# 13.5v cutoff circuit

Discussion in 'The Projects Forum' started by cecil-p, Jan 2, 2009.

1. ### cecil-p Thread Starter New Member

Jan 1, 2009
4
0
hi, i am searching for a diagram or some information on a project.

i am trying to build a discharger to discharge an 18v battery down to 13.5v and cut off at that point. i was told to use a relay inline with the resistor bank that will be carrying the load, and to also use a resistor to power the coil in the relay sized to give the coil a low voltage and release at the 13.5v. is this a safe way to make a self-powered discharger or is there a way to make it more of a solid state unit? i plan to discharge at about .9A with ceramic resistors in parallel (4x 80Ω 10watt). thanks for any information.

2. ### SgtWookie Expert

Jul 17, 2007
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Do you want to discharge it faster? You could use some automotive bulbs (6v and 12v in series) for a more constant discharge rate. You could also use a voltage regulator like an LM317 rigged as a constant current source to drain the battery.

The problem with using fixed resistors is that as the voltage drops, the current through them will decrease. I=E/R, or Current = Voltage / Resistance. Since the resistance in fixed resistors won't change, and the voltage drops as the battery discharges over time, your current drops correspondingly.

Incandescent bulbs have a non-linear resistance characteristic; as the filament heats up, it increases in resistance; as the voltage across the bulb drops, the filament cools and decreases in resistance.

Using LM317 or LM350 voltage regulators as constant current sources isn't such a big deal, but keeping them cool with that kind of power dissipation could be a challenge. 0.9A at 18v is 16.2 Watts of power dissipation required.

The automotive bulb idea would require matching the wattage between the 6v and 12v bulbs. The 6v bulbs would have to be 1/2 the wattage of the 12v bulbs so that they would have even power dissipation.

Are these forklift batteries, or what is the construction and ampere-hour rating of them?

3. ### SgtWookie Expert

Jul 17, 2007
22,182
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I just threw something together, it's really kind of messy but it works in simulation.

The LM317 provides a regulated 10v output. It could be replaced by a 7810 regulator, then R4 and R7 would not be needed.

R1 and R2 set the LM339 inverting input reference level to approximately 5v.

R3 sets the trip level. When the battery voltage causes the wiper output of the pot to fall below the voltage on the LM339's inverting input, the output of the LM339 is pulled from 10v to 0v, which also turns off Q1, a power MOSFET.

Q1 has an 18 Ohm resistor connected from the drain to the battery's + output (simulated by V1). R9 keeps the MOSFET turned off in case something happens to R8.

C1 helps suppress transients on the voltage supply. R5 provides hysteresis to the input of the LM339 comparator; this keeps it from switching on and off when the inputs are just a few mV's apart.

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4. ### cecil-p Thread Starter New Member

Jan 1, 2009
4
0
thank you sgtwookie,the batteries are just dewalt 18v ni-cd batteries (15 sub-c cells in series) at about 2400 mah. i am trying to make a self contained unit so i can set it on a battery and let it discharge the rest of the way before it gets charged again. i am only looking at a low discharge rate because they will be mostly discharged by the time i do use this circuit. i was thinking of the resistors because of the less amount of space required and hpoefully less heat dissipation with them being in parallel. would this diagram still be used if resistors were used instead of auto lamps?

5. ### SgtWookie Expert

Jul 17, 2007
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Sure, it would still work if you used resistors - it would just take longer to discharge the battery, because the resistance will be constant.

If you used resistors that totalled 18 Ohms, you would start drawing 1A when the battery was fully charged. By the time the battery voltage was down to 14V, the current would be 0.778 Amperes.

Note that after the circuit stops the heavy discharge cycle, it will continue to draw about 25mA (0.025A) from the battery, since the circuit is powered by the battery.

There is no indication that it's operating either. It's just a "quick and dirty".

6. ### cecil-p Thread Starter New Member

Jan 1, 2009
4
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i am also having trouble figuring out what the "Q1" is.

7. ### SgtWookie Expert

Jul 17, 2007
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"Q1" is an N-channel power MOSFET. Basically, it is a voltage controlled switch.

You don't have to use that exact same MOSFET; an IRF520 or IRFZ24 would also work, along with many others.

8. ### cecil-p Thread Starter New Member

Jan 1, 2009
4
0
i will give it a try and let you know if it works. it might take me a while due to low level of experience that i have. i do thank you for your time and talent.

9. ### welcomb New Member

Feb 18, 2009
5
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Hi sgtwookie, I'm trying to build a cut-off for a single Li-Po cell. From what I've gathered from the LM339 datasheet, a single cell should be able to power the IC directly (min 2V) and I can use the 1.25V off the LM319 as the bias voltage. By replacing R3 with a 51k and 43k resistor divider, I should get a cut off at about 2.7V. However, with my limited knowledge of electronics, I'm not sure how the values of R6, R8 and R9 will affect it, though my guess is it won't. Also would I have to change value of the R5 hysteresis resistor?

10. ### italo New Member

Nov 20, 2005
205
1
since your load is constant as 4x80Ω the discharge rate will not ever be constant at any set current of .9amps

11. ### italo New Member

Nov 20, 2005
205
1
I looked at wookie design is good there except for too many parts to achieve the results I think the idea is there execpt that use the lm317 as a current source until the lm339 triggers then the lm317 is configured as a voltage source of 18v or whatever.