# Could use help on Diff Amp calculation

Discussion in 'Homework Help' started by laguna92651, Mar 22, 2010.

1. ### laguna92651 Thread Starter Active Member

Mar 29, 2008
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I posted this earlier with no response, I could really use some direction on this.

How do I calculate the max peak to peak output swing and the max input voltage that can be applied without any distortion.

Rid>=20k
±10v

I'm not sure what I need to calculate and what assumptions to make to calculate.

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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For single ended output
Au=Rc/( 2*(Re+re)) = 2.4KΩ/ 2*(28Ω+10.4Ω)=2.4KΩ/76.8Ω=31[V/V]

And the voltage output swing for negative voltage is cut-off of a Q3.
And for positive swing cut-off if a Q4.
max input voltage = ( max peak to peak output swing)/Au

3. ### laguna92651 Thread Starter Active Member

Mar 29, 2008
101
0
Thanks for the reply,what is the re=10.4Ω, where did that come from? Is Au a the max output peak to peak or the gain? Thanks

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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re is the dynamic emitter resistance dependent upon the DC emitter current value and device temperature - plus some constants.

re is typically calculated from the relationship 26/Ie(mA)Ω

5. ### laguna92651 Thread Starter Active Member

Mar 29, 2008
101
0
How is this the max peak to peak output?

For single ended output
Au=Rc/( 2*(Re+re)) = 2.4KΩ/ 2*(28Ω+10.4Ω)=2.4KΩ/76.8Ω=31[V/V]

This looks like a gain, so it is relative? I assume the max output is before distortion occurs, so it is an absolute number?

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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To calculate max negative swing remove Q3 from the circuit and calculate the output voltage.
For max positive voltage remove Q4 and again calculate the output voltage.
And Au is a voltage gain.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I'd be interested to see how the re=10.4Ω value was determined. I'm not sure how one could easily predict the current source value (& subsequently re) from an analysis of the circuit - without reference to some device parameters.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,434
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The output current of a current mirror is given:

$Io=\frac{Vt}{R2}*In\frac{Ic2}{Io}$

So, for R2=12.4Ω ; Vt=26mV; and IR3=(20V-0.7V)/320Ω=60.3mA
And from datasheet, we can read the "exact" value for VBE voltage at Ic = 60mA.
And β=200.
http://www.iele.polsl.pl/elenota/ON_Semiconductor/p2n2222a-d.pdf
Vbe≈0.73V and finally IR3=60.21mA so Ic2≈β/(β+2)*IR3=59.61mA.

And to find the output current I start the first iteration by assuming Io=50mA
26mV/12.4Ω=2.096mA

Io = 2.096mA*In(59.61mA/50mA) = 368.613uA (1)
Io = 2.096*In(59.61mA/368.613uA) = 10.6638mA (2)
Io = 2.096*In(59.61mA/10.6638mA) = 3.6084mA (3)
Io = 2.096*In(59.61mA/ 3.6084mA) = 5.8804mA (4)
Io = 2.096*In(59.61mA/5.8804mA) = 4.8565mA (5)
Io = 2.096*In(59.61mA/4.8565mA) = 5.2576mA (6)
Io = 2.096*In(59.61mA/5.2576mA) = 5.0912mA (7)
Io = 2.096*In(59.61mA/5.0912mA) = 5.1586mA (8)
Io = 2.096*In(59.61mA/ 5.1586mA) = 5.1311mA (9)
Io = 2.096*In(59.61mA/5.1311mA) = 5.1423mA (10)

So in the end Io ≈ 5.1mA

Therefore
re=26mV/2.55mA 10.1Ω ( previously i calculate re for 2.5mA)

And spice results are Ic2=59.74mA and Io=5.91mA .

And I skip the fact that current mirror is very poor design.

Last edited: Feb 13, 2018
9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
Thanks Jony130,

I'd not seen that done before. Much appreciated for your effort.

I wonder if the original 'question' entertained the idea that this degree of deduction was required on the part of the student.

Cheers,

t_n_k