12v Switch

[dave21]

New here so I say hello.


Some advice on car 12v please.


I want to fit a dual volt and amp meter in my dash, and have one with the appropriate shunt. My expectations are, when the engine is off it will show the state of the battery. When the engine is on it will show the active volts and amps?


I assume I can wire to the cigarette lighter socket with is always on?


As I don’t want the meter ‘always on’, I’d believe I can fit a simple on/off switch on the positive side. However, would a common accessory switch do the job or do I need a beefy high amp type?


Thanks, Dave

 

[telectrix]

just fit a voltmeter to the battery with a 2A fuse in line. the meter will show battery volts ( 12. summat ) when engine off, and 14.2V or thereabouts with engine running.

 

[Deservices]

The ammeter, if I remember correct, needs to be in series with the load you want to measure. So if you wanted normal running current would have to break into the main battery feed. Start up amps could be a lot have a look at the max output of your battery could be over 150 amp. Which is why they have large main cables. If you connected to cig lighter, it would show current draw from whatever is plugged into cig lighter. Althou I may just be talking rubbish lol.

 

[telectrix]

think OP isa confused between ammeter and voltmeter.

 

[tazz]

No you right, the shunt is usually a 0.5 ohm resistor, and needs to be in series with the main positive cable to the battery, the amp meter then measure the voltage across the shunt, usually +1v to -1v which is proportional to the current being drawn, and shown in Amps

 

[UKMeterman]

Tazz,
What voltage drop are you expecting at say 10 amps with your 0.5ohm shunt...?

It may be better to use a hall effect device and a bit of electronics Electric Vehicle Battery Monitor

 

[tazz]

Please read the post ....its not what is better....its what the op has...

 

[NickD]

No you right, the shunt is usually a 0.5 ohm resistor, and needs to be in series with the main positive cable to the battery

Don't believe so. Starter motor draw from the battery is many tens of amps, depending on engine size, compression (diesels draw shedloads), oil gloopiness etc. Call it 40 amps ballpark though this will vary massively. You can't get 40 amps through a 0.5 ohm resistor off of 12V, and if you could it would dissipate 800 watts. Bang.

 

[tazz]

Don't believe so. Starter motor draw from the battery is many tens of amps, depending on engine size, compression (diesels draw shedloads), oil gloopiness etc. Call it 40 amps ballpark though this will vary massively. You can't get 40 amps through a 0.5 ohm resistor off of 12V, and if you could it would dissipate 800 watts. Bang.

IT is a shunt...rated at 150 amps...very common on cars and boats for measuring amps...been used for years....google before making rash comments, or I`ll set Glenn on you....!!!

 

[NickD]

Why do you want the meters? If you just like gizmos then fair play to you. But you almost certainly already have a battery voltage indicator of sorts, i.e. the battery warning light, which comes on when the battery voltage when being charged by the alternator drops significantly below 14.4V. When the battery voltage is being kept at this level then it is being charged / kept topped up by the alternator so any current in the main battery lead will be going in not coming out. I'm not sure this is actually the current you want to be measuring. If you really want to measure some current somewhere I'd go for the Hall effect sensor someone mentioned or some other sort of external current clamp which works at DC.

 

[NickD]

IT is a shunt...rated at 150 amps...very common on cars and boats for measuring amps...been used for years....google before making rash comments, or I`ll set Glenn on you....!!!

Do you mean milliohms or something? If what I've put is wrong, someone has rescinded Ohm's Law.

 

[tazz]

Nick, a shunt is made of carbon steel, able to withstand high currents, but to measure high currents is not practical......so the shunt has a resistance to its total load. ie 0.5 ohm....the voltage across the the shunt is measures in mV, which is proportional to the current being drawn....full 100amp = 1v forward, therefore 500mV = 50amp, and so on....even tho the amp meter shows amps, it actually measures mVolts

 

[NickD]

100 amps through a 0.5ohm resistor entails voltage drop across it of V = IxR = 100 x 0.5 = 50 volts.

100 amps through a 0.5milliohm resistor entails voltage drop across it of V = IxR = 100 x 0.5x10^(-3) = 50mV.

You mean 0.5 milliohms, dude.

 

[tazz]

This may help

 

[NickD]

Sorry, that comes up as blank to me.

Shunt (electrical) - Wikipedia, the free encyclopedia

"In this case the shunt, a manganin resistor of accurately known resistance, is placed in series with the load so that all of the current to be measured will flow through it. In order not to disrupt the circuit, the resistance of the shunt is normally very small. The voltage drop across the shunt is proportional to the current flowing through it and since its resistance is known, a voltmeter connected across the shunt can be scaled to directly display the current value.
Shunts are rated by maximum current and voltage drop at that current. For example, a 500 A, 75 mV shunt would have a resistance of 0.15 milliohms, a maximum allowable current of 500 amps and at that current the voltage drop would be 75millivolts. "

Similarly a 100A, 75mV shunt would have a resistance of 0.75 milliohms. A 100A, 750mV shunt would have a resistance of 7.5 milliohms.