Aoe question : easy transistor circuit...

Thread Starter

Zeeus

Joined Apr 17, 2019
459
Hi...Please take a look at image

I think q3 will not switch if R3 is 100 ohms because current through R2 is about 4.3mA and if R3 is 100 then current through R3 is 7mA so there is current entering the base of Q3...Correct??

I made the circuit..same input signal, same R1...Vcc is 12v...R2 is 4.7k....and load is an LED in series with 3k...I used 20 ohms as R3 and I was not expecting the load (LED) to turn on but it does

What is wrong??

Thanks
 

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sghioto

Joined Dec 31, 2017
1,038
I was not expecting the load (LED) to turn on but it does
You are correct it should not, which means you need to check the wiring and components.
Make sure the transistors are the correct ones and installed correctly.
R3 is what is known as reverse bias on Q3. With a value of 20 ohms Q3 cannot conduct.
SG
 

Thread Starter

Zeeus

Joined Apr 17, 2019
459
You are correct it should not, which means you need to check the wiring and components.
Make sure the transistors are the correct ones and installed correctly.
R3 is what is known as reverse bias on Q3. With a value of 20 ohms Q3 cannot conduct.
SG
Yeah maybe made a mistake..Will re check tonight

Thank you
 

WBahn

Joined Mar 31, 2012
24,850
Hi...Please take a look at image

I think q3 will not switch if R3 is 100 ohms because current through R2 is about 4.3mA and if R3 is 100 then current through R3 is 7mA so there is current entering the base of Q3...Correct??
Not really. The current in R2 and R3 is going to be very close to the same unless there is a significant base current in Q3.

Plus, the current in R2 will always be MORE than the current in R3.

I made the circuit..same input signal, same R1...Vcc is 12v...R2 is 4.7k....and load is an LED in series with 3k...I used 20 ohms as R3 and I was not expecting the load (LED) to turn on but it does

What is wrong??

Thanks
There are some assumptions that go into their claim (and they probably state -- or at least imply -- them somewhere).

You need to get enough current through R3 to pull the base of Q3 down by about 0.6 V to 0.7 V. But that current (plus the base current) has to flow through R2, which is going to produce a voltage drop across it and the most it can be is 14.3 V (and more like 14 V -- do you see why?).

Since I'm taking your word that this is not school work (and I do think that that is the case), the let's see if we can find the smallest value of R3 in terms of R2 and the key voltages.

We'll ignore the effect of the Q3 base current.

Vcc = Vce2 + Vr2 + Vr3
Vr3 = I·R3 = Vbe3
Vr2 = I·R2

All of the EE is in the above equations. Everything else is pure math.

I = Vbe3 / R3

Vr2 = Vbe3·(R2/R3)

Vcc = Vce2 + Vbe3·(R2/R3) + Vbe3
Vcc - Vce2 = Vbe3( 1 + (R2/R3))
(Vcc - Vce2) / Vbe3 = 1 + (R2/R3)
[(Vcc - Vce2) / Vbe3] - 1 = R2 / R3
[Vcc - (Vce2 + Vbe3)] / Vbe3 = R2 / R3
R3 = R2 · Vbe3 / [Vcc - (Vce2 + Vbe3)]

If R2 = 3.3 kΩ, Vcc = 15 V, Vce2 = 0.2 V, and Vbe3 = 0.7 V, then the corresponding value of R3 would be about 164 Ω.

For the values you used, it will even be higher.

If R2 = 4.7 kΩ, Vcc = 12 V, Vce2 = 0.2 V, and Vbe3 = 0.7 V, then the corresponding value of R3 would be about 296 Ω.

You can move it around a little bit by assuming slightly different voltages, but not enough to get you anywhere close to 20 Ω.

Check your circuit carefully. Be sure that it really is a 20 Ω resistor and not a higher multiplier. It's possible that even 200 Ω will be enough to get the LED to light up, but only if Vbe for Q3 is down in the 0.5 V range, which s probably a bit much to ask for.
 
Last edited:

WBahn

Joined Mar 31, 2012
24,850
hehe...Think finals is over...

It is a waste of time trying to learn the circuit? How to learn transistor circuits and never forget?
Why would it be a waste of time?

However, you want to focus on learning how to analyze circuits using fundamental concepts and solid analysis techniques and not try to memorize the specifics about individual circuits.
 

WBahn

Joined Mar 31, 2012
24,850
cmon guys...Beginner here trying to learn...why so much negative feedback?
Patience, patience. Not everyone here is just sitting around waiting to answer your questions. Some people won't see your post for days after you make it. I was in the middle of making a detailed response when I actually had to step away for several hours to do things other than sit here answering your questions.

If I didn't respond quickly enough for you, I'm sorry. I have a life (such as it is and what there is of it). I'll go try to help someone that doesn't need such a rapid turnaround. Good luck.
 

Thread Starter

Zeeus

Joined Apr 17, 2019
459
You need to get enough current through R3 to pull the base of Q3 down by about 0.6 V to 0.7 V. But that current (plus the base current) has to flow through R2, which is going to produce a voltage drop across it and the most it can be is 14.3 V (and more like 14 V -- do you see why?).

I can not see unfortunately...
We'll ignore the effect of the Q3 base current. (why should we?)

Vcc = Vce2 + Vr2 + Vr3
Vr3 = I·R3 = Vbe3
Vr2 = I·R2

All of the EE is in the above equations. Everything else is pure math.

I = Vbe3 / R3

Vr2 = Vbe3·(R2/R3)

Vcc = Vce2 + Vbe3·(R2/R3) + Vbe3
Vcc - Vce2 = Vbe3( 1 + (R2/R3))
(Vcc - Vce2) / Vbe3 = 1 + (R2/R3)
[(Vcc - Vce2) / Vbe3] - 1 = R2 / R3
[Vcc - (Vce2 + Vbe3)] / Vbe3 = R2 / R3
R3 = R2 · Vbe3 / [Vcc - (Vce2 + Vbe3)]

If R2 = 3.3 kΩ, Vcc = 15 V, Vce3 = 0.2 V, and Vbe3 = 0.7 V, then the corresponding value of R3 would be about 164 Ω.

For the values you used, it will even be higher.

If R2 = 4.7 kΩ, Vcc = 12 V, Vce3 = 0.2 V, and Vbe3 = 0.7 V, then the corresponding value of R3 would be about 296 Ω.

You can move it around a little bit by assuming slightly different voltages, but not enough to get you anywhere close to 20 Ω.
If I understand what you did (I hope I did) then 296 will make Q3 saturated yeah? Looking for R3 that will make Q3 off even when input signal is applied
 

Thread Starter

Zeeus

Joined Apr 17, 2019
459
Patience, patience. Not everyone here is just sitting around waiting to answer your questions. Some people won't see your post for days after you make it. I was in the middle of making a detailed response when I actually had to step away for several hours to do things other than sit here answering your questions.

If I didn't respond quickly enough for you, I'm sorry. I have a life (such as it is and what there is of it). I'll go try to help someone that doesn't need such a rapid turnaround. Good luck.
Yeah right, I am sorry..Although, post a thread and you might be amazed at how many people will view it in 20mins
 

WBahn

Joined Mar 31, 2012
24,850
I can not see unfortunately...
It's those pesky fundamentals.

The total voltage from Vcc to ground has to obey Kirchhoff's Voltage Law, so

Vcc = Vce2 + Vr2 + Vr3

This means that the largest voltage drop we can get across R2 is

Vr2 = Vcc - Vce2 - Vr3

Because R3 is in parallel with the base-emitter junction of Q3, we have

Vr3 = Vbe3

If Q3 is conducting, then Vbe3 is going to be about 0.6 V to 0.7 V, perhaps a bit less since it's not being asked to conduct very much current. This is assuming that you are using a fairly typical silicon PNP small signal transistor.

Vce2 is the collector-emitter voltage of the NPN transistor and the lowest it can get is likely in the 0.2 V to 0.3 V range. Perhaps a bit more or a bit less.

We often assume that the saturation voltage is 0 V in order to do quick calculations.

If I understand what you did (I hope I did) then 296 will make Q3 saturated yeah? Looking for R3 that will make Q3 off even when input signal is applied
Pretty much the opposite. We are looking for the value of R that will prevent Q3 from turning on at all -- that very different from Q3 being in saturation (also known as "fully on" or something similar).

The more current we pull through R3, the greater Vbe3 will be and we need to get it up into that 0.6 V range before the transistor will start conducting much current. But the only way to get more current through R3 is to pull the other end of R2 lower and lower by turning Q2 on harder and harder. But the absolute best we can do is to pull the collector of Q2 all the way to ground. That would put about 11.4 V across R2 when Q3 is on which means that there is about 2.4 mA of current in R2 (Vcc = 12 V and R2 = 4.7 kΩ). Aside from the slight base current in Q3, all of that current has to be flowing in R3 (they are effectively in series). Since we have to have at least 0.6 V across R3 with 2.4 mA of current flowing in it, that means that the resistance can't be any less about 250 Ω. The 296 Ω came from using a Vce2 of 0.2 V instead of 0 V.

With 100 Ω for R3 in the original circuit, we would need 5 mA of current to even get to 0.5 V for Vbe3 (which is probably not enough) and 5 mA through R2 would require 16.5 V across it. That would, in turn, require that the collector of Q2 be at -2 V, which is not possible.
 

Thread Starter

Zeeus

Joined Apr 17, 2019
459
It's those pesky fundamentals.

Since we have to have at least 0.6 V across R3 with 2.4 mA of current flowing in it, that means that the resistance can't be any less about 250 Ω. The 296 Ω came from using a Vce2 of 0.2 V instead of 0 V.

.
Thank you again...who says one has to go to school to learn?

Yeah so not opposite then because if resistance is less than 250 (how did he write the ohm?) then it should not turn on because A will be more than 2.4...

Actually just tried it again..Made a mistake the first time..It does not turn on with 20 @(ohm)..with 300@ it turns on...170@ does not turn on
 
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