Thanks, that's exactly what I wasn't sure about. So if the voltage at the node connecting the diodes is 0.7V, I would need to "step down" 0.7V to get to the V terminal right? So the voltage at V is 0 V?gneill said:
v1 = 0Eltahawy said:
Yeah it looks like v2 in (b) can't be 0 with both the diodes on because then the current I would be negative, which isn't consistent with the Constant Voltage Drop Model.rude man said:
You mean because two diodes have the same anode voltage ("shared node"), their cathode voltages have to be the same too?izelkay said:
"With diode 1 (left) off and diode 2 (right) on, the voltage drop across diode 1 is still 0.7 V because of the shared node with diode 2, but the current would be 0 A which isn't consistent with the model."rude man said:
The anode voltage of diode 1 is not 0.7V. Diode 1 is off so how can its anode be 0.7V more positive than its cathode?izelkay said:
I was under the impression that it'd be something like this:rude man said:
gneill said:
Nope. You've missed out the -5V source in the second term.izelkay said:
Hm okay if I move the 0.7 V source below that 5k resistor, would I need to "step up" by 0.7 V from the -5 V source?gneill said:
Your picture is correct.izelkay said:I was under the impression that it'd be something like this:
Even though the diode's off I thought the anode's voltage would still be 0.7 V because of the other diode's anode, and its cathode would be 0 V because of the ground. If this isn't correct what would it be?
Think I got it now. The current through the 2nd diode is 0.62 mA, and the cathode voltage of the 2nd diode is -1.2 - 0.7 = -1.9 V. (Solved for the anode voltage above).rude man said:
You got it 100%!izelkay said:
The CVD model is a simplified mathematical representation of a diode circuit that assumes a constant voltage drop across the diode, typically around 0.7 volts for silicon diodes and 0.3 volts for germanium diodes.
The CVD model is an approximation and may not accurately reflect the behavior of a real diode circuit in all cases. It is most useful for analyzing simple diode circuits with low currents.
The CVD model does not take into account factors such as temperature, reverse breakdown voltage, and leakage current, which can significantly affect the behavior of a diode circuit. It is also not suitable for analyzing circuits with high frequencies or high currents.
To calculate the voltage drop across a diode using the CVD model, simply subtract the CVD value (typically 0.7V for silicon diodes) from the voltage at the anode of the diode. This will give you an approximate value for the voltage at the cathode of the diode.
No, the CVD model is only suitable for simple diode circuits with low currents. For more complex circuits, a more detailed model that takes into account the non-linear behavior of diodes is required.