+/- 8V power supply - Page 1
Forum section: electronics / beginners

author: ARGLaser
author_rank: Contributor
timestamp: November 11, 2012, 11:59:11 pm
content: Hey everyone,I have a laser power meter project that I'm working on and the laser head requires a 3-8.5V power source. The problem I have is that it needs + - and ground.Most people hook them up like this, but I don't like this since it needs two batteries and the head isn't rated for 9.5V+/- (with a new 9V battery)I am also hooking mine up to an arduino so I can log the data and graph it. I plan to put it in a project box with a plug for a 9V wall wart or 9V battery, is there a way to hook this up so that I have a + and - power source? (Or use the arduino 5V/3.3V source?)I am still rather new to electronics (Just got my first OP amp PCB made a few days ago) and I don't understand much more than the basics, so please have patience with me if this is a stupid question.-ARG

author: Unknown
author_rank: Unknown
timestamp: November 12, 2012, 12:59:55 am
content: You could split the 9V from a battery or floating (not grounded) wall wart in two with something like a resistive divider (if the current is low) and ground this 'virtual ground'. Something likethis. This would give you approximately +/- 4.7 V for a full battery. This is with the 3 - 8.5 V spec.

author: ARGLaser
author_rank: Contributor
timestamp: November 12, 2012, 01:19:33 am
content: Thanks for the reply, that's exactly what I needed, I will look into it some more, thanks!

author: peter.mitchell
author_rank: Super Contributor
timestamp: November 12, 2012, 07:37:40 am
content: Be very careful though, as a virtual ground circuit cannot provide much current without getting unbalanced.

author: Strada916
author_rank: Frequent Contributor
timestamp: November 12, 2012, 11:28:26 am
content: You could use two exact (not smp ones) wall warts and connect them back as per your diagram. DO NOT use switch mode units unless you know they are isolated from mains supply.

author: Zero999
author_rank: Super Contributor
timestamp: November 12, 2012, 11:45:56 am
content: Quote from: Strada916 on November 12, 2012, 11:28:26 amOr an AC wall wart an a voltage doubler.The virtual ground would work and it can be boosted by adding a couple of transistors

author: ARGLaser
author_rank: Contributor
timestamp: November 12, 2012, 06:28:51 pm
content: Quote from: peter.mitchell on November 12, 2012, 07:37:40 amI looked at that link from alm, and the circuit seems stable with an OP amp (from the schematic below). The meter head should draw less than 1mA, will that be enough to make the circuit unstable?

author: Jeff1946
author_rank: Regular Contributor
timestamp: November 12, 2012, 08:15:10 pm
content: Why not use two 9V batteries with a couple of diodes (1N4001 or similar) in series with each of the batteries? Each diode would drop the voltage about 0.7 V.

author: ptricks
author_rank: Frequent Contributor
timestamp: November 12, 2012, 10:10:57 pm
content: I stick to a center tapped transformer whenever I want a dual supply and have an AC outlet.If on battery you just need a rail inverter part,Linear has a lot to choose from.http://www.linear.com/

author: ARGLaser
author_rank: Contributor
timestamp: November 13, 2012, 02:30:33 am
content: Quote from: Jeff1946 on November 12, 2012, 08:15:10 pmMy project box is designed for a single 9V battery, so I would like to avoid using two if possible. I think I have it worked out though.

author: T4P
author_rank: Super Contributor
timestamp: November 13, 2012, 05:40:06 am
content: It can also be the ICL7660S or the TC1044, these are better solutions, rather than a opamp divider

author: ARGLaser
author_rank: Contributor
timestamp: November 13, 2012, 07:05:37 am
content: Thanks! That's probably what I'll end up using since it will take up less room on my PCB.

author: T4P
author_rank: Super Contributor
timestamp: November 13, 2012, 08:08:05 am
content: You need roughly 2 Si diodes to drop about 1.2V though, unless you want to strap a schkotty and a Si diode

author: Zero999
author_rank: Super Contributor
timestamp: November 13, 2012, 10:36:15 am
content: I wouldn't worry about it, a single silicon diode in series with the battery should do. The voltage is likely to sag under load anyway.