PHY405F 2009 EXPERIMENT 6 SIMPLE TRANSISTOR CIRCUITS

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1 PHY405F 2009 EXPERIMENT 6 SIMPLE TRANSISTOR CIRCUITS Due Date (NOTE CHANGE): Thursday, Nov 12 5 pm; Late penalty in effect! Most active electronic devices are based on the transistor as the fundamental solid state component. We want to help you develop some understanding of its characteristics. The most common form of transistor is called a Bipolar Junction Transistor (BJT). There are two types of BJT, NPN and PNP, with different circuit symbols. The letters refer to the solid state properties of the layers of semiconductor material used to make the transistor. Most transistors used today are NPN because they are the easier type to manufacture from silicon. The leads are labelled base (B), collector (C) and emitter (E). The three leads which must be connected correctly in any circuit before switching on any power supplies. A wrongly connected transistor will be damaged instantly. You will find the appropriate locations of the three wire leads on the plastic package by consulting the spec sheet of the transistor you use. WHAT TO DO: 1. Testing transistors. Select an NPN and a PNP transistor pair such as the 2N3904 and the 2N3906. You will need a basic multimeter. Set a digital multimeter to diode test ; an analogue multimeter to a low resistance range.

2 Test each pair of leads in both directions (six tests in total). The base-emitter (BE) junction should behave like a diode and conduct one way only. The base-collector (BC) junction should behave like a diode and conduct one way only. The collector-emitter (CE) should not conduct either way. The diagram shows how the junctions behave in an NPN transistor. The diodes are reversed in a PNP transistor but the same test procedure can be used. Record all your data. 2. The transistor switch. The transistor is basically an amplifier of current. A small current injected between the base and the emitter encourages a larger current to flow from the collector to the emitter. The phenomenon can be easily illustrated by constructing the circuit below. Connect an NPN transistor into the circuit shown above which uses the transistor as a switch. The supply voltage is not critical, anything between 5 and 12V is suitable. The LED should light when the switch is pressed and go out when the switch is released. Repeat the experiment using a PNP transistor in the same circuit but reverse the LED and the supply voltage. For both experiments measure the current into the base and collector and the current out of the emitter. Also record the base-emitter and collector-emitter voltages in the two states. Increase the value of the 10k resistor until the LED no longer switches. Deduce the maximum current gain of the transistor. 3. The Common Emitter AC Amplifier The basic style of a BJT common emitter amplifier is shown in the figure below. Let us go through the design of the amplifier and the selection of components. I used the following reference

3 Choose the transistor and the voltage supply Vcc I suggest 2N3904 and 12 VDC. The quiescent base voltage (Vb) is determined by the potential divider network formed by the resistors, R1 and R2 and the power supply voltage Vcc and is given as: The supply voltage also determines the collector current, Ic when the transistor is fully "ON" (saturation), Vce = 0. The base current Ib for the transistor is found from the collector current, Ic and the DC current gain beta, β of the transistor. The next steps require the family of characteristic curves for the transistor shown below. The curves are determined experimentally. They consist of plots of the collector current against the collector emitter voltage for range of values of the base current.

4 Lets assume a typical load resistor, RL of 1.2kΩ. We can calculate the collector current (Ic) flowing through the load resistor when the transistor is switched fully "ON" by assuming Vce is near zero. Given Ic, we can find an Re with a voltage drop of say 1V across it. We can plot a load line on the characteristic curves. The point "A" on the collector current vertical axis of the characteristic curves occurs when Vce = 0 and IC=9.2 ma. Now when the transistor is switched fully "OFF", there is no voltage drop across either resistor Re or RL as no current is flowing through them. Then the voltage drop across the transistor, Vce is equal to the supply voltage, Vcc. This is point "B" on the horizontal axis of the characteristic curves. Generally, the operating or Q-point of the amplifier is selected to be half-way along the load line so the collector current will be given as half of 9.2mA. Therefore, at Q, IC= 4.6mA. The DC load line produces a straight line equation whose slope is given as: -1/(RL + Re). It crosses the vertical Ic axis at a point equal to Vcc/(RL + Re). The actual position of the Q-point on the DC load line is determined by the mean value of Ib. The collector current of the transistor, Ic is also equal to the product of the DC gain beta and the base current (β x Ib). If we assume a (minimum) β value for the transistor of say 100, the base current Ib flowing into the transistor will be given as: Resistors, R1 and R2 are chosen to give a quiescent base current of 92uA. The current flowing through the potential divider circuit has to be large compared to the actual base current Ib. Assume a value of 10 times Ib flowing through the resistor R2 (rule of thumb). Then the value of R2 is given by: If the current flowing through resistor R2 is 10 times the value of the base current, then the current flowing through resistor R1 must be 11 times the value of the base current at 1012uA or 1mA. The voltage across resistor R1 is equal to Vcc - 1.7v (Vre = 0.7 for silicon transistors) which equals 10.3V, therefore R1 is given as:

5 So, for our example above, the preferred values of the resistors chosen to give a tolerance of 5% are: Then, our original Common Emitter circuit above can triumphantly be rewritten to include the values of the components that we have just calculated. In the old days, hours were spent on these kinds of calculations. Try to reproduce the results using electronic workbench. Build the circuit and measure the quiescent (DC) operating voltages and currents. Do they differ from those calculated? If so, what erroneous assumption may have been made? What value of C1 should therefore be selected to avoid any attenuation of signals above 100 Hz? Measure the input impedance of the circuit? What should the value of C2 be to bypass AC signals above 100 Hz around the resistor Re? If C2 is included in the load line calculation, is the resulting AC load line different from the DC load line we calculated? Measure the AC gain of the circuit above 100 Hz and the greatest sinusoidal input voltage that can be amplified without any distortion of the output signal. Measure the effect of removing C2 on the gain of the circuit?