Old Time Radio Engineers

[Nowhere-Man]

I figured it might be a good reality trip to try and get my head around how the old radio engineers of the Thirties and Forties did their diagnostics. I chanced upon a book that was written in the hungry Thirties and then updated to 1943. It's less technical than my sixties Ham book but more practical from the actual engineering perspective.

Did you know, a lot of repairmen at that time relied on a Milliameter with a full-scale deflection of 0.005 Amp? Much of their work was done reading Milliamp deflections. According to the book, it was typical for the said Milliameter to have a resistance of 100 Ohms and deflection of the full 0.005 Amp at 5 volts. The book then launches into a method for testing voltages. Shunt resistors are wired in series with the circuit where the voltage might be expected to be, say, 50 volts. Then, it seems you would divide 50 by 5 to get 10, multiply the 100 Ohms of the Milliameter and then insert a shunt resistor of 900 Ohms. Presumably then if the device shows a deflection of the full 0.005 Amp. you can assume the circuit voltage will be 50 volts.

Then there is this device called a Galvanometer. This is used to read the Ohms of an unknown resistor. The said Galvanometer is connected to circuit with a large resistance cable shunted across the probes and fixed to a ruler. The centre probe is slid along the resistance wire till it reads 0 and that then is the point of the two resistances. If you know the resistance of the total resistance wire, you can then use the ruler to calculate the proportion.

Maybe it would be fun to rig this test up if I can find a Galvanometer online.

Might be worth while getting a copy of the Radio Engineers vest pocket book as well as that's said to be really good.

[Terry]

Shunt resistors are in PARALLEL with the meter to increase the current required for FSD.

Resistors in SERIES increase the voltage required for FSD.

The system of switching resistors in series or parallel is used in all analogue multimeter's, the values matching the FSD of the movement used.

[ppppenguin]

Resistors in SERIES increase the voltage required for FSD.

And are often know as multipliers.

The device that was described as a galvanometer is actually a simple form of a wheatstone bridge. The galvanometer itslef is simply a sensitive meter that is used to indicate the null or balance point of the bridge.

[Crackle]

I figured it might be a good reality trip to try and get my head around how the old radio engineers of the Thirties and Forties did their diagnostics. I chanced upon a book that was written in the hungry Thirties and then updated to 1943. It's less technical than my sixties Ham book but more practical from the actual engineering perspective.

Did you know, a lot of repairmen at that time relied on a Milliameter with a full-scale deflection of 0.005 Amp? Much of their work was done reading Milliamp deflections. According to the book, it was typical for the said Milliameter to have a resistance of 100 Ohms and deflection of the full 0.005 Amp at 5 volts. The book then launches into a method for testing voltages. Shunt resistors are wired in series with the circuit where the voltage might be expected to be, say, 50 volts. Then, it seems you would divide 50 by 5 to get 10, multiply the 100 Ohms of the Milliameter and then insert a shunt resistor of 900 Ohms. Presumably then if the device shows a deflection of the full 0.005 Amp. you can assume the circuit voltage will be 50 volts.

as Terry says

Series resistors would be used with the ammeter to measure volts.
Shunt resistors would be used to enable the meter to be able to read larger current values.

Ohm's law states V = I x R
Where I is the amps ( I forget why he used I)
so
100 ohms x 0.005 amps = 0.5 volts for full scale deflection, not 5 volts as you said.

Another useful little formula to always remember is in a DC circuit,  Amps x Volts = Watts.
(an AC circuit is similar but other factors influence the power.)

Substituting for amps in the 1st formulae above gives Volts squared, divided by Resistance = watts.

These 3 little formulae are very useful in all forms of electrical work, understanding them is in my view essential.


Mike

[Murphyv310]

Hi.
There were many "multimeter's" in the loose term available prior to WW2. I have a perfectly good AVO7 from 1944 and it is a derivative of the "AVO Multimeter" from 1923.
Many of the bigger service departments would be well equipped with AVO's and other test equipment form Signal generators and oscilloscopes. A trip to the manufacturers the likes of Ekco & Murphy would be full of test equipment, the engineers would be highly trained in being able to read and understand the results. At the other end of the scale would be the small independent shop, often it would have one meter shared by two or three engineers, the most used item would be the wet finger!
Practical Wireless and other electronic journals would often have constructional details of simple multimeter's, they served a basic purpose but were inaccurate mainly due to skimping on multiplier resistors by using 20% carbon versions. These meters would be usually be found in the home of the hobbyist!
After the war and increased sales of Radio & TV even the small repair shop would sport an AVO7 or Multiminor for the field man, there were plenty of ex WD meters about, also there were VVM's and standard multimeter's by Taylor and other manufacturers.
Advance made a decent range of signal generators and good secondhand Cossor Scopes were available.
There was of course the Valve testers of the time, The AVO Twin panel tester goes back before the war and can still hold its own today, its utterly reliable with only the bases being its weakness.

The amount of Pre War test equipment that still exists is incredible, often due to engineers insistence that the job cannot be done properly without it!

[Nowhere-Man]

I figured it might be a good reality trip to try and get my head around how the old radio engineers of the Thirties and Forties did their diagnostics. I chanced upon a book that was written in the hungry Thirties and then updated to 1943. It's less technical than my sixties Ham book but more practical from the actual engineering perspective.

Did you know, a lot of repairmen at that time relied on a Milliameter with a full-scale deflection of 0.005 Amp? Much of their work was done reading Milliamp deflections. According to the book, it was typical for the said Milliameter to have a resistance of 100 Ohms and deflection of the full 0.005 Amp at 5 volts. The book then launches into a method for testing voltages. Shunt resistors are wired in series with the circuit where the voltage might be expected to be, say, 50 volts. Then, it seems you would divide 50 by 5 to get 10, multiply the 100 Ohms of the Milliameter and then insert a shunt resistor of 900 Ohms. Presumably then if the device shows a deflection of the full 0.005 Amp. you can assume the circuit voltage will be 50 volts.

as Terry says

Series resistors would be used with the ammeter to measure volts.
Shunt resistors would be used to enable the meter to be able to read larger current values.

Ohm's law states V = I x R
Where I is the amps ( I forget why he used I)
so
100 ohms x 0.005 amps = 0.5 volts for full scale deflection, not 5 volts as you said.

Another useful little formula to always remember is in a DC circuit,  Amps x Volts = Watts.
(an AC circuit is similar but other factors influence the power.)

Substituting for amps in the 1st formulae above gives Volts squared, divided by Resistance = watts.

These 3 little formulae are very useful in all forms of electrical work, understanding them is in my view essential.


Mike

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.

[Nowhere-Man]

Hi.
There were many "multimeter's" in the loose term available prior to WW2. I have a perfectly good AVO7 from 1944 and it is a derivative of the "AVO Multimeter" from 1923.
Many of the bigger service departments would be well equipped with AVO's and other test equipment form Signal generators and oscilloscopes. A trip to the manufacturers the likes of Ekco & Murphy would be full of test equipment, the engineers would be highly trained in being able to read and understand the results. At the other end of the scale would be the small independent shop, often it would have one meter shared by two or three engineers, the most used item would be the wet finger!
Practical Wireless and other electronic journals would often have constructional details of simple multimeter's, they served a basic purpose but were inaccurate mainly due to skimping on multiplier resistors by using 20% carbon versions. These meters would be usually be found in the home of the hobbyist!
After the war and increased sales of Radio & TV even the small repair shop would sport an AVO7 or Multiminor for the field man, there were plenty of ex WD meters about, also there were VVM's and standard multimeter's by Taylor and other manufacturers.
Advance made a decent range of signal generators and good secondhand Cossor Scopes were available.
There was of course the Valve testers of the time, The AVO Twin panel tester goes back before the war and can still hold its own today, its utterly reliable with only the bases being its weakness.

The amount of Pre War test equipment that still exists is incredible, often due to engineers insistence that the job cannot be done properly without it!

That's interesting, Trevor. There was also some mention of an AvoMinor. as well as the Avo 7 and Avo 8. Maybe I can acquire some of these.

Have you heard of a split anode adapter? What had me wondering about that was, how do you go on with the wide ranges of tube bases? Most of the examples refer to screen tubes, diode detectors or pentodes. This is all well and good in 1940 but, as we know, tube bases and pins continued to develop. Have you ever used a split anode adaptor? Myself, I'd just try and get at the anode pin itself from below the chassis although access on the last set was problematic.

All in all, it's interesting to weigh up a more diagnostic approach in fault finding. My other book was far more theoretical and scientific with a view to construction. This latter is more service and repair. I left it last night at the point of how and when to remove grid wires on tubes. A small part of that refers, of course, to A.F. coupling where you might have a leaky positive voltage on the grid. Or perhaps direct connection of a grid to the bias battery. Not to mention the dreaded damp finger on anode test..

[Nowhere-Man]

I figured it might be a good reality trip to try and get my head around how the old radio engineers of the Thirties and Forties did their diagnostics. I chanced upon a book that was written in the hungry Thirties and then updated to 1943. It's less technical than my sixties Ham book but more practical from the actual engineering perspective.

Did you know, a lot of repairmen at that time relied on a Milliameter with a full-scale deflection of 0.005 Amp? Much of their work was done reading Milliamp deflections. According to the book, it was typical for the said Milliameter to have a resistance of 100 Ohms and deflection of the full 0.005 Amp at 5 volts. The book then launches into a method for testing voltages. Shunt resistors are wired in series with the circuit where the voltage might be expected to be, say, 50 volts. Then, it seems you would divide 50 by 5 to get 10, multiply the 100 Ohms of the Milliameter and then insert a shunt resistor of 900 Ohms. Presumably then if the device shows a deflection of the full 0.005 Amp. you can assume the circuit voltage will be 50 volts.

as Terry says

Series resistors would be used with the ammeter to measure volts.
Shunt resistors would be used to enable the meter to be able to read larger current values.

Ohm's law states V = I x R
Where I is the amps ( I forget why he used I)
so
100 ohms x 0.005 amps = 0.5 volts for full scale deflection, not 5 volts as you said.

Another useful little formula to always remember is in a DC circuit,  Amps x Volts = Watts.
(an AC circuit is similar but other factors influence the power.)

Substituting for amps in the 1st formulae above gives Volts squared, divided by Resistance = watts.

These 3 little formulae are very useful in all forms of electrical work, understanding them is in my view essential.


Mike

Correction. I'll have to check the book again but I'm sure it stated 5 volts. I agree it doesn't add up if you multiply the ohms and current. Not sure why it says 5 volts but I'm pretty sure it does as I memorised the calculation and it all adds up the same as the book by dividing 5 into 50 which gives the 1000 ohms. Funny that! If I used 0.5 volts it wouldn't have been the same as the 1000 Ohms the book gives. I have no explanation for now till I recheck the figures given. Strange.

[ppppenguin]

....but that then means at full scale deflection, the 900 Ohm resistor is taking the remaining current.

I think you mean remaining voltage. The same current is flowing though the resistor and the meter.

There's something very odd going on here. A 5mA meter might typically have a resistance of 50 ohms. Hence 0.005V*50ohms = 0.25V across it for full scale deflection.

To make this 5mA meter into a 50V FSD voltmeter we need a total resistance of: 50V/0.005A = 10000 ohms. Deduct the 50 ohms of the meter and you get 9950 ohms. Unless you're making something very accurate you can forget about the meter's 50 ohms here.

[Mark Hennessy]

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