# single stage BJT differential amplifier

#### stevy123

Joined Nov 19, 2007
61
If anyone can help me with thsi it would be greatly appriciated.

#### S_lannan

Joined Jun 20, 2007
247
ok. i'll walk you through it.
i assume you have been able to tackle the other 'simple' transistor circuits.

if the ccs is to provide 2ma. we shall devise a stiff voltage source with r1 and r2. we can safely say the current coming of off this source will be 2ma / 100.

once we have established that we can calculate vre by analysing the base emitter loop in the ccs.

now that we know vre lets establish a current through it using ohms law.
i = v/r.

since we have a ccs providing 2ma and perfectly matched transistors we can safely assume 1 ma through each transistor in the diff amp.

so i'd say let each collector sit at 1/2 vcc.

so v=ir

once we have that we have satisfied the dc biasing conditions for the amplifier.

run this all in spice and find the voltage gain and output impedance.

#### stevy123

Joined Nov 19, 2007
61
Hi S_lannan

Thanks for your reply, yes i have worked out other transistor circuits before but none that are connected up like this.

What sort of values would you choose for the resistors as im still not sure how to calculate them.

Regards
Steve

#### S_lannan

Joined Jun 20, 2007
247
First off we need to calculate the resistors for the constant current source.

The general idea here is to keep the current going from vee to ground in this network about lets say 10 times greater than the current going through the base of the ccs.

for the hell of it i'll just say r1=2.5k and r2=10k.
and that means this voltage divider gives the base of Qs -12 volts.

so if there is -12 volts at the base of Qs there will be -11.3 volts at the emitter because of the vbe voltage drop.

that means there will be around 2.3 volts across the emitter resistor.

i want a current of 2ma going out of the current source so i select the resistor using r=v/i
r = 2.3v / 2ma = 1150 ohms.

now that we have established a working current source that delivers 2ma to the differential pair we can safely assume each transistor will take 1ma.

i want the collector to sit at roughly 1/2 vcc. so i would like the collector resistor to drop around 7.5v. I know the collector current will be 1ma so...

r=v/i
r = 7.5v / 1ma = 7500 ohms

#### Ron H

Joined Apr 14, 2005
7,012
You don't have to use Spice to get the gain. If you are familiar with the diode equation, you will know that, at ~25C, the dynamic emitter resistance re~.026/Ie. Since Ie=1mA, re~26 ohms. From the input to the noninverting collector output, the diff amp looks like an emitter follower with re=26 ohms, driving a common-base amplifier with re=26 ohms. If you use the 7.5k load resistor that S_lannan suggested, you should be able to calculate the mid-band gain. If you don't, you weren't paying attention in class.
BTW, in his calculations, S_lannan said
so if there is -12 volts at the base of Qs there will be -11.3 volts at the emitter because of the vbe voltage drop.
. I'm sure he meant to say
there will be -12.7 volts at the emitter
.

#### S_lannan

Joined Jun 20, 2007
247
oops.