Dual Voltage Regulator / IC

Thread Starter

DC_Kid

Joined Feb 25, 2008
1,072
in some other threads i mentioned i was using a Intersil 7660 to get some -V to my op-amps. this device requires several components and in basic form is not a regulator (output varies with input, etc).

was wondering if there is a fairly inexpensive -+9v regulator (IC) that has Vin of around +14vDC. output current total is low, less than 1A.

it doesnt have to be -+9v, but it has to be greater than +5v and less than 0v, etc.

any suggestions?
 

DonQ

Joined May 6, 2009
321
Use the 14V, send it to a linear regulator at the voltage you need. There are lots of them around.

Also send the 14V to a polarity inverter. The 7660 can be wired to do this, but I think there is a 7662 that may do it better. I'll let you do the research.

Once you get the voltage inverted (yes it will not quite be the full 14V and will vary), then send it to a negative voltage linear regulator.

Some numbers to look for on the linear regulators are things like 7809, 7909, LM317... There are lots more...

It might work for you to do what I do to get analog supplies for op-amps where I only have +5V. Just use a MAX232 or something else from that family. They are made for serial communications, but one of the things they do is to create bipolar voltages. Generally, they are around +/- 8.5V or so. You could still regulate this if you needed to, but even with a low dropout regulator you couldn't go much above +/- 7V or so.
 

beenthere

Joined Apr 20, 2004
15,819
Use the 7662 - the 7660 is limited to 10 volts.

Life is simplified with a center tapped transformer and a bridge, so you have positive and negative voltages to begin with. That way you can have +/- 5, 6, 8, or 9 volts.
 

Audioguru

Joined Dec 20, 2007
11,248
Any opamp circuit can use a single positive supply if it is biased correctly. Then a negative supply is not needed.

Maybe if you posted your circuit we can show you how easy it is to bias the opamp so the negative supply is not needed.
 

Thread Starter

DC_Kid

Joined Feb 25, 2008
1,072
let me answer all the suggestions.

1. i dont have a schematic to share because i jumped right in with PCB Layout in DipTrace. schematic is currently hand drawn on paper.
2. amp is a INA114 (http://focus.ti.com/lit/ds/symlink/ina114.pdf). i have a precision current resistor 0.05ohm across the amp inputs and then to gnd (gnd on the minus V input). gain is set so that output is +5v when 1A flows. the amp output wont swing fully to the rails so i need more than +5v and less than 0v on -V and +V to the amp. the amp inputs are biased to gnd via a resistor for the required nA bias. i need to see +5v when 1A flows. and when no current flows (0A) i would like output to be zero.
3. i could simply use a +9v regulator to feed +9v to amp and to the 7660, then -9v from 7660 to the amp. but this means more parts, or i can do as DonQ says, flip the +14v to -14v and send that to a -v regulator, but to protect the 7660 i think a regulator 1st would be better. i was looking to simplify things, etc.
4. ICL7662 is obsolete since 1/2006. 7660 is its replacement.
 
Last edited:

Thread Starter

DC_Kid

Joined Feb 25, 2008
1,072
thanks all for the replies.

after some careful thinking about the problem (and my project) i found a way to simplify.

1. i have decided to modularize segments of my main board (there are 16 indentical "mini" circuits on the board). i'll move each of these "mini" circuits onto smaller boards and interface them to main board via card-edge sockets (i guess its now more like a backplane with interface cards). this also makes for a better business model when this project goes to market.

2. the counter IC's i am using have Vdd max of +15v, so i'll swap out my 7805 for a 7809 and use the 7660 in simplest form to get the -V needed. the board will now run on +-9V.
 
It's a sticky problem. There are push-pull converter circuits that produce +/- rails with one switcher IC and a split-secondary transformer, but the regulation on one side is always poor if the loads are asymmetric, as they invariably are.

Inverters (buck-boost topology) are the standard solution for decent -ve supply regulation from a +ve input. The downside is they produce noise on both the input and output, compared to the buck topology that has a quiet output and a noisy input. This means it's best not to tap the regulated +ve rail as the input supply for the -ve regulator unless you can put up with the extra noise on the regulated +ve rail.
 

Thread Starter

DC_Kid

Joined Feb 25, 2008
1,072
It's a sticky problem. There are push-pull converter circuits that produce +/- rails with one switcher IC and a split-secondary transformer, but the regulation on one side is always poor if the loads are asymmetric, as they invariably are.

Inverters (buck-boost topology) are the standard solution for decent -ve supply regulation from a +ve input. The downside is they produce noise on both the input and output, compared to the buck topology that has a quiet output and a noisy input. This means it's best not to tap the regulated +ve rail as the input supply for the -ve regulator unless you can put up with the extra noise on the regulated +ve rail.
funny you mention noise, i just got my digi o-scope out to look at some points on the 7660. i guess one alternative is to feed the 7660 +v from a 2nd regulator which would "isolate" the +v rail from the 7660 noise. i'll try and grab a few pics.
 

Thread Starter

DC_Kid

Joined Feb 25, 2008
1,072
here are two pics of 7660 on the scope. it was driving a LED, Vcc was 4.7vDC @ 20mA. simplest form of 7660 as a voltage inverter. one note is that my 10uF caps are ceramic surface mounts that i soldered some wire to so they could work on my breadboard.

1st one is probe on pin 8 (Vcc). some ripple there. as i moved the probe down Vcc away from pin 8 the ripple slowly went away. at approx 2" away from pin 8 the ripple was gone (on this scale, etc).

the 2nd is on the output pin 5. bigger ripple there. odd that the output was only -2.3v ? i was expecting ~ -4. any idea why output was so low?
 

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