Tube Amps / Music Electronics
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|5/26/2004 5:33 AM|
||5Y3GT rectifier - I can't understand the circuit diagram|
Hello all, probably a stupid question, but I'm having trouble understanding the circuit diagram of the power supply. Can someone tell me what voltage each of the secondaries is (i.e. in terms of their positions on the schematic - like 6.3V at the bottom secondary) also it seems to me (I have very little experience with valve rectifiers) that there is a closed loop connected to the filaments of the 5y3gt, so my knowledge tells me that no current should flow - it would help my understanding greatly if someone could explain it to me in solid state terms, i.e. bridge/full wave diode rectifier. (sorry to bring it down to that level!) also what is the output of the rectifier for - is it to provide a high voltage DC supply?
Thankyou in anticipation, sorry for the long post.
Cheers, Josh - Australia
|5/26/2004 8:48 AM|
||Re: 5Y3GT rectifier - I can't understand the circuit dia|
Josh, think of the 5Y3 as two diodes with a common cathode. The usual configuration is a full-wave circuit fed from a centre-tapped transformer secondary winding. The centre-tap is grounded and each half of the secondary feeds a plate of the rectifier tube. The two halves of the rectified voltage are summed at the tube cathode.
The "closed loop" feeding the filaments is called a heater, or filament winding. Tubes need a heater to emit electrons. This can be directly from the filament or sometimes there's an actual separate cathode emitting element around the heater that heats up and releases the electrons. This is called an "indirectly heated cathode".
With most tubes the first digit(s) indicate the required filament/heater (the terms are interchangable but the latter is more descriptive of function). So with a 5Y3 there is a need to apply 5 volts to heat up the cathode. The electrons released are attracted to an anode(plate) whenever it is positively charged and repelled during any part of the cycle where the plate is negative.
So the quick way to understand this tube rectifier is to think of it as two diodes connected common cathode, that need a heater voltage to work.
Tubes need a high voltage plate supply to function properly. This is why tube circuits deal with much higher voltages than transistors. Correspondingly, currents are much lower. This is why when transistors came into common use they quickly were adapted to power circuits much different than that of traditional tube layouts. Quasi-complementary outputs are neither practical nor cost-effective with tubes, since they deal with low voltages and much higher currents than tubes normally withstand.
The extra kicker is that tube rectifiers have a much greater forward voltage drop than solid state devices. This drop is usually in the tens of volts and increases with stronger current demand. This is the "sag" phenomenon referred to by ampheads. When a tube-rectified amp is driven hard the plate voltages drop somewhat on power peaks, adding an extra compression to the sound. This appears as a looser, more "bluesy" sound. Solid state rectifiers have only a trivial sag that would take a university physics lab to measure and could never be heard by the human ear. This makes the amp's dynamic response sound much "tighter" - a faster slew rate with no compression, if you like.
Hope this is useful...
|5/26/2004 10:07 AM|
Thanks Wild Bill, that was some excellent information.
I think I understand now! What I was most confused about was how the high voltage was transferred to the other valves when there didnít seem to be a circuit.
Please correct me if I am wrong, but my understanding is that the high DC voltage is carried to one side of the cathode/heater, there is a potential difference of 5 volts across the cathode due to the wire passing through the 5V secondary winding of the transformer on the other side of the cathode. This means that there is a circuit because there is a high potential difference between the high voltage DC and the centre tap ground.
I think my biggest problem has been trying to analyse it from a conventional current point of view, rather than an electron flow point of view.
Thanks again for your help and as I said, if Iíve got this totally wrong, let me know!!!
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