Old telcom power supply, advice needed.

[Noxx]

Hello,
I recently acquired an old Telcom power supply, able to deliver to following current:

+12V, max 6A
-12V, max 6A
and
+5V max 80A

Now that is a lot of amperage lol.

My question is,
how do I limit the current to be drawn ? Let say I run a deplating cell, I'm not sure it would be a good idea to shoot 80 amps, the solution would get hot in no time...

Also, since I don't know much in electricity, is there a possibility for the PSU to blow up ? (exept shortage)

Thanks

 

[Harold_V]

Current (amperage) will be limited by the circuit. Unless you have a dead short, you won't use the full amperage. Current flow is a funciton of resistance and voltage.

Harold

 

[Noxx]

Dead short I meant sorry...

Ok I understand thanks.

 

[lazersteve]

Noxx,

Harold is correct, with a constant voltage supply the resistance of the cell will determine the current flow up to the maximum available from the supply. Of course if the cell resistance goes too low the fuse in the power supply will blow.

Take care to limit the current flowing through the cell by putting a fuse in line with the output, for example a 20 amp fuse will blow at 20 A and stop any more current from flowing than you want. Be sure your power leads going to the cell can handle 1.5 times the rated amperage you are going to allow. If you expect the cell to draw 5 amps use wire capable of 7.5 amps ( AWG 12).

Here's a table of AWG amperage ratings:

Wire Ratings

Here's an example:

Let say your electrolyte plus leads resistance was 50 ohms, this means the current flow (amps ) with the 5V regulated supply would be:

E=IR (VOLTS=AMP x OHMS); E/R = I; E/I=R

I= 5 v/50 OHMS = 0.1 AMPS

you will draw 80 amps if your total resistance is

5 v / 80 A = 0.0625 OHMS

If your total resistance is 1 OHM, you will draw:

5 v / 1 Ohm = 5 Amps

Determine the total resistance with a multimeter, set to measure resistance, hooked up across the cathode and anode leads without the power supply in place. Note: The resistance will change as the electrolyte heats up and as it gets more dissolved ions in it.


Steve

 

[Noxx]

Oh thanks, very useful information.

I will follow your advices, I don't want to blow anything.

If I understand correctly, if I want to limit the amperage I must correct the resistance accordingly with the following law:

V = RI

where
V is Voltage
R is Resistance (ohm)
I is amperage

So in my case:

5/R = I

The formula is the same, only the terms differ.

Thanks for all the info.

I also suppose that resistance will decrease as it heats up ?