How does a resistor work in opamp scematic with transistors?
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The BJT starts to turn on when Vshunt- falls below Vshunt+. This BJT action is an attempt to make the op-amp input voltages (V+ and V-) the same. This is negative feedback in action.
In other words, the voltage across the 15k resistor (in the emitter) is at 0V when the shunt voltages are the same (i.e. no current flowing thru the shunt) and, as Vshunt- starts to fall relative to Vshunt+, the two op-amp inputs are forced equal by the BJT dumping current thru its emitter resistor. The BJT has no-choice because it is contained within the op-amp's negative feedback loop.
In other words, the voltage across the 15k resistor (in the emitter) is at 0V when the shunt voltages are the same (i.e. no current flowing thru the shunt) and, as Vshunt- starts to fall relative to Vshunt+, the two op-amp inputs are forced equal by the BJT dumping current thru its emitter resistor. The BJT has no-choice because it is contained within the op-amp's negative feedback loop.
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