The amplifier shown in figure consists of two NPN transistors and powered by two DC voltage sources.

VBE1, 2 = 0.7 V; HFE1 = hFE2 = 200; HFE1 = 200 hfe2 hie1 = 120, 2 = 5 kOhm

1 - What are the areas of operation of the two transistors? Justify

For this i would say Q1 and Q2 are in the ative region.what should I start by analyzing?Should o determine the base current of Q1 then de colector current of Q1.
And after that how can i determine VCEQ1 and VCEQ2?

Thanks

 

1 - What are the areas of operation of the two transistors?

Can you tell me why do you think that both BJT work in the active region?

For this i would say Q1 and Q2 are in the ative region.what should I start by analysing?Should o determine the base current of Q1 then de colector current of Q1.
And after that how can i determine VCEQ1 and VCEQ2?

Yes, start your analysis be finding Ib and Ic currents

 

Because i have found values for Ic >0 mA.Therefore the transistor Q1 is not in the cut/off region.
The problem is that for determing VCEQ1 i need to know VCEQ2.
For doing that i will have to assume that the colector current of Q1 is equal to the emiter current of Q2.Then assuming that Q2 is also in the ative region i can find VCEQ2?Is that it?

Thanks

 

Because i have found values for Ic >0 mA.Therefore the transistor Q1 is not in the cut/off region.
The problem is that for determing VCEQ1 i need to know VCEQ2.Do i have to make a mesh involving only VCEQ2 and then use that value to find VCEQ1?
For doing that i will have to assume that the colector current of Q1 is equal to the emiter current of Q2.Then assuming that Q2 is also in the ative region i can find VCEQ2?Is that it?

Thanks

It's possible to find Q1 base current and therefore deduce Q1 emitter voltage. If you know Q2 base is at 5V and VBE=0.7V then surely you can deduce VE for Q2. Further, since Q1 & Q2 are directly connected at the collector and emitter respectively, you may then find VCE for Q1.

 

Because i have found values for Ic >0 mA.Therefore the transistor Q1 is not in the cut/off region.

But in saturation Ic current is also greater than zero.

The problem is that for determing VCEQ1 i need to know VCEQ2.Do i have to make a mesh involving only VCEQ2 and then use that value to find VCEQ1?
For doing that i will have to assume that the colector current of Q1 is equal to the emiter current of Q2.Then assuming that Q2 is also in the ative region i can find VCEQ2?Is that it?

Thanks

But why you want to start you solution by finding Vce1 and Vce2?
This is the last thing that you need to do. BJT is a base control current source. So first you need to find Q1 base current and Q1 collector current. Next find Ic2 current. And after you find all the current. Then you can start to find Vce voltage for Q1 and Q2.

 

But in saturation Ic current is also greater than zero.


But why you want to start you solution by finding Vce1 and Vce2?
This is the last thing that you need to do. BJT is a base control current source. So first you need to find Q1 base current and Q1 collector current. Next find Ic2 current. And after you find all the current. Then you can start to find Vce voltage for Q1 and Q2.

Lets suppose that ICQ1=1 mA(looking at the resistors attached to the base of Q1 ib should be somehting like 5 micro A or something like that)

Now for determing VCEQ2 could i do the following:

VCEQ2=VCQ2-VEQ2=(10 V-(1kOhm)*(1 mA))-(5V -0.7 V)=4,7 V

VCEQ1=10 V-RC*IC-VCEQ2-RE*IE=10 V-(1kOhm)(1mA)-(4,7 V)-(50 Ohm)(1mA)=4,25 V

Would this be correct?

 

Lets suppose that ICQ1=1 mA(looking at the resistors attached to the base of Q1 ib should be somehting like 5 micro A or something like that)

Now for determing VCEQ2 could i do the following:

VCEQ2=VCQ2-VEQ2=(10 V-(1kOhm)*(1 mA))-(5V -0.7 V)=4,7 V

VCEQ1=10 V-RC*IC-VCEQ2-RE*IE=10 V-(1kOhm)(1mA)-(4,7 V)-(50 Ohm)(1mA)=4,25 V

Would this be correct?

Looks good, by Ic current is larger then 1mA.
And also we can find Vce voltage much easier

Ve2 = Vc1 = V2 - Vbe2 = 5V - 0.7V = 4.3V

and Vc2 = Vcc - Ic2*R3 and

finally

Vce2 = Vc2 - Ve2

And now Vce1

Vc1 = Ve2 = 4.3V

Ve1 = Ie*Re1

Vce1 = Vc1 - Ve1

 

The amplifier shown in figure consists of two NPN transistors and powered by two DC voltage sources.

VBE1, 2 = 0.7 V; HFE1 = hFE2 = 200; HFE1 = 200 hfe2 hie1 = 120, 2 = 5 kOhm

1 - What are the areas of operation of the two transistors? Justify

For this i would say Q1 and Q2 are in the ative region.what should I start by analyzing?Should o determine the base current of Q1 then de colector current of Q1.
And after that how can i determine VCEQ1 and VCEQ2?

Thanks

But how can you say that Q1 is in the active region when the base voltage is ~-9V?

Look at how V1 is connected.

Both transistors are in backwards, so while they will still act as NPN transistors, they won't be very good ones.

The R3 resistor will have -10V at the top and about 4.3V at the bottom, so about 14.3V across 1kΩ yielding about 14.3mA, even if it all has to come from the 5V source through the base (though the voltage drop will probably be noticeably more than 0.7V in that case).

But this means that the lowest that the other side of the Q2 (the emitter acting as collector) can be is 4.3V, which is also the Q1 collector (acting as emitter) voltage. Thus both the collector and emitter of Q1 are well above the -9V base voltage and that transistor is in cutoff.

So Q1 is cutoff and Q2 is in saturation with all of the Q2 current coming from the base.

 

I'm guessing that the polarity of V1 is drawn opposite of what was intended.

 

Funny how the supposedly trained mind fails to recognize an obvious blunder.

I just saw this as a typical cascode amplifier bias design problem with the 10V source polarized as one would expect it to be.

Good spot WBahn.

 

Lets suppose that ICQ1=1 mA(looking at the resistors attached to the base of Q1 ib should be somehting like 5 micro A or something like that)

Provided the 10V supply were correctly orientated ...

Think again - about 11 uA would be nearer the mark for the base current on Q1. Why guess when all the necessary data is there to resolve the matter.

 

Yep, we all have a tendency to see what we expect to see, which is why the novice sometimes has the advantage in situations where things are just a little bit out of kilter with what is normally encountered.

It's also we we are always harping on posters to post there problems and there work, because it is often the case that they are seeing what they want to see and not what is actually there.

 

I'm guessing that the polarity of V1 is drawn opposite of what was intended.

More than likely, that is the case.

I've given problems in which the problem, as stated (due to a mistake or a typo on my part), either couldn't be solved or the solution made no rational sense. I'm always amazed at how many students will read the problem and not see the mistake and use the data I had intended. I'm not amazed by the number of students that just blindly plug in whatever numbers are there and underline an obviously nonsensical answer and move on. The number of students that catch the mistake is, unfortunately, pretty rare -- but they get extra credit.