23-08-2020, 07:25 PM
Like Mark, I wasn't pleased to see them perpetuating the very poor current controlled model. It's bad mainly because the current gain of a transistor is very badly defined. The exact value must not be relied on when doing sensible design. They should also have said "hfe" or "beta" rather than gain because that's rather ambiguous.
At least this time they have specified that the base current is peak, rather than making you guess if they really mean RMS. So 10uA peak base current multiplied by gain of 100 is 1mA. Without a signal the collector will sit at 10V - 2mA x 1.5k = 10V -3V = 7V. 1mA peak signal will change the collector voltage by 1mA x 1.5k = 1.5V. This is an inverting amplifer so when the base current increases by 10uA the collector current increases hence the collector voltage will fall by 1.5V. On the other half cycle when the base current drops by 10uA peak the collector voltage will rise by 1.5V.
So the peak collector voltage will be 7V + 1.5V = 8.5V.
At least this time they have specified that the base current is peak, rather than making you guess if they really mean RMS. So 10uA peak base current multiplied by gain of 100 is 1mA. Without a signal the collector will sit at 10V - 2mA x 1.5k = 10V -3V = 7V. 1mA peak signal will change the collector voltage by 1mA x 1.5k = 1.5V. This is an inverting amplifer so when the base current increases by 10uA the collector current increases hence the collector voltage will fall by 1.5V. On the other half cycle when the base current drops by 10uA peak the collector voltage will rise by 1.5V.
So the peak collector voltage will be 7V + 1.5V = 8.5V.
www.borinsky.co.uk Jeffrey Borinsky www.becg.tv
