reducing amps

[29a]

Hi all,
Due to a batt failure i'm gona have to run at 48V 24A for a while

I see lots of threads on increasing current, but nothing on reducing current.

Edit:- sorry found it


BUT

Am i right if i snip one of the shunt wires i will be at 26A
Is that close enougth for 8Ah 3C Nicd

 

[Lowell]

For a quick fix, desolder one of the shunt wires. If you have the parts available, try the adjustable current mod as posted in the controller thread.

 

[fechter]

Am i right if i snip one of the shunt wires i will be at 26A
Is that close enougth for 8Ah 3C Nicd

Right.

Desoldering one end of one shunt wire as Lowell suggests is a good idea, since you could solder it back later. Snipping is a bit harder to fix. The shunt wires are apparently made of a special alloy, and might be hard to replace.

 

[29a]

Could i disconect one of the shunt wires and place a on /off switch in series with it to give a 26 or 35 amp limit
If so should it be a 20A rated switch

 

[Matt Gruber]

Could i disconect one of the shunt wires and place a on /off switch in series with it to give a 26 or 35 amp limit
If so should it be a 20A rated switch

i'd wager u won't get 35.
but,
why not try it? i'd be impressed if u get over 30.
TIP: use SHORT HEAVY wires.if u use 8 or 10 and just a few inches, well, u just might get 34.9 LOL

 

[fechter]

Could i disconect one of the shunt wires and place a on /off switch in series with it to give a 26 or 35 amp limit
If so should it be a 20A rated switch

Like Matt's trying to point out, the resistance of the wires and switch will be much higher than the shunt wire.

What might work is to snip or remove one shunt wire completely and install a pair of wires to a switch like you suggested. Just guessing, about one foot of a pair of 10ga wires and a hefty switch (10A minimum) will be about equal to one shunt wire.

A shorter pair of 12ga wires would be equivalent. You might be able to install a switch on the end plate of the controller and have all the wires inside the box.

The four shunt wires together are around 1 milliohm. Each wire would be about 4 milliohms.

I have no idea what the resistance of a typical switch would be. You could change the length of the wires to get the desired resistance.

 

[29a]

The four shunt wires together are around 1 milliohm. Each wire would be about 4 milliohms.

Am i correct that each shunt is 0.00025 ohms, each shunt wire passes
8.7 A , so 2.175 A per milliohm so each shunt passes 2.175 millivolt

Any suggestions for calibration (i don't have DB but i do have clamp on ampprobe and 1/10 ohm multimeter)

 

[fechter]

Not quite.

There are 4 shunt wires in parallel. The combined wires have a resistance of about 1 milliohm. At 35 amps, the voltage drop across the shunt would be 35 mV.

For resistors in parallel, the total resistance
Rt= 1/(1/R1+1/R2+1/R3+1/R4)

The resistance of each individual shunt wire would be around 4 milliohms.

If you snip one, you'd want the switch+wire resistance to be around 4 milliohms to get the right limit when the switch was on.

You can calculate the resistance wire based on gauge and length. The resistance of the switch is harder to estimate, but it will be more than zero.

Your clamp on meter will only work if it is an AC/DC type.
Otherwise calibration will be difficult. I use a 100 amp shunt and analog meter. A DrainBrain would be great.

The only other way to calibrate would be if you had a constant current power supply to put a steady 1 or 10 amps throught the shunt and then measure the voltage drop across it. My meter is not sensitive enough to get a good reading with 1 amp of current (it only reads to the nearest tenth of a millivolt).

 

[Matt Gruber]

i'd ride it, note the amps, if it's too high, add wire.
why is this hard?

 

[29a]

:wink: