Transistors. Lesson #9 Chapter 4. BME 372 Electronics I J.Schesser


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1 Transistors Lesson #9 hapter 4 252
2 JT egions of Operation Saturation Active i amps i =50 ma 40 ma 30 ma 20 ma 10 ma 0 ma v volts utoff There are three regions of operation for a JT: 1. utoff egion where no current flows in the collector or base circuits. When a transistor is in this state, we say that the transistor is OFF 2. Active egion where amplification takes place 3. Saturation egion where the maximum current flows in the collector for a given load. 253
3 JT egions of Operation Saturation i amps i =50 µa 0 µa v volts 40 µa Active 30 µa 20 µa 10 µa utoff 1. utoff: both junctions are reverse biased, i = 0, v <V FThreshold, i =0, v =V 2. Active: mitterase junction forward biased, ollector ase junction reverse biased, i >0, v V FThreshold, i = βi >0, v =V i, where is the series (load) resistor in the collector circuit. 3. Saturation: both junctions are forward biased, i >0, v V Threshold, v = V Saturation 0 < v, i =(V V Saturation )/, i < βi, where is the series (load) resistor in the collector circuit. 254
4 Large Scale D Models For each of these regions, we can define an analysis model to aide in the D operation of the transistor 255
5 Large Scale D Models utoff egion npn pnp V < V FThreshold 0.6 V V < V FThreshold 0.6 V V > V FThreshold 0.6 V V > V FThreshold 0.6 V 256
6 Large Scale D Models Active egion npn pnp β β 0.7V I 0.7V I > 0; V 0.7V > V FThreshold > 0; V 0.7V > V FThreshold = b ; V > 0.2 V = b ; V < 0.2 V 257
7 Large Scale D Models Saturation egion npn 0.7V 0.2V I pnp 0.7V 0.2V I > 0; V 0.7V > V FThreshold 0 < < β ; V 0.2 V > 0; V 0.7V > V FThreshold 0 < < β ; V 0.2 V 258
8 An xample 200k This circuit uses one D source to set the Q point. The term, ias point, is synonymous with Q point. Therefore we call this circuit a selfbias circuit V β = 100 V Let s first assume that this circuit is in cutoff. We will use the model for cutoff 259
9 An xample If this is in cutoff, then V = 15V. ut this can t be since to be in cutoff V < V FThreshold k β = 100 V V 200k 1k =0 =0 V V 260
10 An xample 200k 1k β = 100 Now let s assume that this circuit is in the saturation region. And we will use the model for the saturation region V V 261
11 200k An xample If this is in the saturation region, then i = (15 0.7) / 200k =71.5m A i = (15 0.2) / 1k =14.8 m A β =100 * 71.5m = 7.15 ma < i NOT OK!!!! β = 100 V V 200k 0.2V V I<b ic 0.7V V 262
12 An xample 200k 1k β = 100 Now let s assume that this circuit is in the active region. And we will use the model for the active region V V 263
13 200k 1k β = 100 An xample If this is in the active region, then i = (15 0.7) / 200k =71.5m A i = 100 * 71.5m A = 7.15 ma V =15 (7.15 m * 1k) = 7.85 V > 0.2V OK!!!! V V 200k 0.7V I=βIc V V 264
14 Another Selfias ircuit xample 1 10k 2 V V 5k 1k β = 300 I This is another selfbias circuit which uses 4 resistors to set the Qpoint. Let s redraw this circuit to see how to analyze it! 265
15 Another Selfias ircuit xample Use Thevenin s equivalent on the redrawn circuit to yield another simpler circuit to analyze 1 10k 1 β = V V 5k 1k β = 300 I 10k 2 V V 5k 1k I 266
16 Another Selfias ircuit xample Use Thevenin s equivalent on the redrawn circuit to yield another simpler circuit to analyze 1 10k 2 β = 300 V V 5k 1k I V TH TH V β = 300 V 1k I 267
17 V TH Another Selfias ircuit xample Now let s assume the active region, replace the transistor with the active region model, and analyze the circuit. First, the Thevenin s quivalent: TH β V V I I V β = 300 V TH 1k I = = I TH = βi = V TH V V T (1 β ) = 4.24mA; I I V I I V TH TH = = V = 15 = 5V k = 5 10 = 3.33k Next, applying KVL to the ase And ollector ircuits: = I TH = 4.25mA V = 3.33k (301)1 k ( I I ) = 14.1µ A V = m 1k 4.25m 1k = 6.51V = I TH (1 β ) I 268
18 Homework Probs. 4.26, 4.27, 4.28, 4.29, 4.30, 4.33, 4.34,
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