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This is a circuit I made recently; a differential bjt amplifier (without any signal, only DC).

So I was trying to see, what happens when I change the voltage difference between the base of Q1 and base of Q2. Q3 represents constant current source.

According to the example shown here: https://www.youtube.com/watch?v=mejPNuPAHBY , with changing the Vbe drop across Q1, the I_e1 should change inversely proportional to I_e2. As the I_e1 increases, I_e2 decreases.

Well, I tried to achieve that, but it didn't happen as said. If 5M potentiometer would be removed, the both of I_e would have a value of approx. 550uA. As I connected the potentiometer into the circuit and changed it, I_e1 was decreasing but I_e2 didn't change a bit (pictures given down below).

  • Where is the glitch in this circuit, as differential pair doesn't behave as it should?

schematic

simulate this circuit – Schematic created using CircuitLab

Without potentiometer

enter image description here

With potentiometer

enter image description here

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  • \$\begingroup\$ Your youtube video comes up as "not available". Either way, I recommend that you simulate your circuit, being able to quickly change parameters makes it very fast to understand how something works in a matter of minutes. "Ooooh, an op-amp's DC gain is heavily related to the bias current" \$\endgroup\$ – Harry Svensson Oct 28 '17 at 12:14
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    \$\begingroup\$ You cannot have two GNDs at wildly different voltage levels. That's against the spirit of a GND. Rethink your circuit by using one GND and change your voltage sources accordingly. You will find out what's wrong that way. \$\endgroup\$ – Janka Oct 28 '17 at 12:24
  • \$\begingroup\$ @Janka Did that. Only thing that changes is that both of I_e are now 450uA when both Vbe are equal, other stays unchanged. \$\endgroup\$ – Keno Oct 28 '17 at 12:53
  • \$\begingroup\$ Is your minus terminal from +5V PSU is connected the minus of a +30V PSU? Or not? \$\endgroup\$ – G36 Oct 28 '17 at 13:15
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    \$\begingroup\$ This circuit will have well defined Q3 Vce voltage obrazki.elektroda.pl/5454591100_1509210687.png \$\endgroup\$ – G36 Oct 28 '17 at 17:12
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To get a better understanding of how differential BJT amplifier work.

You must treat the BJT as a Voltage controlled (Vbe) current source (Ic).

So to be able to test this it try to build this simple circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

And as a name suggest the differential amplifier "react" on voltage difference.

$$V_{ID} = V_A - V_B$$

And if this \$V_{ID}\$ is larger then \$150\textrm{mV}\$ all the \$I_{EE}\$ current will flow through \$Q_1\$.

And if \$V_A - V_B = 0\textrm{V}\$ (\$R_X = 0\Omega\$) the \$I_{EE}\$ current will split equally between two BJT's.

So to change \$ V_{ID}\$ voltage all you need to do is to change the\$R_X\$ resistor value between \$0\Omega\$ to \$220\Omega\$

Also notice that \$V_{ID} = V_{BE1} + V_{EB2}\$ (from KVL loop). And this means that as \$V_{BE1}\$ increase the \$V_{BE2}\$ must decrease by the same amount.

enter image description here

For example, if \$V_{BE1}\$ changes from 0.60V to 0.61V the \$V_{BE2}\$ will drop by 0.01V from 0.6V to 0.59V.

And sometimes we can treat the differential amplifier as a cascade connection of the two "basic" BJT amplifiers. The first one is CC (Common Collector) and the second stage is CB (Common base) stage.

You can modify the original circuit by adding two separate voltage divider.

One for Q1 and the second one for Q2 So you can "set" \$V_A\$ and \$V_B\$ separately.

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Notice how your circuit is configured, especially where the grounds are.

You actually have three entirely separate circuits.

schematic

simulate this circuit – Schematic created using CircuitLab

If ground for 30V rail is the same ground as the 5V Ground, circuit 3 does nothing. It your 30V power supply is floating relative to the 5V supply, that is also a problem.

You need to rethink how this circuit is supposed to work.

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  • \$\begingroup\$ So, now I removed 5V rail and supplied voltage to both of bases from 30V rail (so only one ground remains). I figured out where the problem is - the constant current source; the voltage across should be approx. 15V but when I measured it there was barely 1V across it. This is probably the problem but I don't know how to fix this. Any ideas? \$\endgroup\$ – Keno Oct 28 '17 at 14:08
  • \$\begingroup\$ @Keno did you actually read this answer.... As long as you have ground connected in the middle of your circuit you have three separate circuits. \$\endgroup\$ – Trevor_G Oct 28 '17 at 14:15
  • \$\begingroup\$ Yes, and I also commented that I removed that ground and also 5V rail. \$\endgroup\$ – Keno Oct 28 '17 at 15:01
  • \$\begingroup\$ @Keno read your comment, you said nothing about removing the ground. You say "the constant current source; the voltage across should be approx. 15V but when I measured it there was barely 1V across it" You need to me more specific. The voltage across where? \$\endgroup\$ – Trevor_G Oct 28 '17 at 15:10
  • \$\begingroup\$ Collector to ground voltage of the current source. \$\endgroup\$ – Keno Oct 28 '17 at 15:33

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