Discussion in 'General Electronics Chat' started by rc_rs_ss, Aug 10, 2011.

1. rc_rs_ss Thread Starter New Member

Aug 10, 2011
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Hey Everyone- I have been researching on ways to send out an electric signal to trip a relay to shut off my battery charger. I did find some simple circuits and I did the math for what I would need and could not find big enough zener diodes and resistors.

I need this circuit to send a 8.2 or 10Volt 15mA signal when batteries reach 235 to 240 volts.

The circuit I found consisted of a Zener diode, resistor and LED which i substituted my relay for. I could not find a 125.46 Watt 8.2V Zener diode or a 3.5 Watt 14953.33 ohm resistor.
Any help on this would be great Thanks!

Nov 16, 2007
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3. rc_rs_ss Thread Starter New Member

Aug 10, 2011
5
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Battery+(Resistor)(Zener Diode)(Relay Input)(Battery-)
That is the circuit I found- Pretty basic. If there is any way I could make it work for my 235 Volt target it would be perfect. I am open to any other circuits I can solder up in my free time.
Others use an arduino and I have not taken the time to learn the programming part.

4. #12 Expert

Nov 30, 2010
18,078
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www.mouser.com
588-GW10J15K0E \$4.84 plus shipping.
It seems like the last part should be "Oh-E" but it looks like "zero-E"

If you just put this resistor in series with the relay, it will deliver 15ma when the batteries get to 235 volts and there is 10 volts on the relay coil. Problems include the fact that relays do not act at exactly the voltage they are rated for. You need to have something more accurate switch the relay on, and it isn't a zener diode. This requires an intelligent circuit, but not so intelligent as a micro-computer.

5. Ron H AAC Fanatic!

Apr 14, 2005
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In other words, rc_rs_ss, don't waste your money on this resistor.

6. #12 Expert

Nov 30, 2010
18,078
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True. Your basic idea is flawed.

Aug 10, 2011
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8. Ron H AAC Fanatic!

Apr 14, 2005
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Why do you need an 8.2 or 10Volt 15mA signal when the batteries are charged?
I'm trying to get you to give us the big picture.

9. SgtWookie Expert

Jul 17, 2007
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Do you have a low voltage DC power supply available along with this charger?

If so, you could use a comparator circuit to compare the battery voltage to a reference voltage, and have the comparator trip when the desired battery voltage was reached.

You really should let us know what kind of battery/batteries you are charging; how many cells per battery, and how many batteries in series are connected. For example, 25 8v 4 cell batteries or 50 4v 2 cell batteries would work for your numbers.

Last edited: Aug 11, 2011
10. rc_rs_ss Thread Starter New Member

Aug 10, 2011
5
0

The signal will be coming from the Zener diode to Trip my relay to shut off AC Power on my charger. And there is no low voltage DC power supply- I suppose I could rig something like that up-have to research that more-.

11. rc_rs_ss Thread Starter New Member

Aug 10, 2011
5
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Batteries being charged are 28-12 Volt NImh cells. I need to charge this pack to only 70% which give me the 235 Volts. My charger does not have an automatic shutdown and overcharging the batteries is dangerous. I am also putting some thermistors in the pack to trip my relay when it gets warm during charging.

Based on my figures I would need a 3.56 Watt 15Kohm resistor and a 125 Watt 8.2 Volt Zener Diode to shoot 8.2 Volts and .0156 Amps to My Relay Input rated:::
Control Voltage Range 3-32 Vdc
Typical Input Current 3.4mA @ 5 Vdc, 20mA @ 28Vdc
Nominal Input Impedance 1500 Ohms

In this case I can Only get a 10 Watt rated 15Kohm Resistor, and the Zener I need is also not available. So, the 10 Watt resistor changes the Zener Diode specs I need, Right? Or am I confused about something here?
My interpretation of the circuit is: The power going through the resistor will not be great enough until it reaches 8.2 volts(the zener voltage) which then will push 8.2V 15mA current through the Zener and then to the Relay. The resistor will lower the voltage from 235 down to 8.2 correct?
Am I confused?