I understand what it means for the transistor to be biased for Class A, AB, and B with regards to the location of their DC operating points (\$I_{C_Q}\$ and \$V_{CE_Q}\$ are found by finding the point of intersection of the DC load line and the transistor characteristic curve).

For the Class C amplifier, the sources I've read say that the transistor must be biased beyond cutoff, but then I don't understand how this can be since there is no characteristic curve beyond the cutoff:

Transistor characteristic curve with load line

Also, does this mean this type of biasing forces a current (and with an opposite direction with respect to normal operation) through the transistor?

  • \$\begingroup\$ You are not allowed to continue the linear line to negative collector currents because - in reality - there is no negative current.The condition "beyond cut-off" can be visualized in the transfer characteristic only (Ic=f(Vbe)) with Ic=0 for Vbe below cut-off.. \$\endgroup\$
    – LvW
    Aug 7, 2014 at 6:24
  • \$\begingroup\$ I only copied that image from a Google search result, but I think I've seen it somewhere else also... So is it misleading then? Is the Q-point for Class C and Class B the same? \$\endgroup\$
    – Phil S
    Aug 7, 2014 at 13:56
  • \$\begingroup\$ Phil S: According to the graph, at point C the collector current as well as the base current would be negative! Is this a realistic approach? Beyond cut-off simply means that Vbe<0 (for npn devices). \$\endgroup\$
    – LvW
    Aug 7, 2014 at 16:00
  • \$\begingroup\$ So \$V_{CE_Q} and I_{C_Q}\$ for Class C amplifiers are actually located at the dot for Class B, and what makes it biased "beyond cut-off" is a negative \$V_{BE_Q}\$? \$\endgroup\$
    – Phil S
    Aug 8, 2014 at 11:02
  • \$\begingroup\$ Yes - Ib and Ic are zero and Vbe<0. As a result - only 50% of the input signal are used to "open" one of the transistors. \$\endgroup\$
    – LvW
    Aug 8, 2014 at 11:58

2 Answers 2


First, remember that the load line drawing solves a particular set of equations. Where the lines cross gives the the operating point for that combination of power supply, load resistor, and transistor base current.

Second, it's correct that there is no characteristic curve for the BJT that goes through the region you circled. The reason is conservation of energy. If the BJT operated in that region, it would mean that the BJT was delivering energy to the circuit, rather than taking energy provided by the power supply and turning it into heat. Since a BJT doesn't contain a reserve of energy that can be released in steady-state conditions, it simply can't operate in that region.

There is, however, probably a small region right near the origin where the transistor characteristic curves do pass through quadrant IV of the graph. Consider this circuit:


simulate this circuit – Schematic created using CircuitLab

This is essentially taking the resistor/power-supply load line and moving it down for the case where Vcc goes to 0. In this case, the base-collector junction will be forward biased and some power from the base bias supply will be delivered to the load resistor. And the load line will give a solution in quadrant IV, but very close to the origin.

If we were talking about a MOSFET instead of a BJT, even this solution would not be possible, since there's no way for current to transfer from the gate side to the drain side of the FET.

  • \$\begingroup\$ Doesn't this circuit operate in the second quadrant? The equation for the load-line is \$V_CE=-I_CR_1\$ meaning \$V_{CE}\$ is negative when there is a collector current? \$\endgroup\$
    – Phil S
    Aug 8, 2014 at 10:49
  • \$\begingroup\$ @PhilS, \$V_E\$ is 0, because that's where I put the ground reference. If I1 is positive, current will flow through the base-collector junction to R1, and \$V_{CE}\$ will be positive. But \$I_C\$ will be negative (current flowing out of the collector). So on OP's chart, that's quadrant IV. \$\endgroup\$
    – The Photon
    Aug 8, 2014 at 15:40
  • \$\begingroup\$ Is this the region you're referring to? Is it correct to say that at that region, the CB junction is forward biased, and at the point where the characteristic curves intersect the x-axis, the CB junction switches from forward to reverse bias? \$\endgroup\$
    – Phil S
    Aug 9, 2014 at 2:02
  • \$\begingroup\$ @PhilS, yes, I think that's right. \$\endgroup\$
    – The Photon
    Aug 9, 2014 at 3:33

A typical class c amplifier is typically used in RF amplification of a carrier wave: -

enter image description here

The reason the amplifier operates fine is that the tuned circuit in the output will largely reproduce a decent sinewave even when the transistor is biased significantly against 50% of the sinewave at the input.

In fact you could, with a decent tuned circuit apply short pulses of current or voltage at exactly the right time to keep the tuned circuit output looking fairly reasonably sinusoidal. A class c amp is doing similar and is biased so that with no input signal, there is zero collector current.

  • \$\begingroup\$ What do you mean by "..biased significantly against 50% of the sinewave .."? \$\endgroup\$
    – sherrellbc
    Aug 7, 2014 at 12:53
  • \$\begingroup\$ @sherrellbc it means that for the lower half of the input sinewave, the base emitter junction will be reverse biased and only start to conduct (and therefore amplify) as the input rises through the positive half of its cycle. \$\endgroup\$
    – Andy aka
    Aug 7, 2014 at 12:57

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