BJT differential amp questions

Thread Starter

Pavlova

Joined Oct 14, 2008
1
I am using this site for reference : http://www.ecircuitcenter.com/Circuits/BJT_Diffamp1/BJT_Diffamp1.htm

and I just wanted to know the formula they used for differential gain, they got an answer of 20 V/V, is that mean the gain is 20dB of the circuit?
Also, I'm wondering what VCM, VS, VCC, VDD, VC1 and VC2 should be to get a gain of 20dB?

I am fairly new to this so some of these questions may seem stupid, but I really need help.

Cheers
 

hgmjr

Joined Jan 28, 2005
9,029
It appears that the author of the article has designed the circuit to operate on 2 milliamps of current into the two emitters of the differential pair. He does this by setting the bases of both transistors to 0 volts and then he calculates the resistance needed to produce 2 milliamps from the -15 Volt supply. The value he calculates is 7.2Kohms. He has included the equation that he uses to make this calculation.

Next he assumes that the 2 milliamps is split equally between the two transistors when the two base voltages are both at 0 volts.

He takes the 1 milliamp flowing in the emitter of one of the transistors and divides by 25 millivolts (derived from the diode expression) to calculate the transconductance of the transistor. Transconductance has the units of Amps/Volt. That means that when you multiply transconductance by a resistance in Ohms, you get Volts/Volt.

He then uses the value of collector resistance, 1Kohm, together with the calculated transconductance to determine the gain of one of the transistor stages.

hgmjr
 
Top