Service manual

STP 11-25R13-SM-TG
P - 2
a. A shorted Ql would cause excessive current to flow in the circuit, which may or may not cause
several associated malfunctions. At any rate the circuit would lose its ability to regulate the output
voltage. If the circuit were to continue to operate, the output voltage would be too high.
b. If D1 were to open, Rl would get very little current. Since it would drop very little voltage, the
base of Ql would get a much higher positive potential. Ql would approach saturation and the output
voltage would be nearly equal to the input.
c. A shorted Rl would cause the same net result, a high output voltage. An additional symptom
would be that the Zener diode would probably burn up.
d. A shorted Dl would cause a loss of the forward bias on Ql. Thus, Ql would approach cutoff and
the output voltage would drop to nearly zero. As with any short, a burnout might occur. In this case,
resistor Rl would be the most likely component to burn out.
e. An open Rl would also cause Ql to cut off, which would cause the output voltage to drop to
nearly zero.
P-4. Amplified Series Voltage Regulators.
a. If Dl were to open (Figure P-3), Rl would lose most of its current flow and would decrease to a
much lower value.
Figure P-3. Amplified Series Voltage Regulator
(1) Q2 and R2 would also lose most of their currents and their voltage drops would also decrease.
Since Dl is in series with these components, Dl would have to drop most of the input voltage; the base of
Ql is more positive than before.
(2) This results in an increase in the forward bias of Ql. Thus, E
Q1
decreases and so the load
device receives a larger share of the input voltage (the output voltage increases).
b. On the other hand, suppose that Dl shorts. The emitter of Q2 would be at a more negative
potential than before. This would be enough to cause Q2 to saturate and E
Q2
would decrease
substantially.
(1) Resistor Rl would have to drop all the input voltage by itself and so it might burn out if it has to
dissipate too much power.
(2) When Q2 saturates, its collector voltage decreases; and this causes the forward bias of Ql to
decrease because its base is less positive than before. A decrease in the forward bias of Ql would cause
it to drop a greater share of the input voltage. As a result, the output voltage would decrease.