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Scaling a Voltmeter (& an Ammeter)

A representation for our real 250 mV Voltmeter

In the previous post on Electrical Theory, “The Load affects the Process. (Electrical Theory and More)“, I introduced the idea that the load can affect the circuit. To show that we used a real life meter and datasheet and circuit designed to make things look really bad.   Tonight’s post is also “real theory” and we will be talking more about the same multi-meter but with a whole lot more functions.  If you are not familiar with the basic concepts taught in previous posts including Ohms law, please you the Category menu and choose Electricity to see a list of the previous posts to be able to understand this post.

There is a term in electrical theory called “duality”.  As you will find as we switch from measuring Voltage to measuring current, everything is the exact opposite.  So hold on to your hat, the ride is going to be fun.  It will be helpful for you to again look at the Simpson 260-8 datasheet pdf.  Also this is a link to Simpson’s website with many pictures of the meter.

When we left the last post, we said that on the 250mV scale the meter would draw 50μA of current to actually measure 250mV.  One way to understand the meter is to combine an ideal voltmeter with a resistor.  The resistor represents the input resistance (impedance) of the meter so now we can easily understand its effect on the circuit.

The meter scaled to read 1 V.

All of that is fine and dandy if we only had to measure 250mV, but this meter has many ranges of Voltages it can display.  The next higher scale is 1V.  To have the meter scaled to read one Volt, we simply add a resistor in series with our “real Voltmeter” and create a Voltage divider.  0.75 V will be dropped across the scaling resistor and 0.25 V will be dropped across the meter movement.  50μA will continue to flow in the meter circuit.

The meter has other ranges:  Specifically it has 2.5V, 10V, 50V, 250V, 500V, and 1000V ranges.  How do we obtain those?  We just keep adding series resistors.  I will leave the calculations up to you and will provide the answers in about a week.  (You do really want in the fun don’t you?)  Seriously, the best way to understand this is to push yourself to do it.   Voltage dividers are used in almost every electronic circuit we will work with so this is a real-life problem.

An Ammeter inserted into a circuit.

Voltmeters are used by measuring across, in parallel to, a circuit element. Ammeters must be inserted in series.  An ideal Voltmeter would draw no current so it has infinite resistance,  An ideal Ammeter would create no Voltage drop so it has zero resistance.  (Starting to feel like an old Twilight Zone episode in a parallel universe? 🙂 ).  Our actual meter will have a 250mV drop if measuring a 50uA current on the 50uA range because again it has a 5K internal resistance (see the picture below).  In the circuit I designed, the meter would display 25μA because the 250mV source would “see” a load of 10K (5K meter and 5K real load.)  Using Ohm’s law, 0.25V/10,000 Ω = 0.000025 A or 25μA.

A representation for our Real Ammeter.

Because circuits must be opened up to insert the Ammeter, in real life Amperage measurements are not taken as often as Voltage measurements.   The above circuit was designed purposely to act badly to prove the point.  If I had used a 50V source and a 1 MΩ resistor the circuit would draw 50μA and inserting a 5KΩ meter in series would have very little effect.


Now, how do we obtain the other current ranges for our meter?  The meter is advertised to have ranges of 1mA, 100mA, 500mA, &10A.  You might guess that since everything else has been exactly the opposite of the Voltmeter, we would want to put the scaling resistor in parallel to the meter.  If you guessed that you guessed right.

Our meter scaled for 1 mA current.

The final circuit is designed to shunt 950 μA around the meter movement.  950 + 50 μA = 1000μA or 1mA.  The shunt resistor value can be calculated a couple of different ways.  the way I chose to calculate it was 0.25V/.000950A.  However you could calculate it by using the parallel resistance formula with some algebra manipulations.


Now we have a few more ranges to calculate.  100mA, 500mA and 10A.   We just need to keep adding resistors in parallel.   Please note the Simpson datasheet says the voltage drop will be different values for different ranges.  We will assume that we will maintain a voltage drop of 250mV.  My guess is they may be adding some series resistors as well as parallel, because the values of actual resistors will be very hard to obtain.  Again I will provide my results of these calculations in about a week.

One final point.  NEVER NEVER NEVER connect an Ammeter across a Voltage source.  The Ammeter is designed to have a low input impedance and will look like a short circuit to the Voltage Source.  The meter may be damaged if you connect it across the source.   You may even damage the source.

We now know enough to start tackling some simple Op-Amp circuits and get even more real.   The next electrical post will be about some equipment you will need to buy to start building and testing.  I will try to point you toward good and the least expensive as possible sources.


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