Terminator3
Advanced Member level 3
For example, we repairing some FET-based oscillator.
Biasing resistors used: 30 Ohm on drain and 400 Ohm on source. Supply voltage is 2v, so i guess that I=2v/430Ohm=0.0046=0.5mA
Then i measure Vgs, for example it is -1v. Then I=1/400=0.0025A. So the Q point is -1V / 0.0025A, and DC load up to 0.5mA... I know the numbers are wrong, but how about idea?
Biasing resistors used: 30 Ohm on drain and 400 Ohm on source. Supply voltage is 2v, so i guess that I=2v/430Ohm=0.0046=0.5mA
Then i measure Vgs, for example it is -1v. Then I=1/400=0.0025A. So the Q point is -1V / 0.0025A, and DC load up to 0.5mA... I know the numbers are wrong, but how about idea?