The Path of Least resistance?

[Germaine Williams]

Hi, and thank you in advance. The circuit below represents a real life design that we're experiencing problems with, two 12v 28AH batteries are joined together for a 24v supply, however the switching of this supply via relays (inappropriately marked R2,R3) is only wired to one of the two batteries. The load when the system is switched on but not being used is 440ma total. After 6 hours of the system being on standby it develops a fault where by dimming the load up (r1 = dimmer attached to incandescent lights) sends the circuit into a fault where the relays continuously disengage and try and reengage.

I think what is happening is that when the batteries are full and the load is low (R1 = 170 Ohms) the current is happy to flow through the relays as there's a PD of 13.7v and a resistance of 40 Ohms , however after several hours this PD drops to 12v and once we increase the load to full R1 = 1.2 Ohms, therefore the path of least resistance is to bypass the relays and go through the load, thus the relays fail which disconnects the load and the cycle starts again.

If anyone could offer any advice on if my theory has any grounding, it would be greatly appreciated, I'm really not sure how to apply the basic pd divider equations once you introduce a second battery, or why battery 1 which is more than adequate (should be able to provide 87 Amps for 7 minutes and the load is 20 amps max) wouldn't be able to deal with these loads.

Regards
G

 

[ericgibbs]

hi,
I suspect the internal resistance of the two batteries are higher than you believe and its that which causing the redirection of the current paths thru your network.
Is it possible to connect the two 12V relays in series across the full 24V supply.??
E

 

[Germaine Williams]

Thanks for the swift reply. I'll try and measure the internal battery resistance as described online. Could you explain how a high internal resistance could cause this behaviour? Limiting max current supply or making the relays a more attractive return path? I'm probably going to get rid of the relays as they aren't helping but I'm just trying to better understand why this behaviour is happening.

Thanks again
G

 

[jpanhalt]

Have you measured the voltage drop across the low resistance load? If it is <<24V, then the difference between 24V and what you actually measure has to happen somewhere.

John

 

[Germaine Williams]

Hi john so you think I should measure the voltage drop across the load and compare it to the two batteries? The difference being power that is taking a different path?

 

[jpanhalt]

There are no different paths for the current unless you change the circuit.

Since your load is 1.2Ω and R2 and R3 in parallel are equivalent to 40 Ω, then most of the current is going through the load. For the time being, ignore R2 and R3. 24 V across 1.2Ω will produce 20 amperes of current. Conversely, if you have 20 A of current across a 1.2 Ω resistor, the voltage drop across that resistor should be 24 V.

What Eric suggested and I concur with is that your batteries cannot supply 20 A to that load. Higher internal resistance of the batteries than you show in your model would be a likely explanation. In other words, if your batteries really have 0.01 Ω internal resistance each, then the voltage drop due to internal resistance across the batteries in series when producing 20 A would be just 20(0.01x2) = 0.4 V and the voltage across the load would be reduced by that amount. It is quite possible that the batteries are not identical. That is, one may have a much higher internal resistance than the other. Higher internal resistance will be evident as lower voltage across the load.

So, measure the voltage drop across your load. While you are at it, measure the voltage across each battery while the load is connected.

John

 

[AnalogKid]

The schematic does not match the description. In the schematic, two batteries are in series with a 1.2 ohm load, for about 20 A of loop current.

ak

 

[Germaine Williams]

Hi Guys, So if my Maths is correct the internal battery resistance is 0.032 Ohms per battery I had estimated the load and it's obviously going to be constantly changing. The Voltage across the load is 23.6V and each each battery is 12.3v, therefore there is 1 volt being dissipated somewhere. My Load when everything is on is actually closer to 11.2A does that help shed any more light on this? I feel a bit more confused than when I started.

Regards
G

 

[Colin]

When you are measuring the current, the resistance of your multimeter is adding to the resistance of the circuit and producing a false reading.

 

[rfranzk]

Hello All, What size or gauge of wire are you using?? The wire between batteries in series and the load wiring could be source of voltage drop if it is not large enough to carry the current.. I would try measuring voltage drop at each node in the circuit to isolate either a single high resistance point or a sum of several resistances causing a voltage drop.

 

[ChrisP58]

Yes, double check your wire size, length and integrity of all connections. When current is flowing, check the voltage drop accross every element in the loop.

What are the relays for? Does the current flow through the relay contcts? If so, look at how much voltage drop you might have there.

Also, what are the specs of the relays? Typical 12DC volt relays will engage with about 10VDC and stay engaged till the voltage falls to about 2 or 3 volts. If they are going on and off as you describe, there must be something else going on.

 

[mbarazeen]

your relays are connected to only one battery marked as V1, make sure they are seperately wired to te battery terminals. You can use smaller wire size for relays in this case as the standard for control circuits.