Making an ohmmeter from a voltmeter?

[yhelothar]

I have the LM3914 LED Bar driver to use as a voltmeter. It works by having a series of comparators that each activate at incrementing voltages.

I also have a current regulator with a rotary dial switch that can switch between four different current settings.

So here I have a known voltage and a known current. I could calculate the resistance easily with ohm's law.

Is there a way to divide the voltage with the known current in a circuit some way so the LED bar graph would act as an ohmmeter instead of a voltmeter?

 

[silicon]

I have the LM3914 LED Bar driver to use as a voltmeter. It works by having a series of comparators that each activate and incrementing voltages.

I also have a current regulator with a rotary dial switch that can switch between four different current settings.

So here I have a known voltage and a known current. I could calculate the resistance easily with ohm's law.

Is there a way to divide the voltage with the known current in a circuit some way so the LED bar graph would act as an ohmmeter instead of a voltmeter?

digital ohmmeters are very cheap, why not spend the $10-$15 and get one of these?

 

[yhelothar]

digital ohmmeters are very cheap, why not spend the $10-$15 and get one of these?

I'm making a device that would specifically benefit from its own ohmmeter that's integrated into its circuit into a single portable device.

 

[serpretetsky]

I have the LM3914 LED Bar driver to use as a voltmeter. It works by having a series of comparators that each activate at incrementing voltages.

I also have a current regulator with a rotary dial switch that can switch between four different current settings.

So here I have a known voltage and a known current. I could calculate the resistance easily with ohm's law.

Is there a way to divide the voltage with the known current in a circuit some way so the LED bar graph would act as an ohmmeter instead of a voltmeter?

So you have a known current, and a variable voltage (that you can measure).

if you can set your current to a known value, all you are looking to do is scale the voltage shown by some fixed number.

Why try to do some electronic math when you can just rescale your units on the actual LED display. So scale your LED light bar to have voltage shown on one side, and resistance shown on the other side.

For example: pretend your current regular is set to 0.25 amps and here is your LED bar scale
5V-
4V-
3V-
2V-
1V-
0V-

V=IR
V= 0.25R
R = 4*V

(assuming 0.25 amps)
5v-20ohms
4v-16ohms
3v-12ohms
2v-8 ohms
1v-4 ohms
0v-0 ohms

I dont know with what current you're planning on setting and I dont know what resistances your planning to test. Small sensitive electronics may get burned out by 0.25 amps, for example. On the other hand, setting very small currents will require you to measure the very small voltages, and i'm not sure if the chip you are using is precise enough. Good luck!

 

[yhelothar]

So you have a known current, and a variable voltage (that you can measure).

if you can set your current to a known value, all you are looking to do is scale the voltage shown by some fixed number.

Why try to do some electronic math when you can just rescale your units on the actual LED display. So scale your LED light bar to have voltage shown on one side, and resistance shown on the other side.

For example: pretend your current regular is set to 0.25 amps and here is your LED bar scale
5V-
4V-
3V-
2V-
1V-
0V-

V=IR
V= 0.25R
R = 4*V

(assuming 0.25 amps)
5v-20ohms
4v-16ohms
3v-12ohms
2v-8 ohms
1v-4 ohms
0v-0 ohms

I dont know with what current you're planning on setting and I dont know what resistances your planning to test. Small sensitive electronics may get burned out by 0.25 amps, for example. On the other hand, setting very small currents will require you to measure the very small voltages, and i'm not sure if the chip you are using is precise enough. Good luck!

That would work great if I were only using one current setting. But I have 4 current settings from 0.5mA to 2mA. I need the LED Bar to consistently display the resistance. So say if 2KOhm was activated 2 LEDs at 1mA, it still needs to activate 2 LEDs at 2mA. As a voltmeter, there would be twice the voltage at 2mA over 1mA, so there would be 4 LEDs activated at 2mA.

 

[yhelothar]

I just thought of a solution. Instead of using a 1 pole switch to control the current, I can use a 2 pole switch and use the 2nd circuit to add on opamps for voltage gain on the smaller currents. 2mA would have no gain, 1.5mA would have 1.33x gain, 1mA 2x gain, etc.

 

[Paperdoc]

You're going at this the wrong way. Common commercial ohmeters use a fixed voltage supply and measure the current flowing in the circuit, displayed on a scale to read directly in ohms. Why? Two major reasons:

1. Most devices have a limit on what voltage they can tolerate.So ohmeters are designed to use small voltages and measure small currents. If you start out with a current supply that will raise the voltage to whatever is necessary to make that current flow through the device, it may destroy the device.

2. The current source itself has an upper limit on the voltage it can supply to force that current through. With a high-impedance device it's likely the voltage will not be high enough. You will not get the current flow you thought you were going to get, and the voltage you are reading and converting to ohms in fact is just the limiting voltage of the current source. You get the wrong answer.

 

[yhelothar]

You're going at this the wrong way. Common commercial ohmeters use a fixed voltage supply and measure the current flowing in the circuit, displayed on a scale to read directly in ohms. Why? Two major reasons:

1. Most devices have a limit on what voltage they can tolerate.So ohmeters are designed to use small voltages and measure small currents. If you start out with a current supply that will raise the voltage to whatever is necessary to make that current flow through the device, it may destroy the device.

2. The current source itself has an upper limit on the voltage it can supply to force that current through. With a high-impedance device it's likely the voltage will not be high enough. You will not get the current flow you thought you were going to get, and the voltage you are reading and converting to ohms in fact is just the limiting voltage of the current source. You get the wrong answer.

The voltage will be limited by a 9V battery. If it goes over, then the meter just maxes out. I only need the ohmmeter to work in a very limited range (1-6KOhm)

 

[soccerballtux]

most ohmetters drive a super duper low current with high resolution voltage measurement
I think

or maybe the other way around, a voltage with super high res current detection

you can use a fet don't need a switch. Just perform calculation scale with the drain to source voltage accounted for.

 

[phasseshifter]

digital ohmmeters are very cheap, why not spend the $10-$15 and get one of these?

i am with this guy get a cheap digital meter 10 dollars pull it to pieces hot wire its switch to read yor resistance setting and fit it to your unit that yo want it for..the lm3914 is a an led voltage divider in its self but you may want to google for a schem but i cannot find one i have asked further and awaiting a return...from another sorce...also check this page out http://www.penseeprofonde.org/images/e/e9/LM3914.pdf

 

[phasseshifter]

We published an insulation tester in May 1996 that measured resistance in Giga ohms and displayed using an LM3915. The LM3915 is essentially the same as the LM3914 except for its logarithmic scale instead of linear. The basic circuit arrangement could be used for ohms measurement with a lower test voltage. For back copies www.siliconchip.com.au

 

[Jeff7]

I have the LM3914 LED Bar driver to use as a voltmeter. It works by having a series of comparators that each activate at incrementing voltages.

I also have a current regulator with a rotary dial switch that can switch between four different current settings.

So here I have a known voltage and a known current. I could calculate the resistance easily with ohm's law.

Is there a way to divide the voltage with the known current in a circuit some way so the LED bar graph would act as an ohmmeter instead of a voltmeter?

1) Generate a voltage on the outputs.
2) Push the current through a shunt resistor.
3) Op-amp across the shunt resistor to multiply its voltage drop.
4) Display voltage on the bar graph.



Or something like that. I'm pretty sure an op-amp would be useful here though.




* - Not an expert in op-amps by any means. But I've seen some really nifty circuits that use them, and they sound like terribly useful little devices.