# Next step electronics 3; A simple BJT transistor bias circuit

The first step in every transistor based amplifier design is to define its bias circuitry. Perhaps the most important parameter in the schematic is the DC current in the collector of the transistor, because it is directly proportional to its small-signal gain. In this post a simple bias method for bipolar junction transistors (BJT) is being shown, where the collector current is being set with relative accuracy.

The most unpredictable parameter of a BJT is its current gain (hfe). On the other hand, the supply voltage and the resistor values are, in general, constant (of course they can deviate from the nominal values, but not as much as hfe). So, we want to set the value of the collector current with respect to these predetermined parameters.

Let’s say we want a collector current of 5 mA with a supply voltage (VDD) equal to 5 Volts. In addition, the larger the voltage difference between the collector and emitter the better, in order to avoid saturation. Therefore, we can pick a collector voltage of 4 Volts and an emitter voltage of 1 Volt. That means

$R_1 = \frac{5-4}{5*10^{-3}} = 200 \Omega$

and

$R_4= \frac{1}{5*10^{-3}} = 200 \Omega$

assuming that the base current is much smaller than the collector current and, hence, the emitter current is approximately equal to the collector current.

We now have to determine $R_2$ and $R_3$ using the assumptions that we have made. For this amplifier, we are going to use the BF199 RF transistor. If you take a look at its datasheet, you will notice that its on base-emitter voltage is between 0.77 and 0.9 Volts. Maybe you will notice that they are not referred to these collector current and collector-emitter voltage conditions, but if it is operated in its forward-active region (and it will, since it is an amplifier), then it must lie somewhere in this interval. Let us assume a base-emitter voltage of 0.8 Volts. In that case, the voltage at its base is 1.8 Volts, because we have assumed a 1 Volt emitter voltage. However, this voltage depends on the transistor’s base current, which depends on the unpredictable parameter hfe, because

$V_B = I_{R_3} * R_3 = (I_{R_2} - I_{B}) * R_3$

That means that if we want to make the base voltage independent of the base current we must make $I_{R_2}$ much larger than $I_B$. Again, if we look at the transistors datasheet, we are guaranteed a minimum hfe of 40, so it would be adequate to also set $I_{R_2}$ to 5 mA, which means that

$R_2 = \frac{5-1.8}{5*10^{-3}} =640 \Omega$

and also

$R_3 = \frac{1.8}{5*10^{-3}} =360 \Omega$

Building the circuit on LTspice and running the simulation we get the following quite satisfactory results.