Below is an email exchange I had with Lee Heart on a modification to his Batt-Bridge.
On 1/4/2011 7:19 PM, David Nelson wrote:
Hi Lee,
I was wondering how hard it would be to have some sort of analog
display like a typical auto ammeter or galvanometer with zero at the
center to measure the difference between the two halves of a battery
pack?
That's even easier than the existing circuit. View with a fixed width font:
HTML:
________/\/\__________
| R1 |
__|__+ |
___ upper half of pack |
| - __ <
|________/ \_________\< R3
| \__/ meter /< trimpot
__|__+ <
___ lower half of pack |
| - |
|________/\/\__________|
R2
Use a zero-center analog meter. R1 and R2 are equal values, about 10 times higher than R3, and limit the meter current in case R3 is turned all the way to one end. R3 is a trimpot to set the balance. Adjust it so the meter reads zero when the two half packs are at exactly the same voltage.
The meter should be something like +/-50 microamps full scale, to allow the resistors to be large values and so not draw much current or dissipate much power. Set the pot *first* and then connect the meter.
Another idea I had was to have some sort of comparator circuit
that reads the two halves and outputs a + or - voltage reading to a
display. I would also want this device to be powered directly from the
pack it is measuring and not rely on the 12V system. I really like
that the CycleAnalyst runs entirely off the pack since it won't put an
imbalance on the pack.
That's not hard, either. Connect the LEDs of an optocoupler with a series resistor across each half-pack. Connect the transistor output of the two optocouplers in series. Pick the series resistors for each LED so the voltage between the transistors is half of the power supply when the two half-packs balance. It will then swing up and down as the pack's voltages are different.
With either circuit, you'll have to pick your values experimentally, based on the specific parts you use.
=======================================
Thanks Lee, I think I followed that. Just to be sure see notes below.
On Wed, Jan 5, 2011 at 10:08 AM, Lee Hart <
[email protected]> wrote:
> On 1/4/2011 7:19 PM, David Nelson wrote:
>>
>> Hi Lee,
>>
>> I was wondering how hard it would be to have some sort of analog
>> display like a typical auto ammeter or galvanometer with zero at the
>> center to measure the difference between the two halves of a battery
>> pack?
>
> That's even easier than the existing circuit. View with a fixed width font:
>[see circuit diagram above]
> Use a zero-center analog meter. R1 and R2 are equal values, about 10 times
> higher than R3, and limit the meter current in case R3 is turned all the way
> to one end. R3 is a trimpot to set the balance. Adjust it so the meter reads
> zero when the two half packs are at exactly the same voltage.
>
> The meter should be something like +/-50 microamps full scale, to allow the
> resistors to be large values and so not draw much current or dissipate much
> power. Set the pot *first* and then connect the meter.
Ok so the meter is a microamp meter with an internal shunt and is
connected to the wiper of the trimpot. The meter will indirectly
measure voltage difference where the meter scale would be some
constant times the current through the meter depending on R1, R2, R3
and the total pack voltage. So if the parts were selected such that at
a pack voltage of 64V a full swing of 50uA would mean say 2V
difference that with a pack voltage sagging down to 55V a full swing
of 50uA might be something like 2.2V (not calculated, just a guess).
>> Another idea I had was to have some sort of comparator circuit
>> that reads the two halves and outputs a + or - voltage reading to a
>> display. I would also want this device to be powered directly from the
>> pack it is measuring and not rely on the 12V system. I really like
>> that the CycleAnalyst runs entirely off the pack since it won't put an
>> imbalance on the pack.
>
> That's not hard, either. Connect the LEDs of an optocoupler with a series
> resistor across each half-pack. Connect the transistor output of the two
> optocouplers in series. Pick the series resistors for each LED so the
> voltage between the transistors is half of the power supply when the two
> half-packs balance. It will then swing up and down as the pack's voltages
> are different.
>
> With either circuit, you'll have to pick your values experimentally, based
> on the specific parts you use.
So the advantage with this circuit would be that the output would be
the true voltage difference between the two halves regardless of the
changes in total pack voltage assuming that the device was
built/calibrated properly, right? This circuit will take me a little
longer to study out and understand. It is times like this I wish I had
spent more time in my Physical Electronics classes to understand what
really was going on and how everything inter-related.
================================
On 1/6/2011 1:21 AM, David Nelson wrote:
Thanks Lee, I think I followed that. Just to be sure see notes below.
Ok so the meter is a microamp meter with an internal shunt
Yes, though these meters have no internal shunt (and don't need one). Their coil is many turns of very fine wire, having 2K or so total resistance. Full scale is therefore about 50 microamps at 0.1 volts.
and is connected to the wiper of the trimpot.
Yes... meter between center tap of pack and wiper of the trimpot. It directly measures the voltage difference between these two points (with full scale being a 0.1v difference. Too sensitive; but you fix that with R1 and R2.
So if the parts were selected such that at a pack voltage of 64V a full swing of 50uA would mean say 2V difference that with a pack voltage sagging down to 55V a full swing of 50uA might be something like 2.2V (not calculated, just a guess).
Yes; that's the idea.
To calculate it, start with the equivalent circuit at the wiper. It is a voltage of half the pack, with a series resistance of the parallel combination of the upper and lower resistances (R1 + R3/2 and and R2 + R3/2). Since R1=R2 and R3 is split exactly in half at balance, this is just Req = (R1 + R3/2)/2.
Suppose we want a 2v imbalance to produce full-scale on the meter, and the meter's resistance is 2000 ohms. When one half of the pack is 2v different, the center tap is 1v different from the voltage on the wiper. To get 50ua to flow with a 1v difference, the total resistance must be Rtotal = 1v / 50ua = 20k. The meter has 2k of this, so we need 18k more.
Thus, Req = 18k, so R1 + R3/2 = 36k and R2 + R3/2 = 36k. You can use any combination that adds up to 72k; two 36k and no pot, or two 35k and a 2k pot, etc.
The pot sets the balance point (compensates for R1 and R2 not being identical values). With perfect 0% resistors you wouldn't need it. But as a practical matter, if you use 5% resistors the pot needs to be at least 5% of their value to be able to compensate for the difference between them.
With precision resistors, the meter can be calibrated in volts difference. (Remember that a 1v difference at the meter means a 2v difference in the two half-pack voltages). If you don't use precision resistors, you can add a second pot in series with the meter to adjust the meter so full scale is exactly a 2v difference (or whatever you want).
So the advantage with this circuit would be that the output would be
the true voltage difference between the two halves regardless of the
changes in total pack voltage assuming that the device was
built/calibrated properly, right? This circuit will take me a little
longer to study out and understand. It is times like this I wish I had
spent more time in my Physical Electronics classes to understand what
really was going on and how everything inter-related.
The meter circuit measures true voltage difference, regardless of the pack voltages.
The circuit with the optocouplers has an isolated output, but won't be as accurate because the optocouplers are not precision components. Their characteristics change with temperature, with age, and each part is a little different from the others. But it should work "good enough" for spotting a cell in danger.
Warning... remember that this is a very simple circuit; it can't spot a problem is TWO cells are both low in voltage, one in the upper half and one in the lower half of the pack.
--
Lee A. Hart | Ring the bells that still can ring
814 8th Ave N | Forget the perfect offering
Sartell MN 56377 | There is a crack in everything
leeahart
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