Question

Use IS=10-15 A, BF=100, BR=5 for all problems except for those involving a commercial part (e.g. Problem 6, 2N3904) for both hand calculation and simulation. You are asked to calculate VBE / VEB using the IS given. EB junction turn on voltage is 0.0258*In(IC/IS). So do NOT make the 0.7V assumption in design. For analysis, you can use 0.7V approximation for turn-on.

Answer 1

Vp Vp + RC 5V 100 2+ D1 NSCW100 2- 1+ RB DIO_0 Q1 2N3904 DIO_0 6.8k + OV .tran 0 100m 0

Vidio_0 Transistor is.in cut-off model. -D10-0-OV and-Re=100-ohm 49.960907ms, OV V(2) 1.0mV 0.6mV+ 0.2mv- -0.2mV- -0.6mV+ -1.0mV- -4.99999V- 4.999971- 4.999951- 4.99993V- 4.99991V 4.999891 4.99987 74.40 V- 74.3014 74.2001 74.10 V- 74.000V 73.90 V- 5.005V- 49.960907ms,4.9999259V V[ 2,24 49.960907ms,74.144177V 5.000V- 4.994V- 5.0051 49.960907ms,5V Vivp) 5.000V+ 49.960907ms,5V 4.994V+ Oms 10ms 20ms 30ms 40ms 50ms 60ms 70ms 80ms 90ms 100ms

V[2) 41.8mV- 50ms,40.746309mV Transistor in saturation mode, Dio_o=5V, RC = 220 ohm 40.7mV- V12+,2.) 50ms, 3.3006014V 39.6mV. 3.3040V- 3.3026V- 3.3012V- 3.2998V- 3.2984- 3.2970V 5.005V- 50ms,5V 5.000V- 4.994V- 5.005V Vidio 0) 50ms,5V 5.000V- 4.9941+ Oms 10ms 20ms 30ms 40ms 50ms 60ms 70ms 80ms 90ms 100ms

V(2) 4.99998V- 49.960907ms,4.9999259V Transistor in off mode, Dio_O =OV, RC=220 ohm -4.99993- V[22,2) 49.960907ms,74.144177V -4.99987V- 74.400V 74.30 V- 74.20 V 74.10 V- 74.000V+ 73.90 V- 5.005V 49.960907ms,5V 5.000V- Vidio 0) 4.994V 1.0mV- 0.6mv- 49.960907ms,OV 0.2mV- -0.2mv- -0.6mV- -1.0mV- Oms 10ms 20ms 30ms 40ms 50ms 60ms 70ms 80ms 90ms 100ms

Vp Vp + 5V RC 220 2+ D1 NSCW100 2- 1+ RB DIO_O Q1 2N3904 DIO_0 6.8k OV 1- .tran 0 100m 0

V(2) 57.4mV- 50ms,56.32703mV Transistor is in saturation mode. Dio_0=5V, RC = 100 ahm 56.3mV+ 55.2mV- 3.5189V- V12-.2.) 50ms, 3.5148607V 3.5151V- 3.5112V- 5.005V 50ms,5V 5.000V- 4.994V- 5.005V- Vidio O) 50ms,5V 5.000V. 4.9941+ Oms 10ms 20ms 30ms 40ms 50ms 60ms 70ms 80ms 90ms 100ms