Old Time Radio Engineers

[Nowhere-Man]

I think you misunderstood what I said about the 5 volts. Having said that, it's probably the way I explained myself. According to the paragraph, the meter used as a milliameter had a full scale deflection of 0.005 Amp at 5 volts. The resistance of the meter without any added is 100 Ohms. Therefore, if the meter reads the full five milliamps you know you have 5 volts across the meter. However, the voltage in the circuit may need to to be more than 5 volts. You may wish to test a terminal that's designated as 50 volts. In order to do that, a series resistor needs to be used but it has to be calculated. Divide 5 into 50 to get 10 and you can multiply the 100 ohms you already have. Therefore, 1000 Ohms but you need to allow for the 100 Ohms we already have so 900 Ohms in series ought to allow for a 50 volt reading. The meter will still read 0.005 Amp but that then means at full scale deflection, the 900 Ohm resistor is taking the remaining current.

This is the same system that is used with ammeter shunts on boats. The ammeter itself doesn't take more than, say, 100 milliamps but the current in an engine circuit will be at least 100 amps max. Therefore, a large, high watt, low resistance shunt is used in parallel with whatever the meter resistance may be. This is like a long bar with terminals that is placed in the alternator + battery negative line.

I assume this method is explained because back then voltmeters were expensive.

It is easy to confuse yourself when reading text books. Especially as the overwhelming majority of textbooks are written in a way that makes perfect sense if you already know the subject, but if you've not had a lesson or a lecture before attempting to read the textbook, then they can actually do more harm than good. Think of textbooks as references rather than teaching aids. That said, there are good books out there that fall into the latter category, but finding them is the problem...

Another trend I've noticed is that many early books are very theory-heavy, and might present page after page of equations and similar, but hardly any explanation of what is going on in words. In some cases, it's obvious to me that the authors didn't really have a grasp on the subject matter and were hiding behind the maths. Some books feel like they are deliberately trying to exclude people from their "gang" if they can't do the maths. But there are many different adult learning styles out there, and it's wrong to exclude many of them - some of the best engineers I know failed O-level maths.

I say the above as someone who teaches for a living. I realise that not all will agree.

Anyway, to the problem.

The first thing to understand is that a moving coil meter requires current to move the pointer. That's true, whether you're using it to measure current or voltage.

The next point is that the coil of wire will have some resistance.

As a result of these two, then yes, a certain voltage will cause full scale deflection, but that's just Ohm's law in action. But do not think of the meter as a voltage-driven thing, because it is not. Things will make much more sense if you can make that mental leap.

Let's work with the example you give - 0.005 amps FSD. Let's start by using sensible units - let's call it 5mA, please.

And let's re-state that it has a resistance of 100Ω.

Before going further, we ought to say that 5mA through a 100Ω resistor gives a voltage drop of 0.5V, not 5V. Crackle was right about that - it's possible that your book is wrong.

To measure 0 to 5mA, the meter alone is all we need. But, to measure 0 to 50mA, we need to divide the total current using a shunt. You can use a water analogy here - think about an island in a wide river, and think about how the water will flow either side of this island, and re-combine afterwards. If the island is in the middle of the river, then the water will divide itself equally either side of the island. However, if the island is much nearer to one of the banks of the river, then most of the water will flow past the island on the other side, leaving a relatively small amount passing between the island and the nearest bank.

So in the case of your meter, we need to place a resistor in parallel with the meter movement. The incoming current will divide itself between the resistor and the movement. The exact ratio of this depends on the value of the added resistor. If we used 100Ω then the current will divide equally, meaning that if we passed 10mA through the contraption, 5mA would flow in the resistor, and 5mA would flow in the meter - which gives an FSD of 10mA.

To get 50mA FSD, we need to choose a resistor that passes 45mA, leaving 5mA for the meter. That might very well be 11.1Ω - as an exercise, prove that to yourself.

To get 500mA FSD, we need a resistor that passes 495mA, again leaving 5mA for the meter. That's 1Ω

To measure voltage, a series resistor is needed. This is often called a multiplier. The resistor reduces the current that might otherwise flow in a circuit - a current that might damage the meter if not reduced.

Let's say we want 5V FSD. In this example, we know that we want 5mA to flow in the meter coil. How do we get 5mA flowing in a circuit that is powered from 5V? We need a total resistance of 1kΩ (Ohm's law: 5V over 5mA equals 1000).

We know that the meter has a resistance of 100Ω, so we need to add a resistor in series that gives a total of 1000Ω. I reckon that's 900Ω.

How about 50V? Again, we know that we have to get 5mA flowing through the coil of the meter. How to draw 5mA from a 50V supply? How about 10kΩ? So, 10,000Ω minus 100Ω is 9.9kΩ.

Or 500V? Well, that'll be a total of 100kΩ, which requires a series resistor of 99.9kΩ.

Having said all that, I'll say the following:

1. Others have already said what I've said. I've just said it again in a slightly different form perhaps.
   
2. Your messages indicate that actually, you understand most of it, and the apparent confusion is just springing from your exact choice of terminology. Sometimes, that's harder to grasp than the underlying technology.
   
3. I'm in a rush, so haven't proof-read the above properly. Minor mistakes expected. I will edit later if required.
   

Mark

Cheers for that, Mark. I think I need to re-emphasise that what sometimes happens is I may refer to a textbook without having the book on me. Therefore, the way I'd explained it was definitely confusing. To avoid any frustration, I'd ask you all to bear in mind I may possibly be on limited time in the library and it's quite possible I can slip up over a detail.

O.K. so the book did use 5 volts as an example but the resistance was an actual circuit resistance of 1000 Ohms total. The current max deflection was 5 Milliamps. So here it is: 0.005 * 1000 R = 5 volts.

The confusion appeared because I stated 100 Ohms resistance I think. Well, this is apparently because the author then goes on to state the resistance of the meter itself is 100 Ohms (but it's in circuit with 900 Ohms).

Now, Crackle correctly pointed out the voltage didn't correspond and it would really have helped if the book had simply started off with the following: 0.5 volt, 100 R = 0.5 That would have prevented me getting mixed up.

So, I suggest it can be done this way:

(1) Voltage 0.5, Resistance 100 Ohms, Current 0.005 or 5 Milliamps

(2) Voltage 5 volts, Resistance 1000 Ohms, Current 0.005 5 Milliamps

After this, I figured for me it's easier to just go up in steps of 10 so:

10 * 1000 Ohms = 10000 Ohms and thus 10000 * 0.005 = 50 volts (the author subtracts 100 Ohms from the meter itself to give 9900 Ohms circuit resistance.

I can't really go beyond what I've posted if someone disagrees with the figures but the book is EveryMan's Wireless Book and we're on page 40, chapter testing Instruments. It may be possible to access it on the net.

I should add finally, given I've never really had to do it the old way, some of this use of plug and socket resistances is a bit new. In normal life I'd just use my digital meter but I thought it would be a good idea to try and look at how things needed to be done in the Thirties.

Apologies again for any confusion here.