HK A500 power amp
Hi MR2, I had a similar problem with an A500 that was troublesome because of noise in the preamp section. I really did not want to deal with totally rebuilding it because it has 5 mini tubes and the circuitry of various tone controls, filters, etc, etc. So, I decided to just chuck the whole front end and turn it into a stereo amp. So after cutting all the connections to the preamp I was left with the 12ax7 input/voltage amplifier and the 12au7 phase splitter, plus one other tube for keeping the cathode bias voltage of the power tubes. The A500 uses 3 minitubes to set the 7355 power tubes bias by taking their cathode voltage and having it power the filaments in series. Originally, it was the phono tubes and the preamp tube. But, since I am only keeping the power amp section which only needs a 12ax7 and 12au7 I had to add another tube, in this case I just put a 12at7 in the socket to achieve the current drop needed for the 4 7355 outputs. The rest of the circuit stayed the same. I also removed the front panel and replaced it with a nice piece of Oak. The end result is a very nice sounding PP amp that is super quiet since the front end tubes are DC powered. I just thought you might consider an easy alternative. good luck with your project! 808
Hi MR2, I had a similar problem with an A500 that was troublesome because of noise in the preamp section. I really did not want to deal with totally rebuilding it because it has 5 mini tubes and the circuitry of various tone controls, filters, etc, etc. So, I decided to just chuck the whole front end and turn it into a stereo amp. So after cutting all the connections to the preamp I was left with the 12ax7 input/voltage amplifier and the 12au7 phase splitter, plus one other tube for keeping the cathode bias voltage of the power tubes. The A500 uses 3 minitubes to set the 7355 power tubes bias by taking their cathode voltage and having it power the filaments in series. Originally, it was the phono tubes and the preamp tube. But, since I am only keeping the power amp section which only needs a 12ax7 and 12au7 I had to add another tube, in this case I just put a 12at7 in the socket to achieve the current drop needed for the 4 7355 outputs. The rest of the circuit stayed the same. I also removed the front panel and replaced it with a nice piece of Oak. The end result is a very nice sounding PP amp that is super quiet since the front end tubes are DC powered. I just thought you might consider an easy alternative. good luck with your project! 808
That 8.636V RMS is the signal voltage coming out of the 6DJ8 plate going to the EL84 grid that is necessary to get the amp to full output power to the speaker. It's equal to 12.2V peak, which is also the grid-to-cathode (bias) voltage of the EL84s.
8.646Vrms * 1.414 = 12.2Vpeak
The EL84 can only accept a peak signal swing equal to its grid-to-cathode voltage before the EL84 grid starts to draw significant current. Once that happens, the EL84 can't amplify any more, so full output is reached. (Simplified -- I'm limiting this to the driver being capable of voltage amplification only.)
EDIT: The 47k mentioned is the plate load resistor for the 6DJ8 input stage.
Nice! What is the DC voltage at the 5751 cathodes (pins 3 and 8)?
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You have 333R resistors as the cathode loads for your 5751.
Even with 1.5mA of plate current going through each 5751 triode (which would be a lot), you'd have a grid bias of only 0.5V. With that low of a bias voltage, it's very likely the 5751 will be drawing grid current the whole time. I'd try to get that up to over 1V. (Or maybe the input transformer makes that unnecessary? Comments?)
With 0.5V grid bias, and open loop gain of let's say 50x, your 5751 can only deliver 25V peak maximum to the EL34 grids. That's not enough to drive the EL34s to full power plus extra gain to drive the negative feedback loops. Remember that the tube can only accept peak voltage signal input equal to its grid-to-cathode voltage (actually a bit less in real life, due to grid current effects).
With a 5751 or 12AX7 common cathode stage, you'll usually see a 1.5k or 2.2k resistor used in the cathode. Using a 1.5k cathode resistor, and assuming 0.75mA plate current through the 5751 (ballpark estimate), that would give your 5751 a grid bias of -1.1V. That might be clear of grid current.
That 1.1V would allow up to 1V peak signal to be applied to the 5751 grid. With gain of 50x, that means up to 50V peak output signal can be delivered from the 5751 plate (anode). That covers the 25V to 30V needed to drive the EL34, plus only another 5dB of gain for negative feedback. That's not a lot of gain available for negative feedback.
Usually you want a minimum of 3x the gain needed to drive the outputs to full power before feedback is applied. Then you can reduce the gain of the circuit by 3x using (beneficial) negative feedback. That 3x reduction in gain represents -10dB of negative feedback. That's not a whole lot for a pentode amplifier.
To get that 3x the input voltage needed to drive your EL34s to full output open loop, you'd need the driver to supply at least 75V peak from its plate.
If your 5751 has a gain of 50, then you'll need it to accept an input voltage of 1.5V. That means the grid bias of the 5751 will need to be at least 1.6V (grid to cathode). 2V would be better. Headroom is always a good thing.
It would help if you can take voltages and update your schematic with those.
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Here is the feedback configuration I was talking about earlier, now that the server issues have been resolved. This allows you to easily control the operating point of the first stage relatively independently of the global feedback connections.
Values are about what I would recommend as a starting point with the 5751, you might find the higher mu of the 12AX7A advantageous here as a side note.
Values are about what I would recommend as a starting point with the 5751, you might find the higher mu of the 12AX7A advantageous here as a side note.
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Yes, this book is not very helpful for designing a feedback loop. The most helpful info I found is this one.
Amplifier Compensation.
Thanks everyone for all the help so far. I've attached an updated schematic including the power supply. I know the zener regulator is not the best but I want to try it anyway. I have a couple of Tom Christiansen's older Maida boards but I'm not sure I can solder SMD type capacitors successfully. And I prefer not to use transistors if I can get away with it.
Thanks again, Kevin
Thanks again, Kevin
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You have the feedback inserted in the wrong part of the circuit. The feedback should be inserted at the junction of the 1.5k and 100 ohm resistors. You have it inserted at the junction of the 1.5k resistor and 5751 cathode in the schematic.
Also, isn't 47uF kind of small for the cathode bypass capacitor? I'd think 100uF would be more optimal.
In the power supply, I'm not sure how much ripple reduction you're going to get from a 47R resistor with a 50uF capacitor. I assume the 5751 tubes are drawing about 1mA each, correct? If so, a 1k ohm resistor will only drop 4V if it's feeding the 5751 plates. That would give you a quieter plate supply for the 5751's.
Also in the plate supply for the 5751's, for a B+ of 380V I think you can use a 450VDC rated electrolytic cap. In that case, consider raising the capacitance to 100uF or 220uF. Those values are available in 450VDC rated parts.
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Also, isn't 47uF kind of small for the cathode bypass capacitor? I'd think 100uF would be more optimal.
In the power supply, I'm not sure how much ripple reduction you're going to get from a 47R resistor with a 50uF capacitor. I assume the 5751 tubes are drawing about 1mA each, correct? If so, a 1k ohm resistor will only drop 4V if it's feeding the 5751 plates. That would give you a quieter plate supply for the 5751's.
Also in the plate supply for the 5751's, for a B+ of 380V I think you can use a 450VDC rated electrolytic cap. In that case, consider raising the capacitance to 100uF or 220uF. Those values are available in 450VDC rated parts.
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Not sure where the 220K resistors came from should be 2.2K or less. As Rongon points out the feedback is connected to the wrong place.
Off the cuff 47uF should give a -3dB corner of approx 8Hz considering the thevenin equivalent of the cathode resistor and 5751 internal cathode resistance, so 100uF wouldn't hurt.
Off the cuff 47uF should give a -3dB corner of approx 8Hz considering the thevenin equivalent of the cathode resistor and 5751 internal cathode resistance, so 100uF wouldn't hurt.
Sorry, the feedback connection was a drawing error as was the missing decimal for the feedback resistor. The 47uf bypass should be fine. The 50uf cap in the plate supply for 5751 should be fine. It is there to shunt any noise from the diode separating EL34 B+
from 5751 B+. As with any circuit this is a starting point and adjustments might need to be made once the circuit is working. I do however want to get it right in case anyone might use the circuit in the future.
from 5751 B+. As with any circuit this is a starting point and adjustments might need to be made once the circuit is working. I do however want to get it right in case anyone might use the circuit in the future.
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I see that you changed the cathode bias resistors on the EL34s.
I think you will find that 270 Ohms runs them too hot (too high an idle current). 470 Ohms is the datasheet recommended value.
Also
Many of the application notes recommend a common 470 Ohms to feed the screens of the EL34 pair. Note that in push pull, as one sides screen current increases the other side decreases and then they swap for the opposite "half cycle".
The total screen current is relatively constant so a common resistor works well to maintain constant screen voltage.
I would add this even if you are running the screens from a separate regulated supply.
Cheers,
Ian
I think you will find that 270 Ohms runs them too hot (too high an idle current). 470 Ohms is the datasheet recommended value.
Also
Many of the application notes recommend a common 470 Ohms to feed the screens of the EL34 pair. Note that in push pull, as one sides screen current increases the other side decreases and then they swap for the opposite "half cycle".
The total screen current is relatively constant so a common resistor works well to maintain constant screen voltage.
I would add this even if you are running the screens from a separate regulated supply.
Cheers,
Ian
Hey Guys,
I'm using a zener calculator to figure out the regulator for the screens. The limiting resistor is set for about 45ma. At that current to get 360 volts for the screens I'm using six 60 volt zener diodes. The total wattage dissipated is 19.8 across the string. (The calculator output is for a single diode.) That is dissipated over the six zeners isn't it? Like a resistor? 19.8/6 = 3.3 watts across each zener, is that correct? So 5 watt diodes should be safe?
I'm using a zener calculator to figure out the regulator for the screens. The limiting resistor is set for about 45ma. At that current to get 360 volts for the screens I'm using six 60 volt zener diodes. The total wattage dissipated is 19.8 across the string. (The calculator output is for a single diode.) That is dissipated over the six zeners isn't it? Like a resistor? 19.8/6 = 3.3 watts across each zener, is that correct? So 5 watt diodes should be safe?
My foggy understanding is that Kirchoff's Law states that the current drawn through the circuit will be the same at all points in the circuit. Each 60V zener will be passing 19.8mA (let's call it 0.02A for convenience). Power (watts) is volts times current, so...
60V * 0.02A = 1.2W
If I'm right about that, then it might be better to use more lower-voltage zeners. 12V zeners generate less noise than 60V zeners, so...
12V * 0.02A = 0.24W
That means you could use 12V 1W zeners, but you'd have to use 30 of them to get your 360V.
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60V * 0.02A = 1.2W
If I'm right about that, then it might be better to use more lower-voltage zeners. 12V zeners generate less noise than 60V zeners, so...
12V * 0.02A = 0.24W
That means you could use 12V 1W zeners, but you'd have to use 30 of them to get your 360V.
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If you have the space the 30 x 12V zeners will be substantially quieter too. (Think statistical regulator, probably >6dB quieter and possibly more if as I suspect the 12V zeners are quieter to start with than the 60V ones) Noise is not critical here, but it can't hurt either, and likely less expensive as well.
Well I fired up the amp today, literally! The 350 ohm resistor for the zener regulator started to smoke so I shut the amp off. I put a DVOM across it to test for current and turned it on. Then the resistor caught fire. Needless to say a 3 watt wasn't enough. If each zener is dissipating 1.2 watts does that mean the resistor has to be at least 6 x 1.2 = 7.2 watts minimum?
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New Schematic
If anyone is still listening I made some changes to the amp. The EL34's drew too much filament current to be used. So I followed Jim McShane's advice and went with 6L6's. I also went to fixed bias. Problem is I have almost no output. I started by disconnecting the feedback which had no effect. The 5751's have 220 volts across them at 1 ma of current. I've checked everything and its connected properly. With one volt input to the grid of 5751 I have .4 volts AC at the plate. It seems likely that its somehow canceling but I have no idea why? Any pointers would be appreciated, Kevin
If anyone is still listening I made some changes to the amp. The EL34's drew too much filament current to be used. So I followed Jim McShane's advice and went with 6L6's. I also went to fixed bias. Problem is I have almost no output. I started by disconnecting the feedback which had no effect. The 5751's have 220 volts across them at 1 ma of current. I've checked everything and its connected properly. With one volt input to the grid of 5751 I have .4 volts AC at the plate. It seems likely that its somehow canceling but I have no idea why? Any pointers would be appreciated, Kevin
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