This question is referring to npn BJT CE:

If we look at base voltage vs. collector current plot we can see that at approximately 0.7V on base to emitter junction, the BJT starts rapidly conducting.

enter image description here

In a circuit DC BJT is biased so that quiescent point is in the middle of the load line (for class A amplifier).


  • Does DC quiescent point represents reference point for the AC input signal?

  • If we want to normally amplify input signal (without distortions), does the input signal have to vary from 600mV to 750mV (approximately)?

  • \$\begingroup\$ "normally amplify" means what (as opposed to just "amplify"). Without distortion is a myth. \$\endgroup\$
    – Andy aka
    Apr 25, 2017 at 12:29
  • \$\begingroup\$ does the input signal have to vary from 600mV to 750mV The BJT will still follow that DC curve, the AC just causes (small) variations across that curve. Between 600 mV and 750 mV there would be a huge ratio between minimum and maximum current. Assuming that is even possible (in a given circuit) that would mean huge non-linear distortions. A more reasonable choice would be Ic between 10 mA and 30 mA so Vbe would be between 600 mV and 620 mV. \$\endgroup\$ Apr 25, 2017 at 12:35
  • 1
    \$\begingroup\$ find a way to get... No you don't, you'd use a "clever" circuit to make Ic = 20 mA and then couple in the AC signal to the input using a capacitor, see G36's answer. \$\endgroup\$ Apr 25, 2017 at 13:55
  • 1
    \$\begingroup\$ Place a resistor in the emitter, and drop 0.26 volts (10 x kt/q); the distortion drops 10:1 ------ without overt feedback. Drop 2.6 volts, and distortion drops 100:1. Life gets pretty good, and fast and cheap. \$\endgroup\$ Apr 25, 2017 at 16:45
  • 1
    \$\begingroup\$ This was an interesting question especially the second query about the range of the input voltage swing. Unfortuantely the answers aren't very clear and I still don't quite understand how an input voltage can swing 1 volt and still be within the base-emitter voltage range which seems to be about 0.2 volts at most. \$\endgroup\$
    – rhody
    May 8, 2017 at 1:57

3 Answers 3


Does DC quiescent point represents reference point for the AC input signal?

Yes. If you have time try to analysis this simple CE amplifier.

enter image description here

Where Vcc = 12V; Vc = 6V (red plot); Ve = 2V (green plot) ; Vinput = 1V (blue plot).

And we have a ideal BJT (Vbe is fixed and equal to 0.7V, and Beta = 100)

DC quiescent point is Vb = 2.7V; Ve = 2V; Vc = 6V; Ic = 6mA; Ib = 60μA. No the AC input signal voltage will "modulate" our DC operating point the in the rhythm of the input signal.

The AC voltage at input will cause that the base voltage will change from 3.6V to 1.6V in "rhythm" of a AC input signal. Those changes will result that the emitter voltage will also change from 3V to 1V. The emitter resistor (Re) will "convert" this changes in Ve voltage into Ie current. This will result the changes in emitter current from 0.9mA (3V) to 0.3mA (1V). And now this changes in Ie and Ic current will once again be "convert" in to voltage in Rc resistor. And this change in collector current will cause change in VRc voltage, between 9V to 3V. We have a three times larger change in VRc voltage because Rc is three times larger then Re resistor. So the voltage gain is equal to Av = Rc/Re

Also when Vin is at peak Vin = +1V we have 3.7V on the base of the BJT, so the emitter current is equal Ie = 3V/330Ω = 9mA the collector current is also equal 9mA

And collector voltage is equal Vc = Vcc - Ic* Rc = 12V - 9V = 3V

The base current needed to provide 9mA of a collector current is equal:

Ib = 9mA / 100 = 90μA

The base current provided form Rb and Eb is:

I = (Eb - Vb)/Rb = (4.7V - 3.7V) / 33KΩ = 30uA

But the emitter current must be equal to 9mA, and that's requires the base current to be equal 90uA

So we have a situation: Eb delivers 30uA, base needed 90uA, so that extra current (60uA) will (must) be "deliver" by AC voltage source Vin.

So Vin will be deliver (sourcing) 60uA of a current.

For the negative peak swing at Vin (-1V) We have this situation Vb = 1.7V ; Ic = 3mA; Vc = 9V

The base current needed to ensure Ic = 3mA is

Ib = 3mA/100 = 30uA

The Eb delivers I_Rb = (4.7 - 1.7V)/33K = 90uA

But we only need 30uA for the base current to "deliver" 3mA at emitter.

But Eb delivers 90uA so that extra current (60uA) must flow to Vin (must be sink by Vin).

I hope this will help.


The input impedance is:

Rin = RB||(beta+1)Re = 33k||101*330R = 33k||33k = 16.67k

Or Rin = Vin/Iin = 1V/60uA = 16.67k

As you can see everything falls into place.

If we want to normally amplify input signal (without distortions), does the input signal have to vary from 600mV to 750mV (approximately)?

No, the input signal don't have to vary from 600mV to 750mV. Because the input coupling capacitor provide necessary DC-offset (DC level is shift thanks to Cin capacitor). And if Vbe is increased by about 60mV the collector current will increase his value ten times. Also BJT is a highly nonlinear device, so distortion will always occur. And this is why almost no one uses a single CE stage anymore.


1: Yes. At least, the signal at the base; see 2. 2: No. Yes. Sort of.

To elaborate, a BJT must be biased into the appropriate mode of operation, and then your AC signal is typically capacitively coupled into it; this effectively just overlays your AC small signal on top of the DC that the bias network provides.

It should be noted, however, that amplifying 'without distortion' is not going to happen. There's going to be distortion no matter what; the question is how much.

  • \$\begingroup\$ If AC signal of amplitude 10Vpp is applied to the input (to the base), what happens? Does BJT amplifies the signal or it gets destroyed since 10Vpp (or 5Vp) is much greater than 0.7V? \$\endgroup\$
    – Keno
    Apr 25, 2017 at 13:24
  • \$\begingroup\$ If you feed 5V into a BJT directly between its base and emitter, it'll get destroyed. But if you have a proper bias network on it that includes an emitter resistor, it will survive, and amplify the signal based on your bias network conditions. \$\endgroup\$
    – Hearth
    Apr 25, 2017 at 13:52
  • \$\begingroup\$ If Vbe is biased so it equals 0.7V and we apply 5Vp AC signal to input, transistor doesn't get destroyed, right? \$\endgroup\$
    – Keno
    Apr 25, 2017 at 14:50
  • \$\begingroup\$ Like I said, it depends on your bias network. Vbe is a function of the impedance between emitter and ground, the impedance between input and base, and the input voltage. \$\endgroup\$
    – Hearth
    Apr 25, 2017 at 15:21
  • \$\begingroup\$ So AC input signal does not changes the bias network of transistor? \$\endgroup\$
    – Keno
    Apr 25, 2017 at 15:58

Try this


simulate this circuit – Schematic created using CircuitLab

The operating point has 1mA Collector current, because emitter current is ~ 1mA, because we have ~~ 2 volts across R5. We have 2 volts across R1.

To compute gain, we need value for 'reac', slope of the emitter diode. That slope is 1/gm. At 1mA, 'reac' is 26 ohms Gain is $$(R1 || Zload) / (R2 + reac) = 2,000 / 126 = 16$$

assuming a 10MegOhm||10pF scope probe is the load.

Because of the linearizing effect of 100_Ohm in emitter, at 1mA Ie, we can swing nearly +- 100mV before "serious distortion" occurs.

Look up the writings of Willy Sansen, on computing bipolar distortion and the IP2 and IP3 values. You can extend his math, for a linearizing resistor.

Assume the input distortion Intercept Points are 2 volts rms.

What is the input random noise floor? Ignore the large Cmiller, and just use the Collector F3dB; 2K ohm and 20pF = 40 nanosecond, 25Meg Radians, or 4Mhz. Assume our noise density is set by 'rbb' of 2N3904, assumed to be 1Kohm. Noise density is 4 nanoVolts/rtHz. Total input referred noise is 4nV * sqrt(4Mhz) = 4nV * 2,000 = 8 uV rms.

SPDR (spurious free dynamic range) is 2 volts rms / 8uV rms, or 250,000 or 108dB.

====================================================== edit Here is link to Sidney Darlington discussing the bipolar transistor


And this next link gives you insight into bipolar distortion, including the Taylor Series expansion to describe the various coefficients (equation 12)


  • \$\begingroup\$ As long as the transistor doesn't go into saturation or cut-off region, everything is goes normal as the thing concerns A-class amplifier, right? \$\endgroup\$
    – Keno
    Apr 25, 2017 at 19:17
  • \$\begingroup\$ I mean, output voltage must be smaller than power supply voltage (usually Vce= 0.5 * Vcc) and biasing resistors must be biased properly. \$\endgroup\$
    – Keno
    Apr 25, 2017 at 19:20
  • \$\begingroup\$ Expect almost 4 volts peak-peak output swing, until clipping. Run the SPICE distortion analysis on the circuit at 3 (three) volts output PP. \$\endgroup\$ Apr 26, 2017 at 4:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.