Lecture 12: DC Analysis of BJT Circuits.

Transcription

1 Whites, 320 Lecture 12 Page 1 of 9 Lecture 12: D Analysis of JT ircuits. n this lecture we will consider a number of JT circuits and perform the D circuit analysis. For those circuits with an active mode JT, we ll assume that V = 0.7 V (npn) or V = 0.7 V (pnp). xample N12.1 (text example 5.4). ompute the node voltages and currents in the circuit below assuming β = 100. (Fig. 5.34a) We ll assume the device is operating in the active mode, then we ll check this assumption at the end of the problem by calculating the bias of the J and J. f the JT is in the active mode, V = 0.7 V then V 3.3 V = 4 V = 3.3 V and = = = 1 3 R With = α then ma Keith W. Whites

2 Whites, 320 Lecture 12 Page 2 of 9 β = 1 ma=0.99 ma β + 1 onsequently, using KVL 3 3 V = 10 R = = 5.3 V Finally, using KL + =, or = = = 0.01 ma Now we ll check to see if these values mean the JT is in the active mode (as assumed). V = = 1.3 V. This is greater than zero, which means the J is reversed biased. V = 0.7 V. This is greater than zero, which means the J is forward biased. ecause the J is reversed biased and the J is forward biased, the JT is operating in the active mode. Note that in the text, they show a technique for analyzing such circuits right on the circuit diagram in Fig. 5.34c. Very useful. (Fig. 5.34c)

3 Whites, 320 Lecture 12 Page 3 of 9 xample N12.2 (text example 5.5). Repeat the previous example but with V = 6 V. Assuming the JT is operating in the active mode: = α = ma = 1.58 ma kω = 2.57 V = 5.3 V 5.3 V = = 1.6 ma 3.3 kω From the last calculation V = 2.57 V V = 3.43 V. onsequently, the JT is not in the active mode because the J is forward biased. A better assumption is the transistor is operating in the saturation mode. We ll talk more about this later. For now, suffice it to say that in the saturation mode V 0.2 V (see sat Section 5.3.4). Assuming this and reanalyzing the circuit:

4 Whites, 320 Lecture 12 Page 4 of 9 = =0.96 ma 4700 = = 0.64 ma V = = 5.5 V V = 0.2 V sat = 5.3 V 5.3 V = = 1.6 ma 3.3 kω Notice that 0.96 = = This ratio is often called β forced. Observe that it s not equal to 100, as this ratio would be if the transistor were operating in the active mode (see Section 5.3.4). xample N12.3 (text example 5.7). ompute the node voltages and currents in the circuit below assuming β = 100. To begin, we ll assume the pnp transistor is operating in the active mode.

5 Whites, 320 Lecture 12 Page 5 of 9 = =4.65 ma 2000 = = 0.05 ma 0.7 V ma 1 kω= 5.4 V = α = ma = 4.6 ma Now check if the JT is in the active mode: J? Forward biased. J? Reversed biased. So the JT is in the active mode, as originally assumed. xample N12.4 (text exercise D5.25). Determine the largest R that can be used in the circuit below so that the JT remains in the active mode. (This circuit is very similar to the one in the previous example.)

6 Whites, 320 Lecture 12 Page 6 of 9 = 4.65 ma 0.7 V = α = 4.6 ma We ll begin by assuming the JT is operating in the active mode. n the active mode, the J needs to be reversed biased. The lowest voltage across this junction for operation in the active mode is V = 0 V = V = 0 V. Therefore, by KVL 10 + R = 0 or R = = = 2,174 Ω This value of R and smaller is required for the JT to operate in the active mode. xample N12.5 (text example 5.10). Determine the node voltages and currents in the circuit shown below. Assume the JT is operating in the active mode with β = 100.

7 Whites, 320 Lecture 12 Page 7 of 9 First, we ll use Thévenin s theorem to simplify the base circuit 15 V 100 k R TH + 50 k V TH - The Thévenin equivalent resistance and voltage are then R TH = 100 k Ω 50 kω= kω 50 and V TH = 15 = 5 V Using this Thévenin equivalent for the base circuit, the overall circuit is then

8 Whites, 320 Lecture 12 Page 8 of 9 15 V 5 V 33.3 k 5 k V KVL V 3 k V To find the emitter current, we ll apply KVL over the loop shown giving 3 5 = ,000 The quantity of interest is. With = β and = α for a JT in the active mode, we find β β = = = α α β ( β + 1) = β + 1 or ( ) Using this in the KVL equation = ,000( β + 1) With β = 100 then solving this equation we find 12.8 = β + 1 = 1.29 ma. = μa ( ) Next, by KL = = 1.29 mα 12.8 μa= 1.28 ma The node voltages are then

9 Whites, 320 Lecture 12 Page 9 of 9 V = 15 5 kω= 8.6 V V = 3 kω= 3.87 V V = kω= 4.57 V Lastly, let s check if the JT is operating in the active mode. V = V V = = 0.7 V. This is 0.7 V originally assumed for a forward biased J. V = V V = = 4.03 V. This is less than zero, which means the J is reversed biased. Therefore the JT is operating in the active mode, as originally assumed.