HW#4 Solutions
Dr. Du
ME06/EE106 Mechatronics
Problem 1: 3.4b (P92) The following circuits are called clipping circuits. Assume ideal diodes and
sketch the output voltage Vo for two cycles of the input Vi.
Solution: the period of Vi is: T =
2π
ω
=
2π
= 1 (s) ,
2π
the input voltage is shown as:
Since the diode is ideal, the voltage drop across the diode is zero volt.
For Vi > +0.5V, the diode is reversed biased => the diode is OFF. No current will go through the circuit.
Vo = Vi- R x 0 = Vi.
For Vi < +0.5V, the diode is forward biased => the diode is ON. The current will go through the circuit.
Vo = +0.5V.
Input Waveform
Output Waveform
Problem 2 3.13 (P95) In the following circuit, find the minimum Vin required and the resulting voltage
Vout to put the transistor in full saturation. Assume that the beta for the transistor is 100 in full saturation.
Solution: When the min Vin that makes the transistor in full saturation, the following equations are true:
VCE = 0.2 V , VBE = 0.7 V , and
I C = 100 I B
IB
(1)
can be found from the input circuit:
IB =
Vin − VB Vin − (0.7 + Vout )
=
1000
RB
⇒
Vin = 1000 I B + 0.7 + Vout
(2)
I C can be found from the input circuit:
IC =
Vs − VC 5 − (0.2 + Vout )
=
1000
RC
⇒
Vout = 4.8 − 1000 I C
In addition:
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdusjsu.edu
(3)
Vout = I E Rout = ( I B + I C ) Rout = 1000( I B + I C )
Solve about 4 equations, we have: Vin =3.13 V and Vout =2.41V (see the figures below)
Problem 3 3.18 (P96)
[Solution] The upper FET is a p-channel enhancement mode MOSFET and the lower is an n-channel
enhancement mode MOSFET. When Vin = 5V, the upper MOSFET doesn’t conduct (since it needs a
negative voltage to turn on) but the bottom one does, so Vout =0V(ground).
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdusjsu.edu
(4)