CUT OFF CIRCUIT

[scottraynor]

Guys, I'm planning to build a battery discharger for a single sub C type battery. I need an adjustable cut-off circuit that ranges from .9V down to .6Vdc. Any help would be gladly appreciated.

 

[Roff]

What is your discharge rate?

 

[scottraynor]

Ron H, I usually do it 5 to 10Amp discharge rate per cell.

 

[Roff]

How high is the voltage prior to discharging?

 

[scottraynor]

That's around the level of a fully charge sub C cell, usually 1.3 to 1.45V I think!

 

[Roff]

Here's a circuit that should work. I have simulated it, but it has not been built. You can substitute any MOSFET with an Rds(on) of 30 milliohms or less. You'll need a heatsink for the MOSFET, as it dissipates several watts during discharge.

 

[scottraynor]

Ron H, Thanks for your help man... But, I felt so stupid not mentioning that the circuit I'm looking for is to be used as an equalizing tray for a stick pack(6 sub "C" cell connected in series side by side). Sorry about that man.... Need to equalize each cell to a certain level before recharging them again. Hope you get what I mean.

 

[Roff]

I looked at a product description for the **broken link removed**. It seems to me that you can build 6 of these circuits, all powered by the same 9 volt battery, controlled by the same pots for current level and discharge voltage, and started by the same switch. The batteries cannot be connected in series or in parallel. They must be discharged individually. If you use 6 of these circuits, they will all stop discharging at exactly the same voltage, independent of time, which I gather is what you want.

 

[Sebi]

For cell equalizing i use this circuit.

 

[scottraynor]

Ron H:

Yes! you're right... I'll try to finish it this week and inform you later for some updates. Thanks again man...

Sebi:

That's the simpliest equalizing circuit i've ever seen 8) 8) 8) But, unfortunately what i preferred is the one built with cut-off circuit. Thanks man....

 

[Roff]

Scott, build one and test it before you jump in with both feet.

 

[ante]

I use this circuit, one for each cell. The resistor sets the current and the diode sets the final voltage. If you use a “normal” diode the final voltage is 0.6 and with two schottkys 0.7 . These voltages can vary but you can find schottkys with 0.25 – 0.55 per diode. The life of the cells is much longer when discharging the cells one by one.

Ante :roll:

 

[ante]

I use this circuit, one for each cell. The resistor sets the current and the diode sets the final voltage. If you use a “normal” diode the final voltage is 0.6 and with two schottkys 0.7 . These voltages can vary but you can find schottkys with 0.25 – 0.55 per diode. The life of the cells is much longer when discharging the cells one by one.

Ante :roll:

 

[Roff]

I like the simplicity of the diode-resistor circuit. Whether it meets you needs is another question. See the plot below of a simulation of both methods. The diode-resistor discharge starts out at 1 amp, then decays pretty much exponentially. The constant current discharge is 1 amp. The time scale is arbitrary, depending on the battery, and I have no idea how much charge they actually hold. This sim is actually used two 1000 Farad capacitors to simulate the batteries, each charged to 1.5 volts. In the diode-resistor circuit, I used a 0.8 ohm resistor and an MUR120 rectifier, which was a diode that I had a spice model for.
You can change the discharge rate to meet your needs in either circuit. The cutoff voltage is set-and-forget with the constant current circuit. Here, I set it for 0.7 volts.

 

[ante]

Ron,
Interesting test you have made there. I have not tried something like that, but you gave me an idea when using a big cap for battery. I have to see if ISIS will do the work. Many years ago I experimented with a cut off load version but didn’t get it right. The problem I encountered was after cutoff the cell regained power (different for each cell) and this caused the cells to differ even more before they were charged again. Then I discovered that the diode resistor method was the right one for the job. I just leave them on till the next day (sometimes I forget them for days) and they survive and stay equal for a long time.

BDW sorry for the double post

Ante :roll:

 

[Optikon]

Discharge design

RON, in your design, can U1A be used like an opamp as part of the battery current sink? I've never actually used a 393 like this before.. but I've heard rumors... Saves a part if it can..

 

[Roff]

RON, in your design, can U1A be used like an opamp as part of the battery current sink? I've never actually used a 393 like this before.. but I've heard rumors... Saves a part if it can..

I can't remember if I have actually done this in hardware, but I've simulated it in several circuits. In this circuit, R1 and C2 form the dominant pole in the feedback loop, compensating it against oscillation. So long as the loop gain goes below 1 (0 dB) while the phase shift is less than 360 degrees (including the inversion), the loop will be stable. This means that the loop gain has to be below 1 before the other poles (RC time constants) inside the comparator become significant. Since we don't need high closed loop bandwidth or high slew rate, I figured the LM393 (with the MOSFET) would make a passable op amp if I added C2 to stabilize it.
To clarify:
Each RC lowpass node in the loop will contribute 90 degrees of phase shift in the limit. Looking at the schematic of the LM393, it has 3 gain stages including the output transistor. Each of these will contribute a pole at some frequency. We have to roll off the output stage at a low enough frequency so that the loop gain is less than one before the cumulative phase shift of the 3 gain stages is 360 degrees.

 

[Optikon]

RON, in your design, can U1A be used like an opamp as part of the battery current sink? I've never actually used a 393 like this before.. but I've heard rumors... Saves a part if it can..

I can't remember if I have actually done this in hardware, but I've simulated it in several circuits. In this circuit, R1 and C2 form the dominant pole in the feedback loop, compensating it against oscillation. So long as the loop gain goes below 1 (0 dB) while the phase shift is less than 360 degrees (including the inversion), the loop will be stable. This means that the loop gain has to be below 1 before the other poles (RC time constants) inside the comparator become significant. Since we don't need high closed loop bandwidth or high slew rate, I figured the LM393 (with the MOSFET) would make a passable op amp if I added C2 to stabilize it.
To clarify:
Each RC lowpass node in the loop will contribute 90 degrees of phase shift in the limit. Looking at the schematic of the LM393, it has 3 gain stages including the output transistor. Each of these will contribute a pole at some frequency. We have to roll off the output stage at a low enough frequency so that the loop gain is less than one before the cumulative phase shift of the 3 gain stages is 360 degrees.

Would it be better to have Q1 pull down the Iset node to shut off the discharge rather than the gate so that U1A doesnt have to drive C2 directly when it is slamming on/off (Q1 will turn on & off very fast)?

If C2 must be driven by U1A, a 50 Ohm isolation resistor on output of U1A is probably a good thing. What do you think?

 

[Roff]

Would it be better to have Q1 pull down the Iset node to shut off the discharge rather than the gate so that U1A doesnt have to drive C2 directly when it is slamming on/off (Q1 will turn on & off very fast)?

Do we really care about microseconds when deciding when to terminate the discharge of a battery? In any case, Q1 can dump C2 faster than the LM393 can, because the LM393 output sink current is 18ma typical, 6ma min. Q1 will probably typically sink about 100ma while dumping C2, plus you don't have the propagation delay of the LM393. Also, pulling down Iset would leave the feedback loop active, so any Vce(sat) of Q1 or offset voltage of U1A will allow the battery to continue to discharge, if only at a miniscule rate.

If C2 must be driven by U1A, a 50 ohm isolation resistor on output of U1A is probably a good thing. What do you think?

I don't see any point in this. Why would you need it?

 

[scottraynor]

Hi! Ron, can you give me another battery discharger circuit with adjustable cut-off of 5V to 6V and a discharge rate of 25 to 30Amps. I'll be using this to discharge a RC battery pack which has 8 to 9V/3300ma capacity. Hoping for your soonest reply, thanks in advance.