Say i have a PN2222 transistor that i want to integrate into a circuit to amplify a signal with a gain of 100 or so.

From the datasheet of the transistor i can determine that

Vce = 10V and Ic = 150mA for a min gain of 100.

So i need to derive a circuit to satisfy at least those conditions. (Lets say i have a 12 volt power supply as the source.)

How do i come to a decision about the values of the resistors? Is it just a matter of picking them arbitrarily at this point? Is there some algebraic way to solve for the best possible values i could use? Maybe plotting all the different possible values for the resistors or something?

I'm very lost as to how to proceed, and google only seems to return very basic 'intro to circuits' stuff.

Thanks in advance for anyone who replies.

enter image description here

  • \$\begingroup\$ How do you determine those values “from the datasheet”? Sure, you can get a gain of 100 with that, but a gain of 100 does not require such values. \$\endgroup\$ Commented Dec 3, 2018 at 4:13
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    \$\begingroup\$ The ratio between R2 and R3 sets the gain. R1 and R4 divide the 12 volts down to set the base voltage of Q1. The Vbe is.65 volts so that voltage across R3 sets the bias current. For good values R1 should be 100K and R2 about 10K. You do the rest.... \$\endgroup\$
    – user105652
    Commented Dec 3, 2018 at 4:18
  • \$\begingroup\$ Look for design ce amplifier voltage divider bias. \$\endgroup\$ Commented Dec 3, 2018 at 4:29
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    \$\begingroup\$ Possible duplicate and here. \$\endgroup\$ Commented Dec 3, 2018 at 4:38
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    \$\begingroup\$ And/or here. \$\endgroup\$
    – jonk
    Commented Dec 3, 2018 at 5:02

1 Answer 1


This circuit will provide gain of 100, yet with substantial distortion for inputs larger than a few milliVolts input


simulate this circuit – Schematic created using CircuitLab

The values of R2+R3, being approximately beta*R1, causes the Vce to be approximately VDD/2 or 5/2 = 2.5 volts. The voltage across R1 is also 2.5v, thus current thru R1 is 2.5 milliAmps.

The 2.5mA makes the gm be 0.1 amps per volt.

The voltage gain is gm * R1, ignoring all other parasitics acting in parallel with R1 (such are Early Effect and Rload and R2).

The voltage gain, computed as gm * R1, is 100.

Notice the dimensions on gm: amps per volt, and the dimensions on the resistor: volts per amp, cancel each other when multiplied, leaving the gain to be a pure ratio; more properly, delta volts in becomes delta volts out.

Many assumptions are in this circuit. But the DC-feedback of R2, C1, R3 does ensure this circuit will self-bias and let some signal through. In fact, FETs may even work in this circuit.

  • \$\begingroup\$ No emitter resistor for stability? Ok... \$\endgroup\$
    – user105652
    Commented Dec 4, 2018 at 22:25
  • \$\begingroup\$ There is DC feedback. This circuit guarantees a bias point that will respond to a small input signal in a linear fashion. The best bias point requires some guess at the BETA. However, for any functioning bipolar (not horrible leakage), this circuit will adjust and compensate for temperature and VDD and BETA changes. Yes, the operating-point (Vce, Ic) may vary a lot, if BETA changes a lot. But unlike the fixed-base-bias with R-emitter, which can saturate the collector if R-collector is too large, this circuit will not saturate the collector; you always have a linear response. \$\endgroup\$ Commented Dec 5, 2018 at 17:16

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