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#1 |
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diyAudio Member
Join Date: Oct 2002
Location: Atlanta, GA
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My amp will be cycled on and off daily... So I would like to implement a standby mode for my Aleph-x similar to what Nelson said he uses in his amps.
If I recall (guess) correctly, he leaves the power supply powered up and just shuts off the transistors. I was thinking this could be done by shorting the gate to source on Q1 and Q10 (current followers)-(Grey' schematic) and also shorting gate to drain on Q6a (current source mosfet). Turning off the current source mosfet would turn off Q5, Q7, Q2 and Q11 (diff pair and main output mosfets) wouldn't it? If so, this could all be done with a small 3 pole relay, without having to switch a large amount of current each time. Would this cause any type of turn on or off transients that would require output relays? Or should it work fine if a single relay were used to switch them all at the same time? Please let me know if this thinking is correct. Thanks, Jeff |
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#2 |
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diyAudio Member
Join Date: Dec 2002
Location: Illinois, USA
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Shorting the gate to source for Q1 and Q10 could work, but I think I would leave the differential pair powered in standby mode.
Here is another option to spur discussion. You could instead connect the bases of Q3 and Q8 to the positive rail through a 2k resistor in series with the relay switch. That should be enough to turn Q3 and Q8 fully on and shut off the current source. Jeremy |
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#3 |
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diyAudio Member
Join Date: Oct 2002
Location: Atlanta, GA
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If the gate-source were shorted on Q1 and Q10 that effectively shuts off Q2 and Q11 (with the exception of what flows through the output to ground resistors (R1, R4, R44, R45) Would that create a possible relative DC offset problem?
does anyone know what transient behaviour to expect by shutting down the current source to the diff pair? |
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#4 | |
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diyAudio Member
Join Date: Dec 2002
Location: Illinois, USA
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Quote:
In the second case where you turn off the output current sources, turning off the diff pair current source will have only a limited effect on the power consumption, since it runs at only 20mA. You may need to be careful of the increased power supply voltage when turning off the output because you are no longer loading the transformer (or inductors, with an LC supply filter) or passing much current through filter resistors. Jeremy |
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#5 |
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diyAudio Member
Join Date: Oct 2002
Location: Atlanta, GA
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Jeremy,
Thank you for helping me understand a little better, although that does raise one more question in my mind. ![]() Wouldn't leaving the diff pair on, while shutting off the output current sources, pull both the + and - outputs towards the negative (-15 Volt) rail through the output to ground resistors (via the output gain transistors)? If so, this may cause the same DC offset problem (although smaller in magnitude). I guess you would still have some feedback in that case... I wonder if there is a benefit to turning off "everything" vs. just the output current sources. It seems that it would be good to leave the diff pair on to keep them warm, but I specifically remember Nelson saying he turned them off in the production units (can't find the exact thread now) I wonder why he chose this method? Jeff |
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#6 | |
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diyAudio Member
Join Date: Dec 2002
Location: Illinois, USA
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Quote:
And thus, we are back where you started. What is the transient behavior? If you turn of the diff pair CS before the output CS, you may get a loud thump as the feedback shuts down and a differential voltage appears across the speaker. If you turn off the output CS before the diff pair CS, the diff pair controls the thump, then when the diff pair shuts down, the gain transistors should turn off nearly simultaneously without a thump. Jeremy |
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#7 |
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The one and only
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What I do is shut off the bias to both the diff pair and the
positive output stage, otherwise the outputs go to the - or + rails (respectively). The filter capacitance on each of those circuits needs to be balanced so that they shut down and turn on gracefully. |
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#8 |
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diyAudio Member
Join Date: Apr 2004
Location: Erkelenz
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The Aleph X is a 2 stage design. When you shut off both stages were is the standby mode then ?
In my opinion the input diff stage has to keep in normal working conditions and the current of the output stages have to shut off. Regards Ralph |
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#9 |
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diyAudio Member
Join Date: Oct 2003
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Ralph,
It depends on how you define standby. In Nelson's case, standby means the capacitors are still charged, but the amplifier draws no current. If one would like to keep the diff pair on (though I don't see the benefits of doing that), then one would need to shunt the drain resistors of the diff pair, by e.g. switching on the protection BJT's for the -ve rail power FETs. Alternatively, one might also choose to half the current through the diff pairs to reduce the gate voltage for the said power FETs to below their threshold. But one might also define standby as full current consumption and no sound (just to keep the transistors at constant temperature all times), in which case one only needs to shunt the inputs to ground and disconnect the speakers, say via a relay. In my own particular case, I did the same as Nelson. Patrick |
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#10 |
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diyAudio Member
Join Date: Apr 2004
Location: Erkelenz
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Hello EUVL,
but what is the benefit if only the power supply capacitors keep their energy? Even if they are empty they get the whole energy back after switch on nearly immediately. As far as I know, Nelson shut down the bias of the output circuits (and brings the current to zero) in the X600 amp and keep the input amplifier under normal working conditions - and this is for me also the logical way. In standby mode you want to spare energy, but to keep the important parts of the circuit in their working conditions. What you want to reach is to save time after switch on and to reach best operation (and sound) as soon as possible. Of course it is the best way to keep the amplifier totally switched on - but than we need no standby circuit. Yes, all we have to do is to short the input to ground and maybe disconnect the loadspeaker. Regards Ralph |
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