5FP7 project starting with the focusing coil.

[AdrianPH]

Jeffrey, did not get a notification so sorry for the delay.

Not always sure on terminology.
If I stick approx +25 volts DC into the Frame scan coils at 50mA I will get full deflection one way from centre of the tube if I stick -25 volts DC at (-)50mA I will get full deflection in the opposite direction from centre. So I would need 50 Volts peak to peak and the total current peak of 100mA. The resistance of the coil is 511 ohms so actually +/-25.55 Volts DC, the reactance at of the coils comes out at 373mH but at 12.5 Hz is low in comparison, some where around 30 Ohm.

This is where I do not fully know if things will work so I will add a paragraph or three here.

The line scan coils at DC needed 104mA at 6.5 Volts to fully deflect from centre to one side. If I take into account the reactance of the coils at 400 Hz, I calculated the line scan coils would need 14 Volts across the scan coils at a peak current of 204mA it is actually 25 Volts peak to peak with a bit of a spike which is overshoot from the flyback.

The line driver is an EL84 connected to the primary side of a home made 8.8:1 transformer, the HT is 265 Volts during the flyback period the voltage on the Anode pin of the EL84 is going up to 380 Volts and as the sawtooth is applied to the grid the current increases through the valve and the lowest voltage on the Anode is 150 Volts.

So that looks as though I have a peak to peak on the primary of 230 Volts and that matches with the 8.8:1 ratio of the valve, the idea was to keep the anode current change as low as possible so in theory I am changing some 23-24 mA in Anode current.

Back to the Frame scan add in a bit of reactance and I got that the X of the frame scan coil would still be in the region of 512 ohms?
So ignoring the inductance i would still be in the region of 50 volts pk-pk voltage from the secondary of the transformer. I picked 25 VA as a number for the transformer thinking that if I tried any mains transformer at that rating it would be OK as I assumed it was ampere turns that caused saturation and I would have 5 times what I needed at 5 VA with 12.5 Hz.

I can tell I am loosing the plot here and beginning to waffle and not getting to any particular point, so will stop and go cook tea. I may be able to return with a clearer head.

Adrian

[ppppenguin]

The error is that it's not ampere-turns that cause saturation. An unloaded transformer will saturate just as easily as a loaded one. If you take any iron cored transformer and wind up the voltage applied to a winding, at some point the current will start to increase rapidly as the core enters saturation. The voltage at which you get saturation will fall linearly as you reduce frequency.

This is why some US transformers, designed for 60Hz, will saturate and overheat on 50Hz even with correct mains voltage.

Your problem is also made worse by using a sigle ended output stage where the primary of the transformer is also carrying DC.

To calculate the scan power, assume that the load is resistive. As you correctly state, the reactance of the frame coils is a secondary effect at 12.5Hz. You need 50Vp-p which is approx 18V RMS. Using V^2/R gives 18x18/511 = 0.63W average power. In other words, not very much.

[AdrianPH]

OK thanks for that. Not sure why but I tend to look at peak levels as I seem to always work on voltage swings such as the anode voltage swing etc. So I may be able to use a smaller transformer core then I am thinking???

Anyway what I was thinking is that I have a 220 V primary transformer with two 24 Volt secondary windings, so I can wire them up to give 48 Volt output in normal rms values it is normal 40VA transformer.

If I wired the frame scan coils across the 48 Volt section and the 220 Volt primary between EL84 anode and 250 Volts HT, I was thinking that to get my 50 Volts peak to peak I would need a voltage swing on the anode of 230 Volts, if the flyback voltage manages to get the anode Voltage above HT as in the line drive then I could be working with an anode swing from 135 Volts to 365 Volts even though the sweep rate is 12.5Hz the flyback time is the same as the line stage at 0.23mS. Hence hoping I still get some higher voltage on the anode.

If it does show some promise I will then look to winding a transformer for it. If it shows no promise then I may have to look at using the transformer to provide a + /- 25 Volt spilt rail supply using a valve/semiconductor hybrid circuit to drive the scan coils between rails to get the required swing.

The last option that I do not think would work, was to wind some more frame scan coils with lots more turns but finer wire, getting the peak current down into the 15mA range, this would require two scan coils of approx 11,000 turns each of 38awg wire and the physical size of the coils could be prohibitive for me to hand make. The total resistance would be around 9.4K ohms and online calculators suggest the total inductance would be 23 H !

Think it is time to play over the next few days.

Adrian

[AdrianPH]

I have another what is most likely a stupid question, but as I do not mind making a fool of myself here goes: -

On the line stage as said the HT is normally 265 Volts, I am seeing the anode volts rise above HT to 380 Volts and then I get the sawtooth waveform dropping the anode volts to 150 Volts. is the current flowing as voltage drops from 380 to 265 volts provided from the stored energy in the coil or is it still out of the power supply and through the valve?, I am wondering what the current is through the valve is it providing 204mA/8:8 = 23mA or is it just half of that and only provide current for the 265 volts down to 150 Volts so it is 102/8.8 = 11.5mA? Plus quiescent of course.

Not sure if I explained that correctly or how i could measure it on a scope?

Adrian

[ppppenguin]

Anyway what I was thinking is that I have a 220 V primary transformer with two 24 Volt secondary windings, so I can wire them up to give 48 Volt output in normal rms values it is normal 40VA transformer.

The core will saturate badly at 12.5Hz. As I've said before, to get 50Vp-p (approx 18V RMS) at 12.5Hz you need a nominal secondary voltage of at least 72V on a nominal 50Hz tansformer. That's before even thinking about the DC.

Going for a higher VA rating doesn't help.

[AdrianPH]

OK I have had a sleep and time to try and digest but need to do more reading and playing before it sinks in and I may start to understand this better.  I wish I had a variable frequency ac power supply to play with and measure things, as I only seem to learn by trying and seeing the results, but I do not have such gear.

I found an article in the early hours and had a read of it, it is on audio OPT transformers and it talked about power de rating as in the following:-

I give an example: suppose you select an OPT which can handle 50 Watts at 30 Hz.
Then this transformer can handle 50/2 = 25 Watt at 30/1.414 = 21.2 Hz.
Or the transformer can handle 50*2 = 100 Watt at 30*1.414 = 42.4 Hz.
The rule behind all this is: “the power capability doubles when the frequency is a
factor 1.414 larger. The power capability halves when the lowest frequency is divided
by 1.414 (square root 2)”.

This reduces the VA of the transformer by a factor of 16 which is much, much more than I thought it would.
Somewhere else that the voltage ratio was dependant on the turns ratio.

I was thinking that if I applied the above for a 50 Hz transformer to get to 12.5 Hz
50Hz/1.414 = 35.35Hz. VA = 20
35.35Hz/1.414 = 25Hz. VA = 10
25Hz/1.414 = 17.67Hz. VA = 5
17.67/1.414 = 12.5Hz. VA= 2.5
Current wise 40 VA at 230 Volts means the primary current could normally be up to 173mA rms.
de rate to 2.5 watts and it drops to 11mA.

I want a peak secondary current of 100mA which is 35.4mA rms the primary current would be 35.4 x (48/230) = 7.4mA
I need 50 Volts peak to peak which as you say is 18 Volts ish rms, the turns ratio is 230/48 so the primary voltage required is 86 Volts rms, Things I do not know is if i could apply the same rules for sawtooth as one does for sine wave.  So I generally thought the transformer may still have a chance? No No I had not considered any standing dc current.

I need to try something to progress the project in some way, so I may as well build the EL84 stage and rewind a transformer, provide an air-gap by separating the E and I laminations and stick as much copper in there as I can.  Not sure what else to try.

Adrian

[ppppenguin]

For the 0.5W or so of power you need for V scan why not do it push-pull with a double triode such as ECC82. Then there's no DC in the transformer.

Sawtooth simply means there are higher frequency components in the signal as comapred to a sine. Quite a lot higher in the flyback.

But as I keep on trying to stress, its not the power rating of the transformer that's problem, it's that the core will saturate at low frequencies unless you derate the voltage.

[AdrianPH]

For some reason in my mind I had discounted push pull, I think because of the flyback pulse, I was relying on that to get the spot to the left hand side of the screen and I am unsure of what a push pull stage would do, or how the pulses in both triodes would effect the spot etc.  But it is something I can certainly try, I am assuming biasing the valves near cut off??

Also please do not think I am trying to prove you wrong or disagreeing with what you are saying to me, you are obviously way more technical/knowledgable in this field than myself.  But I have always found it hard to just take things as read or believe what I am told without it being shown to me in a practical way.  I was a repair engineer, for 40 years, I did not have to understand how a circuit worked, just how to find a faulty component, in fact the bosses did not want highly technical staff, they would have to pay a lot more! So I have to rely on trying something, if it does not work dig into it until it makes sense.

I am still finding it difficult to see why you are saying voltage is the driving factor, this is because I see valves as a current source, so if I bias a valve to give 10mA it cares not if the voltage on it's anode is 100 or 200 volts, or if the load resistance is 500 or 5K ohms within reason.  I can understand if the  transformer was on the mains supply or some really low impedance source at 12.5 Hz then the current would be just limited by the resistance of the windings.  lastly I am past the 60 mark the grey matter is not as quick to understand unknown territory so please do bear with me.


Adrian

[ppppenguin]

Do you have a variac? Find a transformer and apply variable voltage 50Hz to it. Use a voltmeter and ammeter to measure the voltage and current. Plot the result. As you raise the voltage above a certain level the core will start to saturate and the current will rise alarmingly. A very practical demo of saturation.

[AdrianPH]

Hello Jeffrey;

No unfortunately I do not have a variac.

The following was described to me once if making mains transformers without details of the core saturation, the numbers are going off memory.

If I had a variac, I could wind say a few hundred turns on a bobbin and slowly bring the variac volts up until I saw a knee in the current drawn and then back off a little, so that if I had 250 turns on the bobbin and just below the knee was 30 Volts then use the formula of 250/30 = 8.33 turns per volt, so for 240 Volt mains input it would be 2000 turns and for secondary use 8.33 x 1.04 to allow for some loss. So for 30 Volts output use 30 x 8.33 x 1.04 = 260 turns per volt. But how that would apply to lower or higher frequencies was not mentioned.
I have not come across a reasonably priced fully isolating variac, so never picked one up.

Adrian