gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
230602-1627 EDT
Jraef:
I disagree with what you described.
An ideal semiconductor diode has a reverse impedance of infinity, and a forward impedance of 0 ohms.
A real world semiconductor diode has a reverse impedance very close to infinity up to some breakdown voltage. So it is quite similar to the ideal diode.
In the forward direction a diode has a small voltage drop that depends on the magnitude of forward current. In a useful working range. This is around 1 V for a silicon diode. Too much forward current will cause excessive diode heating and failure. The size of the semiconductor chip and its thermal conductivity, and heat sinking will determine current rating.
If the load on the diode is a pure resistance, then the output voltage across the load resistance will be close to a 1/2 sine wave.
If the load is a parallel combination of a resistor and capacitor, then depending upon the component values of the load you will get some stored energy on the capacitor ( means a voltage on the capacitor at the start of the next cycle ), and capacitor voltage will not drop to 0. The diode forward drop is still in the 1 V range, and on the next cycle the diode won't start to conduct until the source voltage exceeds the capacitor voltage by the diode drop. Thus, forward conduction is less than a half cycle. Make the capacitor big enough, and forward conduction will occur for a short time, and voltage ripple on the capacitor will be very small.
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Jraef:
I disagree with what you described.
An ideal semiconductor diode has a reverse impedance of infinity, and a forward impedance of 0 ohms.
A real world semiconductor diode has a reverse impedance very close to infinity up to some breakdown voltage. So it is quite similar to the ideal diode.
In the forward direction a diode has a small voltage drop that depends on the magnitude of forward current. In a useful working range. This is around 1 V for a silicon diode. Too much forward current will cause excessive diode heating and failure. The size of the semiconductor chip and its thermal conductivity, and heat sinking will determine current rating.
If the load on the diode is a pure resistance, then the output voltage across the load resistance will be close to a 1/2 sine wave.
If the load is a parallel combination of a resistor and capacitor, then depending upon the component values of the load you will get some stored energy on the capacitor ( means a voltage on the capacitor at the start of the next cycle ), and capacitor voltage will not drop to 0. The diode forward drop is still in the 1 V range, and on the next cycle the diode won't start to conduct until the source voltage exceeds the capacitor voltage by the diode drop. Thus, forward conduction is less than a half cycle. Make the capacitor big enough, and forward conduction will occur for a short time, and voltage ripple on the capacitor will be very small.
..