sincere advise
please read some elect. books about schotky diodes bec of grave misconceptions about them.
Hi Christer,
Yes, indeed, the reverse current is virtually non-existing and that is why normally you don't find a Trr number on their data sheets. But they DO switch off very sharp and generate di/dt-induced bursts. The sharp reverse current turn off in other rectifiers generates the hf in those, so there you want a soft recovery from reverse current back to non-conduction. It doesn't make much difference whether the di/dt is generated by forward current switch-off as in a schottky or by reverse current recovery in a "normal" diode, the transient causes the hf bursts. And you don't want that in your audio stuff. At least this is my understanding of it.
Jan Didden
Yes, indeed, the reverse current is virtually non-existing and that is why normally you don't find a Trr number on their data sheets. But they DO switch off very sharp and generate di/dt-induced bursts. The sharp reverse current turn off in other rectifiers generates the hf in those, so there you want a soft recovery from reverse current back to non-conduction. It doesn't make much difference whether the di/dt is generated by forward current switch-off as in a schottky or by reverse current recovery in a "normal" diode, the transient causes the hf bursts. And you don't want that in your audio stuff. At least this is my understanding of it.
Jan Didden
sagarverma said:
Thanks for the advice. Since I have so many different books, it would help if you could mention one to me, preferable with chapter? Or maybe - even better - explain the grave misconceptions?
Jan Didden
janneman said:
i dont think that this is happening with a 50-100hz sine wave , maybe with a square wave .....
but i agree with u that there is no advantage in using fast diodes where u can use a regular ones .
Jan,
maybe I am lost somewhere, but in an ordinary 50/60 Hz PSU, as we are discussing, it is the circuit itself that will make the forward current switch off softly. The diode conducts as long as the voltage from the transformer is higher than the (increasing) voltage over the cap (for simplicity I neglect the voltage drop over the diode).The current is mainly restricted by the ESR of the cap and other resistances and it will decrease gradually until the transformer voltage drops below the cap voltage. At the point where the diode turns off, the voltages are practically identical and no current will flow. Hence, there is no forward current to turn off abruptly, and the Schottky diode won't cause any trouble. An ordinary diode on the other hand would have a stored charge that causes a reverse current spike after turn off. That is, as far as I can see, the dI/dt of the forward current before and at turn off is dictated by the rest of the circuit, not by the diode.
maybe I am lost somewhere, but in an ordinary 50/60 Hz PSU, as we are discussing, it is the circuit itself that will make the forward current switch off softly. The diode conducts as long as the voltage from the transformer is higher than the (increasing) voltage over the cap (for simplicity I neglect the voltage drop over the diode).The current is mainly restricted by the ESR of the cap and other resistances and it will decrease gradually until the transformer voltage drops below the cap voltage. At the point where the diode turns off, the voltages are practically identical and no current will flow. Hence, there is no forward current to turn off abruptly, and the Schottky diode won't cause any trouble. An ordinary diode on the other hand would have a stored charge that causes a reverse current spike after turn off. That is, as far as I can see, the dI/dt of the forward current before and at turn off is dictated by the rest of the circuit, not by the diode.
you can observe the phenomena Jan describes with a decent 'scope using the AC input -- there can be many microvolts of RF ripple which, because of their comparatively short wave-length -- propogate themselves all over the place. it's not just the diode, it's the transformer leakage inductance, interwinding capacitance, etc.
with one very inexpensive, beastly torroid bought off EBay I experienced zero, zip, nada, niente RF ripple on the DC power rails using MUR860's. with another high quality EI transformer from a cannibalized HP meter you could detect the ripple with an RFI sniffer on the HP 3585a spectrum analyzer. it pays to measure.
by the way, Analog Devices has reprinted their amusing tale of ADC problems in the teeth of RFI.
with one very inexpensive, beastly torroid bought off EBay I experienced zero, zip, nada, niente RF ripple on the DC power rails using MUR860's. with another high quality EI transformer from a cannibalized HP meter you could detect the ripple with an RFI sniffer on the HP 3585a spectrum analyzer. it pays to measure.
by the way, Analog Devices has reprinted their amusing tale of ADC problems in the teeth of RFI.
Christer said:
Hmmm. Yes. Sounds logical. Seems I have learned something again. Thanks.
Jan Didden
This gets interesting. As the voltage across the diode (forward) gets smaller the current gets smaller also by virtue of the forward characteristic curve, isn't it? So, if there is load current, AND the voltage approaches zero, the only source of the load current is ... the reservoir capacitor!? Should be easily simulated.
Jan Didden
Jan Didden
Charge is stored in p-n junction capacitance?
This in-built p-n capacitance of a si diode creates an voltage divider with power supply capacitance. Several pF or less versus some mF?
janneman said:
If the di/dt is high (for example 1kHz square), the load transition from the diode to the capacitor is very fast. The load current and the transition time isn't only 50/60 Hz related.
Regards,
Milan
Yes understand, but with 50/60Hz the voltage dv/dt is rather low so presumably the di/dt is also very low, also in view of the diode forward curve where the current falls off anyway as the forward voltage goes to zero. So the forward di/dt will not be very high.
However, the REVERSE di/dt would be very high independent of the rectified signal frequency so for that you would want a soft recovery. His point was that since there is no reverse current in schottkys there is no di/dt in that respect, IIUC.
Jan Didden
However, the REVERSE di/dt would be very high independent of the rectified signal frequency so for that you would want a soft recovery. His point was that since there is no reverse current in schottkys there is no di/dt in that respect, IIUC.
Jan Didden
darkfenriz said:
I think for the rates of change we talk about here the reservoir cap is virtually invisible due to wiring L and its own ESL.
Jan Didden
moamps said:
Yes, but the only interesting case is if this causes the diode to stop conducting. That could only happen if the cap is charging and the load current is in the positive half cycle of the square wave and switches over to the negative half. However, that only mean that the load current drops to either a lower value or all the way down to zero. But, this will not cause any increase in capacitor voltage (maybe slightly due to ESR) so even though it may cause a quick change in charge current, it will not suddenly cause the diode to stop conducting. That will only happen when the transformer voltage drops to the cap voltage. Or did I go wrong somewhere?
darkfenriz said:
No, the reverse current is mainly due to recombination, except for Schottky diodes where it is almost only due to the junction capacitance. It's in the paper I linked to previously.
janneman said:
So, to wrap up at least some parts of this discussion, I did a quick sim. See diagram below. A simple single phase rectifier, 10V AC 50Hz into 1000uF and a 15 Ohms load....
... and the graph. Of note:
The yellow curve is the voltage across the diode (right scale). We see that when Vdiode goes through zero, Idiode also goes through zero (green curve, left scale). This is at 186.7mS.
But we see that before that point (at 186.0mS) Idiode can no longer support the load current (blue, left scale) so the capacitor, which has been charging until now (red curve, left scale) reverses current and starts to supply the load current. In fact, after 186.7mS, Iload and Icap are equal but of opposite direction (red and blue).
So indeed the current turn off in the forward direction is gradual and there is an equally gradual transfer of the load current from diode to cap just before the Vdiode=0 point. So the forward current turn off will not cause any hf bursts as seems.
Jan Didden
The yellow curve is the voltage across the diode (right scale). We see that when Vdiode goes through zero, Idiode also goes through zero (green curve, left scale). This is at 186.7mS.
But we see that before that point (at 186.0mS) Idiode can no longer support the load current (blue, left scale) so the capacitor, which has been charging until now (red curve, left scale) reverses current and starts to supply the load current. In fact, after 186.7mS, Iload and Icap are equal but of opposite direction (red and blue).
So indeed the current turn off in the forward direction is gradual and there is an equally gradual transfer of the load current from diode to cap just before the Vdiode=0 point. So the forward current turn off will not cause any hf bursts as seems.
Jan Didden