I see in the guides to standard Bipolar Transistor circuits (emitter-follower, common emitter amplifier), they always say: Choose a quiescent current of 1mA.

How is this value chosen? I look at the specification sheets and I am not sure what I am looking for specifically. From my understanding it should be based on the output characteristics of the transistor so that clipping does not occur and maximum symmetrical swing is obtained. However, I have to use the 2N3904 transistor, and the spec sheets don't have the output characteristics. I also have to drive a load of 100k ohms. Does choosing a quiescent current depend on the load?

  • \$\begingroup\$ One thing I would definitely not do first when designing a transistor amplifier is: Choose a quiescent current of 1mA. That's my cunning plan and don't assume I'm Baldrick LOL. \$\endgroup\$ – Andy aka Oct 16 '19 at 11:46
  • \$\begingroup\$ That 1mA is just an example / starting point. The most optimum biasing current depends on the transistor and what you want to do with it. For example, high frequency transistors often have a biasing current value that will result in the highest bandwidth. Another example: for an Audio stage driving a speaker of 8 ohms, a 1 mA biasing current is extremely low considering that more than 1 A can flow in normal operation. \$\endgroup\$ – Bimpelrekkie Oct 16 '19 at 11:49
  • \$\begingroup\$ let Ic*Rc = V+/2 then let Ib = Ic/ hFE (100 ish). Vb will be <600 mV with a current limiting Bias from Vref with R divider Ohm’Law then AC couple In faster than input swing V/ T(=RC ) slew rate. Add Re and choose gain = Rc/Re \$\endgroup\$ – Tony Stewart EE75 Oct 16 '19 at 12:03
  • \$\begingroup\$ If you are stuck using a specific BJT (because that's already been decided for you), then you can find the range of collector currents over which it is useful (typically, this is two to three orders of magnitude range -- so you have many options, still.) It's possible you have a specific battery system (button battery or 9 V battery) and this may, taking into account the entire circuit functions, also suggest some limiting boundaries. (A teacher could demand the value.) But if you were to write down a list of specs, I think others may help you re-orient yourself with respect to this question. \$\endgroup\$ – jonk Oct 16 '19 at 20:01
  • \$\begingroup\$ @Andyaka I also, wouldn't firstly choose Icq. But these are steps we have to follow, so I just trying to figure out the justification behind this assumption. \$\endgroup\$ – Edmund Blackadder Oct 22 '19 at 2:22

This circuit will give you 1MHz bandwidth, and drive a 100Kohm load with less than 10% error because of the load resistor.

Your gain (notice I made this YOUR circuit) will be 10K||13K / (1K + 26 ohms) where the 26 ohms is the small_signal gradient of the diode equation for certain diode dopings, also useful for transistor transconductance thinking.


simulate this circuit – Schematic created using CircuitLab

What dos this schematic illustrate?

--- the 100K load resistor is 10x the Collector resistor, thus there is less than 1 dB gain error (1dB = 12% voltage ratio)

--- the bandwidth (at the collector), with NO ALLOWANCE for any Cload across the 100Kohm, is approx. 10Kohm in parallel with 10picoFarad, the Cob of 2N3904 (from my memory). In truth we should include the DC_biasing feedback resistor of 13K to be in parallel with the 10K. Ignoring that for now, as we approximate the collector TAU to arrive at the bandwidth, we have 10K ohm * 10pF = 100 nanosecond. Since this is a 1-pole approximation, just invert that and divide by 6.28 to have bandwidth being 1.59MHz.

--- the gain is roughly Rcollector/Remitter (10Kohm/1Kohm); for more accuracy, read the 2nd paragraph of this answer

--- input resistance? so you might compute loading effects on any sensor or voltage source such as a prior stage? R4 || R5 || (Beta * R3)

--- input impedance? Cmiller will make the Cin be (1 + Av) * 10pF; you can cascode this circuit, and reduce that to (1 + 1) * 10pF

--- distortion? if I recall rightly, 4 milliVolts across the emitter-base will produce 10% distortion of collector current; however there is a linearization resistor in the emitter, of value 1,000 ohms. The ratio of 1,000 / 26 is the distortion-improvement ratio, which is 39:1. Thus the distortion for 4 milliVolts input (across EB and Remitter) will be 10% / 39 or 0.25%.

Again this is for 4 milliVolts input. For fun, set up a SPICE simulation, alter the Vin to be 4mVPP, include .fourier behavior, and examine the result.

  • \$\begingroup\$ Thanks for the feedback but I not really looking for a solution, I am trying to understand why a value of 1mA is chosen. Does this come from the datasheet and if so, what specifically am I looking at to find this value. What if I choose something like 10 mA, how would that affect the result. \$\endgroup\$ – Edmund Blackadder Oct 22 '19 at 2:20
  • \$\begingroup\$ given your load of 100,000 ohms, at 1mA the voltage would be 100 volts. Since that is a very large voltage, we can assume only a small portion of the collector current will go thru the Rload, and the rest will continue thru the collector resistor. This, most of the current remaining in the collector resistor, is important for low or moderate distortion, and I assume that is also one of your goals. \$\endgroup\$ – analogsystemsrf Oct 25 '19 at 10:48

1mA is just a typical and convenient value for the collector current.

The higher the chosen collector current, the lower can be the value of the collector resistor to achieve mid-supply dc biasing. Since the collector resistor nominally defines the output resistance of a common emitter amplifier it means that a lower value collector resistor results in less reduction in the amplifiers gain due to the potential divider formed between the collector resistor (output resistance) and the load resistance which is often the input resistance of the following stage.

The drawback of a higher chosen collector current is that more power is dissipated in the amplifier and so choosing the collector current is a trade off between gain and power dissipated (power drawn from supply).

  • \$\begingroup\$ 1mA is a convenient value for a small signal amplifier. Now imagine the collector current in a common emitter power amplifier. 1mA isn't going to cut it for a 1 watt output amplifier unless you use an awful high voltage power supply. \$\endgroup\$ – JRE May 7 at 15:39

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