# Determining whether or not a BJT is in forward active

Hi, I'm having some trouble determining whether or not a BJT (used in an amplifier topology for example) is in forward active, given that we dont know what the base-emitter and base-collector voltages are (we might know how they are related). My textbook likes to argue: assume it is in forward active and then verify that the conditions for forward active mode hold true. This doesn't seem like the way you prove statements (show that true implies true) - rather we should assume that the equations for forward active mode do not hold and then show that the transistor is not in forward active. However, if we assume the equations for forward active mode do not hold, we have no further relations to relate the base-emitters and base-collector voltages to determine the mode of operation.....

dlgoff
Gold Member
Hi, I'm having some trouble determining whether or not a BJT (used in an amplifier topology for example) is in forward active, given that we dont know what the base-emitter and base-collector voltages are (we might know how they are related).

That's what volt meters are for. It's easy to measure these voltages and determine if it is forward or reverse biased.

My textbook likes to argue: assume it is in forward active and then verify that the conditions for forward active mode hold true

Without the whole context of your text, I believe the bold I added in your statement is not meant to be proved through transistor model equations. By assuming forward bias, you know the base-emitter junction voltage will be between 0.5 and 0.7 volts for NPN transistors. vk6kro
If you have no idea of the base emitter voltage of a BJT transistor, then you also have no idea if it is operating in linear mode or not.

However, philosophy aside, you can get this information quite easily. This is a practical subject and we can get predictable results.

If the transistor has a voltage of around +0.6 to +0.8 volts between base and emitter, then it is possible that the transistor is operating linearly. Outside that small range, the transistor will either be turned off (not drawing collector current) or it will be drawing all available collector current (ie it is saturated) or enough to place the transistor in danger of being destroyed.

The critical factor is that the collector circuit should have a source of current available.

If it has, then an increase in base current will be amplified in the transistor to give an increase in collector current.
This current amplification is fairly linear and predictable, while the relationship between the input voltage and collector current is far from linear or continuous.

Thank you for answering. That makes sense if I had physically built the circuit - I could then just let something happen, some voltages and currents will be established in the nodes, and then I can probe the base-emitter to see whether or not I am in forward active. However I don't see how it's possible to solve this sort of thing on paper (like in the form of a textbook problem). Do we need to use iteration and if so how do we do it?

If the transistor has a voltage of around +0.6 to +0.8 volts between base and emitter, then it is possible that the transistor is operating linearly. Outside that small range, the transistor will either be turned off (not drawing collector current) or it will be drawing all available collector current (ie it is saturated) or enough to place the transistor in danger of being destroyed.

I don't think that is quite correct.

You can run BJTs in reverse, by passing current from the base into the collector. Then, the voltage on the emitter ([let's talk npn's;] above the voltage on the base, which is ~0.7V above the collector) would flow down to the collector.

The Hfe in reverse bias is less than that in normal bias, but you can definitely have the voltage on the emitter higher than the base, if there is current flowing from the base into the collector. It is a symmetrical junction (npn or pnp) after all. The reduction in Hfe and is because you have a bigger np junction on one side than the other.

AlephZero
Homework Helper
However I don't see how it's possible to solve this sort of thing on paper (like in the form of a textbook problem). Do we need to use iteration and if so how do we do it?

If you use a general equivalent circuit for the transistor that would include both linear and nonlinear operation, then you won't be able to solve it "mathematically". Of course you can solve it numerically, which is what programs like SPICE are for.

But you can often get a to solution "on paper" by doing what your textbook says. There is a general point about solving any sort of complicated problem here. You can do anything you like to "guess" something that might be a solution, and then prove your "guess" really is a solution.

That's how your "textbook method" works. You guess the circuit might be operatiing linearly, and work out the consequences of that guess. Then finally, you show that all the currents and voltages in the circuit really are consistent with linear operation.

If it turns out they are not consistent, then your guess was wrong, the circuit is not operating linearly, and you have to start again with some different assumptions.

vk6kro
Thank you for answering. That makes sense if I had physically built the circuit - I could then just let something happen, some voltages and currents will be established in the nodes, and then I can probe the base-emitter to see whether or not I am in forward active. However I don't see how it's possible to solve this sort of thing on paper (like in the form of a textbook problem). Do we need to use iteration and if so how do we do it?

Getting the correct voltage across the base emitter junction is easier than it might seem.

This is because the voltage across this junction, or any silicon diode, will be about 0.6 volts if you send a reasonable level of current through it. You make the current reasonable by selecting a suitable series resistor.

For example, suppose you want a collector current of 10 mA and you know the transistor has a gain of 100, then the base current has to be 100 μA. ( 10 mA / 100 = 100 μA).
If you have a 10 volt supply, and assuming a drop of 0.6 volts across the B-E junction then a series resistor from the supply to the base will have 9.4 volts across it and 100 μA flowing in it.
So, it should be 94000 ohms. ( 9.4 volts / 0.0001 amps = 94 K ohms).
So, you might use an 82 K or a 100 K resistor because you can't normally buy 94 K resistors.

You can do this type of approximation or you could use a simulator like the excellent, free program called LTSpice. You would still need to adjust the final circuit to suit the actual components, but you would get quite close with a simulator.

dlgoff