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Electrical Engineering - The voltage across the emitter resistor in a Voltage divider Common Emitter Transistor
Expert: cleggsan - 10/29/2009
Question
How do we go about calculating the voltage across the emitter resistor? I have seen that it should be 1/10 the supply voltage for good Q point stabalisation. I would like to find out why is it 1/10 the the supply voltage. It would be great if you can show the working for this. Thank you in advance.
Answer
Can I assume you are talking about a Class A biased linear amplifier? If so, the correct biasing it to put the voltage at the collector at the midway between the max and min swing. The 1/10th rule is an approximation or rule of thumb - like so many things in the engineering world. It is a place to start, but is not a hard fast requirement. In fact, for very small signals and high gain where keeping noise low is necessary the bias may be tilted way down low to keep emitter current low for the sake of keeping noise low.
But back to best Q point. Assume the supply voltage is 10v. Assume the collector resistor is 10k and the emitter resistor is 1k. This is a very common case where the gain would be 10. So, the bias would be to put the collector voltage at 5.5 volts. This point allows the output swing to go from 5.5 to 10 up and 5.5 to 1 down for a total swing range of 9v p - p. The emitter voltage under no signal would be .55v.
Does it make sense??
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