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.

20>=Ad<=30 (single ended output)
Rid>=20k
±10v

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

 

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

 

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

 

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

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

 

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?

 

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.

 

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.

 

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.

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.

 

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