# Digital Voltmeters

1. Feb 16, 2010

### kusiobache

How do digital volt meters work?

If I connect a Digital voltmeter to a battery, how does it measure the voltage. And i'm assuming it measures terminal voltage, not emf?

This question's probably simple but it's been bugging me for awhile. All I can think of is that the DVM produces (or measures maybe?) a current, and then because of the internal resistor it measures the drop in voltage? (which would explain why the DVM has to be in parallel). Then the drop in voltage somehow = the voltage of the battery? (that's that part that would make the least sense to me, if the rest of my theory is right).

But I'm probably way off, so can someone explain it to me?

2. Feb 16, 2010

### Staff: Mentor

For measuring voltage, DVMs use high-input-impedance voltage followers/amplifiers. These will basically have an input structure similar to CMOS Opamps. Are you familiar with those? If not, wikipedia.org is a reasonable starting point.

3. Feb 16, 2010

### kusiobache

I'll Wikipedia what you said and see where it gets me. Otherwise, just looking up digital voltmeters on Wikipedia (which I did a few times before) is pointless.

EDIT: So the amplifiers have a high input resistance and low output resistance, and since they are similar to CMOS Opamps they can amplify voltage. That's what I gathered/ already knew

Well then, does the DVM use the resistance and ohms law to calculate the voltage or something? Or maybe the followers/amplifiers act as comparators somehow? Or is it something else.

I kindof already know what the parts in a DVM DO, I just don't quite know HOW they work together to determine a voltage. (unless it's like a comparator or something...).

Last edited: Feb 16, 2010
4. Feb 16, 2010

### vk6kro

A digital voltmeter reads the voltage at its terminals. Whether this is the output of a power source or just the voltage drop across a resistor, doesn't matter.

Internally, they are like this:

http://dl.dropbox.com/u/4222062/digital%20voltmeter.PNG [Broken]

Those skinny rectangles at the left are precision resistors. The input is at the left of this drawing.

Without the switch, the amplifier will produce enough output to give a reading of, say, 1999 on the display, with 199.9 mV input.

So, if the switch is set to the top of the top resistor, 199.9 mV input will produce a display of 199.9 on the display. The decimal point is added to make the display easy to read.
If the input is 60.5 mV, the display will be 60.5.

Now, the resistor divider is arranged so that on the next step down, 1.999 volts input to the terminals will produce 199.9 mV at the switch position. The amplifier works as before, but the decimal point on the display is moved to give a proper display. So, the display now shows 1.265 for 1.265 volts input.

Moving the switch further down, reduces the input voltage accurately to 1/100 th or 1/1000 th of the actual input value. Each time the amplifier gets a voltage input between zero and 199.9 mV, but the decimal point is moved to reflect the switch position.

The resistance of the resistors in the resistor string adds up to 1 megohm or 10 megohms depending on the meter. In most circuits, the small current drawn by this very high resistance has little effect on the circuit operation.

Last edited by a moderator: May 4, 2017
5. Feb 16, 2010

### kusiobache

Thank You! I was looking for a way to +rep you or something, but I don't see one. And also, thank you for the diagram!

Last edited by a moderator: May 4, 2017
6. Feb 16, 2010

### Mike_In_Plano

Extra - Most voltmeters have either a resistance-per-volt rating or an input resistance rating.

The input resistance of old fashioned types was determined by the scale you were on and the resistance-per-volt rating of the meter.

i.e. if you had a 10K-per-volt meter and you were on the 5 volt scale, the input resistance was 50K-ohm.

Modern meters usually have a fixed resistance that does not change with scale - typically 10 Megohm.

- Mike