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4hv.org :: Forums :: General Science and Electronics
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need help with cap bank charger

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Pinkamena
Tue Apr 17 2012, 04:21PM Print
Pinkamena Registered Member #4237 Joined: Tue Nov 29 2011, 02:49PM
Location:
Posts: 117
Good day!

I am trying to design a charger for my capacitor bank. Right now, it's simply rectifying 240VDC, smoothing it out with a large capacitor, and connected straight to my cap bank.
The problem is, my cap bank is rated for 300V max, and I'm charging it using 240VDC rectified So if I'm not careful, I'll destroy my caps by overcharging them.
"No problem!" I thought, I can just cut off the charger before it reaches 300V. I was first thinking of using a regular old manual switch, but I'm afraid I'll forget to turn it off one day and overcharge the caps.
So I was thinking of some way to automate the shutoff, by comparing the voltage of the caps to a reference voltage, lets say 270V to keep it on the safe side. So when the caps reaches the reference voltage, a relay will shut off the charger. Question is, how can this be made? I have tried using op-amps to compare the voltage, but that didn't work well... So have you guys got some advices?
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Thomas W
Tue Apr 17 2012, 05:45PM
Thomas W Registered Member #3324 Joined: Sun Oct 17 2010, 06:57PM
Location:
Posts: 1276
did you use a potential divider before feeding it into the opamp...
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Pinkamena
Tue Apr 17 2012, 08:05PM
Pinkamena Registered Member #4237 Joined: Tue Nov 29 2011, 02:49PM
Location:
Posts: 117
Yes. Here is the circuit. There must obviously be something very wrong, but I do not see what. Any help would be appreciated!
1334693117 4237 FT137189 Untitled
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Sulaiman
Tue Apr 17 2012, 08:18PM
Sulaiman Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
A low-tech method is to use a relay, say 12Vdc coil
and a string of zener or TVS diodes in series with the coil.
The string of diodes + relay coil goes across the cap bank.
Use the normally-closed contact in series with the charging circuit,
ideally on the ac side.
The relay contacts wil initially be closed allowing the cap bank to charge
when the cap bank voltage reaches (zener voltage + relay pull-in voltage)
the charging will stop and the cap bank will slowly discharge.
When the cap bank voltage falls below (zener voltage + relay drop-out voltage)
the contacts will close again.
Thus the cap bank will maintain it's voltage +/- a few volts.

The zener/tvs diodes have to carry the relay current so a small/sensitive relay is best.

Have some resistance in series with the charging circuit
(on the ac or dc side) to limit 'inrush' current.
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Saz43
Tue Apr 17 2012, 08:33PM
Saz43 Registered Member #1525 Joined: Mon Jun 09 2008, 12:16AM
Location: America
Posts: 294
Can your op-amp source enough current to trigger your relay? Most op-amps are open collector and need a pull up resistor in order to source current.
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Pinkamena
Tue Apr 17 2012, 09:57PM
Pinkamena Registered Member #4237 Joined: Tue Nov 29 2011, 02:49PM
Location:
Posts: 117
Saz43 wrote ...

Can your op-amp source enough current to trigger your relay? Most op-amps are open collector and need a pull up resistor in order to source current.
I don't know, but that's not where the problem lies. It lies in the fact that the op-amp inputs doesn't get a steady voltage. I thought it would, since the AC has been rectified and smoothed with the 1mF capacitor. If you look at the picture I've attatched, you can see the voltage the two inputs get.
Here is the circuit if you want to play around with it.

Sulaiman wrote ...

A low-tech method is to use a relay, say 12Vdc coil
and a string of zener or TVS diodes in series with the coil.
The string of diodes + relay coil goes across the cap bank.
Use the normally-closed contact in series with the charging circuit,
ideally on the ac side.
The relay contacts wil initially be closed allowing the cap bank to charge
when the cap bank voltage reaches (zener voltage + relay pull-in voltage)
the charging will stop and the cap bank will slowly discharge.
When the cap bank voltage falls below (zener voltage + relay drop-out voltage)
the contacts will close again.
Thus the cap bank will maintain it's voltage +/- a few volts.

The zener/tvs diodes have to carry the relay current so a small/sensitive relay is best.

Have some resistance in series with the charging circuit
(on the ac or dc side) to limit 'inrush' current.
Any chance you can make me a drawing or give me a link to somewhere I can read more about this?
1334699858 4237 FT137189 Untitled
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Pinkamena
Wed Apr 18 2012, 07:18PM
Pinkamena Registered Member #4237 Joined: Tue Nov 29 2011, 02:49PM
Location:
Posts: 117
I tried Sulaimans tip, and it worked. Thanks!
I'm just gonna put the circuit and a description of it here, in case someone searches for a charging circuit.

The 1kOhm resistor is a potentiometer (0-35kOhm) that allows for a variation in when the charging stops. The 1,8kOhm resistor is constant. The 98mF cap is my cap bank, and the 1mF cap is for smoothing the voltage from the rectifier.
There's two SPDT switches. One on-off-on for either charging, keeping the charge, or discharging (throug the 35 or 100 Ohm load). The other, an on-on switch is for choosing between two different loads. I found that when charging to a low voltage, like 50V, the relay would use too long time to turn on, and thus getting letting the cap bank charge to over 50V, and giving the relay coil too much current. The 100 ohm load makes the charging go slower.
The Zener diode has a zener voltage of 10V. The diode next to it is just for show so one can see when it's done charging.
11373482
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Forty
Thu Apr 26 2012, 07:01PM
Forty Registered Member #3888 Joined: Sun May 15 2011, 09:50PM
Location: Erie, PA
Posts: 649
A question and a suggestion:

When determining the values of the resistors for the voltage divider (not the ratio, I imagine that is ~zener voltage divided by target voltage), would the correct process be to find the current required by relay to switch, divide that by the hfe of the transistor, and then set that current equal to the target voltage divided by the sum of the two resistors? I'm using the same design (thanks pink and sulaiman) for a zvs charger where the NC connects 12v to the gate supply and the NO connects the gates to ground via two diodes.

You might want to consider adding a freewheeling diode and a capacitor across the relay coil to (respectively) protect the transistor and prevent rapid relay switching. Also adding a fixed resistor to your 1k trim -or- making the 1k fixed and the 1.8k variable could prevent the accidental case where your capacitor/charging voltage is connected directly to the zener and transistor base which could possibly fry them (unless I'm misunderstanding something, which could be probable.) Also, do you really need the smoothing capacitor still? Seems like your bank will do all the smoothing you need and the extra 1mF discharging through the load resistor will just keep your SCR latched (if that's what you intend to discharge your bank with anyway)

Oh and I guess a bonus question b/c it's not worth starting a thread about: Why does a battery connected to a mosfet, + to drain - to source, cause the fet to heat up instantly? or is it because the gate is left floating?
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Pinkamena
Fri Apr 27 2012, 10:03AM
Pinkamena Registered Member #4237 Joined: Tue Nov 29 2011, 02:49PM
Location:
Posts: 117
Well, I'm sure that you're right about the process to find the two values for the resistors. But since I am not completely certain in my own knowledge of the math going into these things, I just used the falstad sim and played around with the resistor values until I found a pair that gave enough current to the relay.

What is a freewheeling diode? And I will definitively add a capacitor, thanks! The smoothing capacitor isn't needed now, you're right.
I have added a low-current fuse between the zener diode and the voltage divider, just in case what you mentioned will happen.

Regarding my way of bank discharging, it will consist of one IGBT, connected to 9 SCRs. The IGBT turns voltage on and off, and the SCRs decide what coil the current go to.
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Forty
Fri Apr 27 2012, 02:11PM
Forty Registered Member #3888 Joined: Sun May 15 2011, 09:50PM
Location: Erie, PA
Posts: 649
The free wheeling diode is just any old diode (common 1n400x family will do) that is connected antiparallel (cathode to positive side, anode to ground) across the coil. When the magnetic field of the coil collapses it produces a backwards emf that will then get shorted out through the diode rather than traveling back into your circuit. You'll want to put them across your coilgun coils as well (but those ones will need to be higher pulse current rated, the exact value I'm not quite certain how to calculate but the first negative peak in Barry's RLC sim should give you a rough idea.)

With your sim, try increasing your divider resistor values by 10x and seeing if that still works. Because if you charge up to 300v with the current ones, the power dissipated in them will go up by a factor of 36, which at the least will be about 32W.
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