Let's say there's a basic differential amp constructed with two npn BJTs. And a current source is working as the tail. circuit

The task I have is to amplify a small AC signal of around 1mV p-p, referenced to ground. Can I just connect this signal to one of the inputs, and ground the other input to observe an amplified differential signal at the outputs?

I was taught that when Vin is smaller than the base-emitter saturation voltage (~0.6V) the BJT is not active. Does that means my small AC signal is not going to work? (need a DC bias?)

I also read that the current source is biasing the circuit by drawing Ie (and therefore Ic). Then the Vbe "becomes" 0.6V consequently.

So at the end of the day, what parameters are needed for a BJT to enter active mode? Is the current source alone correctly biasing the transistors so my 1mV signal is getting correctly amplified?

Any help is greatly appreciated!

  • \$\begingroup\$ If you are using an ideal current source then the shared emitter voltage will be adjusted to whatever is needed and you don't need to show that "lower rail" as it is entirely irrelevant to the circuit. (You only need the upper end of the current sink, as shown.) But the main point here is that using an ideal component makes the realistic question about a saturated BJT moot, given the signal magnitude you indicated. \$\endgroup\$
    – jonk
    Jun 18, 2020 at 23:19
  • 1
    \$\begingroup\$ It should be noted that the circuit as shown will NOT work. This is because the base node has no ground reference. You must not connect any source BETWEEN both base terminals. As to your question: Yes, you can ground one base and connect Vin to the other base. In this case, the circuit works as a common collector-common base combination. \$\endgroup\$
    – LvW
    Jun 19, 2020 at 6:30
  • \$\begingroup\$ Yes - assuming ground is something like halfway between the supply rails drawn... \$\endgroup\$
    – user16324
    Jun 19, 2020 at 10:46

1 Answer 1


Tie the tail to -9 volts.

Tie the top of the 2 collector resistors to +9v.

Notice this requires a total of 18 volts. Call the midpoint "ground"

Now tie both bases to GND thru 1Kohm resistors, to supply the base current.

Set the tail current to 2 milliamps (thus approximately 1ma per transistor).

Put 4,700 ohm resistors in each collector, with the top ends to +9v supply.

With 4.7 volts across each of the two collector resistors, we see each transistor has about 4.3 volts collector-base. And the common emitter node is about -0.6 volts. Approximately.

You might have fun varying the tail current, in factors of 2:1 or 4:1 or 8:1 or 10:1, and using log-linear plot paper to examine the logarithmic behavior of that emitter voltage as the current widely changes.

Apply 1mvpp to either base. Unless that 1mv is precisely centered around ground, you may want to insert a series 100uf between the sin generator and the base. Use a much faster frequency than 1 Hertz.

What gain? for small signals (and 1mv qualifies as small signal), the delta_Iout / delta_Vbase for each of the 2 transistors is 1/26 ohms if operating at 1ma.

When you wiggle 1 base at 1mvpp, the shared emitters wiggle at 1/2 of that Vin or 0.5mv; and the other base should be almost still (should be << 1 millivolt); the changing base current causes this undriven base to move.

Your gain, looking at either output collector, will be 4700 / (26 + 26) , approximately. The predictability depends on how close to exact abrupt junction the transistor emitter-base junction is manufactured.

For additional learning, perform a SPICE model of this, and examine the THREE input voltages (the driven input, the common_emitters, and the undriven input). Also examine how closely the two collector voltages are exact opposites (or are not exactly opposite); above 10,000,000 Hertz input, you will be significant imbalances; as you understand the causes of imbalances, circuit design houses will be interested in hiring you.


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