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An emitter follower with 12V on the base with have about 11.3V at the emitter, which is sufficient to operate the LM7805. The means the collector emitter voltage will be about 0.7V but that's not a problem..................
Yes, a diode does need to be added to protect the transistor on shutoff. But the diode drops I was referring to at the collector are for something else.
An NPN can only turn on if the base is at a higher voltage than the emitter. Higher by the amount of B-E voltage drop (about 0.7V). This is why NPN's are used for low-side switching, not high-side switching like being done here. If the NPN is "full on" then the emitter voltage is only slightly below the collector voltage. So almost 12V. This means the base has to be at least 12.7V, which won't happen in your setup. And definitely won't happen with R1 and R2 creating a voltage divider.
So my thinking was to drop some voltage at the collector with diodes so that the emitter is low enough for the base to work.
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The 100uF cap will indeed discharge through the voltage regulator when the NPN is turned off, with a time interval depending upon the load, but until it's discharged, the base-emitter will be momentarily reverse-biased with a voltage greater then the typical Vbe breakdown voltage. Thus it is a danger to the NPN and could damage the junction, which is why you need the diode.....................
The 100μF smoothing cap (or whatever value) will discharge through the voltage regulator and is not a danger to the NPN - no need for a protective diode. No harm, though, if you feel the need to add one anyway. .
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I stand corrected.The 100uF cap will indeed discharge through the voltage regulator when the NPN is turned off, with a time interval depending upon the load, but until it's discharged, the base-emitter will be momentarily reverse-biased with a voltage greater then the typical Vbe breakdown voltage. Thus it is a danger to the NPN and could damage the junction, which is why you need the diode.