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In Reply to: RE: shunt regulated upgraded ParamourII 45 project starts today posted by Doc B. on May 22, 2007 at 10:04:17
I have a question in general. Is it really necessary to burn off the same current in the shunt as the output stage in a Class A SE parafeed stage? For example, for a stage biased at 55ma, with a plate choke, would a ccs to feed the shunt regulator set 75ma work? Or would you need to set it 110ma? I have seen many state that the shunt should shunt the same amount of current as the load, but never a reason why for a class a stage.
Thanks,
Chris
The "rule of thumb" comes about from series feed power stages, where all of the signal current flows through the power supply. The peak signal current, normally supplied from the last capacitor of the power supply, must then come from the shunt regulator shunting less current - essentially, the output stage and shunt regulator act like a class-A push-pull stage, swapping current back and forth.
For all circuits, the shunt regulator must be able to supply the peak current demanded from the circuit being regulated. In parafeed, the power supply signal currents are small except at the lowest frequencies. For 2 watts output at 30Hz with a 3000 ohm load (the upgrade Paramour output stage) the current through the 40 henry inductor is 14.5mA peak. Assuming this is the worst case, that is the minimum shunt reg current. For driver and preamp circuits, the peak signal current is usually no more than 20% of the quiescent current.
Of course there is always some virtue in operating circuits (such as shunt regulators) with some current overhead. So we don't really know how well this low current reg/parafeed combo will work - hence the experiment. Based on other experiments with shunt regulators over the years, we expect great things - but listening will tell.
Thanks for the explaination. Would you be so kind as to share the formula for determining the current swing through the choke?
Thanks,
Chris
Sure, but you probably already know them!
Voltage at the plate of the tube comes from:
** Power equals voltage squared over R
R is the OPT impedance, 3000 ohms in this case. Power is a bit less than two watts, so voltage is the square root of 2*3000 or 77.5 volts. That is RMS voltage, so multiply be 1.414 to get peak plate voltage of 109.5 volts.
Since the plate choke is at constant voltage at the top, and plate voltage below, apply Ohm's law to get the current in the choke:
Current equals voltage over impedance
Impedance is the inductive reactance, 2 * pi * frequency * inductance. The BH-6 plate choke is 40 henries, so at 30Hz this is 7540 ohms, and peak current is 109.5/7540 or 14.5mA.
Thanks again. You were right, I did know them. It helps to know how exactly to apply them though.
Chris
That idea of equal current levels is just a rule of thumb. We are shunting what is left of the available power transformer current capacity beyond what the driver and output stages draw, about 13 mA. The 45 is drawing about 36 mA and the driver a little under 4mA. And of course there is a C4S feeding the shunt reg and the driver and the output stages, and a C4S loading the driver stage as in the stock ParamourII. So there is a little more current drawn by the bias circuits for those active loads.
.
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