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Why Aren't There More Aspies Into Electronics?

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Dr-David-Banner
On ‎11‎/‎03‎/‎2016 at 9:16 PM, Pinky and his brain said:

The problem is that 50uF is no longer a standard value. Today Capacitors and resistors follow a decade standard. It's either going to be E12, E24, or E96 standard. There are other standards too, but they are not used much. You can see the decades here: E decades

So you have to settle with either a 47uF or a 56uF. But that's not a big problem, because the precision on capacitors is usually +-20%. So they will always be a little off the value anyway. I think you can feel quite safe just using a 47uF or 56uF instead.

On Farnell, Mouser or Digikey you can find many different caps, with a value of either 47 or 56 uF and voltage rating between 400 and 450 Volts. Remember a higher voltage rating is not a problem, only a too low rating is a problem.

Here is a link to a search on Farnell: Farnell caps

:)

 

Cheers. I'm not too great at finding capacitor values online. I'll check it out.

By the way I finally discovered the solution to my question (or part of it). I noted in my HAM book a few days ago that cathode bypass capacitors perform a very simple function. They attenuate the A.C. signal amplitude and that essentially stops an A.C. voltage drop across the cathode. Thus, there is the big bias resistor between cathode and the steel chassis. That does have a D.C. drop which can be tapped to bias the grid. And the cathode bypass capacitor then chops any A.C. so the D.C. bias is stable. That's all there is to it. I suppose the capacitor virtually shorts out high frequency A.C.

Should you know how to deduce the resultant reactance of two opposing reactances in parallel mathematically please let me know. By that I mean, a capacitor and an inductance. The series explanation I grasped easily but the formula for parallel wasn't clear. It has the following:

-XC XL
----------
XC-XL    (XC is capacitive reactance and XL inductive reactance)

You can see the minus sign at the very top? If it were two basic resistances (not reactances) you'd just multiply the top digits and add the lower before division. So, anyone know what the minus sign means at the top?


 

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Dr-David-Banner
On ‎13‎/‎03‎/‎2016 at 1:31 AM, Little Guy said:

I was into electronics when I was young and built crystal sets and then tube sets. I got kind of lost once transistors came along. Built short wave radios but never got Morse code due to tone deaf. Still today in the US, Morse code is no longer a requirement so I got my general ham license after 50 years of first studying the ARRL books. Later, I got back into electronics and started fixing and building computers.

Today though programming fulfills the same need to use my logic pathways in the brain. I program web applications for a living now sometimes several thousand lines all out of my head.:wacko:

That's the book I have and it's a pretty good one.

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Pinky and his brain

@Dr-David-Banner

Well if you're talking about the current, that runs through the inductor and the capacitor, then those will have to be subtracted, because they work in the opposite direction so to speak. They try to cancel each other out.

If we're talking about combined impedance, then the following formula should work:  Z= (XC * XL) / (XC - XL)

But if the frequency is higher than the resonance frequency of the circuit, then the formula has to be adjusted.

The resonance frequency is:  f = 1 / (2 * pi * (sq-root of L * C))

If frequency is above resonance frequency, then Z = (XC * XL) / (XL - XC)

The reason XC and XL has to switch place, is because below res. freq. the load is inductive, and above res. freq. the load appears capacitive. 

 

**since it's hard to make formulas in here, I just want to explain what the different symbols mean in the above text.

/ = divide

* = multiply

f = frequency

All values must be in their basic units. ( Ohm, Henry, Farad and Hertz)

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Dr-David-Banner
20 hours ago, Pinky and his brain said:

@Dr-David-Banner

Well if you're talking about the current, that runs through the inductor and the capacitor, then those will have to be subtracted, because they work in the opposite direction so to speak. They try to cancel each other out.

If we're talking about combined impedance, then the following formula should work:  Z= (XC * XL) / (XC - XL)

But if the frequency is higher than the resonance frequency of the circuit, then the formula has to be adjusted.

The resonance frequency is:  f = 1 / (2 * pi * (sq-root of L * C))

If frequency is above resonance frequency, then Z = (XC * XL) / (XL - XC)

The reason XC and XL has to switch place, is because below res. freq. the load is inductive, and above res. freq. the load appears capacitive. 

 

**since it's hard to make formulas in here, I just want to explain what the different symbols mean in the above text.

/ = divide

* = multiply

f = frequency

All values must be in their basic units. ( Ohm, Henry, Farad and Hertz)

So, it seems as if basically the simple way is to just multiply the XC XL and then add and divide (as you would with two resistances in parallel). What had me confused was why the book had a minus sign such as here where I duplicate it : -XC XL
                                                                                                           ---------
                                                                                                          XC+XL
The resonant frequency formula I understood as fortunately there were examples in the book. I hopefully recall the frequency is expressed in Megacycles, the capacity as micromicrofarads and then use the square root and pi to finally divide by 1000000. However I always seem to keep forgetting these sums!!

It does state that due to the phase difference of the two reactances, XL tends to be positive and XC negative. Like I said, I was a bit confused by the parallel example.

Thanks for the assistance.

 

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Dr-David-Banner

Just to add, I took a bit of time to some simple experiments with bulbs and filaments.

There are two ways the battery set tube heaters were powered in the Forties and Thirties, Either the 2 volt battery was put in parallel with the filament or the filaments were wired in series. In the latter case, the voltage would drop across the filaments. This method was popular with the mains transformerless receivers.

It would seem my set has parallel filament connection. Each tube has two pins which are the heater pins. My Thirties workshop book recommends a simple test where you wire the filaments in series with a small lamp and battery. From this I gather at least one of my tubes seems O.K.

Here is the odd bit. The filament, as in light bulbs it seems, is merely a thin strand which will get hot when connected to 2 volts. I have to make sure each pin is getting the voltage directly and that 2 volts potential difference is evident throughout. I believe it is only 1 watt per filament. Two large torch batteries will give me 3 volts in series so then I can add 1 Ohm and ensure 2 volts is the final voltage.

Once the 120 volts DC HT current is applied to the circuit, electrons should flow through the tubes and act as a Low Frequency amplifier for RF and AF signals. This is not a Superheterodyne set but a grid leak regenerative feedback design. I've no idea when it last worked but can only guess maybe the early sixties before being stored and fporgotten.



 

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Dr-David-Banner

The light turned on yesterday. One tiny step forwards and hopefully a good omen. I renewed a lamp holder wire in case the original was shorting and that circuit is now O.K.

For some reason the negative leads on these very old battery powered sets always have some resistors between them. The L.T. (2 volts) negative wire tends to be grounded to the chassis (no PCB boards back in those days) but the H.T. negative lead has a potential divider between it and the L.T. negative. This intermediate junction can be used to tap off grid bias for the tubes.

The engineering side of this activity is kind of different to theory. You need to check the basic stuff like locating the aerial lead, the H.T. leads, the speaker wires and even the RF earth wire. It only takes one broken connection to get no response.

I'll be testing that battery eliminator transformer soon at a safe distance as well.

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Dr-David-Banner

O.K. so it now seems the 1940's battery eliminator is fixed. The pic here shows some of the work done, including repair of the transformer primary.

Originally the metal frame on which the unit was built was covered in grease and dirt (as were the wires). This boiled down to a big cleaning job and some disassembly. Then most of the wiring re-soldered. I'm a bit short of decent wire at the moment so the thick multi-strand I used was a headache to solder. It's really far easier to use thinner wire although I tend to be wary of poor quality wire as is churned out today.

The black bladed device you can see at the back is an old selenium rectifier as there were no silicon diodes in the Forties. It works on the same principle of a P/N junction and only allows positive voltage amplitudes to pass and thereby allow direct current.

The breadboard allows for shunt resistances. Somewhere I ought to have a 2 volts D.C. tapping.

The main D.C. output today measured 130 volts.

This is now the power supply for me Thirties radio set. At some point the whole thing will be connected up and I'll be attempting to get some stations.

It still looks a bit grubby but far better than at the start when it was almost black with dust and grease.

Metal rectifiers fell out of favour with engineers and are considered dangerous but I think I tend to get my kicks from the buzz of it all. I did clean the metal rectifier, of course.

Note how this whole device is metallic. Should a live wire ever contact the chassis you'd surely know about it when holding the case. Maybe people weren't so health and safety geared In the war years.

I plan on making this all much safer and maybe coating the chassis.

HTELIMIN.jpg

Edited by Dr-David-Banner

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Dr-David-Banner

One thing I'm doing lately is carrying out basic experiments with a 9 volt battery and resistors or even lamps. Or I may cut off the plug of a transformer, identify the polarity of the 2 wires and then make some us of it.

For example, as I said earlier, I need a source of just 2 volts at the moment. This is for the filaments of the 4 electron tubes in my Thirties set. I had some confusion earlier over the current and watts required here but, after some revision. I find that usually neither the current or watts is all that big. A total current of just 1 amp can be normal for a heater circuit although it's probably a touch more in a battery set. In the case of my set the heaters are connected in parallel with the battery so I expect the current to divide as the resistances are in parallel.

The unit shown above seems to have no 2 volt tapping. So, I've taken a modern 9 volt mains transformer and used some resistors to tap off just over 2 volts. Now I just need to test this 2 volts can deliver enough watts to power my circuit. I tend to use small bulbs and a milliameter to gauge this. If it works the transformer will deliver the 2 volts I need without having to worry about batteries going flat as I test the receiver. This is a pretty hefty unit that can deliver 20 watts so it ought to be up to the task at hand.

The unit shown above is kind of scary. The case that goes on top is metallic so I dread to think what would happen if it ever became live. They had no Earth Wire or RCD back in those days.

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Pinky and his brain
2 hours ago, Dr-David-Banner said:

One thing I'm doing lately is carrying out basic experiments with a 9 volt battery and resistors or even lamps. Or I may cut off the plug of a transformer, identify the polarity of the 2 wires and then make some us of it.

For example, as I said earlier, I need a source of just 2 volts at the moment. This is for the filaments of the 4 electron tubes in my Thirties set. I had some confusion earlier over the current and watts required here but, after some revision. I find that usually neither the current or watts is all that big. A total current of just 1 amp can be normal for a heater circuit although it's probably a touch more in a battery set. In the case of my set the heaters are connected in parallel with the battery so I expect the current to divide as the resistances are in parallel.

The unit shown above seems to have no 2 volt tapping. So, I've taken a modern 9 volt mains transformer and used some resistors to tap off just over 2 volts. Now I just need to test this 2 volts can deliver enough watts to power my circuit. I tend to use small bulbs and a milliameter to gauge this. If it works the transformer will deliver the 2 volts I need without having to worry about batteries going flat as I test the receiver. This is a pretty hefty unit that can deliver 20 watts so it ought to be up to the task at hand.

The unit shown above is kind of scary. The case that goes on top is metallic so I dread to think what would happen if it ever became live. They had no Earth Wire or RCD back in those days.

Having the heaters in parallel, will double the current, not divide it in half. Putting two equal resistors in parallel, will give you half the resistance. And half the resistance will double the current.

If the two tubes are the same, then you can put them in series. That would then require 4 Volts at 1 Amp. But this will only work with identical tubes.

But I agree, that the new transformer is the right thing to use. You just need to get the right regulator circuit made. :)

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Dr-David-Banner
On ‎15‎/‎04‎/‎2016 at 7:00 PM, Pinky and his brain said:

Having the heaters in parallel, will double the current, not divide it in half. Putting two equal resistors in parallel, will give you half the resistance. And half the resistance will double the current.

If the two tubes are the same, then you can put them in series. That would then require 4 Volts at 1 Amp. But this will only work with identical tubes.

But I agree, that the new transformer is the right thing to use. You just need to get the right regulator circuit made. :)

Well, the heater circuit supply idea turned out to be a bit of a flop and I'll share with you what I did and what happened.

You're right about the current but I didn't quite figure it out till after I'd experimented.

The reason I finally dismissed my transformer idea is there was a lack of watts. Sure, I got a potential divider and I did get the 2.9 volts my heater circuit needs. However, this 2.9 volts in practice wouldn't light up a 3 volt flashlight bulb (I know 3 volts is a bit high here). Not even the 6 volt voltage divider would do it. The main 9 volt supply would light it up very brightly but further along the voltage divider, there was too much loss of power.

I added another 9 volt transformer in parallel and found the voltage rose a bit throughout.

I conclude probably the wattages of my resistors at 7 watts were absorbing a lot of power (even though the transformer is rated at 20 watts at 2 amps). Yes, the current ought to be the same throughout with resistances in series and I had mine in the positive line.

I then tried putting two 1.5 volt batteries in parallel and, in that case, the bulb lit up dimly. However the case may be, it was that heat around a filament I need and it has to happen at low voltage.

Back to the drawing board:

Here is where I screwed up a bit: On revising my books, the fact is a filament circuit even in a battery set doesn't draw a lot of current or need a lot of watts. 1 watt for the entire filament circuit is the norm. The current will be around 200 milliamps.

Thus, now I have four 1.5 volt batteries in parallel. I tested my 3 volt bulb out of curiosity and it lit up dimly with a current draw of 40 milliamps. I made a very crude wooden battery holder using springs (salvaged from a junk pump) and a length of metallic strip. Now, I have a more realistic L.T. Filament battery. True, it's 0.5 volt too low.

I think it's a good idea to do these crude experiments before risking actual tubes. My feeling is I've been lacking actual experience seeing how things work.

So finally here is what must be done:

Test all resistances. Make sure all the anodes are in the H.T. circuit as well as screens. Check there is a hefty resistance between the positive and negative supply. Use a series dim bulb tester when powering up.

Once the filaments are hot and giving off electrons, the 100 volt H.T. must be applied.

Should I be unable to get this set to work, I can consoled myself I gave it my best shot and at least I got the H.T. Unit to work.

I'm pretty worked up about it all now mainly out of suspense.

P.S. I was pleased to find that all my tube pins register 1.5 volts so it seems this circuit is functional. Later I may check what current is being drawn.

 

Edited by Dr-David-Banner

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