# 12v from battery to 5v

1. Jan 13, 2009

### lowspeedchase

Hi all...

I plan on using the following device: "[PLAIN [Broken] Manual.pdf"]http://www.pc-control.co.uk/images/Wasp [Broken] Manual.pdf[/URL] to monitor 12v lines in a jeep and respond with various actions.

A simple example:

Monitor the 12v+ line for the vehicles brights, when it gets hot (hot meaning a current flowing through the line), enable a set of auxiliary lights.

The issue is that the board I am going to purchase, per the aforementioned link, can monitor up to 5v for input.

The solution i have come to is to attach a "+5V Fixed-Voltage Regulator 7805" between the 12v line and the input on the board.

I have a couple of questions though that I hope some of you experts can provide some guidance on:

Do I need to use a diode on the positive line to prevent current from flowing the wrong way?

Should I use a capacitor to reduce the current below 5v? If so what would you recommend?

Lastly, I read this line in the manual for the board that confused me:
"As a rough guide the most important thing is to ensure that the source impedance (Rs) of the analogue signal (VA) is less than 10k"
Do I need to pay attention to that in this automotive environment?

Any input or direction you guys (and gals) could provide would be greatly appreciated!!

Chase

Last edited by a moderator: May 3, 2017
2. Jan 13, 2009

### mgb_phys

If you just want to detect lights on/off you don't really need an analogue input a simple digital input would do. You could use this to monitor the voltage to the lights (so you can them turn off if the battery gets low)

The easiest way is just to make a voltage divider out of a pair of resistors.
Put say a 2K resistor from the 12V line to the device and a 1K resistor from that point to ground - the input to the device will now be at 1/3 of the voltage of the 12v line (ie a maximum of 4V). You then simply multiply the answer by 3 in your software.
You will be wasting power through the resistors but only 12V/3K = 4mA which is tiny compared to what the lights are using.

The 10k input impedance says that this is a rather crappy analogue sensor. Ideally the analogue input should take no current at all, a decent digital meter would have 10M input impedance. This is why you have to use relatively low values for the two resistors, 1K rather than 100K.

3. Jan 13, 2009

### yungman

Don't worry about the source impedance. THis is the circuit you can build:

It is important to have D2 to protect the regulator. One thing concerning is the power disipation of the regulator. you drop 12-5=7V across the 7805, it you draw 1A, that will be 7W!!! I would either put 5 diodes in series for D1 to help dropping 3.5V to lower the power by half. Or if you want to, buy the 78SR05 switching regulator equivalent instead. I think the company making this is Power Trend.

4. Jan 13, 2009

### lowspeedchase

Thank you so much guys, you answered all my questions!

Chase

5. Jan 13, 2009

### mgb_phys

Just out of interest - is there any reason not to just use a simple relay?

6. Jan 13, 2009

### lowspeedchase

The lights themselves are powered/engaged by a simple relay. This board is being used because it can control it's outputs from a computer via USB. The goal is to have complex rule-sets followed based upon a myriad of inputs ranging from temperature to ambient light to simple on/off on the 12v lines.

Thanks again for the help!

Chase

7. Jan 13, 2009

### mgb_phys

8. Jan 13, 2009

### lowspeedchase

Indeed. That product was in fact considered; however, the whole system is being driven by a touch-screen lcd so the computer was going to be necessary already.

Additionally the computer would be controlling audio as well as receiving information over cellular internet and a gps radio signal.

Chase

9. Jan 13, 2009