I decided to try a couple, very simple, experiments and take a few measurments just so I could boil it all down and try to digest it properly.
It is all based on the mathematical (fact/principle) that on a 120vac line, a 1/2 wave rectifier will produce an 'effective' equivalent of 84vac.... Right?
Ok I got that.. and I believe it.
But after putting in the rectifier I had the gnawing feeling that the 'effective' results must be lower only because the radio seemed to have dimmer lamps and lower volume than it did with the ballast tube in.
1.)
So I first measured everything with the ballast tube back in place as normal:
Of course the radio volume sounded normal and acceptable.
The AC voltage is right around 84vac measured right after the ballast tube at the begining of the series filament string.
AND...the current through the heaters is measuring 330ma as measured on the AC-amp scale of my digital "Extech true RMS" multimeter.
( doing a little ohms law that computes to 254 ohms as the actual load resistance after the heaters are hot)
2.)
Then I removed the ballast tube and replaced it with an adjustable DC power supply adjusted to 84 VDC.
The radio volume seemed to sound just the same as with the ballast tube.
So I measured the current through the heaters again on my DC-amp scale and it also read 330ma.
( so far so good)
3.)
I placed the 1/2 wave rectifier diode in place of the ballast tube... (in theory this should be producing plusating DC at an 'effective' 84vac 'equivalent'... right?
( but the volume still seemed somewhat lower than with the ballast tube ...just as I had noticed before)
(Now..I cannot really measure this voltage properly for all the reasons you guy's pointed out.) .. but I thought that perhaps I can try to derive it.
So, I measured the current ... first on my DC-amp meter as 210ma. ...and using the 254 ohms as the load that calculates to only 53 vdc
So..then I tried, as an experiment, to measure the current on my AC-amp meter and it measures 250ma. Now using the 254 ohm load this calculates to 63vac.
So I don't know what conclusions to draw from all this yet.
My general observations are:
(theory)=Voltage as the electrmotive force is what pushes current through a load.
And for a given fixed load the greater that voltage the greater the current through the load will be... right?
Now:
If the 1/2 wave rectifier were really producing 84v it would be able to "push" 330ma of current through that 254 ohm heater-string load... right?
But as it is only "pushing" about 210ma into that load which tells me that it must not actually be anywhere close to 84volts.
It is supposed to be generating 84v AC "equivalent energy".
..But it just doesn't seem to be producing that "84v equivalent" energy compared to EITHER that of the normal 84vAC or that of the 84vDC comparison basis.
By the way, when I try to measure the pulsating 1/2 wave voltage with my 'true rms" multimeter... it actually reads 53v on the DC scale and 63v on the AC scale... exactly as calculated from the current/resitance calculations above.
In the interests of getting good "scientific" test results...along the way...( concerned that the heater string may be a serious inductive load screwing up my measurments) ...I even tried replacing the heater-string load with a purely resistive load of 330 ohms (close to 254 ohms) to see if there were any significant differences in my readings and there were not.
So what do you "learned" guys think is actually going on here?
Did I do the experiment incorrectly?
Is it that I am not properly able to correctly measure the 1/2 wave current this way either?
The only flaw I see in your measurements is that, if the voltage is low for whatever reason, the filament resistance will be lower than what you measured at 84 volts. Your resistor experiment covers this case though.
Did you verify that the input voltage driving the diode is 120 with the diode and radio hooked up and running?
The 84 volts (.707) number assumes that the input waveform is a pure sine wave. If it is not, the number is different. The last time I looked at the line on my scope, it was pretty bad looking. FWIW it seems that it would have to be pretty screwed up to explain this much difference, but ituition is not always very reliable.
::REF: Replacing a ballast tube with a 1/2 wave rectifier:
::(I don't have any real problem because the diode does work "good enough".. but)
::
::I have been musing on all the thoughts and facts you guys have, so patiently, offered and which we've all tossed around in these threads lately.
::
::I decided to try a couple, very simple, experiments and take a few measurments just so I could boil it all down and try to digest it properly.
::
::It is all based on the mathematical (fact/principle) that on a 120vac line, a 1/2 wave rectifier will produce an 'effective' equivalent of 84vac.... Right?
::Ok I got that.. and I believe it.
::
::But after putting in the rectifier I had the gnawing feeling that the 'effective' results must be lower only because the radio seemed to have dimmer lamps and lower volume than it did with the ballast tube in.
::
::1.)
::So I first measured everything with the ballast tube back in place as normal:
::Of course the radio volume sounded normal and acceptable.
::The AC voltage is right around 84vac measured right after the ballast tube at the begining of the series filament string.
::AND...the current through the heaters is measuring 330ma as measured on the AC-amp scale of my digital "Extech true RMS" multimeter.
:: ( doing a little ohms law that computes to 254 ohms as the actual load resistance after the heaters are hot)
::
::2.)
::Then I removed the ballast tube and replaced it with an adjustable DC power supply adjusted to 84 VDC.
::The radio volume seemed to sound just the same as with the ballast tube.
::So I measured the current through the heaters again on my DC-amp scale and it also read 330ma.
::
::( so far so good)
::
::3.)
::I placed the 1/2 wave rectifier diode in place of the ballast tube... (in theory this should be producing plusating DC at an 'effective' 84vac 'equivalent'... right?
::( but the volume still seemed somewhat lower than with the ballast tube ...just as I had noticed before)
::(Now..I cannot really measure this voltage properly for all the reasons you guy's pointed out.) .. but I thought that perhaps I can try to derive it.
::So, I measured the current ... first on my DC-amp meter as 210ma. ...and using the 254 ohms as the load that calculates to only 53 vdc
::So..then I tried, as an experiment, to measure the current on my AC-amp meter and it measures 250ma. Now using the 254 ohm load this calculates to 63vac.
::
::So I don't know what conclusions to draw from all this yet.
::My general observations are:
::
::(theory)=Voltage as the electrmotive force is what pushes current through a load.
::And for a given fixed load the greater that voltage the greater the current through the load will be... right?
::
::Now:
::If the 1/2 wave rectifier were really producing 84v it would be able to "push" 330ma of current through that 254 ohm heater-string load... right?
::But as it is only "pushing" about 210ma into that load which tells me that it must not actually be anywhere close to 84volts.
::
::It is supposed to be generating 84v AC "equivalent energy".
::..But it just doesn't seem to be producing that "84v equivalent" energy compared to EITHER that of the normal 84vAC or that of the 84vDC comparison basis.
::
::By the way, when I try to measure the pulsating 1/2 wave voltage with my 'true rms" multimeter... it actually reads 53v on the DC scale and 63v on the AC scale... exactly as calculated from the current/resitance calculations above.
::
::In the interests of getting good "scientific" test results...along the way...( concerned that the heater string may be a serious inductive load screwing up my measurments) ...I even tried replacing the heater-string load with a purely resistive load of 330 ohms (close to 254 ohms) to see if there were any significant differences in my readings and there were not.
::
::So what do you "learned" guys think is actually going on here?
::Did I do the experiment incorrectly?
::Is it that I am not properly able to correctly measure the 1/2 wave current this way either?
::
::
::
::
::
::
Without a diode, power = V^2/R. With the diode, power delivered to the resistor will be halved. The only way for that to happen is for the rms voltage to be 0.7V: (0.7V)^2 = 0.5V^2. You can also use calculus to compute the average value of voltage squared over a a whole cycle (1/60 sec).
I can see one problem with the experiment. Your "true rms" meter will not give the correct reading, amps or volts, when measuring an asymmetric wave form, which a half-wave output from a diode is. Except for a very sophisticated DMM, the meter will not be able to tell whether the half-wave is 100% above zero or 50-50, or whatever; and it makes a difference for computing rms. Therefore, the reading will be low. I have spoken to an engineer at Fluke about all this, since calling a meter "true rms" is a little misleading.
So what is a "true rms" meter good for? Really only when measuring voltages or currents that include harmonics. All the harmonics will add up to a symmetrical waveform. A square wave or triangular wave that is symmetric about the zero axis will be measured correctly by a true rms meter.
Now, I can't explain why the filaments are dimmer with the diode than with the ballast tube. Tony suggested that the forward drop through the diode is cutting the voltage significantly. The forward drop through a 1N400x silicon diode will only be about 1.0V at rated current (1A); and your peak current with the diode should be less than 1A.
Here is a theory: perhaps with your half-wave supply, the heaters are flickering, and their temperature is dropping off slightly during half of the cycle. Weird.
Your analysis of the .707 thing assumes that the signal is symmetrical in amplitude and time so that the energy in the upper and lower parts of the signal are the same. Then when you half wave rectify it, the power will drop in half. If there is any asymmetry in the signal then the answer is different.
Tony
:Here, for everybody's info, is the math behind why a diode cuts rms voltage to 70% of input rms voltage:
:
:Without a diode, power = V^2/R. With the diode, power delivered to the resistor will be halved. The only way for that to happen is for the rms voltage to be 0.7V: (0.7V)^2 = 0.5V^2. You can also use calculus to compute the average value of voltage squared over a a whole cycle (1/60 sec).
:
:I can see one problem with the experiment. Your "true rms" meter will not give the correct reading, amps or volts, when measuring an asymmetric wave form, which a half-wave output from a diode is. Except for a very sophisticated DMM, the meter will not be able to tell whether the half-wave is 100% above zero or 50-50, or whatever; and it makes a difference for computing rms. Therefore, the reading will be low. I have spoken to an engineer at Fluke about all this, since calling a meter "true rms" is a little misleading.
:
:So what is a "true rms" meter good for? Really only when measuring voltages or currents that include harmonics. All the harmonics will add up to a symmetrical waveform. A square wave or triangular wave that is symmetric about the zero axis will be measured correctly by a true rms meter.
:
:Now, I can't explain why the filaments are dimmer with the diode than with the ballast tube. Tony suggested that the forward drop through the diode is cutting the voltage significantly. The forward drop through a 1N400x silicon diode will only be about 1.0V at rated current (1A); and your peak current with the diode should be less than 1A.
:
:Here is a theory: perhaps with your half-wave supply, the heaters are flickering, and their temperature is dropping off slightly during half of the cycle. Weird.
Now if you take 120 volts AC and rectify it with half wave rectification, and you don't filter it, you are effectively cutting the total reaction time in half (the time the power reacts on any one device. If you don't filter this half wave rectified current, you'll effectively have an apparent voltage of 60 volts (perhaps even less)--half of 120, if this voltage has a load on it. There may be some peaks around 84 volts, but they are too small to ever react on the tube heaters, just as the peaks in AC voltage are too small to actually drive light bulbs up to 160 volts (otherwise light bulbs which are sold rated at 120 volts would burn out). This is why the tubes are dim in your half wave rectifier circuit. The only way you can brighten them is by installing an electrolytic condenser across the tube heaters so as to filter the voltage which is being supplied to them. Larger condensers will retain more of each half wave, and will cause the tubes to be brighter. Select a value accordingly.
It does not matter what theoretical voltage may be available from an unfiltered half wave rectifier. Tube heaters can't react that fast. All that matters is the apparent voltage. You get that apparent voltage by measuring across the tube heaters while they are being supplied by the half wave rectifier.
If the voltage still doesn't come up even after installing a large electrolytic, then you may have a high resistance diode. This would be obvious because it would get rather warm with such a load imposed upon it.
Anyone who has ever worked with tube radios (and we all have) knows that when the electrolytics fail in a radio (not from shorting, but from drying up), the voltages drop quite low. About half of the radios available to us from this era utilize half wave rectification. Experience from such radios will give you an idea as to what is actually going to happen in a loaded half wave rectified circuit.
I'm sure that some of what I've written in here may not make perfect sense. I tried, though I can't always pull everything out of my brain accurately. All I can tell you is to try this and see what happens. Wire a diode in series with the filaments. Observe polarity across the filament string (after the diode, not across the AC line). Install an appropriate electrolytic across the filament string (not the AC line) observing correct polarity. I don't exactly expect everyone to use this method of ballast replacement, but as long as we're going to discuss it, we may as well get it working correctly. It can.
Thomas
The Peak voltage of the 120 volt line is 170 volts. The peak voltage of the half (or full) wave rectified line is still 170 volts (minus the diode drop). You are correct in your description of what happens when you half wave rectify it. The negative half of the cycle is removed, chopped off. And you also describe correctly what will happen if you add a capacitor.
My only problem remains with the 84 volts. That was supposed to be without a capacitor and Dougs argument about the power being cut in half makes sense to me. But this leads to an RMS voltage of 84 which seems to be wrong.
The half-wave pulse will be rich in high-frequency harmonics, all off which must add together to produce the predicted rms current thru the string.
However, the inductive impedance for the higher frequency components will be much higher than for a pure 60-Hz waveform, which is what you get with a ballast resistor.
Just a hunch.
Thomas
:My hunch is: INSTALL AN ELECTROLYTIC ACROSS THE FILAMENTS.
:
:Thomas
I'd be curious about any results you get from the same tests though.
I had the 1/2 wave diode (84v) feeding my string of filaments at ~300ma and I (then believing that the voltage might actually be low)..added a 15uf 160vdc cap across the filament string which boosed it nicely to (80vdc as read on my multimeter) however after a few hours of running the cap got blisteringly hot and started to swell on the top.
I looked at the waveform and saw 120v ripple riding on about 20-30vdc base. It didn't seem, from the scope waveform, that the ripple was being smoothed out very much if at all!!
Well I removed that cap and found a higher voltage rating one. This one was 22uf at 200vdc.
That kicked up the DC even more (~90vdc) and the waveform still showed big 120v ripple now riding on about 50vdc base. Well this cap was getting hot too... so I was afraid to leave it in.
So I got an even higher rated cap... a 22uf @450vdc...
Now it wasn't getting so warm as quickly... but not really staying cool either. And the waveform still looked bad. ~120v ripple riding on 50vdc base line.
So I decided to do a bench test.
I got a 330 ohm wire-wound 20 watt resistor and fed it with the 1/2 wave rectifier.(84v...right?)
It drew about 250ma as expected.
Nice clean 1/2 wave pulses riding on zero dc-volt base line.
Ok.... so then I put the 22uf 200vdc cap across it and the same thing... getting hot very fast and the waveform was 120v ripple riding on a base line of about 50vdc. Again I used an even higher voltge 450vdc cap... but it didn't seem like that would stay cool very long either... and that high ripple was killing me wondering why.
So I decided to try to understand a little bit why.
So I reduced the load greatly.
I went back to a 15uf 160vdc cap but this time I tried a small 4watt christmas-tree type lightbulb load instead of that 330 ohm resistor.
Wow! What a difference... the waveform smoothed right out to a very small ripple and the capacitor stayed nice and cool.
So I started increasing the load again and sure enough the cap started getting hot and the ripple got worse.
So I started increasing the capcitance to like 200uf @ 160vdc... and THEN it started to get better... the ripple was reduced a bunch and the cap stayed cool. But of course NOW the voltage was way up there in outer space... like 140-160vdc or something....
Remember all I originally wanted to do (thinking at that time that I wasn't really getting 84volts) was to boost that 1/2 wave diode output a little bit. Like maybe 15-20 volts or so. But the smaller uf-value caps get too hot and the larger-uf caps boost the voltage much too high.
So...I don't fully understand why the caps get so hot.
My guess is that high ripple riding on the 40 or so vdc base line all adds up to too much voltage for the dielectric to handle... ??
What do you think?
Please set up that simple bench test and let me know.
Peter
So
The problem seems to be that the 84 volts thing is wrong. The true RMS meter seems to be reading correctly. Matlab definitely knows where zero volts is.
Now I just have to figure out why the 84 volts thing is wrong. Dougs argument sounds correct to me. This must be a trick question :).
Put in an electrolytic. You'll have bright filaments and you can listen to your radio.
Thomas
The fact that the peaks have little effect is exactly what RMS is trying to address. The line is 170 peak, but we don't call it that. We call it 120 which is RMS. The question seemed to revolve around what the RMS value of a half wave rectified line was.
Adding the cap will bring it up, so, if thats all you want to do, then go for it. I was just trying to understand what was going on so that I would get it right next time. For me, that means making the theory and the practice agree.
Tony
http://www.tpub.com/content/neets/14179/css/14179_175.htm
http://www.eece.ksu.edu/~starret/589/man/E.html
The correct RMS voltage is 84. Doug is correct.
The meter is probably lying to you.
Tony
:Sorry all, but all this RMS stuff is not just theory. I think I found the problem. I have Matlab, a mathematics program that we use here at work. I gave it the 170 volt peak sine wave and asked it to calculate the RMS value. It gives back 120. Then I half wave rectified it by zeroing out the part of the signal below zero and asked it for the RMS value of that. It gave back 66 (Actually 65.something).
:
:The problem seems to be that the 84 volts thing is wrong. The true RMS meter seems to be reading correctly. Matlab definitely knows where zero volts is.
:
:Now I just have to figure out why the 84 volts thing is wrong. Dougs argument sounds correct to me. This must be a trick question :).
:
:
I've suggested this advertising slogan to Fluke: "If it works, it must be a Fluke."
:Never mind. I made the same mistake the meter is making. Reading the "help" for the RMS function says it is only valid for a symmetrical waveform. You have to do it differently for the general waveform.
:
:The correct RMS voltage is 84. Doug is correct.
I just "assumed" that it was a Matlab supplied function.
My mistake.
:Tony, that's rich. Now I'm contemplating a class-action suit against both Matlab and Fluke! :>)
:
:I've suggested this advertising slogan to Fluke: "If it works, it must be a Fluke."
:
::Never mind. I made the same mistake the meter is making. Reading the "help" for the RMS function says it is only valid for a symmetrical waveform. You have to do it differently for the general waveform.
::
::The correct RMS voltage is 84. Doug is correct.
:
Ok you guys!... You are all great!! I've been reading every single word from all of you...thanks.
I think we can all rest a little easier after I explain the following:
I got to thinking that all the scientists and mathematicians can't be wrong... so it's probably just my metering thing and that you probably can't accurately measure the current like this either.
So..to determine "EXACTLY" what is going on here I decided to "go to the butcher" and get the meat!!!
The only real way to determine "effective" results is to measure that actual work it can perform... RIGHT?
Aha.. of course that's right!... and what is that? ...Watts or HEAT right???
So I took a 1000 ohm 7watt wire-wound resistor and taped it down to my bench. Then I taped my digital temperature probe to it.
Then I performed a comparison test.
I used 84vDC and waited for the resistor to max out at 333 degrees F
I measured the AC line voltage as 119vAC going into the recitfier.
Then I used the output of the 1/2 wave rectifier into the same resistor and watched the temp as it maxed out at 323 degrees F
Wow!!! very close, only TEN degrees less!
So I then repeated this test about 5 times both ways for repeatability. Fine.
I used a 1 amp diode and a 3 amp diode and even a 6 amper...All produced the same results.
So 323 degrees is about 97% of 333 degrees. That's pretty good to prove they are VERY close.
(Close enough for government work!!... as they say)
So if I "perceived" that the volume on my radio was somewhat lower using the rectifier... it must NOT have been very much indeed.
I think this test proves that I simply cannot use these meters as test tools to measure the cuurent anymore than I can the voltage of the 1/2 wave signal.
AND... It restores our confidence in Joe Ohm or Larry Watt or Fig Newton or whoever did these original calculations ... No?
Nice work.
So, why doesn't the diode work the same as the ballast?
I am assuming that the ballast is only connected to the filament string (and the pilot lamp) right?
Tony
:
:"IF YOU WANT MEAT... GO TO THE BUTCHER"
:
:Ok you guys!... You are all great!! I've been reading every single word from all of you...thanks.
:
:I think we can all rest a little easier after I explain the following:
:
:I got to thinking that all the scientists and mathematicians can't be wrong... so it's probably just my metering thing and that you probably can't accurately measure the current like this either.
:
:So..to determine "EXACTLY" what is going on here I decided to "go to the butcher" and get the meat!!!
:
:The only real way to determine "effective" results is to measure that actual work it can perform... RIGHT?
:
:Aha.. of course that's right!... and what is that? ...Watts or HEAT right???
:
:So I took a 1000 ohm 7watt wire-wound resistor and taped it down to my bench. Then I taped my digital temperature probe to it.
:
:Then I performed a comparison test.
:I used 84vDC and waited for the resistor to max out at 333 degrees F
:
:I measured the AC line voltage as 119vAC going into the recitfier.
:Then I used the output of the 1/2 wave rectifier into the same resistor and watched the temp as it maxed out at 323 degrees F
:
:Wow!!! very close, only TEN degrees less!
:
:So I then repeated this test about 5 times both ways for repeatability. Fine.
:I used a 1 amp diode and a 3 amp diode and even a 6 amper...All produced the same results.
:
:So 323 degrees is about 97% of 333 degrees. That's pretty good to prove they are VERY close.
:
:(Close enough for government work!!... as they say)
:
:So if I "perceived" that the volume on my radio was somewhat lower using the rectifier... it must NOT have been very much indeed.
:
:I think this test proves that I simply cannot use these meters as test tools to measure the cuurent anymore than I can the voltage of the 1/2 wave signal.
:
:AND... It restores our confidence in Joe Ohm or Larry Watt or Fig Newton or whoever did these original calculations ... No?
:
:
:
:
:
I think it must work the same Tony.
I'm now seeing that my "perception" of it not being as loud must have been flawed. That's all I can think now when faced with the evidence... lol
Because ...the proof was not only "in the pudding" but in the temperature of the pudding!
From the 1N400x datasheet, the diode drop is 1V at full load (1A); for a silicon diode, it would be 0.7V at minimal load. Let's assume that it's, on average for this case, a 0.85V drop through the diode.
So, if 120V ac through an ideal diode (no drop) puts out a theoretical 85V rms, then with the diode drop let's call it 85 - 0.85 = 84.15V rms. So the proportional difference in power, which is related to the square of the ratio of voltages, will be:
(85/84.15)^2 = (1.0101)^2 = 1.02
So, the diode drop will account for about a 2% drop in power delivered to the heater string.
Peter's observation of a 3% difference in temperature of the resistor isn't directly related to a corresponding difference in power disipated. This is because the heat transfer from the resistor will be related the temp differential between the resistor and ambient air (conduction and convection) and to the resistor's absolute temp (radiation).
The most accurate way to measure the power disipated by the test resistor would be to run a calorimetric experiment: dunk the resistor in an insulated cup of water, and measure the time to heat the bath temp by a certain number of degrees.
What Peter's experiment seems to show is that what we observe is affected by what we expect to observe. And, in this case, it's not helped by DMMs that are labeled "true RMS," but don't tell you that they measure true rms only under special conditions.
Onward and upward.
I'd be curious about any results you get from the same tests though.
I had the 1/2 wave diode (84v) feeding my string of filaments at ~300ma and I (then believing that the voltage might actually be low)..added a 15uf 160vdc cap across the filament string which boosed it nicely to (80vdc as read on my multimeter) however after a few hours of running the cap got blisteringly hot and started to swell on the top.
I looked at the waveform and saw 120v ripple riding on about 20-30vdc base. It didn't seem, from the scope waveform, that the ripple was being smoothed out very much if at all!!
Well I removed that cap and found a higher voltage rating one. This one was 22uf at 200vdc.
That kicked up the DC even more (~90vdc) and the waveform still showed big 120v ripple now riding on about 50vdc base. Well this cap was getting hot too... so I was afraid to leave it in.
So I got an even higher rated cap... a 22uf @450vdc...
Now it wasn't getting so warm as quickly... but not really staying cool either. And the waveform still looked bad. ~120v ripple riding on 50vdc base line.
So I decided to do a bench test.
I got a 330 ohm wire-wound 20 watt resistor and fed it with the 1/2 wave rectifier.(84v...right?)
It drew about 250ma as expected.
Nice clean 1/2 wave pulses riding on zero dc-volt base line.
Ok.... so then I put the 22uf 200vdc cap across it and the same thing... getting hot very fast and the waveform was 120v ripple riding on a base line of about 50vdc. Again I used an even higher voltge 450vdc cap... but it didn't seem like that would stay cool very long either... and that high ripple was killing me wondering why.
So I decided to try to understand a little bit why.
So I reduced the load greatly.
I went back to a 15uf 160vdc cap but this time I tried a small 4watt christmas-tree type lightbulb load instead of that 330 ohm resistor.
Wow! What a difference... the waveform smoothed right out to a very small ripple and the capacitor stayed nice and cool.
So I started increasing the load again and sure enough the cap started getting hot and the ripple got worse.
So I started increasing the capcitance to like 200uf @ 160vdc... and THEN it started to get better... the ripple was reduced a bunch and the cap stayed cool. But of course NOW the voltage was way up there in outer space... like 140-160vdc or something....
Remember all I originally wanted to do (thinking at that time that I wasn't really getting 84volts) was to boost that 1/2 wave diode output a little bit. Like maybe 15-20 volts or so. But the smaller uf-value caps get too hot and the larger-uf caps boost the voltage much too high.
So...I don't fully understand why the caps get so hot.
My guess is that high ripple riding on the 40 or so vdc base line all adds up to too much voltage for the dielectric to handle... ??
What do you think?
Please set up that simple bench test and let me know.
Peter
The current through a capacitor will depend upon the capacitive impedance. This, in turn, varies with the applied frequency. In fact, the impedance of a capacitor is inversely proportional to the freq.
For example, at 60Hz a cap will conduct more current than at 50Hz.
Now, a half-wave sinusoid coming out of a diode is not 60Hz. Yes, there will be a strong 60Hz component, but there will be mucho higher frequency harmonics. These higher freq components, as well as the 60Hz fundamental, will go through that cap like shelled corn through a goose - the cap will present less and less impedance as the freq increases.
Result - too much current through the cap, and it overheats.
By the way, when you have, say, a 450V electrolytic cap, it's good for 450V DC. Now, if you slap 60Hz, 120Hz...across it, it's no longer good for 450V.
But when you slapped the big load (the filament string) across the cap, then the cap saw a huge ripple and that's when the current thru the cap goes sky high.
Maybe when you use a small condenser on a large load, the condenser is being fully charged, but then nearly fully discharged again for each cycle. When you use a condenser in a radio circuit, the condenser, compared with the load, is rather large. It maintains nearly all of its charge during its entire operation. This is why you get smooth DC. Perhaps a condenser builds up heat each time it is charged, but since it maintains most of its charge in the B circuit of a radio, the heat is probably very small. Since it is constantly being nearly discharged in the filament circuit, maybe it's building up heat. A condenser doesn't actually have current flow "through" it, but current is flowing into and out of it. If the condenser is constantly being fully charged and nearly fully discharged, a lot of current is flowing into and out of it.
Now here's my problem. Sean, the guy with the Airline radio, installed two 10 MFD condensers back to back, and now they don't heat up, but they're still undersized because they only raise the voltage up to that required by the tube filaments, and not up to a full 120 volts. Also, if you install an undersized cap in the B circuit of a radio, it doesn't heat up. You get a hum because the cap isn't maintaining a large enough charge.....but it doesn't heat up.
Your theories sound better. I guess we can still use this idea if we use nonpolarized caps, though.
Thomas
humm.. now I'm curious and I wonder what caps are used say in a good bench supply like a 500ma 150vdc power supply design?
peter
Also, if you install an undersized cap in the B circuit of a radio, it doesn't heat up. You get a hum because the cap isn't maintaining a large enough charge.....but it doesn't heat up.
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:Your theories sound better. I guess we can still use this idea if we use nonpolarized caps, though.
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:Thomas
T.
Electrolytics are used all the time in things like train transformers and other high load applications. They don't usually heat up. You guys have your theories and I have mine, but in reality I don't see why an electrolytic would heat up if it's being fed pure direct current, even if the current is pulsating at 60 cycles a second. I've never tried this idea with tube filaments, though, so perhaps I have something to learn. I've used small capacitance electrolytics in B filter circuits, and they don't heat up even though the radio hums. It just seems odd that they'd heat up for the diode-tube filament set-up. ...And then, how they don't heat up if you put two back to back. Putting them back to back means that they can handle alternating current. There shouldn't be alternating current of the reversing polarity kind after a diode. There certainly isn't after a tube diode. There shouldn't be any after a solid state diode, either.
Well, you all figure it out and let me know what you come up with because it puzzles me.
Thomas