In a switching DC-DC voltage converter, the oscillatory nature of the switching circuit generates a small "ripple" effect in the output voltage which is supposed to be minimized via careful design of the overall circuit. The output current of this type of converter typically flows through a diode into the rest of the system. The voltage measured at the cathode of this diode will exhibit the aforementioned ripple.
Measuring ripple frequency would determine if a diode were open in a bridge rectifier circuit because the ripple frequency is normally twice the input frequency in a functioning full wave bridge rectifier. If one diode were open, the ripple frequency would only be the input frequency. Note: This is true for single phase or bi-phase operation. Three phase operation is more complex, but still doable - You would expect three times input frequency in normal state, and two times (asymmetric) with one open diode.
if you reverse the diode in a half wave rectifier, you would expect the A- Ripple to increase B- output to be less filtered C- out put polarity to be reversed D- output voltage to decrease
when is ripple fator minimum
it is nothing but ac ripple neutralizer
zener diode :zener diode operates under reverse bias voltageideal diode :ideal diode operates under forward bias voltage
replace the diode
actually a diode is forward biased by at least 0,7 volts.so what it means is that when it is removed that 0,7 voltage drop on the diode is eliminated thus increasing ripple voltage
A: Then the phase that the diode is remove will not work or Rather get an output
Measuring ripple frequency would determine if a diode were open in a bridge rectifier circuit because the ripple frequency is normally twice the input frequency in a functioning full wave bridge rectifier. If one diode were open, the ripple frequency would only be the input frequency. Note: This is true for single phase or bi-phase operation. Three phase operation is more complex, but still doable - You would expect three times input frequency in normal state, and two times (asymmetric) with one open diode.
if you reverse the diode in a half wave rectifier, you would expect the A- Ripple to increase B- output to be less filtered C- out put polarity to be reversed D- output voltage to decrease
It depends on how the diode is damaged. There are generally two cases. One, the diode is shorted, and conducts with a low impedance in both directions. The other, the diode is open, and does not conduct, having a high impedance, in both directions. The effect depends on the particular circuit. In a power supply, a shorted diode will often blow the fuse, while an open diode will result in no output, or in high ripple voltage output. Is it possible that diode has normal voltage output but wrong current,meaning low mA?
you can't have half wave rectified polyphase AC power, unless you are using only one of the phases. however with only one diode from each phase of three 50 Hz phases the ripple frequency is 150 Hz, with two making a full three phase bridge the ripple frequency is 300 Hz. perhaps that was what you meant.
ripple
it clips all the bottom voltage off the sine wave,to get steady dc voltage you will need to clean up the ripple to get good voltage regulation <><><> A diode does not change AC to DC, but it does allow alternating current to flow in only one direction. A diode (or often 2 or 4 diodes) with a filter can be used to rectify AC into DC.
The effective resistance of the capacitor reduces the ripple current through the capacitor making it less effective in its function of smoothing the voltage. But if the capacitor filter is fed by a transformer and diodes, the resistance of the transformer exceeds that of the capacitor.
Removing the freewheeling diode in a single-phase semi-converter can lead to several adverse effects. Without the diode, the inductive load will not have a path for the current to continue flowing when the SCRs are turned off, potentially causing voltage spikes and damaging the circuit components. Additionally, the efficiency of the converter may decrease, leading to increased ripple in the output voltage and a less stable operation. Overall, the absence of the freewheeling diode compromises the converter's performance and reliability.
I'm not sure you understand what you're asking. A diode will have a voltage drop of ~.5-.7 volts. If you put a diode and resistor in series, the voltage across the diode will be .5 - .7 volts, and the voltage drop across the resistor will be (supply voltage - diode voltage drop). If you are trying to rectify to DC, you need at the least a half wave rectifier (two diodes), and some system to remove the ripple. The rectifier simply chops the AC waveform, so for the part of the supply sine wave that is ~.5 or less, the output of the rectifier will be zero. The top part of the sign wave will show up on the output of the rectifier, but will be slightly smaller (due to the voltage drop across the diode). You'll need to get ride of this rippling for true DC. One fairly easy way to do this is to use a zener diode. It will attempt to keep the voltage drop across it the same, so purchase a 1.2 volt zener diode. The problem with this is the zener diode will saturate if you have too heavy a load. What I've done on simple projects is to use a zener diode to bias the base to collector voltage of a transistor, with the emitter acting as the output (an NPN BJT usually). I also used a fairly large capacitor to help minimize the ripple as well, although this may not be necessary for you.