Old Time Radio Engineers

[Mark Hennessy]

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?

To save confusion and possible frustration, please can we restrict this thread to meters only? In fact, if no-one objects, I'm thinking of renaming it so that we can be clear.

No problem at all if you want to discuss valves, but let's have another thread for that.

And this is a UK site, so we will call them valves. To me, a tube base is the rear of a CRT. I was fixing an oscilloscope at the weekend, and had to unplug the tube base connector as part of that.

Thanks in advance,

Mark

[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.

[Nowhere-Man]

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?

To save confusion and possible frustration, please can we restrict this thread to meters only? In fact, if no-one objects, I'm thinking of renaming it so that we can be clear.

No problem at all if you want to discuss valves, but let's have another thread for that.

And this is a UK site, so we will call them valves. To me, a tube base is the rear of a CRT. I was fixing an oscilloscope at the weekend, and had to unplug the tube base connector as part of that.

Thanks in advance,

Mark

Hmmm, I definitely didn't want to wind up with a thread on meters. I really don't know a great deal about the workings of a test meter except when I might need to wire an ammeter shunt on a boat. Unless someone else wants to start a thread on moving coil design for test meters?

I don't object to someone pointing out a mistake I made because in some ways it's good for me to dig out the textbook and go over it again in more detail. However, so long as hopefully we are now clearer than before, it would be nice to move on back to the main topic: The way old time service engineers did their repairs and diagnostics. In fact, I have before me some pics of an Avominor, A pocket meter from 8 to 240 volts, a Wristlet Voltmeter up to 100 volts, a Rotameter and large Ferranti Radio Service Tester.

[Nowhere-Man]

O.K. Mark, I just read your comments on voltage and series resistors - I'll probably print out the page here at the library and then stick it in my textbook for reference. I hope you understand I'm not being rude when I seem to delay responding directly to explanations because sometimes I have to do this later on. Let's say I get about 4 or 5 posts but only have an hour and bit, plus e-mails to sort through.... Anyway, thanks for your help and patience.

I had a pretty late night trying to advance further because now there is some similar information on testing resistors. This bit was tricky. The fundamental point that gave me pause for thought is that, back in the early Forties, the radio engineers weren't as spoiled as we are. Checking if a resistor was O.K. wasn't simply a case of sticking your DMM on, say, 200 K, and probing a resistor. Times were hard and testing equipment was costly so pocket Milliameters were vital. 

Yes, as you said in your post, current limitations were an issue. The book states (from memory):

1000 R (up to 30 Milliamps on tester)
10000 R (up to 10 Milliamps on tester)
100,000 R (up to 5 Milliamps).

So, for the sheer hell of it, this is what I did in various imaginary scenarios:

With the aid of a calculator, I imagined I needed to test a 20,000 R resistor. The voltage according to the book was 100 volts. According to my calculation, the meter ought to deflect to 0.005 or 5 Milliamps because 100 divided by 20,000 is 0.005.

However, here is the punch line: Suppose the resistor is knackered? Suppose it's only able to offer a resistance of a mere 10,000 R? Or suppose the resistance is so low you could get 500 milliamps kicked out. Well, apparently the done thing was to start off with a much smaller voltage and work out the whole scenario through proportion. I won't risk giving any examples in case I make a typo or mistake and jar a few nerves but basically I plotted out a few expected current readings at, say, 30 volts. Then imagined the resistor was damaged and how that could be detected by proportion.

Of course, why bother? All I can say is maybe it gives a kind of perspective on how different things were back then. Not only did most people wear a hat on the street but a fair few seemingly relied upon a pocket milliameter. Even worse, no calculator so they probably did it all on paper!

Lastly, any mistakes or typos here, please all be aware it may happen and if you point it out, I'll go over it again!

Finally, how much is a pocket milliameter of the Forties era? As well as Avo 7's and 8's?

[Mark Hennessy]

O.K. Mark, I just read your comments on voltage and series resistors - I'll probably print out the page here at the library and then stick it in my textbook for reference. I hope you understand I'm not being rude when I seem to delay responding directly to explanations because sometimes I have to do this later on. Let's say I get about 4 or 5 posts but only have an hour and bit, plus e-mails to sort through.... Anyway, thanks for your help and patience.

Hi NM,

No problem at all - I understand the time pressure at the library and everything.

It reminds me of the days before having broadband at home, where you'd set aside a couple of hours for the computer. First, dial up to download the messages and grab all the web pages you wanted to read. Then, once off-line, respond to all the emails and read all the web pages that you got earlier. Then dial up again to get the messages sent and follow any new thoughts. It was very different.

I held off getting broadband for as long as possible because I knew it would take over. And it certainly does. But of course, you can't be without it!

I had a pretty late night trying to advance further because now there is some similar information on testing resistors. This bit was tricky. The fundamental point that gave me pause for thought is that, back in the early Forties, the radio engineers weren't as spoiled as we are. Checking if a resistor was O.K. wasn't simply a case of sticking your DMM on, say, 200 K, and probing a resistor. Times were hard and testing equipment was costly so pocket Milliameters were vital. 

Yes, as you said in your post, current limitations were an issue. The book states (from memory):

1000 R (up to 30 Milliamps on tester)
10000 R (up to 10 Milliamps on tester)
100,000 R (up to 5 Milliamps).

So, for the sheer hell of it, this is what I did in various imaginary scenarios:

With the aid of a calculator, I imagined I needed to test a 20,000 R resistor. The voltage according to the book was 100 volts. According to my calculation, the meter ought to deflect to 0.005 or 5 Milliamps because 100 divided by 20,000 is 0.005.

Absolutely right; 100V will cause 5mA to flow in a 20k resistor.

A movement that has a 5mA full-scale deflection is pretty insensitive, and as you can see, to measure high resistors, you'd need a big set of batteries!

In the famous AVO 8, the FSD is 50 micro-amps. Pretty sure that's right - mine are up in the attic. At least, the most sensitive current range is 50µA, which is pretty handy occasionally.

So now, to measure 100k, we only need around 5V. There is a 15V battery in an AVO 8 for the high-ohms range...

However, here is the punch line: Suppose the resistor is knackered? Suppose it's only able to offer a resistance of a mere 10,000 R? Or suppose the resistance is so low you could get 500 milliamps kicked out. Well, apparently the done thing was to start off with a much smaller voltage and work out the whole scenario through proportion. I won't risk giving any examples in case I make a typo or mistake and jar a few nerves but basically I plotted out a few expected current readings at, say, 30 volts. Then imagined the resistor was damaged and how that could be detected by proportion.

This is a good "thought experiment" to go through

Luckily, resistors tend to fail high.

But even so, the principle is a good one: with a moving-coil meter, which is at constant risk of being damaged via overload*, then yes, you start on the least-sensitive range, and work your way up.

* OK, some meters have overload protection built in, but it's best not to stress them needlessly.

The scenario you're describing suggests that you have a variable voltage source, and by knowing the voltage, and being able to read the current from the meter, then you're able to apply Ohm's law to determine the resistance. That's good - it works. Of course, it depends on how accurately you can determine the voltage. In effect, you have to make 2 measurements in order to determine resistance, so you've got double the opportunity for errors to creep in... Also, there is that risk of damage.

As you might know, a "real" multimeter doesn't work that way. With that, you short the probes together, and the needle goes over to the right hand end of the scale. You can adjust this with a pre-set resistor on the front of the meter. Having done that, there's nothing a resistor can do - whatever the value - to cause the needle to go beyond the right-hand end stop.

The down-side of this approach is that the scale is non-linear. But it's a scheme that served us well for many decades.

A better scheme might be to use current source rather than a voltage source. Then, we pass a known current through the unknown resistance. If we choose the currents sensibly, then we get sensible voltage readings. Imagine the current is 1mA - so if the resistor was 1kΩ, then we'd see 1V.

But the trouble with a current source is that it requires some electronics. A couple of transistors, perhaps. No problem in a digital meter, where there's already plenty of electronics (and that's how most DMMs work), but it's hardly ever seen with analogue meters...

Of course, why bother? All I can say is maybe it gives a kind of perspective on how different things were back then. Not only did most people wear a hat on the street but a fair few seemingly relied upon a pocket milliameter. Even worse, no calculator so they probably did it all on paper!

To be honest, I've no problem with thinking back in time. It helps to give us an understanding of how we got to where we are today, and it's humbling to think how much was achieved with so little. Slide rules rather than calculators - and much more reliance on mental arithmetic (a skill I really admire). I always try to estimate the answer before hitting "enter" on the calculator, just so I know that I didn't make a typo when entering it into the calculator. Try persuading a kid today to do that!

Keep at it!

Mark

[Yorkie]

In a former life, during my career at British Gas, we were still using slide rules into the mid 70s. Calculators were classed as 'capital' rather then revenue items and needed a capital application and business case to be made, in the same way as buying a vehicle, say.  However,  other items from stationers were 'revenue' and needed no approval, so when we needed calculators, we just asked for a receipt for a 'stapler' or a dozen box files.  Anyone who's worked in a large outfit, be it in the public or private sector will be familiar with all the ruses used to 'oil the wheels of industry'.

Never in our wildest dream could we have imagined that a pocket calculator can now be had for 60p at Tesco.

As to 'most people wearing a hat in the street', yes, and they raised it ('doffed' it) when a lady walked by! On the front page of Practical Wireless in the mid 1950s, not only did men at workbenches were a suit and a tie, as often as not, it was a bow tie, and of course, the obligatory briar pipe in one hand and a soldering iron in the other.

How times change.

[Murphyv310]

I still have my slide rule and use it too.
I showed it to my 10 year old grand daughter and she was quite fascinated how it worked.

[Refugee]

Absolutely right; 100V will cause 5mA to flow in a 100k resistor.

Oops!
I have spotted a mistake in the calculations.
It aught to read 1mA and not 5mA.

[ppppenguin]

I still have the slide rule I used at school. I still just about remember how to use it. Not many people younger than me (I'm 59) will have used a slide rule to do real work.

Which reminds me of a story......

God had finished his creation and was mighty pleased with himself. He told all the animals to be fruitful, to go forth and multiply. So all the animals obeyed god and walked, swam, flew to do you know what. Apart from a pair of small snakes who were looking very worried and confused. They said to god that they didn't know how to go forth and multiply. God was perplexed, then the penny dropped. He got one of his angels to cut down some trees and fashion them into rustic tables. The small snakes were delighted. As they said: "We're adders, so to multiply we need log tables!"

[Mark Hennessy]

Absolutely right; 100V will cause 5mA to flow in a 100k resistor.

Oops!
I have spotted a mistake in the calculations.
It aught to read 1mA and not 5mA.

Oops indeed - thanks for pointing that out! Clearly a brain-fart, but no idea how I missed it when I proof-read. I'll correct the original post to save confusion, but happy to leave this here to show that I'm human