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Wiring pot as voltage divider across polarities

Hi

Is it possible to wire a pot as a voltage divider so that it gives 0v DC at its mid point of resistance, and positive voltage one way, and negative the other?

I have an input to a synthesiser that takes -5vDC to +5vDC and i would like to build a simple control for it.
 
Hi

Is it possible to wire a pot as a voltage divider so that it gives 0v DC at its mid point of resistance, and positive voltage one way, and negative the other?

I have an input to a synthesiser that takes -5vDC to +5vDC and i would like to build a simple control for it.
Yep. You need a dual power supply, with +5V, 0V and -5V rails. Then you connect the pot beween +V and -V. You could use a small centre-tapped transformer with half-wave rectification and large reservoir caps on both rails, then a 7805 regulator for the +5V, and a 7905 for the -5V.
 
Hi again!

I already have both a -5v and +5v power rail to tap into. So does that mean i can just connect the pot across those two rails and it would work like that?
 
Yep again. That should be no problem. :)
Consider the logic - with the pot wiper at the 'top', it's directly connected to +5V. With the wiper at the 'bottom', it's directly connected to -5V, so 1/2 way between these, it must be 0V.
You're good-to-go. ;)
 

hevans1944

Hop - AC8NS
And make sure the pot resistance is high enough not to present an excessive load on the power supply that would destroy the pot.
 
thanks for all your advice..

it turns out that my power rails are actually +14.5 VDC and -14.5 VDC, instead of -5 and +5, which is what i want..

I am still trying to find if there are ±5V power rails on the PCB, but if i cant find them, can anyone suggest a way that i can turn the 14.5 into 5 V?

would it be as simple as using a 7805 and 7905 regulator circuit for the positive and negative 5V respectively?

should i tap both rails? or run both of them off the +14.5VDC rail?
 
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thanks for all your advice..

it turns out that my power rails are actually +14.5 VDC and -14.5 VDC, instead of -5 and +5, which is what i want..

I am still trying to find if there are ±5V power rails, but if i cant find them, can anyone suggest a way that i can turn the 14.5 into 5 V?

would it be as simple as using a 7805 and 7905 regulator circuit for the positive and negative 5V respectively?

should i tap both rails? or run both of them off the +14.5VDC rail?
Yep, 7805 & 7905.

They need to be set up on both rails, the 7805 on the positive and the 7905 on the negative, with all of the usual caps and a protection diode on each.
It's 12:42am here right now and I'm exhausted and on the way to bed, but if nobody else has helped by tomorrow morning I'll draw you up a diagram.

The 7905 datasheet:-
http://www.ti.com.cn/cn/lit/ds/symlink/lm7905.pdf

And the 7805:-
https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf
 
Here's the promised diagram. Take careful note of the 7905 pinouts. They're different to the 7805.
With these values, no protection diodes are really necessary, but if larger caps are used, it's a good idea to add reverse diodes across the input/output of both regulators. With larger caps following the regulators, you'll need those diodes too. A 1N400x is recommended.

And as GPG says, 78L05 and 79L05 regulators are available, useful for currents below 100mA, and fine for this circuit. A quick search for their datasheets will be needed if you go that way - the pinouts are different to these. Sometimes, too, it can be hard to get 79L05s, depending on where you shop.

(It always feels weird drawing this circuit. The 7905 just doesn't 'look' right upside-down.)

Dual rail supply.JPG
 
Thanks so much for that. Really helpful, once again! Ill let you know how i go.

EDIT:
one quick question. Should i use electolytic caps or tantalum?
 
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Thanks. I have to give a talk tomorrow so i probably wont get a chance to work on it til thursday. But ill let you know how i go..
 
Hi

I had some time spare tonight so i had a go on the breadboard. It works great!

The only issue is that the positive rail is giving 4.96V and the negative is giving -5.06V, thus putting off my 0v differential..

Is there a way i can make sure that i get +5v and -5v?

Ive tried reducing them both to 4.90 with a trimpot. And it is a bit better, but still a bit off..

I didn't have any aluminium 0.1uf caps so i used tantalum 0.1uf 35v.. is that alright?
 
I don't know what to say. Did you really expect a perfect 5.00V and -5.00V, from cheap standard regulators? You're asking too much, I'm afraid.

And if you got the voltages to +4.90V and -4.90V and still had a problem, the pot was the cause. You'll be lucky to do much better. You'd need precision regulators and a precision pot to improve the voltages using this method, but they still won't be perfect. I can't recommend anything specific. Temperature will affect the values too, so you'll be forever trimming to keep it right, even when you get it there.

You'll need to go to a higher-tech circuit to get better - perhaps op-amp based.

Edit: The datasheets for the 7805 and 78L05 only claim to be within 4.8V and 5.2V, so you're actually lucky to get the values you have.
 
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Yeah. No worries. I thought that might be the case. Was just checking if it was possible or not.. i put two trimpots in and it will do. I will mostly use it to give CV to the synth filter than frequency, so the slight disparity wont affect it heaps..

Thanks again for all your help
 
You do not need the two regulators. A pot and 2 fixed resistors can be used.

upload_2015-10-7_10-42-41.png
Bob
 

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Hi again.

I have got it generating the right voltage. However i measured a normal control voltage signal and it is around 5-10mA, the 7805 circuit i have built is outputting somewhere around 125mA. Is there a way i can reduce this current? Or should it be fine? It is to generate a control voltage for an analogue synthesizer and i dont want to blow it up by putting a current an order of magnitude larger than what it is used to, if that will damage it.

So, 2 questions: firstly, will 125mA damage a synthesiser if that is the strength of the control voltage current.
And secondly, if that is the case, how can i reduce the current to an acceptable level?
 
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