# battery monitoring circuit

Discussion in 'General Electronics Chat' started by sujinvipin, Sep 6, 2011.

1. ### sujinvipin Thread Starter New Member

Aug 18, 2011
28
0
i would like to cal the fuse rating for the battery monitoring circuit.
i have attached the image.
Vinput=56V battery input and my desired vout to the plc input should be 9V.
i have chosen the resistances to be 5.6K and 1K ohm.
So by calculation i get the max current in the plc to be 6.6mAmps.
Now how would i determine the fuse rating. for eg if i take a fuse rating of 0.5A then would the plc be protected when the battery voltage is 56V.

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2. ### cjdelphi New Member

Mar 26, 2009
272
2
if it's a car question the mod's are not going to be happy lol

3. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,793
771
Boy !! U are going to need very high power resistors and a container to keep them touching anything.

Ever consider getting a DC to DC converter

4. ### SgtWookie Expert

Jul 17, 2007
22,183
1,729
With the resistor values shown, you'd have ~8.5mA current flow through the two resistors, and you'd need a resistor rated 3/4W or higher for the 5.6k resistor. That's quite a bit of power to use just to monitor the voltage of the battery.

At a minimum, use 10x as much resistance to get the current below 1mA, and use a 10nF to 100nF cap from the junction of the resistors to ground to provide a low-impedance input to your PLC's ADC. Keep the ADC sample rate low so that the cap doesn't get drained needlessly. This will enable you to use 1/10 Watt resistors for both the 56k and 10k values. You will not need a fuse, as resistors don't fail shorted. Power dissipation in the 56k resistor will be ~50mW.

If you wish to further reduce the power consumption by the monitoring circuit, you can increase the resistors up to ~8.5 times more; 85k for the lower and 470k for the upper. This will give you roughly 0.1mA current through the two resistors, and ~8.58V at the junction of the two. If you then use a 10nF cap to ground from the junction, it will require about 8mS for the cap to fully charge, or 80mS for a 100nF cap to charge.

I've attached a schematic/simulation of the latter; only the time segment from 40mS to 80mS is shown.

• ###### ADC input RC time.PNG
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Last edited: Sep 7, 2011
5. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,793
771
Shooot ! I missed the milli in the Amperage value.

6. ### eblc1388 Senior Member

Nov 28, 2008
1,542
102
You still need a fuse. A 0.5A or 1A one will do.

This is not needed for the resistors but will protect you from accidents like shorting of the DC 56V via whatever possible means, like using multimeter wrongly on current range while doing a voltage measurement.

7. ### Crispin Member

Jul 4, 2011
88
2
thanks! That answered a question I did not know existed. I was seeing this on my current project (400V -> 3V divider)

/hijack