C-354 choke current calculation.

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C-354 choke current calculation.

Postby 20to20 » Sat Jul 10, 2010 10:57 am

Per the diagram, 435v. - 415v = 20v drop through 56R DC choke = 357ma. If the replacements are rated 200ma., what else has to be factored to make it work?
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Postby mesherm » Sat Jul 10, 2010 1:59 pm

56 ohms just is the straight DC resistance. Thats the lowest the impedance will ever get. The amount of DC ripple current determines the actual impedance.
If you don't already have it then download PSUD2. Its great for designing power supplies and seeing how component changes affect output.


http://www.duncanamps.com/psud2/
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Postby 20to20 » Sat Jul 10, 2010 2:41 pm

mesherm wrote:56 ohms just is the straight DC resistance. Thats the lowest the impedance will ever get. The amount of DC ripple current determines the actual impedance.
If you don't already have it then download PSUD2. Its great for designing power supplies and seeing how component changes affect output.


http://www.duncanamps.com/psud2/


Mike,

Thanks, got it, will install it shortly. Will I also have to add the individual rectifier tube data files or is that built in? Without data files or built-in data do you just supply the pertinant specs manually if known?

To the choke question, I've not actually measured the voltage across the choke to confirm the difference ( I should, I will, I promise!) and wonder if the real world drop is actually 20v. But if it is, I still have this picture in my mind that 435v DC (-) 415v DC is 20v. and the current is higher than spec. Can Inductance on top of DC impede the normal DC current?

To your point about ripple, does that make it dangerous to the choke to have too much Cap. at the rectifier before the the choke? If the filtering is already high, would that effectively reduce the choke to a meer resistor placing it at the mercy of the total circuit draw? (I suppose we wouldn't even use it if the cap. input filter was 100%)

The reason I'm trying to get this resolved is that knowing the choke is red-lined already, I was deciding on how to procede in dividing the output power to each channel and adding a second choke to take the strain off. I think I understand that paralleling the choke is not all there is to it, that would lower the Z. Any branching to split the load would have to include a second cap with the second choke. Yes?

Thanks for the PSUD2 link.
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Postby EWBrown » Sun Jul 11, 2010 6:26 am

The higher voltage (435V) side of the C354 choke, connected to the rectifier cathode, will have a higher 120Hz ripple voltage/current so the choke's 1.5H of inductance will significantly reduce that to a more manageable level.

The choke's DC resistance is about 56 ohms, and its "AC resistance" (reactance) at 120Hz is about 1130 ohms:

Using the formula of XL = 2*PI*F*L or 6.28*120*1.5 = 1130 ohms equivalent

If you were to add a second "branch" to the B+ filtering, just add a second choke, connect one end to the rectifier cathode, and then duplicate the filtering caps and resistors on the "downstream" side of the choke. That would halve the current load for each choke. Of course, if you're using an original sized ST70 chassis, this will be like trying to cram ten pounds of "you know what" into a five pound bag... 9^0

To do this "absolutely" correctly, the 6.8K, 2W resistor should then be replaced with two separate 13K, 2W resistors, and the 22K, 1W resistor should also be replaced with two separate 43K, 1W resistors, that would keep their respective voltage drops correct, for the two individual channels Don't skimp on the resistor power ratings, it's best to have some safety margin.

HTH

/ed B
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Postby 20to20 » Sun Jul 11, 2010 8:19 am

Ed, Thanks
The higher voltage (435V) side of the C354 choke, connected to the rectifier cathode, will have a higher 120Hz ripple voltage/current so the choke's 1.5H of inductance will significantly reduce that to a more manageable level.

The choke's DC resistance is about 56 ohms, and its "AC resistance" (reactance) at 120Hz is about 1130 ohms:

Using the formula of XL = 2*PI*F*L or 6.28*120*1.5 = 1130 ohms equivalent


It's obvious that it's impossible to simply use a DC voltage drop across a choke to establish the current through it, if the schematic voltage prints are correct. It would be interesting to see what the DC would be at the input Cap. if the choke was subbed with a 56R res. I guess it's time to really dig into and learn how to use PSUD, Mike turned me onto. I started building this circuit in the program but kept getting messages about my voltages going negative at 5 cycles... and couldn't correct that with value changes. Must start over, there.

To do this "absolutely" correctly, the 6.8K, 2W resistor should then be replaced with two separate 13K, 2W resistors, and the 22K, 1W resistor should also be replaced with two separate 43K, 1W resistors, that would keep their respective voltage drops correct, for the two individual channels Don't skimp on the resistor power ratings, it's best to have some safety margin.


I was hoping to draw up a solution that could let each branch share the PC board load puting it back in series, since it's so much smaller, in order to simplify this "operation." I'll be starting with a stripped chassis and I'll add the second choke under the other OPT then move the biasing components. I can see needing to add a second equal value cap after the choke, to have two equal LC filters, and send that power to the OP tubes, but thought I could rejoin the supply from each branch back to the PC board circuit.

Thanks Much!
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Postby EWBrown » Sun Jul 11, 2010 2:28 pm

The input capacitor (before the choke) should not be increased in value, as that can damage the rectifier tube. SS rectifiers can work into huge input caps, but tube rectifiers cannot. Good 5AR5/GZ34s are expensive, and stay away from the J/J and Chinese "Shuguang" or "Sino" made 5AR4s, they generally aren't quite up to full ST-70 current handling capabilities.

There is a neat "trick" to effectively raise the rectifier tube PIV ratings, and that is to connect an SS rectifier in series with each plate (cathode to the tubes plates), what that does is to effectively increase the Peak Inverse Voltage rating for the tube rectifier, but it will not increase the current handling capability.

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Postby 20to20 » Sun Jul 11, 2010 3:45 pm

EWBrown wrote:The input capacitor (before the choke) should not be increased in value, as that can damage the rectifier tube. SS rectifiers can work into huge input caps, but tube rectifiers cannot. Good 5AR5/GZ34s are expensive, and stay away from the J/J and Chinese "Shuguang" or "Sino" made 5AR4s, they generally aren't quite up to full ST-70 current handling capabilities.

There is a neat "trick" to effectively raise the rectifier tube PIV ratings, and that is to connect an SS rectifier in series with each plate (cathode to the tubes plates), what that does is to effectively increase the Peak Inverse Voltage rating for the tube rectifier, but it will not increase the current handling capability.

/ed B


I really don't want to modify the the PS any more than necessary to take the strain off the choke. I'd like to stay loyal to the original circuit AMAP. Since a 1.5H 300ma choke is about impossible to find, or any specs higher that would fit under the chassis, I believe a second C-354 is the way to go and since that would double the effeciveness, I thought I'd swap the 30uf down to 20uf and make the output cap 30uf. Not sure at this point if 3H + 20uf is equivalent to the stock. More Aljabya to figur'.
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Postby 20to20 » Mon Jul 12, 2010 8:29 am

If you were to add a second "branch" to the B+ filtering, just add a second choke, connect one end to the rectifier cathode, and then duplicate the filtering caps and resistors on the "downstream" side of the choke. That would halve the current load for each choke. Of course, if you're using an original sized ST70 chassis, this will be like trying to cram ten pounds of "you know what" into a five pound bag... 9^0


PSUD seems to be working OK. The high current swing during 5 cycles of start-up it's warning me about is probably unavoidable. So it looks like splitting the circuit and keeping C1 @ 30uf is best and just adding the choke and a second 20uf, output each branch through a 13K R then rejoining the PC board is the least intrusive circuit mod. The extra 20uf. cap will be easier to position, too. This scheme seems to provide slightly cleaner B+, too.

What I'd eventually like to establish is the actual operational current through the choke when the output is driving into AB and two tubes are in cutoff instead of just using steady state idle current to estimate choke current. Is it substantially different? I suppose a quick and dirty, back of the envelope, answer would be somewhere between 100ma and 200ma. and closer to the middle of that, depending on how close to max power you get. How close to class B does a '70 get @max power, short of clipping?
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Postby dcgillespie » Mon Jul 12, 2010 5:58 pm

Assuming that your ST-70 is biased and loaded correctly as Hafler intended, and each amplifier is in fact developing 35 watts RMS into an external load, then each output stage would be pulling about 200.4 ma from the power supply to produce that power output. To that of course would be added the current draw of the AF amplifier and phase splitter stages. Because the stock supply cannot develop that much current without a significant B+ voltage drop when both channels are driven, the unit realisticly only produces about 30 watts RMS per channel from the stock supply under a worst case scenario -- such as at 20 Hz with both channels driven.

Additionally, the ST-70 is typical of so many class AB1 designs, where in it produces relatively little of it's power output in true class A operation. Again, assuming correct bias and loading conditions exist, each channel of the ST-70 will commence class B operation at about 4.36 watts RMS power output.

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Postby 20to20 » Mon Jul 12, 2010 9:32 pm

dcgillespie wrote:Assuming that your ST-70 is biased and loaded correctly as Hafler intended, and each amplifier is in fact developing 35 watts RMS into an external load, then each output stage would be pulling about 200.4 ma from the power supply to produce that power output. To that of course would be added the current draw of the AF amplifier and phase splitter stages. Because the stock supply cannot develop that much current without a significant B+ voltage drop when both channels are driven, the unit realisticly only produces about 30 watts RMS per channel from the stock supply under a worst case scenario -- such as at 20 Hz with both channels driven.

Additionally, the ST-70 is typical of so many class AB1 designs, where in it produces relatively little of it's power output in true class A operation. Again, assuming correct bias and loading conditions exist, each channel of the ST-70 will commence class B operation at about 4.36 watts RMS power output.

Dave

then each output stage would be pulling about 200.4 ma from the power supply


Did you mean 100ma. per channel?

The amp is rated for 80W peak for each channel and 35W continuous with only 1 db down @ 20hz. How many watts is 1 db worth?

Assuming all the voltages are as the design calls for and biasing is set as suggested to 50ma, what negative G1 voltage is required to trigger cutoff in a 6CA7/EL34?
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Postby dcgillespie » Mon Jul 12, 2010 10:33 pm

It would be about 200.4 ma per channel under the conditions where I said "and each amplifier is in fact developing 35 watts RMS into an external load". 100 ma per channel is the correct static current under quiescent (no signal) conditions.

Power output in watts at 1db down from a referenced maximum (35 in this case), is for all intents and purposes 78% of the reference, or in this case, 27.3 watts.

In the ST-70's UL output stage, cut-off in each output tube roughly takes place at around -43 vdc with the application of signal, and quiescent B+ levels maintained. This will happen sooner as B+ drops from it's static value.

Note that the amplifier is rated as 35 watts continuous without regard to frequency, although as a high quality audio amplifier, this is understood to be 20-20 kHz. The 1db down rating refers to the power bandwidth -- or the band of frequencies the amplifier can produce within 1 db of rated power, at no more than 1% THD.

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Postby 20to20 » Tue Jul 13, 2010 6:39 am

Dave, I really appreciate your replies. I'm trying to refresh all this basic theory from studies 25 years ago.

It would be about 200.4 ma per channel under the conditions where I said "and each amplifier is in fact developing 35 watts RMS into an external load". 100 ma per channel is the correct static current under quiescent (no signal) conditions.


So this estimate on current supports my figure of @350ma through the choke based on the diagram's print of 20v DC drop across it. My original question was how this could be with a choke rated for 200ma. Your figures are @400ma. Is this additional current supplied to the tubes from the C2 filter alone, and the choke only freshes the filter at @ 200ma?

Power output in watts at 1db down from a referenced maximum (35 in this case), is for all intents and purposes 78% of the reference, or in this case, 27.3 watts.


That makes sense for 1db. Since the log scale for db really takes a bite out power for 3db.

In the ST-70's UL output stage, cut-off in each output tube roughly takes place at around -43 vdc with the application of signal, and quiescent B+ levels maintained. This will happen sooner as B+ drops from it's static value.


I found a data sheet that I think I am interpreting correctly, that shows 0 Ia at about (-)58v with 400v. Va. Does cutoff occur prior to 0ma Ia at some "for all practical purposes" level?

The 1db down rating refers to the power bandwidth -- or the band of frequencies the amplifier can produce within 1 db of rated power, at no more than 1% THD.


That's my understanding.

So, can I go wrong by splitting the supply to the channels with the mod I'm considering?

Thanks Much!
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Postby dcgillespie » Tue Jul 13, 2010 8:24 am

The published voltage figures are values taken at quiescent conditions. From your calculations, there is obviously an error made somewhere in these figures, as the static draw of the amplifier at quiescent conditions is about 215 ma. With a dc winding resistance of about 60 ohms, this means that the drop across the choke at quiescent is about 12.5-13 volts. The "error" in the published voltage across the choke is likely due to the fact that often, it was common that non-critical voltages were rounded in different ways, so that 416 volts becomes 420, and 403 becomes 400 -- so there's your "20 volt drop". The difference is due to the fact that these are general numbers and were typically supplied for basic trouble shooting use only -- not so much for circuit analysis.

As for the current capability of the choke -- no doubt the choke is significantly underrated from an absolute current need standpoint. But consider the design economies that were used for the original price point. It's inclusion was basically to minimize PS hum under low level conditions. Under those conditions, it is fairly effective. Now obviously under full power conditions, it saturates under the heavy current draw and basically becomes a 60 ohm resistor. But who would hear the increased hum created under that condition anyway? At full power, the choke does pass the full current draw -- it just becomes very ineffective at it's job.

Your effort to use two chokes would have two very positive effects. First, it would halve the voltage drop created across a single choke. That means that there is more B+ available, and therefore more power output possible. Secondly, because each choke would remain effective even at full power output, hum modulation of the B+ would be reduced, which would reduce the THD created under full power output conditions. It is a worthwhile improvement.

As for the cut-off voltage of the EL34s, you can get all caught up in the definition of what that value absolutely is, but for our purposes, cut-off is defined as that point where a given tube is no longer drawing any useful current to produce usable power output. For class B operation, this is all happening on the most non-linear portion of the tube's characteristic curves, so approximations rule the day. But based on transfer curves for this tube, -43 volts is a good approximation of where that happens in this setting. As a general rule of thumb, for all practical purposes, class B operation commences when the current draw for one tube starts to exceed the quiescent current draw for both tubes together. By definition then, if the total current draw from the PS is unchanged from quiescent conditions, but the current is now all flowing through one tube, then cut-off has been achieved in the opposite tube, and class B operation has commenced.

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Postby 20to20 » Tue Jul 13, 2010 9:36 am

The published voltage figures are values taken at quiescent conditions. From your calculations, there is obviously an error made somewhere in these figures, as the static draw of the amplifier at quiescent conditions is about 215 ma. With a dc winding resistance of about 60 ohms, this means that the drop across the choke at quiescent is about 12.5-13 volts. The "error" in the published voltage across the choke is likely due to the fact that often, it was common that non-critical voltages were rounded in different ways, so that 416 volts becomes 420, and 403 becomes 400 -- so there's your "20 volt drop". The difference is due to the fact that these are general numbers and were typically supplied for basic trouble shooting use only -- not so much for circuit analysis.


What's most scary about the choke at first read on the ratings is the 400v. to frame max. It seems uncharacteristic of the designers to have this component pushed so close its max current rating and have such a low voltage rating. That said, typically, as with caps, do published choke ratings reflect a working rating and are generally rated at least 50% higher for current/wattage? I have yet to do a touch check for temps on the choke. The wax drip on the bottom cover, after so many years, is not a reliable clue, alone, I'd say.

Thanks, once again...
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