The continued use of series heaters and droppers was done for cheapness and nothing else. Later sets had a diode and dropper, so the effective Vin to the heater chain was 170v. Cheaper dropper and less heat but certainly can't run it on DC.
Non-standard supply voltages, 1966
- Thread starter PBC_1966
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I'm not sure I understand that. Why should a diode result in less heat (for same supply to the heaters) or prevent use on DC (provided supply was of correct polarity!).
Kind Regards, John
Less heat in the dropper resistors, but I agree it would be fine on DC.
A diode in a series heater chain designed specifically for AC would cause the filaments to run at excessive power levels if you then connected the set to DC of the same voltage, since with AC the diode is letting through only every other half-cycle.
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I need to think more carefully/deeply about this! In terms of my initial thinking, if there was just a diode, then, to deliver the same power to the heaters, the current would have to be roughly double what it would have been without the diode (since the 'VA' would only be present for half the time). If, on the other hand, one also added a large capacitor, thereby getting closer to 'continuous DC', then one would get closer to the 'no diode' situation, with, I would have thought, similar VD in the dropper resistors. Indeed, if the capacitor were large enough, one would end up with DC of a voltage appreciably higher than the RMS voltage of the AC, and hence potentially the need for more VD in the droppers.
As I said, I need to think more carefully - so watch this space!
Kind Regards, John
Yes, if you add a capacitor to the output of sufficient size you end up with near ripple-free DC, which, of course, is what was used for the H.T. in valve sets (usually with an extra series resistor/coil and another smoothing capacitor). But the case Winston is talking about is where the series heater chain was connected with a series diode and dropper resistor across the supply with no reservoir/smoothing capacitors involved.
Indeed - and, as I said, if one does (just) that one presumably needs to run considerably more current through the heaters (if they'll take it!) in order to get the same amount of power into the heaters?
Kind Regards, John
Adding a diode will result in roughly half the rms voltage, so the series dropper resistor has to drop less voltage to reduce the voltage applied to the heaters to the right level. They will then take their design current.
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So a TV designed to run on 240 volt AC would also run on 110 volt DC so what's the problem?
Yes - but, in the absence of a large capacitor, their "design voltage and current" for only half the time (every other half-cycle) - which would presumably mean that the heaters would not get enough power (averaged over whole cycles, or real-world periods of time) to get up to their required operating temperature. ...stick a diode in series with the element of an old-fashioned "one bar electric fire" and the element would probably not even glow.
As far as I can see, the only way one could get the heaters up to operating temperature with an un-smoothed half-wave rectified supply would be to run them at operating voltages (and currents) considerably in excess of normal - if they would tolerate that - so that power averaged over whole cycles would be adequate.
When I have a moment, I need to sit down and do the maths, but I would not be surprised if I find that if one looks at supplying voltage/current for only fractions of a cycle (alternate half cycles or whatever), to get the same (averaged over time) power into the heaters will always involve the same amount of 'lost' power (again, averaged over time) being dissipated in the dropping resistors.
Kind Regards, John
Assuming you're talking about the 'old' ones, in the absence of the diodes we've been discussing, if they worked on 240V AC, the heaters of the valves, and the HT supply, would fall to less than half if one tried to run it on 110V DC. If there were a diode, then it might work for the heaters, but the HT supply would still be very low.
Kind Regards, John
But you need to look at the H.T. side as well as the heater chain.
Yes, the peak voltage of the half-cycles being used to power them will need to be higher than the peak voltage used with un-rectified AC. The idea is to generate the same amount of power in the filaments, and if you're only using half of the sinusoidal waveform to do that, it follows that you need a higher peak voltage (and current) on each half cycle.
With the rectifier, you lose half of the available power from the AC supply without dissipating it as heat as you would need to do to "lose" the same amount of power with a dropper resistor.
Many years ago, before thermostatically controlled soldering irons were in my teeange budget of the time, I made up my own small soldering station with an Antex iron and stand bolted to an ABS project box, on the front of which were two toggle switches and a neon lamp. The second switch was simply wired across a diode placed in series with the live side of the supply so that when I didn't want to use the iron for a few minutes but would need it again soon and didn't want to wait for it to re-heat from almost cold, I could just open the switch to keep it "ticking over" at a lower power. Result: Only every half-cycle actually going toward heating the element, and thus running on about half power (ignoring the tiny voltage drop across the diode and the fact that the element resistance might change slightly with temperature). It would not work with a DC supply, since the iron would still receive full power (minus the small drop across the diode).
That's exactly what I've been saying - and, as I've said, it relies on the filaments being able to cope with that higher peak current.
Indeed so - but, as above, the price you have paid (if nothing else is changed) is that you have also halved the power delivered to the heater filaments. One therefore has to do something to bring the filament power up to what is required and that, in turn, will (subject to my doing the maths!) increase the losses in the dropper resistor.
Yes, many of us (certainly myself) have done things like that but, again, the whole point of such exercises is to decrease the power dissipated in the load, whereas with the valve heater filament, we need to maintain the dissipated power at the level required for it to achieve its operating temperature.
Kind Regards, John
The valves used in these chains are specified for a certain heater current, e.g. in the case of British numbering 300mA for the "P" series or 100mA for the "U" series. Add up the total of the voltages in series for each heater and you then have the total power required for the chain. With a simple dropper resistance, if you must have the current at that specific level and you have only 200-250V AC available as the source, that fixes the total amount of power which must be dissipated and thus determines how much power has to be wasted in the dropper resistor.
But if the total power actually consumed by the filaments is less than half the figure you get by mutliplying the supply voltage by the required heater current in the above, using the diode reduces the losses by removing half of the available power with minimal (near-zero) loss. The dropper then only has to dissipate the difference as wasted power.
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