HELP need -8.5V from -9V

Hi, i fixing a piece of gear that had a dc-dc convertor that put out -8.5v
it was used for the VEE on a LCD display

i cant find one that does -8.5v, so i got one that does -9v

9v is too much, the contrast is off, i need between -8.2v and -8.7v

im sure i cant use a regulator since im at 9v and need only to drop 1/2v

how about a zener diode?

how would i connect it?

i got GND and -9v

i dont know the current draw the the LCD on the VEE line
its not much, my meter shows 3ma
thats not much, is that right?

anyway.. help please

mitch

is duh, a resistor voltage divider?

50ohms and 1kohms

??

mitch

the potential divider will not work I'm affraid, it will not be able to source enough current.

I would suggest you use a low drop out variable regulator, you can get them with drop outs of only a hundred millivolts or so. LM337 might be a good start, don't know if it is a low drop out reg though.

The zener idea would work make sure you build in good enough protection though, you don't want to blow it up!

As you said your load (LCD) consumes 3mA and you wana reduce voltage by 0.5V
Simple way I think is to put a resistor in series of load and source
to select the series resistor
VR = 0.5V
IR = 3mA
R = VR/IR = 0.5/3mA
R = 166.67 ohm you can use a nearby value resistor like 150 ohm or 175 ohm
Let me know if there is any problem with above method

sod a reggy/zener far to lossy,
just put a series DIODE and drop the 0.6V through that, alot less power loss and pretty much any-old diode can be used (as long as it can handle the current)

so 9V - 0.6V = 8.4V

duh a diode in series

sometimes a simple solution blinds...

mitch

how about LM337 so you can set it to anything you want?

panic mode said:

how about LM337 so you can set it to anything you want?

Click to expand...

LM337 has a minimal dropout of 2.5V, so that won't work.