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.
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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
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.
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
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
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
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.
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.
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.
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