knowing that diodes are sufficient for the task helped me make progress. :slight_smile:

Gotcha. That’s a helpful summary. Thanks :+1:

Transistor as current amplifier

Here is a toy problem I’ve whipped up to test if I understand this. I’ve removed the Dlatch to keep things focused. Additionally, this time instead of using the brightness of an LED as measure I’ve made an Astable 555 timer circuit and we are using the current at the collector to drive the frequency Q square wave.

I’ve picked a PN100 tranistor since it has an Hfe of about 100 like the example you gave.

With the base of the resistor can receive a range of 0.045 - 0.05 ma of current.
That means the collector (before R2) has a range of ~0.045 * hfe(100) = ~4.5ma

Here the idea I want to test that I don’t think can possibly be true
If the collector is following the base current, is my PN100 modifying the amount of current that can flow at Pin6 of the 555 timer?

i.e. Is my transistor acting as a resistor in this circuit? Is that what we mean when we talk about transfer of resistance? 5ma at 5v is like… what ~1000ohms?

For me, the 555 timer is NOT important here, it’s just a way to measure the change in impedance because, as bob pointed out," > A LED is usually ON or OFF. ".

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