bjt differential amplifier

Thread Starter

ntetlow

Joined Jul 12, 2019
24
Attached is a simulation of a differential amp with two identical inputs. The output voltage is around 8 volts, with a 5 mA input which would make the gain about 1600. The formula for the gain is RC * gm. Rc is 75k whilst gm is Ic/25mV approx 200uA/25mV. This does not work out at 1600.
Can someone tell me what's wrong here please?
upload_2019-10-5_0-14-42.png
 

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ericgibbs

Joined Jan 29, 2010
8,752
hi,
It appears that the two Base input are identical and in phase, ie: cancellation.
Shift one input by 180 deg.
E
Vb2 Phase shifted 180deg
073 Oct. 03 19.28.gif
 

crutschow

Joined Mar 14, 2008
23,373
As Eric noted, the amp is doing what its name states, amplifying the differential (difference) signal, and with both inputs the same phase, that signal is zero.
 

Thread Starter

ntetlow

Joined Jul 12, 2019
24
hi,
It appears that the two Base input are identical and in phase, ie: cancellation.
Shift one input by 180 deg.
E
Vb2 Phase shifted 180deg
View attachment 187260
I can see that the resulting voltage will be flat but why is it at 8 volts. For some reason my dc operating point analysis is not working so I don't know if 8 volts is the dc voltage point.
 

Thread Starter

ntetlow

Joined Jul 12, 2019
24
hi nt,
What Collector quiescent voltage do you want.?

E
Eric,
firstly I'd like to thank you for your promp replies to what I'm sure must be my annoyingly trivial questions?
I am not worried by the quiescent collector voltage value per se, I just want it to be the correctly calculated one. As far as the voltage is concerned I would like to be able to calculate it from the formula previously quoted.
 

ericgibbs

Joined Jan 29, 2010
8,752
hi nt,
Your questions are not trivial or annoying.:)

So you are expecting a voltage gain of approx 1600, when using a differential +/-10mV driving Input.?

E
 

Jony130

Joined Feb 17, 2009
4,983
The formula for the gain is RC * gm. Rc is 75k whilst gm is Ic/25mV approx 200uA/25mV.
For Iee ≈ 200μA ( tail current ) the collector current will be Ic1 ≈ Ic1 ≈ 100μA
The BJT's transconductance is gm ≈ 100μA/25mV ≈ 4mS hence the voltage gain for a symmetrical output will be around 4mS*75kΩ = 300 [V/V].
And for asymmetric output the will be 150[V/V]. And this is what you will see in simulation.
 
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