How can I convert 12v from a battery to 5v for monitoring in a jeep?

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In summary, the conversation discusses the use of a device to monitor 12v lines in a jeep and respond with various actions. The issue is that the device can only monitor up to 5v for input, so the solution proposed is to attach a "+5V Fixed-Voltage Regulator 7805" between the 12v line and the input on the device. Questions are raised about using a diode to prevent current from flowing the wrong way, using a capacitor to reduce the current below 5v, and the importance of source impedance in an automotive environment. Suggestions are given to use a simple digital input or a voltage divider with resistors. The conversation also touches on using a relay or an embedded computer such as Arduino or Be
  • #1
lowspeedchase
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Hi all...

I plan on using the following device: "[PLAIN Manual.pdf"]http://www.pc-control.co.uk/images/Wasp 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
 
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  • #2
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
Don't worry about the source impedance. THis is the circuit you can build:
2zisge9.jpg


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
mgb_phys said:
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.

yungman said:
Don't worry about the source impedance. THis is the circuit you can build:
2zisge9.jpg


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.

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

Chase
 
  • #5
Just out of interest - is there any reason not to just use a simple relay?
 
  • #6
mgb_phys said:
Just out of interest - is there any reason not to just use a simple relay?

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
 
  • #8
mgb_phys said:
If you are into this sort of stuff have you seen http://en.wikipedia.org/wiki/Arduino
It's an embedded computer (so you don't need a PC)

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

1. How can I convert 12v from a battery to 5v?

The most common way to convert 12v from a battery to 5v is by using a voltage regulator. This is a small electronic component that adjusts the output voltage to a specific level, in this case, 5v.

2. What is the purpose of converting 12v to 5v?

There are many devices that require a 5v power supply, such as microcontrollers, sensors, and LEDs. Converting 12v to 5v allows these devices to be powered by a battery or other power source that provides a higher voltage.

3. Can I use a resistor to convert 12v to 5v?

No, a resistor cannot be used to convert voltage. It only limits the flow of current in a circuit. Using a resistor to reduce voltage would result in a significant amount of power being wasted as heat.

4. How efficient is the conversion of 12v to 5v?

The efficiency of the conversion depends on the type of voltage regulator used. Generally, switching regulators are more efficient than linear regulators, with efficiencies ranging from 70-90% for switching regulators compared to 30-60% for linear regulators.

5. Are there any safety concerns when converting 12v to 5v?

Yes, there are a few safety considerations when converting voltage. It is important to use a voltage regulator that can handle the power requirements of the circuit to avoid overheating. Additionally, proper grounding and insulation should be used to prevent electrical hazards.

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