Author Topic: Measuring DC current out of a battery charger with a 500A 100mV shunt (Read 10951 times)

Jedon · « **on:** January 13, 2011, 01:34:52 PM »

Okay I got my hydro going and the numbers for the raw hydro output look good, 5.26A at 108V DC but I don't think I'm getting all that power into my batteries.
I got an IOTA 13A 55.4V DC battery charger and feed the DC right into that even though it's an AC input charger, I was told it would work and it seems to do something.
I want to measure how much current I'm getting out of the charger. My multimeter can read 20A DC so I tried that but it went blank so it must surge about that?
It seems like the voltage and current when the charger is connected to the battery bank, becomes that of the battery bank?
I think I can just put the shunt on the + side of the charger and measure the voltage drop across that to calculate the outgoing amps, right?
I know I shouldn't be using AC rated breakers but it's what I have for right now so my setup looks like this:
hydro->20A breaker->120V AC socket->battery charger->20A breaker->Batteries ( 8 x 6V golf cart )->inverters
Since I'm getting 12KWh+ per day raw out of the hydro that seems like it should be enough power so I don't have to ever run the generator but I still do so I'm trying to figure out where the issue is, maybe I'm just a power hog?! I have an AC power meter to install that will help with that side.
Thanks!

ghurd · « **Reply #1 on:** January 13, 2011, 01:44:04 PM »

The shunt goes in series with the power wire.

50A will read 10mV.
25A will read 5mV.

10A (500W) will read 2mV.

Depending on the scales in your meter, the margin of error can be pretty high with a 13A charger.
G-

Jedon · « **Reply #2 on:** January 13, 2011, 01:58:35 PM »

Thanks I'll give that a try tomorrow.

Jedon · « **Reply #3 on:** January 13, 2011, 02:09:12 PM »

I'm wondering if the input voltage to the charger is not high enough, it's right at the low end cutoff range, 108V. I'll put bigger jets on the hydro to get some more voltage out of it under load and if that doesn't work I'll lower the field resistance to keep the voltage up under lower flow rates.
This is the charger : http://www.iotaengineering.com/dls5413.htm
it does say 80% efficient so that would lower the 560W down to 450W, then the inverters are 97% so now we are down to 434W max, if the charger is producing less current at the lower voltage,
If it can push 13A into 54.4V, it can only push 11.7 into 49V ( battery bank voltage )? Or it makes 13A at 120V but only 11.7A @ 108? Combine the two for even less power?

ghurd · « **Reply #4 on:** January 13, 2011, 04:23:32 PM »

I have no idea how that one works.

Typical transformer based battery chargers push less amps as the battery voltage increases.
But typical transformer based battery chargers will not work with DC!

Stray thought.
Maybe the battery charger is oscillating.
Do you have a meter with frequency?
Maybe check the HZ on the battery charger's input. If you get a reading that is not as high as the HZ from the Harris should be, then maybe the Iota is switching on and off.
Yes, I know. On paper it should probably not work. Except that a $20 VOM with Hz will read the Hz of a PWM solar controller with a 3A PV and a 115AH battery, and it will do it on the battery terminals.
G-

Jedon · « **Reply #5 on:** January 13, 2011, 05:04:38 PM »

I have this multimeter: http://www.harborfreight.com/ac-dc-digital-multimeter-37772.html
Says it measures Hz to 20K
I wonder if I can/should just disconnect the rectifiers from the Harris and have it send the unregulated AC to the IOTA?

ghurd · « **Reply #6 on:** January 13, 2011, 05:52:01 PM »

That meter is about like the one I use. Maybe exactly.

I do not know anything about IOTA stuff.
Probably not a great idea to try it without some guidance from someone who knows more about them (not me).
G-

joestue · « **Reply #7 on:** January 13, 2011, 06:59:54 PM »

If your meter can read millivolts accuratly just measure the voltage across a length of cable, no need for a shunt.

that battery charger is spec'd for 108vac to 132 vac, which means it would prefer and extra 50 volts on the input.
this is significant because it may be a topology that's locked at up to 45% duty cycle. it may be that high right now even though its putting out about 2/3rd rated output.

Jedon · « **Reply #8 on:** January 13, 2011, 07:38:44 PM »

I don't really understand how the putting DC into an AC charger thing works, so it sounds like you are saying that if it wants 108 to 132V AC then for DC it wants 158V to 182V?

Jedon · « **Reply #9 on:** January 13, 2011, 07:44:06 PM »

I Google'd it and this looks like it explains:
Note that the DC voltage is roughly equivalent to 71% of the peak AC voltage.
http://www.bcae1.com/voltages.htm
So 71% of 158V is 112V and 71% of 183V is 129V so you +50V is pretty good.
Looks like I am really not feeding enough voltage in under load and should lower the field strength on the Harris unit.
Thanks!

joestue · « **Reply #10 on:** January 14, 2011, 10:03:23 AM »

FYI (as a general rule)
Anything that says 85-265 vac will take a dc voltage from about 80 to around 360 volts or so.
if it says auto ranging or auto select, then it has a relay that will switch it from a voltage doubler to a full wave bridge.
in this case it is safer to just bypass the FWB (you'd want to remove it for efficiency reasons anyway) and apply the dc to the filter caps directly. Those types will need about 200 volts or so.

OP's battery charger may work marginally better with the diodes removed.
I suspect it is a standard half bridge inverter.

Jedon · « **Reply #11 on:** January 14, 2011, 01:07:32 PM »

I'll take a picture of the inside of the charger and post it tonight, if there are things I should do to make it work better I'm all for it!

ghurd · « **Reply #12 on:** January 14, 2011, 11:18:07 PM »

Quote from: joestue on January 14, 2011, 10:03:23 AM

OP's battery charger may work marginally better with the diodes removed.
I suspect it is a standard half bridge inverter.

OPs being Other Person's? Meaning mine?
Mine is basically a 10:1 transformer with a rectified output. It has an ammeter on the output. It is center tapped for a 6V battery. And it has a DPDT-CO switch.
Are you suggesting I remove the full wave bridge rectifier from the output of the transformer for increased performance?

joestue · « **Reply #13 on:** January 15, 2011, 08:55:32 AM »

op = original poster

his is a switch mode unit

Jedon · « **Reply #14 on:** January 15, 2011, 08:55:00 PM »

I can't seem to get high enough voltage into the IOTA, is there an electronic way to up the voltage +50V or so? I know I can't use a transformer since that's AC, although the Harris hydro unit does have rectifiers so maybe I could put a transformer before the rectifiers?
Thanks!

Jedon · « **Reply #15 on:** January 16, 2011, 11:54:42 AM »

Here are some pictures of the IOTA charger, any suggestions on how to modify it?

joestue · « **Reply #16 on:** January 18, 2011, 05:48:05 AM »

i'm going to say with 90% confidence that the ac comes in through that common mode filter goes right to the full wave bridge and those three 200 volt caps.

the transformer is a forward converter, that's why there's a 7 or 10 watt ceramic resistor for a snubber, and 6 paralleled current sense resistors.
I would also speculate that the jumpers between the capacitors are for the American 120vac version.
which means one of two things, that is intended to be a voltage doubler, and that cap next to the diode bridge is a 400 volt cap, or all three are 200 volt caps in parallel and the jumpers are just jumpers.

on the output that looks like a standard 6 amp diode for reverse polarity protection. you could potentially bypass it; and the full wave bridge on the input for approximately a 2% improvement in efficiency.
that's about the only thing you can do.

Jedon · « **Reply #17 on:** January 18, 2011, 01:46:16 PM »

Thanks, that doesn't seem very worth it.
Does the output decrease with the input voltage regardless of current?
So it wants 108V AC minimum but I give it ( loaded ) 109V DC ( but that's 71% of 109V AC or 78V? http://www.bcae1.com/voltages.htm ) ? If so then it doesn't seem like it should even be working but it is.
How can I measure the current coming out of the charger? I tried it with my DMM but it backfeeds from the battery bank, ZAP!

ghurd · « **Reply #18 on:** January 18, 2011, 03:47:21 PM »

Quote from: Jedon on January 18, 2011, 01:46:16 PM

How can I measure the current coming out of the charger? I tried it with my DMM but it backfeeds from the battery bank, ZAP!

I had a feeling you were improperly measuring amps from the other thread.
You are not connecting the meter right.
Google how to measure amps with a meter.
G-

Jedon · « **Reply #19 on:** January 18, 2011, 04:49:06 PM »

Ow wow don't I feel like an idiot! I was connecting to the circuit just like when reading voltage! I was configuring the meter right at least.
Time to re-measure.

Jedon · « **Reply #20 on:** January 18, 2011, 04:57:13 PM »

Okay so now I am a bit confused, if I am down at the turbine, and I want to measure current, do I remove the negative lead from the unit and insert the meter between there and the outgoing negative wire?
http://www.sciencebuddies.org/mentoring/project_ideas/Elec_HowToMeasure.shtml

ghurd · « **Reply #21 on:** January 19, 2011, 08:15:15 AM »

Yes, that would work.
Depending on the meter lead terminals, and how the system is physically set up, I usually try to pull out a fuse and measure the current in the gap.

It can be in the Pos or Neg wire. The reading will be the same. What goes out must come back.

Might find this sketch clearer-
http://www.electronics-radio.com/articles/test-methods/meters/how-to-measure-current.php

The turbine will not slow down when measuring amps properly. If anything, it may speed up a microscopic amount.

The "ZAP" may have blown the meter fuse. Best check it before trying the test. Many have a spare fuse inside

Be Carefull!
That 108 VDC loaded will be higher during the time the wire is separated and the meter connected.
G-

Jedon · « **Reply #22 on:** January 19, 2011, 02:41:34 PM »

Thanks! So the red and black wires don't really matter in this case? Is it better to read off of + or -?

Another question, how do I read transformer specs? I poked around on the net but couldn't find a good link.
I have one that reads:
L=500uH 10%
I=40.0Arms
E=3.77Vrms
fnom=60Hz
fsw=20KHZ
HIPOT=4KVrms

So it sounds like it takes some voltage at 60Hz and turns it into 3.77V and can handle 40 amps?
It's a custom unit made by Frost Magnetics so the part numbers didn't do my any good.
Thanks!

TomW · « **Reply #23 on:** January 19, 2011, 02:54:38 PM »

Quote from: Jedon on January 19, 2011, 02:41:34 PM

Thanks! So the red and black wires don't really matter in this case? Is it better to read off of + or -?

Another question, how do I read transformer specs? I poked around on the net but couldn't find a good link.
I have one that reads:
L=500uH 10%
I=40.0Arms
E=3.77Vrms
fnom=60Hz
fsw=20KHZ
HIPOT=4KVrms

So it sounds like it takes some voltage at 60Hz and turns it into 3.77V and can handle 40 amps?
It's a custom unit made by Frost Magnetics so the part numbers didn't do my any good.
Thanks!

That sounds like a choke or inductor rating from here.

Transformers usually have a Primary and Secondary windings ratings.

The uH is microhenries and a rating for a choke or Inductor and probably only has 2 leads where a transformer would have 4 or more.

Just a thought.

Tom

joestue · « **Reply #24 on:** January 19, 2011, 08:16:38 PM »

keep it simple, read the voltage across one foot of wire.

Quote from: Jedon on January 18, 2011, 01:46:16 PM

Thanks, that doesn't seem very worth it.
Does the output decrease with the input voltage regardless of current?
So it wants 108V AC minimum but I give it ( loaded ) 109V DC ( but that's 71% of 109V AC or 78V? http://www.bcae1.com/voltages.htm ) ? If so then it doesn't seem like it should even be working but it is.
How can I measure the current coming out of the charger? I tried it with my DMM but it backfeeds from the battery bank, ZAP!

i'd still bypass the diodes but that's just me.

the output will not follow the input, however there is a lower limit, below which the 55ish volt open circuit voltage will drop.
the fact that it can deliver power to the battery means its working right.
There are a number of reasons it cannot deliver the full rated output below rated input, i can make this as complicated as you want...

Quote

If it can push 13A into 54.4V, it can only push 11.7 into 49V ( battery bank voltage )? Or it makes 13A at 120V but only 11.7A @ 108? Combine the two for even less power?

There is a product limitation (700 watts iirc?). as long as the input voltage is high enough that the duty cycle stays below 45% it will deliver up to 700 watts at whatever voltage. you may even be able to get 700 watts at 45 volts, but it probably is set up to not exceed 13 amps. (is that a blue potentiometer next to the heatsink in the corner? Tweak at your own risk!) 13 amps x 45 volts is 585 watts.

Is this a DC or AC turbine?

Quote

At the power shed ( 750ft from the turbine ) voltage was 109V to the charger and the batteries were are 50.1. Later when the sun was on the 800W of PV the battery voltage went up to 53V and the hydro voltage went up to 114.4V so there is a correlation between the voltage of the batteries and the loaded hydro voltage.

yes, there is a correlation, but the reasons will vary.

Jedon · « **Reply #25 on:** January 20, 2011, 12:04:59 PM »

How sensitive does my DMM need to be to read across the foot of wire? I tried it and got 0, what is the expected range?
I did mess with the blue pot and raised the output voltage a bit up to 56.6V.
I'll measure the current coming out of the charger, that will tell me a lot more about what is going on.
The turbine it turns out, is 3 phase AC rectified to DC.
Thanks!

TomW · « **Reply #26 on:** January 20, 2011, 01:18:45 PM »

it will be in millivolts [mv].

Seems a 1 ft hunk of #10 copper drops 1 millivolt per amp of current flow in the wire. Smaller or larger wire would vary with smaller dropping more and larger less.

Just the nutshell version.

Tom

Jedon · « **Reply #27 on:** January 20, 2011, 01:23:06 PM »

So perhaps using 14g would give me bigger readings?

TomW · « **Reply #28 on:** January 20, 2011, 01:28:06 PM »

Quote from: Jedon on January 20, 2011, 01:23:06 PM

So perhaps using 14g would give me bigger readings?

Indeed.

Smaller wire is higher resistance.

Higher resistance drops more voltage per unit of current.

#10 just happens to work out to about 1 mv per amp and handles 50 amps easily too boot.

Ohms Law is your friend with current shunts.

Tom

Tom

Jedon · « **Reply #29 on:** January 22, 2011, 08:32:34 PM »

Yeah I didn't know it was 3 phase but it is. I measured the right way and now I'm getting:
From hydro at shed: 1.88a at 112.4v ( 211 watts )
Out of IOTA: 3.56a at 50.7v ( 180 watts )
So the charger is being 85% efficient which seems about right?
My guy feeling form our house power usage is that 180 watts seems about what we are getting, about enough to break even most of the time.
Now I can go futz with the alternator and see how much more power I can eek out of it.

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Author Topic: Measuring DC current out of a battery charger with a 500A 100mV shunt (Read 10951 times)