Web Control Power

[Mike P]

Hi, I'm a little rusty on my electronics. I just ordered a Web Control unit and see it operates on 6-9VDC. I would like to power this from a 12VDC Altronix power supply. I forget the formula to figure out how the drop 12vdc down to let's say 8vdc. Any guidance would be great.

Thanks for any help,

Mike

 

[BraveSirRobbin]

Looks like it has a regulator in it, or I might be missing something in the documentation:

WebControl™ requires 9V DC power source not exceeding 12VDC power supply. Current consumption without load is about 170mA in active mode. Due to network activity and BRE activity, WebControl could not get into sleeping mode. With TTL load added on, it may require up to 400mA DC current from the power supply. If you ran into sensor not reading during first installation, please check the power supply voltage with WebControl connected using a meter to make sure the voltage is not too low.

 

[JimS]

You could use a resistor in series and a zener diode in parallel. That works well, especially if the load is low but the current specs are higher than I expected. A 7808 regulator is not much more complicated and a better choice if you need to drop the voltage. The docs make it seem like 12V may be ok.

Here is the first hit on Google for "zener shunt regulator design" that has the calculations

http://www.kennethku...tors_zeners.pdf

 

[Mike P]

Thanks,

After a little more research, I ordered a couple of 7808's. I think this should do the trick. I'll let you know how it goes.

Thanks again,

Mike

 

[JimS]

One thing you need to think about on the 7808 is input and output bypass caps as noted in the data sheet. It can sometimes oscillate if you don't have them..

The other is a heat sink. At almost 500 mA a 4 volt drop gives close to 2 watts of loss. With 65C degrees rise per watt a small heatsink is in order.