Confused by circuits - need some help please!

[peterb]

OK it is actually a cct designed for speed control - with a variable voltage output - not really the most appropriate thing to use in this application.

Most monitors (AFAIK) have a 12 volt input - but it would be wise to check that if you can - either from the monitor case or handbook, or a Google search to the mfr's website.

The reasons why you need to check are:

1. The battery has a 12V output - and if the monitor hada 12 v input, there would be no need for this circuit.

2. If the monitor does have an input voltage of less than 12 volts, feeding it with 12V would cause it to draw excess current - which could be why the fuse is blowing.

The other reason for the fuse blowing could be because you have wired it up incorrectly.

[timtim86]

Why would the 5A fuse be blowing even when the Voltage regulator isn't plugged in? Why is it in parallel? Seems bizarre to me...I'm going to put together a quick circuit diagram so I can show you how it's wired

[timtim86]

Worst circuit diagram ever but I can't remember A-level physics anymore.

thanks for your help Peter

Looking at it, it seems completely ridiculous to me. Why does the fuse complete a circuit with the battery? Is it to make sure that the battery isn't too strong? In which case, do batteries change the amount of current the give out over time? This thing used to bloody work!

[peterb]

The fuse should not be across the battery - it is shorting it out. No wonder it is blowing. DON'T connect the multimeter (set to current) across the battery - lead acid cells will deliver several tens of amps - some hundreds - and with the multimeter set to current, it appears as almost a short circuit.

To measure current it should be in series with teh load - to measure voltage it should be in parallel.

The 5A fuse should be in series with the input to the regulator.

Batterys will deliver the current that the load takes - the only limit is the connecting wiring and the internal resistance of the battery - which for that type of lead acid cell is very low. The Voltage will vary slightly over time, but not significantly unless the battery is deeply discharged (which won't do iyt ant good.

Ohms law states that the current is through a load is proportional to the voltage across it - the constant of prportionality is called risistance -

so V/I=R or I=V/R sop w 12 volt battery connected to a 24 ohm resistro will deliver 12/24 amps which is .5 (or 500 milliamps) double the voltage to 24, and the current flowing through the resistor will be 1 amp.

Power =V*I so doing a bit of substution with ohms law - power =I^2 R or v^2/R. So in the above case - a 24 ohm resistor connected across a 12V battery will consume (or convert to heat (12*12)/24 watts = 6 Watts. Double the Voltage (and the current) and the power quadruples to 24 Watts - the resistor will get very hot if it is a 6 Watt rated component.

Back to the cct - the 5A fuse should be in series with the input to the regulator - the 2A in series with the input to the monitor. The Voltage output of the regulator should be set to the rated voltage of the monitor - but as I (think) I said earlier - this isn't the best type of regulator to use for the job.

[peterb]

BTW - you were lucky that you only drew 5A from the battery - yiou could have drawn much more which could have damaged the mulitimeter - although they usually have internal fuses to protect them from that sort of abuse . The battery may need recharging - a fully charged lead acid cell of that type should have a terminal voltage (use the V setting on the meter) of around 13.2 to 13.8 volts. (Just a thought - some MMs have separate plug socket for measuring current - as well as a switch setting - if the correct socket wasn't selected, you won't get a meaningful reading.)

[Blitzen]

Ok - I can see the pics now (and that is a variable resistor - which I guess is to control the output voltage - or perhaps the overcurrent threshold) - what is the Maplin part number?

It just an old fashioned POT isnt it?