The Bipolar Junction Transistor- Small Signal Characteristics Debapratim Ghosh deba21pratim@gmail.com Electronic Systems Group Department of Electrical Engineering Indian Institute of Technology Bombay 8 September 2012 Debapratim Ghosh Dept. of EE, IIT Bombay 1/13
A Quick Recap, and To-Do Problem Statement Modeling the BJT The BJT as a device A three terminal device- Emitter (E), Base (B) and Collector (C). Looking into each terminal gives us interesting information. An Amplifier using the BJT A Common Emitter (CE) amplifier is a two-port network. i/p CE Amplifier o/p The AC input is given on the B-E side, and output is measured on the C-E side. Forward-active region- B-E forward biased, and C-B reverse-biased Using the Ebers-Moll model in this region Debapratim Ghosh Dept. of EE, IIT Bombay 2/13
The Ebers-Moll Model Problem Statement Modeling the BJT The B-E junction under forward bias is characterized by I E = I ES (e V BE V T 1) (1) Equation(1) shows that the B-E junction resembles a diode. The C-B junction under reverse bias is characterized by I C = αi E = αi ES (e V BE V T 1) = βi B (2) Equation(2) shows that I C is a current-controlled current source. C C B B E E Debapratim Ghosh Dept. of EE, IIT Bombay 3/13
Problem Statement Modeling the BJT The Complete BJT Small Signal Model Q: What is a small signal? A: Once a circuit is biased with DC voltages, an added AC signal small in amplitude not to disturb the bias conditions is called a small signal. B r bb B c b c C i b r b c r b e c b e βi b r o E Resistance r b c can be ignored. Why? At low frequencies, the capacitances c b e and c b c can be ignored. Why? Debapratim Ghosh Dept. of EE, IIT Bombay 4/13
Interpreting the Model Problem Statement Modeling the BJT Once r b c, c b e and c b c are ignored, the model becomes much simpler. B r bb B C r b e i b βi b r o The notation i b indicates the small signal (AC) base current. A similar lowercase notation follows for small signal voltages as well. Here, β is the AC current gain (different from DC β). Define the input resistance r i = r bb +r b e. The output resistance r o is a consequence of the Early effect. E Debapratim Ghosh Dept. of EE, IIT Bombay 5/13
Problem Statement Modeling the BJT Procedure to Obtain the AC Model of a Circuit 1. Do a DC analysis. Compute all DC currents and voltages, e.g. I B, I C, V CE etc. 2. Calculate r b e = V T IB. It is difficult to calculate r bb analytically. 3. If the Early Voltage V A is known, find r o = V A I C. 4. For AC analysis, deactivate all DC sources. Now apply the model to the rest of the circuit. Exercise: For the circuit below, construct the small signal AC circuit. V CC R s R 1 C 1 R C C L V in R 2 R L Debapratim Ghosh Dept. of EE, IIT Bombay 6/13
Experiment- Part 1 In this part, we determine the input resistance of an npn transistor BC547. Use the circuit shown. V CC = 10V 1MΩ 47kΩ 180kΩ 1kΩ 10µF V o V in 10µF 47Ω Debapratim Ghosh Dept. of EE, IIT Bombay 7/13
Steps for Part 1 1. Set the DC operating point (0.6 V BE 0.8V and 4 V CE 6V). 2. Now go for AC analysis. Vary V in to get v be in the range 20 100mV. 3. Clearly, r i = v be V in v be (3) R s where R s = 47kΩ. 4. Vary I B (using the 1MΩ pot) to get different values of r i. Tabulate r i v/s I B. 5. We know that, r i = r bb +r b e = r bb + V T IB. 6. Plot a graph of r i v/s I 1 B to find r bb. Subtract from r i to obtain r b e. r i r bb I B 1 Debapratim Ghosh Dept. of EE, IIT Bombay 8/13
Experiment- Part 2 In this part, we determine the small signal AC current gain (β) of the BC547. 1. Use the same circuit as in Part 1. 2. From the small signal model, we can show that 3. Again, since R C R L, i b = V in v be R s (4) i c = V o R L (5) 4. And of course, β = ic i b. Exercise: The transconductance g m of a BJT is defined as g m = i c (6) IB v be Derive an expression for g m in terms of the device small signal parameters. Also, find g m while you do the experiment. Debapratim Ghosh Dept. of EE, IIT Bombay 9/13
Experiment- Part 3 In this part, we determine the output resistance r o of the BJT. Shown below is the small signal model of the circuit on the output side. V o i c R C r o R L Finding r o becomes easier when the effect of R C is minimized. This will happen when R C. But we can t just open R C - it will disturb the DC bias point! Solution- Use a 5H inductor in series with R C. At DC, it gives zero reactance and at 10 khz, a reactance of over 300kΩ! This nearly creates an open circuit in the R C branch. Debapratim Ghosh Dept. of EE, IIT Bombay 10/13
Experiment- Part 3 (cont d) The effective circuit now becomes simpler. V o i c r o R L 1. First, open circuit R L, i.e R L. Note the output voltage V o as V oc. 2. Now, connect R L and vary it till V o = Voc 2. This will happen when R L = r o. 3. The value of R L now gives the value of r o! 4. The final circuit you need to use is only a small modified version from the one used in Parts 1 and 2. Debapratim Ghosh Dept. of EE, IIT Bombay 11/13
Experiment- Part 3 (cont d) Circuit to be used in this part of the experiment V CC = 10V 1MΩ 47kΩ 180kΩ 5H 1kΩ 10µF V o V in 10µF R L Debapratim Ghosh Dept. of EE, IIT Bombay 12/13
Food for Throught * Study the r i v/s I 1 B plot obtained after doing Part 1 of the experiment. What does the slope of the graph give us? * For the circuit in Part 1, construct the small signal equivalent circuit. Prove that the voltage gain A v = V out /V in < 1, for the given component values. * The point B corresponds to a hypothetical point nside the base, and r bb is the resistance between the base terminal lead and the bulk base material. Why then, don t we have a similar point for the emitter and collector as well, i.e. E and C? * The expected value of r o is usually 40 50kΩ. Compare this with the observed r o. Give reasons for the difference observed in the values. * In an ideal BJT, what values would we expect for the following quantities? (i) r bb, r b e and hence, r i (ii) r o (iii) g m Debapratim Ghosh Dept. of EE, IIT Bombay 13/13