Author Topic: Circuit to provide four adjustable DC outputs  (Read 1640 times)

0 Members and 1 Guest are viewing this topic.

Offline GJS123Topic starter

  • Newbie
  • Posts: 9
Circuit to provide four adjustable DC outputs
« on: June 08, 2015, 03:48:15 pm »
I would like to implement a hardware emulator for a piece of industrial equipment.  The equipment has 4 low voltage, 0-10V DC, isolated monitoring points which represent the machine state.   I will be connecting these to A to D converters to sample them.

So I want the simplest circuit that can provide four adjustable 0-10V DC outputs to use as an emulator.  The current drawn will be virtually zero.  I thought that voltage dividers connected to a single supply would do the trick but then I thought that each one, when adjusted, would change the other output voltages.  Can someone point me in the right direction?

Thanks.
 

Offline flynwill

  • Regular Contributor
  • *
  • Posts: 143
Re: Circuit to provide four adjustable DC outputs
« Reply #1 on: June 08, 2015, 05:31:17 pm »
Yes a 10V power supply and 4 pots will do what you want.  One will effect the other only if the power supply sags, which seems unlikely if your are just running the connections to an A/D converter input.  I would suggest a small unregulated 16V wall wart as the source power, and a programmable regulator to generate your 10V reference.  Pots can probably be between 1 and 10k, you'll want linear taper.

If this is to be a test jig for some control project you might want to consider building something to emulate the machine's operation  -- eg a second microcontroller with a 4-channel D/A converter generating your test signals. Otherwise I can see turning the pots to generate your test stimulus will get old pretty quickly.

Also as a note: Most industrial control sensors use some variation of a current loop (typically 4-20ma) and not voltage for sensor inputs for better noise rejection.  If you google that, there are lots of components to both generate and to properly receive such signals.
 

Offline GJS123Topic starter

  • Newbie
  • Posts: 9
Re: Circuit to provide four adjustable DC outputs
« Reply #2 on: June 09, 2015, 05:45:14 pm »
Thanks for your reply, flynwill.  So do all the tracks of the pots go in series or parallel?  Is it one end of all the tracks to 0V and the other end to +10V?

The ten volts is not very critical so no need for the regulator, I think.  I start by using my bench supply, anyway, then switch to a wall wart when I've checked it works properly.

That's a good point about using a microcontroller to emulate the equipment.  Actually, the equipment is a very large power supply which spends most of the time with the output in a steady state. The monitoring points represent voltage and current output of two seperate DC-DC converters.
 

Offline PSR B1257

  • Frequent Contributor
  • **
  • Posts: 330
  • Country: de
Re: Circuit to provide four adjustable DC outputs
« Reply #3 on: June 09, 2015, 06:21:18 pm »
Quote
So do all the tracks of the pots go in series or parallel?
Parallel (of course) since you want independent output voltages.

Quote
Is it one end of all the tracks to 0V and the other end to +10V?
Sure. And don't forget the other other end  ;)
In theory, there is no difference between theory and practice. But, in practice, there is.
 

Offline flynwill

  • Regular Contributor
  • *
  • Posts: 143
Re: Circuit to provide four adjustable DC outputs
« Reply #4 on: June 10, 2015, 04:20:41 pm »
Yes parallel, to be clear see attached...

If they were in series they won't interact if, as you say, the input impedance on your micro's analog inputs is high, but each would only give it's 1/4 of the 0-10V range (ie 0-2.5V for the first, 2.5-5V for the second, etc).

The pots can probably be 1k,  if you go much lower you need to check the power dissipation and the ratings on the pot (Vsupply^2/R).

If you don't mind a bit of non-linearity (and sounds like you wouldn't), the pots can probably as high in value as the actual input impedance of your analog inputs.  eg if you your analog inputs have a 10k input impedance and you use 10k pots then the voltage with the pot centered will be 4V instead of 5V.
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf