Jump to content
Dr-David-Banner

Why Aren't There More Aspies Into Electronics?

Recommended Posts

Dr-David-Banner

O.K., prepare yourself for a shock. This project has failed. I spent last night p++++d off and reflecting upon how and where I went wrong.

Here is what happened:

I got 1.5 volts right across my filament circuit after connecting voltage. No issues there. There was no H.T. applied to the anodes. My idea was simply to test I had a working heater circuit.

Next test was to verify is amps were being drawn. This tested as problematic. No draw at all.

Finally I figured something wasn't quite right. I re-tested the filaments of a tube in isolation using an old continuity test old engineers used. You simply run an ammeter and a bulb through the two pins in series on the negative line and the bulb will light if the pins are O.K. Well, that didn't happen. It had happened a few days ago but now zilch. I tested the other tubes and likewise dead!!!

No doubt about it the tubes were screwed!

So, the big question to be asked was why and how? I began to suspect that my mistake was to do this test without H.T. also on the anodes but I've never found any information anywhere that that is risky. They do warn about putting on H.T. without the heater voltage but not vice versa.

Some stats: Old battery radio sets like this one averaged about 0.5 Amp in the heater circuit. Watt load was low. With only 1.5 volts I wasn't getting much of a current across a regular filament I had tested. About 0.04 Amp was noted.

Today I tested the draw using just on 1.5 battery on a regular lamp and the current draw remained the same as batteries were added in parallel. I could see no indication of a surge risk. I applied some maths as well and we have basically 4 resistances in parallel so the current should simply divide between each resistance (if it were a series heater circuit current would be the same throughout.)

I can only conclude that either:

(1) Powering filaments in isolation without H.T. also applied can screw the tubes.

Or

(2) The tubes simply blew when actual voltage was applied after 70 years.

I'm seriously down-hearted but mistakes aren't the end of the world if you can find exactly what you did wrong.





 

Edited by Dr-David-Banner

Share this post


Link to post
Share on other sites
Pinky and his brain
On 19/4/2016 at 6:07 PM, Dr-David-Banner said:

O.K., prepare yourself for a shock. This project has failed. I spent last night p++++d off and reflecting upon how and where I went wrong.

Here is what happened:

I got 1.5 volts right across my filament circuit after connecting voltage. No issues there. There was no H.T. applied to the anodes. My idea was simply to test I had a working heater circuit.

Next test was to verify is amps were being drawn. This tested as problematic. No draw at all.

Finally I figured something wasn't quite right. I re-tested the filaments of a tube in isolation using an old continuity test old engineers used. You simply run an ammeter and a bulb through the two pins in series on the negative line and the bulb will light if the pins are O.K. Well, that didn't happen. It had happened a few days ago but now zilch. I tested the other tubes and likewise dead!!!

No doubt about it the tubes were screwed!

So, the big question to be asked was why and how? I began to suspect that my mistake was to do this test without H.T. also on the anodes but I've never found any information anywhere that that is risky. They do warn about putting on H.T. without the heater voltage but not vice versa.

Some stats: Old battery radio sets like this one averaged about 0.5 Amp in the heater circuit. Watt load was low. With only 1.5 volts I wasn't getting much of a current across a regular filament I had tested. About 0.04 Amp was noted.

Today I tested the draw using just on 1.5 battery on a regular lamp and the current draw remained the same as batteries were added in parallel. I could see no indication of a surge risk. I applied some maths as well and we have basically 4 resistances in parallel so the current should simply divide between each resistance (if it were a series heater circuit current would be the same throughout.)

I can only conclude that either:

(1) Powering filaments in isolation without H.T. also applied can screw the tubes.

Or

(2) The tubes simply blew when actual voltage was applied after 70 years.

I'm seriously down-hearted but mistakes aren't the end of the world if you can find exactly what you did wrong.





 

It's option 2.

You cannot destroy a tube by simply powering the heater alone. It's only a heating element. So most likely the tubes were already bad, and what killed them completely, was the last time you applied power to them.

Tubes are quite rugged. You can only kill them by overloading them. Too much voltage will make them arc and short out, too much current will make them melt.

So no, I don't think you destroyed anything. It's always sad when a project goes south, but that's life. Especially when you work with 50-100 year old technology.

 

I think you should try to get hold of a variable power supply, (lab supply) for the next time you make experiments. It doesn't have to be fancy, but it is much easier for you to control your voltage and current level that way. And when you know exactly what is going in, you will no longer have to worry about doing it wrong.

By setting your current limit first, you will always be in safe area. Because even a full short circuit, will never be able to pull more current than the limit you have set. I use that every time I have to power up a circuit for the first time. It saves me a lot of trouble.

A cheap one will cost about £40-60 (new), if you're lucky you might be able to find a used one even cheaper. Just make sure it works before paying for it. They should have both variable voltage from 0-30Volts and variable current limiting from 0-3A. (Typically, there are more than on type) If they don't have current limiting, they are less desirable.
 

Share this post


Link to post
Share on other sites
Dr-David-Banner
On ‎20‎/‎04‎/‎2016 at 7:44 PM, Pinky and his brain said:

It's option 2.

You cannot destroy a tube by simply powering the heater alone. It's only a heating element. So most likely the tubes were already bad, and what killed them completely, was the last time you applied power to them.

Tubes are quite rugged. You can only kill them by overloading them. Too much voltage will make them arc and short out, too much current will make them melt.

So no, I don't think you destroyed anything. It's always sad when a project goes south, but that's life. Especially when you work with 50-100 year old technology.

 

I think you should try to get hold of a variable power supply, (lab supply) for the next time you make experiments. It doesn't have to be fancy, but it is much easier for you to control your voltage and current level that way. And when you know exactly what is going in, you will no longer have to worry about doing it wrong.

By setting your current limit first, you will always be in safe area. Because even a full short circuit, will never be able to pull more current than the limit you have set. I use that every time I have to power up a circuit for the first time. It saves me a lot of trouble.

A cheap one will cost about £40-60 (new), if you're lucky you might be able to find a used one even cheaper. Just make sure it works before paying for it. They should have both variable voltage from 0-30Volts and variable current limiting from 0-3A. (Typically, there are more than on type) If they don't have current limiting, they are less desirable.
 

I appreciate the encouraging words and also the in-depth feedback you provided me throughout this thread. It seems clear to me your experience with electronics, as well as insight, is very deep and impressive.

Anyway, I did quite a bit of searching and a lot of thinking. Even one evening sat next to my German Shepherd, I had the calculator out trying to get a maths angle on it all (while he ate).

First of all, I'm now aware some of my tubes rattled. On close inspection of the output Tetrode I can see some physical damage through the glass where there is disconnected springy wire. The bases had also been loose but I'm not sure if the air vacuum had leaked.

On website forums I find they advise to test the filaments on only one 1.5 volt battery. This is where the maths came in as, to be honest, I can't see any change in current at all with parallel voltage supply (I used more than one). If there are 4 resistances in parallel, the overall sum resistance will be smaller and the current higher as resistance falls. Therefore the current will be divided amongst each resistance as they teach at beginners classes. I can't really see any possible surge with just 1.5 volts. However, I did consider what might happen if the resistance of each filament had been lowered due to age perhaps.

It also seems there are some particular battery tube filaments that are dead specific. Deviate by just a few millivolts and they are supposed to pop. That is a possibility too.

Anyway, I've recovered perspective now. I mean, it could have been much worse. At least I noticed the filaments were faulty before I went on to apply the full H.T.

I've decided to shelve the radio for some time and then will consider substituting new bases and new tubes, maybe from China. I may even need to alter some aspects of the circuit. No need to throw in the towel - it may live to fight another day.

Despite the disappointing conclusion, I seemed to have learned more from this defeat than most of my successes. One lesson I did learn is my basic engineering skills were a bit weak compared with my theory. Doing small tests with 9 volt batteries and 0.125 amp bulbs brought theory a bit closer to home.

Tes.jpg

Share this post


Link to post
Share on other sites
Dr-David-Banner

Lately I've been delving into oscillators and mixer circuits. Where it gets interesting is by comparing it with American circuit designs. For example, some American receivers used pentodes as mixer tubes or even used 2 tubes (oscillator and mixer) which is thought to be more efficient.

In the case of pentodes, they simply injected the oscillator frequency into the same grid as the aerial signal. Then it created the usual I.F.

Some of the theory is now a bit more complex. I've had to delve into resonant frequency calculations a bit. Also the usual Q's.

Share this post


Link to post
Share on other sites
Dr-David-Banner

I must admit I'm still a bit puzzled over these battery electron tubes. I did some fishing around and am told the Cossor tubes are presumed to be especially delicate. The filaments take 2 volts max. The norm is to test the filaments with just one 1.5 flashlight battery and then maybe 2 or 3 in parallel when the main 120 D.C. is applied.

I now have a few simpler projects to work on for the time-being and don't have to mess around using batteries for the filaments as it's all done for you via the transformer.

Still, I'll be blown if I know why the tubes were destroyed. Some time in the future I'll try and find some substitute tubes and try again.

Meantime below is a pic the same as one of the sets I did get going and now it runs well. It's a Fifties model and the speaker is pretty huge. When it runs there's a kind of musty smell and you can see a deep glow around the tubes. Actually this set improved with use. It was really weak the first time I got it to receive and I had to change a few capacitor values to a higher figure.

One basic thing I also learned is how important the Earth RF ground is on these sets. I made a special ground using a metal stake and telephone wire driven into the earth. Connecting this to the set, it pulls in pretty well on LongWave.

Any artists may have a suggestion as to how to re-letter the slats as mine lost their letters - someone cleaned them off at some time. I thought maybe I could do it somehow on a P.C. and make a transfer. I used Perspex to make the slat which fits but I have no lettering. All the scale lamp bulbs are apparently dead as well so will need to be replaced.

hmv1121_1267554.jpg

Edited by Dr-David-Banner

Share this post


Link to post
Share on other sites
Pinky and his brain

I'm still sure the old tubes had a defect before you started your project. Most likely a vacuum leak. If air had leaked into the tube, then the tube will fail once you power it up. Just like a light-bulb, it will look OK when looking at it, but when you power it up, the bulb will blow immediately. The oxygen in the air destroys the filament.

Since you mentioned "Cossor tubes" I found this document online. It has a lot of old tube information from that specific company. (Don't know if you already have it, but here you can find it).

 

https://frank.pocnet.net/sheets/160/suppinfo/Cossor_VM_1935.pdf

 

:)

Share this post


Link to post
Share on other sites
Dr-David-Banner
On ‎15‎/‎06‎/‎2016 at 7:19 PM, Pinky and his brain said:

I'm still sure the old tubes had a defect before you started your project. Most likely a vacuum leak. If air had leaked into the tube, then the tube will fail once you power it up. Just like a light-bulb, it will look OK when looking at it, but when you power it up, the bulb will blow immediately. The oxygen in the air destroys the filament.

Since you mentioned "Cossor tubes" I found this document online. It has a lot of old tube information from that specific company. (Don't know if you already have it, but here you can find it).

 

https://frank.pocnet.net/sheets/160/suppinfo/Cossor_VM_1935.pdf

 

:)

Thanks. Not much time left to answer now as this place is about to close for the weekend.

I guess I'll have to look at finding substitute tubes at some point and try again. I'm still reading a very dated book on radio engineering and it seems the fashion back when it was written was to use transformer coupling in the final stage. The primary would be connected between HT and anode and the secondary would be fed to the grid of the following output tube and the grid bias (which was usually a battery).

I did do the filament tests and before I powered up the set, these tests seemed O.K. However, later I used superglue to try and fix the base better than it was. Still, I just have to accept the tubes were probably on their last legs.

Share this post


Link to post
Share on other sites
Dr-David-Banner

I found a way engineers in the Forties used to test capacitors. There were no digital meters back then and Avometers were used. Anyway you use an AC winding voltage and put a resistor, capacitor and neon bulb in series. Then count the flickers over a minute. Doing this with a known capacitor allows you to apply mathematical proportion to any unknown capacitor. I don't know if you can easily get neon lamps and it didn't say what resistance to use. It does state you can also charge a 1mF capacitor at 10 volts but if it's 0.0001 you can use 100 volts. For HF coils they often used small lamps in series with the winding to test for OC or breaks.

Share this post


Link to post
Share on other sites
Pinky and his brain

@Dr-David-Banner There is a way of calculating the value of a capacitor, if the only tools you have is a voltmeter and a stopwatch.

The formula goes like this:  t / 5 = R x C    The trick is to find a resistor that has a suitable size compared to the capacitor. Because you have to measure the charge time of the capacitor, and if the resistance is too small the capacitor will charge to quickly, and if it's too big it will take forever. But it is possible to find a capacitor value with reasonable precision, with this method.

5t = the time it takes to charge the capacitor to 99% of the battery voltage available. (the reason it's 99% and not 100%, is that a cap has an internal loss, so it will never reach 100%).

R = the resistance in Ohms

C = the capacitance in Farad

And example:

Let's say you have a 9 Volt battery (measures 9 Volt), and you have a 10k resistor, and it takes 10 seconds to reach 99% of 9 Volts (8,91V).

That would then result in:

10 sec / 5 = 2  (so t = 2)  

And t / R = C

 2 / 10,000 = 0.0002 Farads   --->> 0.2 uF or 200 nF.

The only thing you have to be aware of, is the internal resistance of our power source. If the battery has too high internal resistance, then the result will not be precise. In theory you have to add the internal resistance of the power source to the value of R. But in reality, you just have to make sure the source can deliver enough current.

With a 10k resistor and 9 volts, the maximum current possible is 0.9mA. So as long as your power source can deliver 0.9mA, without causing the voltage to drop, you will be fine. Of course you have to calculate the numbers every time. But that's life, people today are spoiled by technology. In the past everything had to be done manually.

But this is a simple way to find a capacitor value, if you don't have a capacitance meter. :)

Edited by Pinky and his brain

Share this post


Link to post
Share on other sites
Dr-David-Banner

I'll have to make a note of that. I kind of like the old electronics books and various ways of testing. A method I found last night was to simply wire a lamp through a primary transformer to check the winding is intact. I'd have loved to have got that radio going as I learned a lot from the battery TRF sets. It's still on hold for a while as all rhe tubes need substitutes. I did manage to repair the HT Eliminator and still try to read up on the engineering theory as there is a lot to learn.

Share this post


Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.


×
×
  • Create New...

Important Information

By using this site, you agree to our Terms of Use.