Help with dropping resistor DC supply question.

[Dialtone]

I have located some universal digital display panel meters that I would like to use for a DC power plant voltage/amperage monitor system.
The meters have an internal regulator for the power supply and state they will need an input of DC 5 to 18 volts. (No current specified, but I assume it is not too great)

The DC system I want to use them on has a nominal operating voltage range of from 45 to 60 Volts DC.

To keep things simple, I would like to employ dropping resistors off the DC plant to supply the operating voltage for the meters. Can someone help me with the resistor values calculations to do this?
Is there a better simple way?

Thanks in advance.

 

[Russlk]

A 10 volt zener diode and 3.6K resistor to the voltage supply should work.

 

[Leftyretro]

I have located some universal digital display panel meters that I would like to use for a DC power plant voltage/amperage monitor system.
The meters have an internal regulator for the power supply and state they will need an input of DC 5 to 18 volts. (No current specified, but I assume it is not too great)

The DC system I want to use them on has a nominal operating voltage range of from 45 to 60 Volts DC.

To keep things simple, I would like to employ dropping resistors off the DC plant to supply the operating voltage for the meters. Can someone help me with the resistor values calculations to do this?
Is there a better simple way?

Thanks in advance.

Well using a small three terminal voltage regulator would be the better method however your 45-60 volt source is too high for many such regulators and would have to be cut down with a voltage divider anyway.

However to calculate dropping resistor value you have to know the current draw for the display in question. Take either a fixed 5volt DC supply and measure the current that the display draws from it. or use 4 series connected AA battery cells and again measure the current draw from the display. We can then go about the proper calculations.

EDIT: Yes the zener diode method would work well and be very simple.
Lefty

 

[Dialtone]

Thanks for the quick responses.
I had forgotten about using a zener also.