Problem with differential amplifier with BJT

Thread Starter

Rastko321

Joined Mar 27, 2017
5
Hi guys,


I do not know to solve this problem. If you know, please help!

For circuit from picture I need to calculate Vin. I have this known data:

Ic1=4*Ic2; Vcc=1.4V; Vref=1.1V; Vces=0.2V; Vbe=0.7V (when transistor conduct); Iee=100μA; β1=β2=100
Rb=10kΩ; Rc=1kΩ; Va1->inf. Va2->inf.

Image in attachment. And thanks in advance!
 

Attachments

WBahn

Joined Mar 31, 2012
25,060
You need to show your best attempt at solving the problem. We can then see if where and if you are going wrong and try to give you hints to help you along.

We won't just work your homework for you.

You might start with determining what Ic1 and Ic2 are.
 

WBahn

Joined Mar 31, 2012
25,060
Hi. Well Ic1 and Ic2 are currents from collector. And my attempt to solve this problem is in attachment. Please help and this is not homework.
But what are the values for Ic1 and Ic2. You know that Iee is 100 uA. You know that Ic1 = 4*Ic2. So what is Ic1 equal to?
 

Thread Starter

Rastko321

Joined Mar 27, 2017
5
I need to calculate Vin when Ic1=4*Ic2, this is condition. Problem will be too easy that we have values of Ic1 and Ic2.

One hint: Q1 & Q2 works in DAR
 
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Bordodynov

Joined May 20, 2015
2,468
The task is not correct. The voltage of the Base-Emitter depends on the current. If we add an additional condition, at what current the Base-Emitter voltage equals 0.7V, then the problem has a simple solution. For example, at 1 mA.
The currents are treated as 4:1 ==> ic (Q2) =100/5= 20uA. ic (Q1)=80uA.
Next, use the exponential law for a semiconductor junction. Find the Base-Emitter voltages for both transistors.
 

Thread Starter

Rastko321

Joined Mar 27, 2017
5
The task is not correct. The voltage of the Base-Emitter depends on the current. If we add an additional condition, at what current the Base-Emitter voltage equals 0.7V, then the problem has a simple solution. For example, at 1 mA.
The currents are treated as 4:1 ==> ic (Q2) =100/5= 20uA. ic (Q1)=80uA.
Next, use the exponential law for a semiconductor junction. Find the Base-Emitter voltages for both transistors.

Can you do for me, please! I do not know exponential law.
 

Bordodynov

Joined May 20, 2015
2,468
I=is*(exp(Vbe/Vt)-1), Vt=KT/q=25.8mV, T=300K=27C.
Vbe=Vt*Ln(I/is+1)~Vt*Ln(I/is), I/is>>1
delta_Vbe=Vt*(ln(ieQ1/is)-ln(ieQ2/is))=Vt*ln(ieQ1/ieQ2) ==> Vb1=REF+delta_Vbe
I'm sorry, the one current I was talking about is not needed.
 

Thread Starter

Rastko321

Joined Mar 27, 2017
5
I=is*(exp(Vbe/Vt)-1), Vt=KT/q=25.8mV, T=300K=27C.
Vbe=Vt*Ln(I/is+1)~Vt*Ln(I/is), I/is>>1
delta_Vbe=Vt*(ln(ieQ1/is)-ln(ieQ2/is))=Vt*ln(ieQ1/ieQ2) ==> Vb1=REF+delta_Vbe
I'm sorry, the one current I was talking about is not needed.
Thanks. But where is Vin?
 

Bordodynov

Joined May 20, 2015
2,468
Of course I could count the answer, but I was already warned and once banned. With respect to the current of the emitter, calculate the current of the base. Under Ohm's law, calculate the voltage drop across the base resistor. You will receive the amount that you did not have.
 

WBahn

Joined Mar 31, 2012
25,060
I need to calculate Vin when Ic1=4*Ic2, this is condition. Problem will be too easy that we have values of Ic1 and Ic2.

One hint: Q1 & Q2 works in DAR
But you are given that Iee, which your "work" shows is equal to Ic1 + Ic2, is equal to 100 uA.

If you can't use this information to figure out what the actual values are of Ic1 and Ic2, then you need to go back and relearn 7th grade algebra.
 

WBahn

Joined Mar 31, 2012
25,060
I need to calculate Vin when Ic1=4*Ic2, this is condition. Problem will be too easy that we have values of Ic1 and Ic2.
The currents are treated as 4:1 ==> ic (Q2) =100/5= 20uA. ic (Q1)=80uA.
Do you plan to follow the TS around throughout their career working every problem for them? You should, since you don't seem to think that there is any need for him to ever figure out how to do anything for himself.
 

WBahn

Joined Mar 31, 2012
25,060
The task is not correct. The voltage of the Base-Emitter depends on the current. If we add an additional condition, at what current the Base-Emitter voltage equals 0.7V, then the problem has a simple solution. For example, at 1 mA.
The currents are treated as 4:1 ==> ic (Q2) =100/5= 20uA. ic (Q1)=80uA.
Next, use the exponential law for a semiconductor junction. Find the Base-Emitter voltages for both transistors.
He doesn't need the exponential law. The problem states Vbe for both transistors as a given, even though the current in them is a factor of four different. So the model that is implied is a constant-Vbe model.

Since the difference in Vbe at this ratio of current is only about 36 mV, it is not an unreasonable model. Even if it were, it's the model he is given to work with. This problem is almost certainly intended as a simplified, introductory exercise for differential pairs.
 

WBahn

Joined Mar 31, 2012
25,060
Can you do for me, please! I do not know exponential law.
You will not learn anything by having others do all the work for you. You have a text book that explains how to do it. You have examples that have been worked for you, both in the text and in class. Yet something hasn't clicked and you can't do it on your own. Seeing yet one more problem worked by someone else is not going to change that. It will only let you copy someone else's work, claim it as your own, think that you've gotten past a hurdle, and then flounder on the exam. YOU need to struggle through the problem so that what you are missing can click.
 

hobbyist

Joined Aug 10, 2008
887
Hi. Well Ic1 and Ic2 are currents from collector. And my attempt to solve this problem is in attachment. Please help and this is not homework.
If this is not homework, then please do not put it in the homework section, as it brings confusion of how to help with your thread, when its homework we need to treat the thread as your personal attempt to work the problem, when its not homework, then your not required to have all the working knowledge on an electronic level, just a knowledge of what your trying to accomplish.

Hope this helps.
 
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