# BJT diff amp output polarity

#### automagp68

Joined Nov 13, 2011
81
Im confused with this simple diff amp circuit i built in my multisim

I was always under the impression that that If you put a positive (aka 0 phase) input on base #1 and a negative (180 phase) on base #2 and took the output from Collector #2 you would see an inversion relative to the input at base #1

When i put my scope on the input and output in my circuit they are in phase. How is that possible considering the drive for the second transistor (base #2 ) is inverted?

So what I'm seeing is

My red line VIN1 as compared to the Blue line VO2 Are in phase.
I dont understand how thats possible?

#### Jony130

Joined Feb 17, 2009
5,228
My red line VIN1 as compared to the Blue line VO2 Are in phase.
I dont understand how thats possible?
But can you explain why you expecting a phase shift ?
Notice that as Vbe1 rising Vbe2 at the same time decreases. This means that the Ic2 collector current also must decrease. If so voltage drop across Rc2 also decreases but Vout is rising because Vout = Vcc - Ic2*Rc2 .

#### automagp68

Joined Nov 13, 2011
81
hmmmmm
Very interesting

As Ib1 goes up Ib2 goes down causing VO to increase not decrease.

Hmmmm ok, for some reason i thought it was opposite.

Been a while since i played Diff amp

#### automagp68

Joined Nov 13, 2011
81
I see your argument as to why they are not invertered but here is a new question

Hos is it possible then for the Ib2 and Vout to BE inverted?

Ib2 is driving the current through RC2 so how is it remotely possible for Ib2 and he drop across RC2 to be inverted like I'm showing

#### AnalogKid

Joined Aug 1, 2013
9,086
Each half of the differential pair is a simple common emitter circuit. By rule, the phase of the collector current is inverted with respect to the phase of the base current. Or, the voltage observed at the collector is inverted with respect to the voltage driving the base (assuming no external phase shifting components. There is inversion from B1 to C1, and inversion from C1 to C2 (the differential amp action), so B1a nd C2 are in phase.

Also, you can eliminate the 180 degree voltage source and ground base 2, and the output will not change in phase. If B1 is 0 degrees, E1 is 0 deg (emitter follower), E2 is 180 deg (differential action of a common current source for both emitters), C2 is 0 deg (Q2 acts as a common base amplifier).

ak

#### Bordodynov

Joined May 20, 2015
2,812
My advice is to change the value of the capacitors C1 and C2 such as the value of 100nF. Even if the input impedance of the oscilloscope still 1M obtained quite a significant phase shift. I do not know whether it mimics your simulator, but in reality the effect.

#### AnalogKid

Joined Aug 1, 2013
9,086
Good catch. C1 and C2 should be 1 uF, not 1 pF.

ak