Assuming the current delivered is within the capabilities of the diode, then while the voltage is applied reverse biased to the diode, the diode voltage will be almost equal to the open-circuit voltage of the Thevenin equivalent ( ie a sinusiodal half cycle ).
22.
If the semiconductor is translucent, the junction becomes the source of light as it is emitted, thus becoming a light-emitting diode, but when the junction is reverse biased no light will be produced by the LED and, on the contrary, the device may also be damaged.
23.
:: : In circuit ( b ), the voltage across where D 1 would have been will be-5 / 3 V . So sticking D 1 back into the circuit will cause D 1 to be reverse biased, which is consisted with the hypothesis that D 1 isn't conducting, so you accept that initial hypothesis.
24.
Consider a positive-going swing : As long as the input is less than the required forward V BE drop ( H " 0.65 V ) of the upper NPN transistor, it will remain off or conduct very little-this is the same as a diode operation as far as the base circuit is concerned, and the output voltage does not follow the input ( the lower PNP transistor is still off because its base-emitter diode is being reverse biased by the positive-going input ).
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