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Topic: 12v to 5v - Voltage divider issues (Read 10212 times)previous topic - next topic

Divinitous

Mar 28, 2013, 03:18 pmLast Edit: Mar 28, 2013, 05:17 pm by Divinitous Reason: 1
Gentlemen,
I'm working on a small project where I'm creating a box to help make things more efficient when servicing machines at work.  First a short bit about the machine I'm working with.  Basically the machine I service is used to find the viscosity of a liquid.  Inside a glass tube the liquid fills up a capillary and two sets of optics detect the amount of time it takes for the minuscus to travel from the upper optic to the lower optic.  Every few months or so the optic stands have to be taken out and the voltage adjusted for both optics to compensate for any wear the optics may have suffered.

The setup that I'm creating will do much more than just this, but this is the part that I'm having an issue with.  The stands will never have a voltage above 11.5v DC so as long as I can get the voltage divider operating properly I am not concerned about an over voltage situation.  My issues lies in the voltage input that I'm getting before the voltage divider.  When I connect the voltage divider circuit the Vin from the source the voltage will drop.  What would normally be a 9.6v drops down to 8.2v.  Connecting the arduino won't affect this, so i know it's not anything with a faulty arduino or bad programming.  I could compensate with some mathematics but the system needs to be operational while I have my setup connected.  Basically I need something to detect the voltage but not create any interference on the source circuit.  Below is my configuration for the voltage divider.  Overall, it works as in the 8.2v is read as 3.3v, but that voltage drop on Vin is gonna be an issue.

Vin -- R1 (220k) -- Vout -- R2 (150k) -- Gnd

fungus

#1
Mar 28, 2013, 04:02 pm
Normally a voltage drop would mean too much current but 220k and 150k is plenty of resistance.

Double-check the resistors with a multimeter to make sure they really are that value (I know you're sure they are, but that's the symptoms...)
No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

Divinitous

#2
Mar 28, 2013, 05:16 pm

Normally a voltage drop would mean too much current but 220k and 150k is plenty of resistance.

Double-check the resistors with a multimeter to make sure they really are that value (I know you're sure they are, but that's the symptoms...)

Yeah, that's something I checked previously.  The values are well within the 5% tolerance.

Any possible alternative to a voltage divider with resistance that i might be able to get away with?
Originally I setup a simple LED voltage meter in a project box and it did a great job. The connector for these stands has 24v for me to work with.

1oldfox

#3
Mar 28, 2013, 05:22 pm
Good morning Divinitus. First a question. Are there other items, circuits, instruments, etc. in your project that depend on the source (Vin) remaining stable and constant. From what you have described so far, voltage isn't the issue. You are wanting to measure *time*, the measure of which can be mathematically massaged to give you the viscosity measure you are looking for. Is my understanding correct? If so, you may want to investigate using OpAmps for the so called sensors rather than a voltage divider. This would transition you out of the *analog* world and take you into the *digital* world. Much easier to measure time.
-oldfox-
If it ain't broke, fix it 'til it is!

cmiyc

#4
Mar 28, 2013, 05:37 pm
What is the source of "Vin"?

Capacitor Expert By Day, Enginerd by night.  ||  Personal Blog: www.baldengineer.com  || Electronics Tutorials for Beginners:  www.addohms.com

Grumpy_Mike

#5
Mar 28, 2013, 05:51 pm
It sounds like the impedance of the Vin is very very high if it is being dragged down by a load of 370K.
You have connected the grounds together haven't you?

Divinitous

#6
Mar 28, 2013, 05:56 pm
Quote from: 1oldfox

Good morning Divinitus. First a question. Are there other items, circuits, instruments, etc. in your project that depend on the source (Vin) remaining stable and constant.

Yes, this optics setup is only a small part of the complete system.  Bascially I need to check the voltage without having to disconnect the setup.  There are times when I may have to run the machine with my arduino attached to the circuit.  So the voltage drop does create an issue.

Quote from: 1oldfox

From what you have described so far, voltage isn't the issue. You are wanting to measure *time*, the measure of which can be mathematically massaged to give you the viscosity measure you are looking for. Is my understanding correct?

No, the optics setup connects to a larger part that does all the timing and calculations.  The machine itself does the viscosity measurements.  I am only looking to create something to detect the voltages.

Quote from: 1oldfox

If so, you may want to investigate using OpAmps for the so called sensors rather than a voltage divider. This would transition you out of the *analog* world and take you into the *digital* world. Much easier to measure time.

This isn't an option for me.  Basically the overall machine detects the voltage drop when the meniscus passes through the optics.  Typical voltage is 9.5v and a drop from this is the machines trigger.  You can see now why the arduino creating a drop would become an issue as the machine would constantly see the voltage drop and cause the machine to error out.

Oldfox,
Maybe I didn't outline everything properly.  The optics stand is connected to a whole machine.  The circuit I'm designing is only to read the voltage so i can service the optics on the stand.

Divinitous

#7
Mar 28, 2013, 05:57 pm

It sounds like the impedance of the Vin is very very high if it is being dragged down by a load of 370K.
You have connected the grounds together haven't you?

Yes, the ground on the connector is connected to the ground on the arduino.  Should this be different?

Divinitous

#8
Mar 28, 2013, 06:04 pm

What is the source of "Vin"?

Basically the connects to the machine with a 4 pin connector.  All voltages are DC.  24v for the stand to operate on, 1 pin for the upper optic voltage, 1 pin for the lower optic voltage (both a max of 12v) and a ground.  The stand itself operates on 24v from a power supply that is also powering other components.  The supply isn't the issue as it was well oversized, it's the circuitry inside the stand and the voltage divider.

DVDdoug

#9
Mar 28, 2013, 06:10 pm

It sounds like the impedance of the Vin is very very high if it is being dragged down by a load of 370K.
Right...   An effective source impedance of around 50K along with your 370K load would another voltage divider, with about 1.4V dropped across the 50K source impedance.

A regular voltmeter/multimeter has very high resistance (maybe 10M) so it doesn't tend to "drag down" the voltage you are measuring.

You can try increasing the resistance values in your voltage divider, or the best solution would be an op-amp buffer.

retrolefty

#10
Mar 28, 2013, 06:10 pm

What is the source of "Vin"?

Basically the connects to the machine with a 4 pin connector.  All voltages are DC.  24v for the stand to operate on, 1 pin for the upper optic voltage, 1 pin for the lower optic voltage (both a max of 12v) and a ground.  The stand itself operates on 24v from a power supply that is also powering other components.  The supply isn't the issue as it was well oversized, it's the circuitry inside the stand and the voltage divider.

If those voltage divider resistors are really the ohms you have stated then the two optic voltage outputs have just too high a source impedance to read like that. You would have to buffer them first with a high input impedance op-amp or some other means to not 'load down' the signal voltage from those two signals.

Lefty

Grumpy_Mike

#11
Mar 28, 2013, 06:11 pm
Quote
The supply isn't the issue as it was well oversized,

Then you are not explaining what your set up really is. Putting a high resistance across a supply like this will not cause it to drop. If you are seeing a voltage drop then something is not as you described it.

Your response also indicates you are not understanding what 1oldfox said.

1oldfox

#12
Mar 28, 2013, 06:20 pm
OK. Check me out on this. You want to measure the voltage at the optic sensors to give you a value of light transmittance for maintenance purposes. As the transmittance drops, that would indicate a *dirty column* for example, and would be time for you to perform your PM's. Your concern is that the Arduino will cause the voltage to drop, giving a false level from the optic sensors. Here, I would again say, look into OpAmps. The perfect solution. You are already hooking up a voltage divider. You just replace it with the OpAmp and associated circuitry. Which, by the way, can be tuned for voltage levels with a potentiometer.
If it ain't broke, fix it 'til it is!

fungus

#13
Mar 28, 2013, 06:56 pm

Any possible alternative to a voltage divider....

In theory: 10 times more resistance in your divider should mean 10 times less voltage drop (more or less).

You could try putting bigger resistors in there, eg. in the megaohm range. The Arduino ADC input has 100 megaohm resistance so there's still some margin to work with.

No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

retrolefty

#14
Mar 28, 2013, 06:58 pm

Any possible alternative to a voltage divider....

In theory: 10 times more resistance in your divider should mean 10 times less voltage drop (more or less).

You could try putting bigger resistors in there, eg. in the megaohm range. The Arduino ADC input has 100 megaohm resistance so there's still some margin to work with.

Yes, but the analogRead() function works best and is recommended being driven by a voltage source of 10K ohms or less, so values read for such a high impedance source are sure to be problematic.

Lefty

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