Okay, so why is it that Vc assumed to be 12v - 0.7v (D1), how about contribution from the 9v battery?? Why is this not affecting the voltage at node C?
Why Vc isn't 9v - 0.7v (D2) = 8.3v?
Even if no current flows, voltage would still be there, no? Infinite resistor would have a voltage across it...A reverse biassed diode drops whatever voltage is available, no current flows.
As no current flows through a reverse biassed diode, it effectively disconnects part of the circuit, a bit like an infinite resistor or insulator.
I believe so, seeing as how potential difference does not actually require current to flow, since electrical fields can exist in a vacuum. Voltage just represents stored energy that could be put to work if it was released.Even if no current flows, voltage would still be there, no? Infinite resistor would have a voltage across it...
Vc is only affected by D1 but not D2, why? There's voltage across both diodes... Even though D2 is off.There is zero current but there is still a voltage gradient and field. The circuit node that the turned off diode is connected to has a very low resistance which sets the current and so sets the voltage gradient across the reverse biased diode.
The voltages are determined by the currents flowing through the components.Vc is only affected by D1 but not D2, why? There's voltage across both diodes... Even though D2 is off.
Wait a sec, so for negative clipper in post #7The voltages are determined by the currents flowing through the components.
If a diode is reverse biassed, no current flows, the voltage across the diode is irrelevant and the reverse biassed diode can be removed from the circuit diagram without changing the analysis.
The minimum output voltage, Vo, will be Vref - Vd.So why is Vo = -Vd + Vref. Wouldn't Vi affect it too?
Vo must be the 3V battery voltage because that is what the output is connected to. How could it be different ?So in this simple circuit, Vo = 3v?
so if R1 = 1k ohm, IR would be = (5v-3v)/1000 = 2 mA, current across R1.The battery is being charged by the 5V source through R1. Charge current = (5V - 3V) / R1
Why not.See, I thought that voltage at that point would be the same as Vo, but apparently not ...
Well then Vo should be 5V-3V = 2V not 3V. Don't you see the contradiction?Why not.
I do not follow your logic and cannot see why you have that opinion.Well then Vo should be 5V-3V = 2V not 3V. Don't you see the contradiction?
Measured relative to what. You show Vo is clearly being measured relative to ground.Where is for Vo Potential difference is measured between one voltage source only, which is 3V...
The definition of an ideal battery is one that maintains its voltage, whatever current is taken from it. Your arrow (Vo) must be at 3V. Quite incidentally, current will flow through R1, given by 2/R1.so if R1 = 1k ohm, IR would be = (5v-3v)/1000 = 2 mA, current across R1.
Voltage across R1 is 5V-3V = 2V
But at this point here:
Isn't the voltage also 5V-3V=2V?
See, I thought that voltage at that point would be the same as Vo, but apparently not ...