A bipolar junction transistor, (BJT) is very versatile. It can be used in many ways, as an amplifier, a switch or an oscillator and many other uses too. Before an input signal is applied its operating conditions need to be set. This is achieved with a suitable bias circuit, some of which I will describe. A bias circuit allows the operating conditions of a transistor to be defined, so that it will operate over a pre-determined range. This is normally achieved by applying a small fixed dc voltage to the input terminals of a transistor.
Bias design can take a mathematical approach or can be simplified using transistor characteristic curves. The characteristic curves predict the performance of a BJT. There are three curves, an input characteristic curve, a transfer characteristic curve and an output characteristic curve. Of these curves, the most useful for amplifier design is the output characteristics curve. The output characteristic curves for a BJT are a graph displaying the output voltages and currents for different input currents. The linear (straight) part of the curve needs is utilized for an amplifier or oscillator. For use as a switch,a transistor is biased at the extremities of the graph, these conditions are known as "cut-off" and "saturation".Output Characteristic Curves
After the initial bend, the curves approximate a straight line. The slope or gradient of each line represents the output impedance, for a particular input base current. So what has all this got to do with biasing ? Take, for example the middle curve. The collector emitter voltage is displayed up to 20 volts. Let's assume that we have a single stage amplifier, working in common emitter mode, and the supply voltage is 10 volts. The output terminal is the collector, the input is the base, where do you set the bias conditions? The answer is anywhere on the flat part of the graph. However, imagine the bias is set so that the collector voltage is 2 volts. What happens if the output signal is 4 volts peak to peak ? Depending on whether the transistor used is a PNP or NPN, then one half cycle will be amplified cleanly, the other cycle will approach the limits of the power supply and will "clip". This is shown below :
The above diagram shows a 4 volt peak to peak waveform with clipping on the positive half cycle. This is caused by setting the bias at a value other than half the supply voltage.
The lower diagram shows the same amplifier, but here the bias is set so that collector voltage is half the value of the supply voltage. Hence, it is a good idea to set the bias for a single stage amplifier to half the supply voltage, as this allows maximum output voltage swing in both directions of an output waveform.
The base emitter voltage, Vbe, for a small signal transistor is typically quoted in many text books as either 0.6 V or 0.7 V Both values are an approximation,and as can be seen from the graph the value of Vbe varies with collector current, device type and temperature. With low base currents of 50uA or less, taking Vbe as 0.6 volts is a reasonable approximation. For higher base currents, and in switching circuits using Vbe as 0.7 V is a better approximation. In large power transistors, Vbe can be even and often be as high as 0.8 or 0.9V.
Simple Bias Circuit
The simplest bias circuit is shown below. It consists only of a fixed bias resistor and load resistor. The BJT is operating in common emitter mode. The dc current gain or beta, hFE is the ratio of dc collector current divided by dc base current. The BJT is a BC107A. The values of Rb and Rc can be determined by either mathematical approach or by using the output characteristic curves for the BC107A.
Quiescent Point (Q-Point)
The point Vo in the diagram above is where the output signal would be taken. For simplicity,the input signal and coupling capacitors have been omitted. For minimum distortion and clipping it is desirable to bias this point to half the supply voltage, 10 volts dc in this case. This is also known as the quiescent point. The ac output signal would then be superimposed on the dc bias voltage.The Q-point is sometimes indicated on the output characteristics curves for a transistor amplifier. The quiescent point also refers to the dc conditions (bias conditions) of a circuit without an input signal.
I have mentioned that setting the Q-point to half the supply voltage is a good idea. It gives a circuit the highest margin for overload. However, any amplifier will clip if the input amplitude exceeds the limit for which the circuit was designed. However, there are certain cases when it is not necessary to bias a stage to half the supply voltage. Examples would be an RF amplifier design where the input signal is in microvolts or millivolts. If the stage had a gain of 200 then the output (assuming a 2mV peak input) would only need to swing up and down 400mV about the Q-point. Hence a stage with a supply voltage of 12 volts could have its Q-point set at 10 volts or even 2 volts without problems. Another example would be a microphone stage where similar low level input signals are involved.
The collector voltage Vc for the simple bias design is 10 volt. The dc current gain, hFE for the BC107A is obtained from the manufacturers data sheets and varies between devices. A typical beta is around 290. Taking a base current of 20uA and reading values direct from the output curves, the collector current, for a collector emitter voltage of 10 volts is around 3.9mA. As hFE= Ic / Ib then a BC107A must have a beta of at least 3.9mA / 20uA = 195 to work with this circuit. Also, the base emitter voltage, Vbe is typically 0.6v. Knowing the above data and using ohm's law , values for Rb and Rc can be determined:
Rb = Vcc - Vbe / Ib = (20-0.6) / 20u = 970k use (1M)
Rc = Vc / Ic = 10 / 3.9m = 2.56K use (2.7K)
Without using the output characteristic curve, values for Rb and Rc can still be calculated. A value for hFE must be estimated first and a desired collector current. As hFE varies in each transistor the value chosen should be the lowest value from the manufacturers data sheets. he equations to use are:
Rc = Vc / Ic
Ib = Ic / hFE
Rb = Vcc - 0.6 / Ib.
Using the example above with Vcc=20 and hFE =195 yields the same values.Temperature Stability
As can be seen both Vq and Iq will vary over a wide range. This is the reason that this circuit is seldom used. It is clear that a different circuit arrangement is needed.
This is similar to the self bias circuit with one difference: the base resistor Rb is returned to the transistor collector instead of the supply voltage. The reason for this is simple; if the transistor used had a high current gain, then the collector voltage would fall. As Rb is connected to the collector then the base current would be reduced to counter the effect. If the transistor had a low value of beta, then the collector voltage would rise. This in turn provides more base current for the transistor to conduct harder and stabilize the q-point. The equations to calculate Rc and Rb follow:-
Rc = Vc / Ic
Rb = Vc - Vbe / Ib
as Ib = Ic / hFE then
Rb = (Vc -Vbe) * hFE/ Ic
Self stabilizing bias example:
A bias circuit is required to bias a transistor to half the supply voltage. A BC107A transistor with hfe of 200 is used and supply voltage, Vcc is 20 volts. The collector current is to be 1mA. The resistor values are:
Rc = Vc / Ic = 10 / 1mA = 10K
Rb = (Vc-Vbe)*hFE / Ic = (10-0.6)*200 /1mA = 1880k a 1.8M resistor is fine here.
Temperature Stability of Self Stabilizing Bias Circuit
This method of biasing is more resilient to changes of temperature as shown in the graph below. It is unlikely that anything you make will be tested under this extreme range of temperatures, however some parts of the world, for example Mongolia have Winters where temperatures plumit to -40° C and Summers that can reach +40° C ! If you live in an extreme climate then the effects of temperature must be taken into consideration. The results below show quiescent collector voltage and currents and can be compared to the simple bias circuit above.
Potential Divider Bias
This is the most widely used biasing scheme in general electronics. For a single stage amplifier this circuit offers the best resilience to temperature changes and variation in device characteristics. The disadvantage is that a couple of extra resistors are required, but this is outweighed by the advantage of excellent stability. The circuit is shown below:
Here R1 and R2 form a potential divider, which will fix the base potential of the transistor. The current through this bias chain is usually set at 10 times greater than the base current required by the transistor. The base emitter voltage drop of the transistor is approximated as 0.6 volt. There will also be a voltage drop across the emitter resistor, Re, this is generally set to about 10% of the supply voltage. The inclusion of this resistor also helps to stabilize the bias: If the temperature increases, then extra collector current will flow. If Ic increases, then so will Ie as Ie= Ib + Ic. The extra current flow through Re increases the voltage drop across this resistor reducing the effective base emitter voltage and therefore stabilizing the collector current. The equations follow:
Rc = Vc / Ic
Ie = Ib + Ic as Ic >> Ib then Ie ~ Ic
Ve = 10% * Vcc
Re= Ve / Ie
Vb = Ve + 0.6
R2 = Vb / 10 * Ib
R1 = Vcc-Vb / 10 * Ib
Using the values of the previous examples a direct comparison of stability can be demonstrated. The values are;
Vcc=20V, Vc=10V, Ic = 1mA, transistor is BC107A with hFE=195
Rc= Vc /Ic = 10 / 1m = 10k
Ve = 10% * 20 = 2V
Re = Ve / Ie= 2 / 1= 2k
Vb = 2+ 0.6 = 2.6V
Ib = Ic / hFE = 1 / 195 =0.005128mA
R2 = Vb / 10* Ib = 2.6 / 0.05128 = 50.7k use 47k
R1 = Vcc-Vb / 10 * Ib = (20-2.6) / 0.05128 = 339.3k use 330k
Using these values and plotting the change in quiescent conditions for Vc and Ic over a temperature range of -50 to +50 celcius is displayed below:
As shown above, this bias circuit offers the best stability against variations in Vc and Ic over a very wide temperature range. As the resistor values used were preferred values, then the quiescent point will be slightly different from the calculated value.