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Registered Member #3943
Joined: Sun Jun 12 2011, 05:24PM
Location: The Shire, UK
Posts: 552
Patrick wrote ...
so do the changes you've made mean? are you using hysteresis? as other have said, and id favor in this situation.
The parts in red are the only changes I have made, like I said I tried hysteresis but it didn't do anything for this circuit. Whilst these mods have fixed all of those issues I was originally having.
btw the capacitor at the bottom right is actually a .22uF, the red box just covers the dot. I didn't really have a plan, I just took a .22uF capacitor and used alligator clips to connect it to different places of the circuit until I found which places worked. The diode in series to the op amps positive input was added incase of any leakage current from the op amp was causing the LED to light up dimly, since I am using an ultra bright LED it doesn't take much current at all for it to start producing light.
The 100k is just a pulldown resistor to stop the positive input of the op amp from floating and picking up noise from the adjacent driver.
The 2.2nF capacitor would only work if it was placed after a base diode.
The 2.2k instead of the 1k was changed becuase according to the datasheet the output transistor can have quite a high voltage drop when currents of more than 8-10mA are flowing through it, so its just there for good measure.
Dr. Dark Current wrote ...
BTW turning the gates off is a very nasty thing to do. This interrupts the current through the series inductor producing a large voltage spike, the avalanche breakdown of the FETs is the only thing which limits the voltage.
A better approach would be to disconnect the driver's ground from circuit ground by means of another FET, and put a diode from this additional FET's drain to the positive votage input rail.
hmm didn't know this, but it seems to be working ok and has survived so far. Could a transorb or some sort of snubber be used instead of an additional fet? Or what about slowing down the turn off via some extra resistance in series with the pull down diodes?
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
Well just leave it as it is, the low voltage FETs can stand a lot of avalanching. It's just not correct in terms of power electronics "rules". If you did this with IGBTs or bipolars, they would die the instant you'd switch the gates off.
A better advice: Disconnect the driver using a relay. Don't forget the "damping" diode. You can use a spare contact on the relay to keep the circuit off until eg. you press a RESET button (simple form of a latch).
Registered Member #3943
Joined: Sun Jun 12 2011, 05:24PM
Location: The Shire, UK
Posts: 552
Dr. Dark Current wrote ...
Well just leave it as it is, the low voltage FETs can stand a lot of avalanching. It's just not correct in terms of power electronics "rules". If you did this with IGBTs or bipolars, they would die the instant you'd switch the gates off.
What about this in parallel with the inductor? since its sort of useful having the driver automatically recharge the bank after discharging, shorting it out doesn't seem to cause any harm either.
Dr. Dark Current wrote ...
A better advice: Disconnect the driver using a relay. Don't forget the "damping" diode. You can use a spare contact on the relay to keep the circuit off until eg. you press a RESET button (simple form of a latch).
Is that in series with the positive rail? And by damping diode do you mean the one in reverse to the relays coil?
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
Alex1M6 wrote ...
Dr. Dark Current wrote ...
A better advice: Disconnect the driver using a relay. Don't forget the "damping" diode. You can use a spare contact on the relay to keep the circuit off until eg. you press a RESET button (simple form of a latch).
Is that in series with the positive rail? And by damping diode do you mean the one in reverse to the relays coil?
It is up to you where you put it, in series with positive or ground rail. The diode goes from the "hot" side of the relay contact to the other rail (I'm sure you can figure out its polarity - so it normally blocks current).
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Capacitor bank charger feedback circuit
