# Comparing two circuits for amplifying a signal to turn on LEDs (NPN vs PNP transistors)

Discussion in 'Analog & Mixed-Signal Design' started by seanspotatobusiness, Sep 28, 2016.

Sep 17, 2016
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Hi, I have two circuit diagrams. One circuit someone else designed using two NPN and one PNP transistors and it amplifies a signal to fade on an arrangement of LEDs. I changed it so the values of the resistors is probably no longer suitable. I don't know why they used a PNP transistor instead of using three NPN transistors so I drew the second circuit diagram which I think would do the same thing but I don't know enough about electronics to be confident. Can someone else advise why to use a PNP transistor?

2. ### AlbertHall Well-Known Member

Jun 4, 2014
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In the circuit with the PNP transistor, when at least one of the inputs is at 2.3V the PNP transistor will be turned fully on and its collector will be within 0.5V of the 15V supply.
In the circuit with only NPN transistors, from the inputs at 2.3V you will lose at least three times a diode forward voltage (input diode plus two base emitter voltages) so the maximum voltage at the emitter of the second transistor will be 0.2V! This is not enough to turn on the third transistor at all. It won't work.

Sep 17, 2016
32
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Hi AlbertHall. Thanks but I'm not sure I understand. I thought I had organised the NPN transistors in a Darlington arrangement so they sequentially amplify the signal? Did I screw it up? I guess I still don't understand transistors as well as I had thought.

4. ### AlbertHall Well-Known Member

Jun 4, 2014
2,266
448
They are connected as a darlington pair but you are taking the output from the emitter and so the output voltage will be less than the input, though more current is available.

5. ### #12 Expert

Nov 30, 2010
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7,331
The way I calculate the second circuit, the voltage on the capacitor will never exceed 0.5 volts. Phooey. No light. The third transistor never comes on.
The first circuit is what I call a double invert or a Sziklai Darlington. That one will apply almost all the 15 volts to the 5k resistor feeding the capacitor. Bingo. Let there be light!

Then we get to the 10 ohm resistors. Not smart to allow so much credence to LED specifications as to expect them all to be 2.90 volts. You are allowing 0.2 volts for the resistor and it should be at least 10% of the power supply. You need to take out one LED in each string and adjust the resistor to 180 ohms.

Then check the current and see if 180 ohms will keep the current down to the safe level for the LEDs.

atferrari likes this.

Sep 17, 2016
32
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Would changing the values of or removing any of the resistors make the first circuit work?

I don't actually know the forward voltage of the LEDs I want to use yet but I guess I can subtract one in series and add another row in parallel and then use a higher value of resistor. Thanks.

7. ### AlbertHall Well-Known Member

Jun 4, 2014
2,266
448
The first circuit works (except, as noted, the LEDs and their resistors).
If you meant the second circuit then, no, changing resistor values will not make it work.

8. ### #12 Expert

Nov 30, 2010
16,693
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Why do you have that capacitor in there? Without that, you need 1200 ohms between the second and third transistor.
With the capacitor, you are creating a 1/2 second delay and starving the third transistor of current with 10,000 ohms.
Why? If you want a delay, you should install it between transistor 1 and transistor 2.

Sep 17, 2016
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Yes, I meant the second circuit! I made some changes so now it looks like this:

According to this circuit simulation, it should work like this: http://everycircuit.com/circuit/6159582916771840

The capacitor is meant to make the light fade on and off.

10. ### AlbertHall Well-Known Member

Jun 4, 2014
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Your simulation does not include the diode at the input. Try it with that diode in place and see what happens.

11. ### Alec_t AAC Fanatic!

Sep 17, 2013
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There is also the matter of the voltage drop across the 1k resistors. This is significant even if you assume an optimistic current gain of ~100 for the transistors.

12. ### #12 Expert

Nov 30, 2010
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After being told 4 times that three base-emitter junctions and a diode in series with 2.3 volts won't work, you insist on using that configuration.

panic mode likes this.

Sep 17, 2016
32
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I only insist on understanding why it won't work. I want to understand the circuits I'm using. I want to see how and why it fails.

14. ### AlbertHall Well-Known Member

Jun 4, 2014
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448
Each diode or base emitter junction needs 0.6V to 0.7V to make it conduct. Your circuit has a diode and three base emitter juntions with just 2.3V available. That's not enough voltage to make them all conduct.

Put the input diode in your simulation and look at the voltages.

Sep 17, 2016
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I did that and indeed, it did not work! However, I measured my PIR sensor outputs and they are 3.3 V. I also have some diodes with a voltage drop of 0.2 V. I will probably be using the PNP version anyway, What I don't understand is why adding more transistors in Darlington pairs causes the final output to be too low when the whole point of Darlington pairs to amplify a signal, not absorb or block it? How do you use a Darlington pair to actually do what it's supposed to do?

16. ### crutschow Expert

Mar 14, 2008
13,498
3,374
A Darlington pair amplifies current, not voltage.
It actually has an AC voltage gain of slightly less than one.
But to turn on (bias) a silicon transistor requires about 0.7V per transistor, so for three transistors in a Darlington triple, 0.7V *3 = 2.1V bias is required to turn them on.

Don't confuse AC gain with DC bias requirements.
They are two different animals.
You have to satisfy the DC bias requirements before you get gain.

17. ### #12 Expert

Nov 30, 2010
16,693
7,331
I'll do the math for you:
0.6V + 0.6V + 0.6V + 0.6V = 2.4V
You need 2.4 volts just for the silicon with nothing left over for current gain.
2.4V is more than 2.3 volts.

Now you've changed the specification to 3.3 volts input and a 0.2 volt diode.
Now you can get 1.3 volts across the last 1k resistor for a current of 1.3 ma.
Using a transistor for a switch provides a gain of 10 to make it switch hard, so you can efficiently supply 13ma.
You want 6 strings of 4 LEDs which require 120 ma
120ma is more than 13 ma.

Would you like to try modifying the second circuit, again? Or would you like to switch to the other circuit?

Last edited: Sep 28, 2016
18. ### crutschow Expert

Mar 14, 2008
13,498
3,374
Albert E supposedly said “The definition of insanity is doing the same thing over and over again, but expecting different results”.
Are we getting there?

19. ### hobbyist Distinguished Member

Aug 10, 2008
773
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Hi,
There is the issue with voltage drops across all the diode junctions in series, so analyzing this circuit it shouldn't work,, so that was my conclusion as well , ,,,,untill I built it,,,,,,,well this video shows the rest of the story.

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