Non-standard supply voltages, 1966

[JohnW2]

If you put a diode in series with the load the RMS voltage becomes 1/root2 of the original. With a 240v supply that becomes approx 170V. Then you design the dropper as if you are using a 170V supply.

All true, and if one reduces the resistance of the dropper appropriately for that reduced RMS voltage, the power supplied to the heaters will, indeed be the same 'as normal', and I now agree that will result in an appreciable reduction in the dissipation (hence heat production) in the dropper.

However, as I've been saying, an issue arises because of the peak current the heater filament will be subjected to in this situation - since the peak voltage (hence peak current, for a resistive load), is sqrt(2) times RMS for 'full AC", but 2 times RMS for half-wave rectified AC.

If one takes, for example a valve with a heater rated at 6.3V (RMS) 0.3A for 'full AC' (or 6.3V 0.3A DC) (i.e. about 1.9W) then if one feeds that with half-waved rectified AC of the same RMS voltage (6.3V) (so as to achieve the same heater power of 1.9W), the peak voltage will be 12.6V, rather than 8.9V, and the peak current 0.6A, rather than 0.42A - which one might expect to potentially have a marked effect on survival of the filament ....

... it's roughly equivalent to taking an incandescent bulb/lamp, designed as 60W at 240V (RMS, full AC), hence 0.25A RMS, and instead trying to run it at the same power by supplying it with every positive (or every negative) half cycle of a 340V RMS supply, the peak current then being around 0.5A, rather than 0.35A with the 'normal' supply.

Kind Regards, John

 

[winston1]

If you put a diode in series with the load the RMS voltage becomes 1/root2 of the original. With a 240v supply that becomes approx 170V. Then you design the dropper as if you are using a 170V supply.

All true, and if one reduces the resistance of the dropper appropriately for that reduced RMS voltage, the power supplied to the heaters will, indeed be the same 'as normal', and I now agree that will result in an appreciable reduction in the dissipation (hence heat production) in the dropper.

However, as I've been saying, an issue arises because of the peak current the heater filament will be subjected to in this situation - since the peak voltage (hence peak current, for a resistive load), is sqrt(2) times RMS for 'full AC", but 2 times RMS for half-wave rectified AC.

If one takes, for example a valve with a heater rated at 6.3V (RMS) 0.3A for 'full AC' (or 6.3V 0.3A DC) (i.e. about 1.9W) then if one feeds that with half-waved rectified AC of the same RMS voltage (6.3V) (so as to achieve the same heater power of 1.9W), the peak voltage will be 12.6V, rather than 8.9V, and the peak current 0.6A, rather than 0.42A - which one might expect to potentially have a marked effect on survival of the filament ....

... it's roughly equivalent to taking an incandescent bulb/lamp, designed as 60W at 240V (RMS, full AC), hence 0.25A RMS, and instead trying to run it at the same power by supplying it with every positive (or every negative) half cycle of a 340V RMS supply, the peak current then being around 0.5A, rather than 0.35A with the 'normal' supply.

Kind Regards, John

If you draw out a sine wave then erase the bottom half below zero the peak does not change.

But even if it did filaments don't break down with large peak voltages (unlike semiconductors). The heating effect depends on the RMS voltage.

 

[bernardgreen]

The peak current when the lamp is supplied by AC via a diode is the same as the peak current when there is no diode. The difference is that there are only 50 peaks all the same polarity per second instead of the 50 positive and 50 negative peaks per second when the lamp is fed AC.

Adding a capacitor to the rectified AC changies things as with a large enough capacitor the lamp will be fed by near constant peak voltage and the continuous current will e the same as the peak current on AC.

 

[JohnW2]

If you draw out a sine wave then erase the bottom half below zero the peak does not change.

That's true if you don't do anything else (i.e. if you just add a diode to a heater chain that was previous working as intended on 'full AC'). However, if you do that, the heaters will only get half of the power that they want (√2 less RMS voltage and √2 less RMS current). As you said, if the supply is 240V RMS, that will fall to about 170V RMS downstream of an added diode.

If, having added the diode, you reduce the dropper resistor so that the RMS voltage (hence also RMS current) supplied to the heaters is the same as it was without the diode (so that the heaters get the correct power), then the peak voltage will be √2 greater than it was without the diode.

But even if it did filaments don't break down with large peak voltages (unlike semiconductors).

Quite so - which is why I have talked about peak currents. For an essentially resistive load, they will also be (for the same heater power {same RMS current and voltage}) √2 times greater than without the diode. As I said, it would be like trying to run an incandescent bulb/lamp intended for 240V (RMS) AC off every other half cycle of a 340V (RMS) supply.

Kind Regards, John

 

[JohnW2]

The peak current when the lamp is supplied by AC via a diode is the same as the peak current when there is no diode.

See what I've just written to winston....

... What you say is true if you only add a diode - but then you will also get only half the required power delivered to to load. If you want the load to get the same power, you have to provide it with the same RMS voltage and current as it had before you added the diode - which would require reducing the dropper resistance such that the load gets √2 more peak voltage (hence √2 more peak current) than it did before you added the diode.

Kind Regards, John

 

[winston1]

As I said, it would be like trying to run an incandescent bulb/lamp intended for 240V (RMS) AC off every other half cycle of a 340V (RMS) supply.

Kind Regards, John

But valve filaments have a much greater thermal lag than incandescent lamps. A lamp "half waving" can be seen to flicker and possible overloads ever cycle. But a valve filament would still appear constant brightness.

A series diode was standard practice towards the end of the valve era and valve life did not seem to be adversely affected. A second diode reversed biassed was also fitted across the heater chain after the dropper diode whose purpose was to take out the fuse if the dropper diode failed short circuit.

 

[JohnW2]

As I said, it would be like trying to run an incandescent bulb/lamp intended for 240V (RMS) AC off every other half cycle of a 340V (RMS) supply.

But valve filaments have a much greater thermal lag than incandescent lamps. A lamp "half waving" can be seen to flicker and possible overloads ever cycle. But a valve filament would still appear constant brightness.

I haven't got a clue as to whether they had greater thermal lags than lamps, but, even if they did, I don't think that's relevant to my point - that, with a diode, they would be subjected to about 41% more peak current than they were 'designed for'. I would have expected that to adversely affect valve life (as I would in the lamp analogy), but maybe I'm wrong.

A series diode was standard practice towards the end of the valve era and valve life did not seem to be adversely affected.

I presume you mean the end of the 'live chassis' era, since some of us were usuing valves long, long after that ... but, yes, I realise that diodes were used. What it did to valve life, I don't know.

A second diode reversed biassed was also fitted across the heater chain after the dropper diode whose purpose was to take out the fuse if the dropper diode failed short circuit.

I don't recall that, but it sounds like a sensible idea.

Kind Regards, John

 

[PBC_1966]

I haven't got a clue as to whether they had greater thermal lags than lamps

Apply 6.3V from your transformer to a 6V flashlight bulb for, say, one second, and what do you see happen? Now apply that 6.3V to the heater of, say, an EF80 for the same period of one second and what do you see?

 

[JohnW2]

I haven't got a clue as to whether they had greater thermal lags than lamps

Apply 6.3V from your transformer to a 6V flashlight bulb for, say, one second, and what do you see happen? Now apply that 6.3V to the heater of, say, an EF80 for the same period of one second and what do you see?

It's a long time since I looked carefully at an EF80, let alone applied voltage to it's filament for one second (a trip to the cellar may be indicated!). However, I suspect what you are talking about the the cathode of the valve, rather than the filament itself, and the cathode clearly will have a much greater thermal mass than the filament of either a valve or a lamp/bulb (and to take time to be heated by the filament), so one would expect it to take longer to start glowing, and to glow for longer after the voltage was removed.

However, as I said, I don't think that thermal mass/lag has got much, if anything, to do with my point about the peak current!

Kind Regards, John

 

[JohnW2]

It's a long time since I looked carefully at an EF80, let alone applied voltage to it's filament for one second (a trip to the cellar may be indicated!).

Trip to cellar, and application of 6.3V to an EF80 heater done

One can see just the upper tip of the cathode, and what one can see takes a couple of seconds to start glowing, and continues to glow for a second or two after disconnection of power, but I can't really convince myself as to whether what I am looking at is just the cathode (with the filament hidden within it) or the filament itself.

Kind Regards, John

 

[PBC_1966]

On most valves you can usually see part of the heater if you look through the envelope at the right angle into the end of the cathode. Another thing to remember is that in many designs the heater is in physical contact with the cathode anyway, so the thermal mass of the cathode is added to that of the filament.

Back on the general topic of DC mains supplies, one other thing I don't think has been mentioned yet is the early arc lamps which continued to be used in many theatres for a surprisingly long time. The early 100/200V DC system still shown as being in service in the Charing Cross area of London in 1966 may well have just been supplying power for the theatres all along and around The Strand by that time.

 

[JohnW2]

On most valves you can usually see part of the heater if you look through the envelope at the right angle into the end of the cathode.

With larger valves, I agree. However, as I said, with an EF80 I really couldn' decide whether or not I was actually seeing the filament, rather than just the cathode. In the second picture, the hotter (yellower) bit towards the bottom of the glowing area could be the filament itself. What do you think? ....

Another thing to remember is that in many designs the heater is in physical contact with the cathode anyway, so the thermal mass of the cathode is added to that of the filament.

That's obviously true, certainly once the cathode starts heating up. Of course, there is only electrical contact between filament and cathode in a 'directly-heated' valve (which can play havoc with some applications, particular audio!).

Kind Regards, John