24-11-2016, 10:37 AM
Hi Andy,
Looking at the "signal present" circuit, it's pretty simple once you spot that the op-amp is simply an inverting amplifier. No idea what the gain is, as I don't see any component values...
(Then I remembered that you emailed me a copy of the service manual - I'll stick it in The Archive in a bit)
This op-amp will amplifier the signal that's coming from the output side of the speaker protection relay. The gain (approx x5) will have been chosen according to the required threshold required to light the LED. At higher levels, the op-amp could well be driven into clipping. That's why the two diodes are there (D20/D21). The negative input of the op-amp is a virtual earth, but only when the op-amp is not clipping. When clipping occurs, the voltage at the input can move considerably, but the diodes clamp that to +/- 0.6V. That's good because it protects the input stage, plus the TL0x series can do strange things when the input stage is overloaded.
After the op-amp is a half-wave rectifier and smoothing capacitor (D23, C40), and a transistor to drive the LED.
If you need more background on how the op-amp is working, this might help: http://www.markhennessy.co.uk/articles/op-amps.htm
Regarding the quiescent settings, that sounds a bit high to me...
Between the bases of the output devices, you need 2 lots of 0.6V, plus whatever is across the current-sharing resistors. The voltage you're monitoring is that minus 0.6 divided in 2 (approx). So perhaps that amounts to 1.72V - in which case, that leaves about half a volt across the current sharing resistors. Those are 0.33 ohms, and there's 2 sets in series, so each resistor sees about a quarter of a volt. Hmm - that suggests a standing current of 0.75A in each transistor. No wonder it gets hot!
Assuming a more reasonable 50mA per transistor, that's 17mV across each of the 0.33 ohm resistors, giving 34mV between the emitters. Between the bases, that'll be something like 1.234V. Knock off the 0.6V of D11 and divide by two to give you what should be across R35: 300mV
Please note that the above is all approximations. 0.6V is a reasonable assumption for a B-E junction and a diode drop, but in practice they will differ slightly.
The best way to set the quiescent current is to monitor the THD while adjusting the current. There should be a certain value that gives the lowest value, but in many cases it can be a bit on the high side - especially with parallel emitter followers like this one. Providing the THD is reasonably low at the value you pick, it'll be fine. For a low-power amplifier, the quiescent dissipation accounts for most of the heat because music has a large peak to mean ratio (or crest factor, as some prefer). This is a bit different, but the fans will help
Hope that makes sense,
Mark
Looking at the "signal present" circuit, it's pretty simple once you spot that the op-amp is simply an inverting amplifier. No idea what the gain is, as I don't see any component values...
(Then I remembered that you emailed me a copy of the service manual - I'll stick it in The Archive in a bit)
This op-amp will amplifier the signal that's coming from the output side of the speaker protection relay. The gain (approx x5) will have been chosen according to the required threshold required to light the LED. At higher levels, the op-amp could well be driven into clipping. That's why the two diodes are there (D20/D21). The negative input of the op-amp is a virtual earth, but only when the op-amp is not clipping. When clipping occurs, the voltage at the input can move considerably, but the diodes clamp that to +/- 0.6V. That's good because it protects the input stage, plus the TL0x series can do strange things when the input stage is overloaded.
After the op-amp is a half-wave rectifier and smoothing capacitor (D23, C40), and a transistor to drive the LED.
If you need more background on how the op-amp is working, this might help: http://www.markhennessy.co.uk/articles/op-amps.htm
Regarding the quiescent settings, that sounds a bit high to me...
Between the bases of the output devices, you need 2 lots of 0.6V, plus whatever is across the current-sharing resistors. The voltage you're monitoring is that minus 0.6 divided in 2 (approx). So perhaps that amounts to 1.72V - in which case, that leaves about half a volt across the current sharing resistors. Those are 0.33 ohms, and there's 2 sets in series, so each resistor sees about a quarter of a volt. Hmm - that suggests a standing current of 0.75A in each transistor. No wonder it gets hot!
Assuming a more reasonable 50mA per transistor, that's 17mV across each of the 0.33 ohm resistors, giving 34mV between the emitters. Between the bases, that'll be something like 1.234V. Knock off the 0.6V of D11 and divide by two to give you what should be across R35: 300mV
Please note that the above is all approximations. 0.6V is a reasonable assumption for a B-E junction and a diode drop, but in practice they will differ slightly.
The best way to set the quiescent current is to monitor the THD while adjusting the current. There should be a certain value that gives the lowest value, but in many cases it can be a bit on the high side - especially with parallel emitter followers like this one. Providing the THD is reasonably low at the value you pick, it'll be fine. For a low-power amplifier, the quiescent dissipation accounts for most of the heat because music has a large peak to mean ratio (or crest factor, as some prefer). This is a bit different, but the fans will help
Hope that makes sense,
Mark
