# Finding a high frequency differential amplifier Transfer Function

Discussion in 'Electronics Homework Help' started by Mobin, Apr 5, 2018.

1. ### Mobin

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May 1, 2017
hi friends,
I have been assigned to analyze this circuit below(it's an improvement to a differential amplifier in order to have better BandWidth):

like a differential amplifier , i tried to find the half-circuit and i came up with this:

in order to find the transfer function i used hybrid-pi model but now I'm stuck with a problem: that there is no connection between the two transistor and the second transistor seems non-functional so how can i find the transfer function? have i done any mistake in a part of solution?

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May 1, 2017

3. ### Ratch

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Mar 10, 2013
What do you want us to do? Do you want the equations for the voltages V1, V2, Vout/2 which you marked in red? Obviously V2 is grounded. A transfer function needs to be defined before is can be calculated (Vout/Vin,Iout,Iin,Vout,Iin,Iout,Vout). Which is it? I have never done double differential amps before, or any differential amps for that matter. But if you give me more info, I will take a swipe at it.

Ratch

4. ### LvW

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Apr 12, 2014
Why do you ignore the current source at the common emitter node of Q1 and Q2.
Yes - in this case, the "second transistor is not-functional".

5. ### Mobin

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May 1, 2017
in ac analysis dc current source would act like open circuit
but i shouldn't make that point grounded, that way the problem could be solved
now i have another problem, with biasing it .
when i write down dc and ac load lines,in order to get maximum swing, it comes up that there is more variables than equations, and also that Q2 bias current is zero!!!!

is that possible?

1,079
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Mar 10, 2013

7. ### Mobin

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May 1, 2017
sorry i missed that
the transfer function i'm looking for is Vout/Vd

8. ### LvW

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146
Apr 12, 2014
For ac analyses, the current source may be replaced by an open circuit - that`s correct.
However, if you want to analyze bias conditions...what do you think: Is the DC current source important or not?

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9. ### Ratch

1,079
330
Mar 10, 2013
Very interesting circuit. Differential amps with a current source have a very small turn on/off range. As you can see from the plot below, ±4 Vt , where Vt = 25.4 mV, is all it takes to switch the paired transistors from ON-OFF to OFF-ON. To stay in the linear region, Vd should be within ±Vt. During operation, the voltages of the transistors from left to right are HIGH-LOW-HIGH-LOW or LOW-HIGH-LOW-HIGH. The two inner transistors from where the voltages are taken switch between HIGH-LOW and LOW-HIGH. The advantage of this double differential configuration is that the output voltage can vary between ±v. A single differential configuration could only vary between 0 to v. That is double the voltage range. The transfer function is 2v/Vt. The higher the supply voltage, the greater the voltage amplification.

Ratch

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10. ### Mobin

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May 1, 2017
that was really useful thanks a lot

11. ### Mobin

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May 1, 2017
of course that is important, and it shows the bias current of both transistors. but the problem is that i can't recognize the relation between bias current of the two transistors. are they biased independently?

Last edited: Apr 12, 2018
12. ### Mobin

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May 1, 2017
is there anybody who could help me with biasing this

13. ### Ratch

1,079
330
Mar 10, 2013
What's the problem? If Vd is zero, all the transistors will be conducting with 1/4 of the current source in each transistor. All the emitter voltages will be at -0.7 volts and all the E-B junctions will be forward biased. If Vcc is greater than 0.7 volts, the the C-B junction will be reversed bias and all the transistors will be operating in the active region.

Ratch

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