Chapter-5 HW

#2- Design a variable voltage power supply capable of supplying (5-15) V at > 250 ma.

- Using a stepdown transformer we take 120VAC to 21.2VAC @ 60Hz. and Voltage Divider

- 21.2VAC = 15Vp               Vp = sqrt(2) Vrms

- Period T = 1/60Hz = 16.7 ms = 0.0167 s

- RC >>T to reduce ripple. Pick RC = 10T = 0.167 s

- Pick total R = R1 + R2 to supply 250mA at 15.V.    R = 15V/0.25A = 60 Ohm

- Select a voltage divider with R1 = (0-40) Ohm and R2=20 Ohm

- Select RC = 0.167 or C =0.167/60Ohm = 2.8 mF

In this design the current supplied to the load will depend on the load impedance! Better solutions
exist!

#3- (a) What happens if the load resistor is shorted. (b)What happens if the load resistor is removed?

(a) If RL =0 the zener would be bypassed and not affected. The 100Ohm would have a 12V drop
with i=12/100 = 120mA. Power = V^2/R = 1.44W. The 100Ohm (1/4W) resistor would be DAMAGED!

(b) VZ= 5.1V
V100 = (12-5.1)V = 6.9 V    Voltage drop across 100Ohm
i = V100 / 100Ohm = 69mA   Current thru 100Ohm

P100 = i V100 = 0.48W DAMAGE  Power across 100Ohm

PZ = i VZ = 0.35 W DAMAGE  Power across Zener

#6-  Design a transformerless 300V power supply with ripple <1% capable of supplying 100mA.

Use a voltage doubler circuit with two diodes.

Vp = sqrt(2) Vrms = 170V

Vout = 2 Vin = 340Vp

C = i / DV f = 0.1/(3/4)(60) = 490uF

#12- 5Vp is applied to the diode circuits. What Vout is expected?

(A) On the positive 1half-cycle the diode and Zener are passing current with a ~0.6V drop.
On the negative half-cycle the diode blocks and the zener limits V to -6.1V. Since V =Vp sin(wt)
the negative half-cycle passes unclipped.

(B) On the positive half-cycle the diode conducts with ~0.6V drop so Vout = Vp-0.6= 4.4V .
On the negative half-cycle the diode clips the signal to virtually zero current.

(C) In the diode clamp the polarized capacitor will charge on the positive half-cycle only.
The voltage at the top of the diode is negative (negative side of the capacitor). The DC level
floats to approximately -Vp+0.6V = -5.4V . Vout = -5.4V + 5 sin(wt).

(D) On the positive half-cycle the first diode does not conduct until Vin > 3V + 0.6V = 3.6V.
Vout rises and then plateaus at +3.6V. then falls back to zero.
On the negative half-cycle the second diode conducts.when Vin < -5V - 0.6V < -5.6V
is available.