aus jd 2703

iTrader: (5)

hey does any one have a wiring diagram on how to make a voltage cut off with a 30amp auto motive relay? i know u can do it just can find a walk through or anything any more?
cheers, Josh P

kufman

iTrader: (6)

What voltage are you looking to cutoff at?

aus jd 2703

iTrader: (5)

above 6 so 6-6.5

kufman

iTrader: (6)

The basic circuit could look like this. The switch thing is supposed to be the relay. The diodes on its coil are too alter the voltage at which it would shutoff. You could also use a resisor, but you would have to experiment to find the right value. The circuit probably needs one more diode to make it better. I will have to think about it more.

kufman

iTrader: (6)

So you would use a 5V coil relay and probably two or 3 diodes in series to make it cutout at a higher voltage. A 5 volt relay will probably stay energized down in to the 3.5 to 4 voltage range. the diodes would raise this up. The circuit does need another diode in it, between the low side of the relay contact and the input to the coil.

kufman

iTrader: (6)

Added diode. May not need D1. Add or subtract D2's and D3's to get the right voltage. Can add a capacitor to get a delay effect so that the system doesn't turn off for momentary dips below the specified voltage.

PitCrew

Don't need D4.
I made one of these using a 30 amp relay from Radio shack. Worked perfect Correct,the coil drops out at like 3.5 volts.

kufman

iTrader: (6)

D4 is only there to prevent load current from going through the push button. Othereise it doesn't have to be there.

Phroz

iTrader: (1)

Anyway to do it with a 0.9v per cell (5.4v) cut off?

kufman

iTrader: (6)

With a couple of diodes in series with the coil you should be able to get around 5.4V. Depends on the relay used. As mentioned above, most 5V coil relays with cutout somewhere between 3.5 and 4 volts. To raise this in approximately 0.6V increments you can add diodes in series with the coil. If a particular relay lets go at 3.5V, adding 3 dioes in series will get you to about 5.3 volts. This isn't an exact science since the diodes voltage drop will depend on the current going through them. You have to look at the current that the coil requires and find that spot on the diodes voltage curve..... or you can just mess around with it until it works.

Jam-ehz

iTrader: (6)

Actually, you can use resistors to calculate exactly what you want your cutoff to be..


Any resistors added to the current coil will increase the mininum voltage it will stay on for.
You'll first need to find the minimum voltage the relay runs at (usually printed on the unit), and the resistance of the relay coil (multimeter can tell you).

Use this formula:
Ra is Resistance you need to add - the desired number
Rc is Resistance of the relay coil
Vc is the desired cutoff voltage
Vr is the normal cutoff of the relay

Ra = Rc * (Vc/Vr) - Rc

In my case, I had a 79.1 Ohm coil in the relay, it cut off at 2.52V. I wanted a 5.4V cutoff.

Ra = 79.1 * (5.4/2.52) - 79.1 = 90.40

So I needed around a 90-Ohm resistor, so I just got a 100Ohm variable resistor, and I dial it in myself. It also offers some flexibility if I wanted a bit lower or higher..

I likewise bought a 6V relay, and used the same formula to make myself a 3.6V 4-cell cutoff unit.

There was one guy's site a while ago that had all of this wicked stuff.. laser timers, and tons of other stuff.. Sanj's Yokomo page.. seems to be gone now.. but at least I kept my spreadsheet that calculates it all for me.. just don't ask me to explain it anymore!

kufman

iTrader: (6)

yup, that works too!

aus jd 2703

iTrader: (5)

jam ehz
which is the always on and the on when a current is passed through the coil ?

Jam-ehz

iTrader: (6)

The always-on is on the left side of the bridge in the pic, not connected to anything. The on-power terminal is depicted on the right side.