Hi all - I'm new here, so hello to everybody.
I'm trying to repair an old 450 watt B&W ASW4000 active subwoofer, so I have been studying the schematic diagrams. The output rails runs at +74V and -74V.
On the main amplifier board, there's a long tail pair / differential amplifier that has a constant current source formed from Q603, a current-setting resistor R603 (current being equal to about Veb divided by R604, or 0.7 V/475 ohms, or about 1.5 mA), two standard diodes D601, D602 to maintain Veb on Q603, a resistor R606, and an "E-822" --constant current diode--- D605 that sets the current in this leg at 8.2 mA.
Evidently, when forward biased, D605 will maintain a constant current of 8.2 mA.
So far as I can figure, if you grounded R606 and omitted the constant current diode D605, the overall 1.5 mA constant current source would still work fine. If so, what's the purpose of the constant current diode D605?
The overall circuit I redrew from is below too.
Thanks for your thoughts!
Joe
I'm trying to repair an old 450 watt B&W ASW4000 active subwoofer, so I have been studying the schematic diagrams. The output rails runs at +74V and -74V.
On the main amplifier board, there's a long tail pair / differential amplifier that has a constant current source formed from Q603, a current-setting resistor R603 (current being equal to about Veb divided by R604, or 0.7 V/475 ohms, or about 1.5 mA), two standard diodes D601, D602 to maintain Veb on Q603, a resistor R606, and an "E-822" --constant current diode--- D605 that sets the current in this leg at 8.2 mA.
Evidently, when forward biased, D605 will maintain a constant current of 8.2 mA.
So far as I can figure, if you grounded R606 and omitted the constant current diode D605, the overall 1.5 mA constant current source would still work fine. If so, what's the purpose of the constant current diode D605?
The overall circuit I redrew from is below too.
Thanks for your thoughts!
Joe
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Welcome to the forum.
You say the rails are +/-74V. A 3k9/0.25W is going to fry itself being fed 74V, I calculate its dissipation as ~1.4W.
With the 8.2mA current source in series, the 3k9 will only dissipate 0.26W so definitely too high for long-term reliability but it will survive quite some time before giving up the ghost. Of course just increasing the 3k9's value will achieve the same thing so I have no idea why they'd want to include a relatively expensive current regulator diode in that spot.
Disclaimer - I worked at B&W when the ASW4000 was being developed though it wasn't my responsibility.
You say the rails are +/-74V. A 3k9/0.25W is going to fry itself being fed 74V, I calculate its dissipation as ~1.4W.
With the 8.2mA current source in series, the 3k9 will only dissipate 0.26W so definitely too high for long-term reliability but it will survive quite some time before giving up the ghost. Of course just increasing the 3k9's value will achieve the same thing so I have no idea why they'd want to include a relatively expensive current regulator diode in that spot.
Disclaimer - I worked at B&W when the ASW4000 was being developed though it wasn't my responsibility.
Hi abraxalito. Thanks for the comment. And super cool that you used to work at B&W, even in that time frame. Did you live and work near Steyning?
Welcome eetrojan!
I'm baffled, too. I do note that the constant current diode likely has very large dynamic impedance, so it will help reject any hum on the supply rail and minimize hum across D001 and D002. Seems like an expensive way to address the issue though...
I'm baffled, too. I do note that the constant current diode likely has very large dynamic impedance, so it will help reject any hum on the supply rail and minimize hum across D001 and D002. Seems like an expensive way to address the issue though...
Thank you!
At the time I lived in Angmering which was the other side of the South Downs from Steyning. But about a 20min scenic commute by car across the Downs.
D605 is power supply noise rejection so that the current in diodes D601, D602 is constant instead of following the supply voltage. Normally a large cap is placed in parallel with the diodes or halfway up the resistor(s) R606. 8.2mA is excessive and it could be as little as 0.2mA. The LTP current maters because this amp uses R604+VR602 to set the DC offset, where a current mirror would automatically minimize it for a wide range of LTP current.
This diode was designed for exactly this sort of circuit to give good supply rejection.
Unfortunately both the diode and 3.9k resistor are being run at 100% of rating, not a good idea.
A resistor/zener circuit is more common as its cheaper. The current should have been around 1mA
Unfortunately both the diode and 3.9k resistor are being run at 100% of rating, not a good idea.
A resistor/zener circuit is more common as its cheaper. The current should have been around 1mA
Follow up question. I'm hearing that it would be bad if the current in diodes D601, D602 followed the power suply voltage, but I just ran across some earlier notes of mine where I was blindly speculating that the designer might have included constant current diode 605 to establish a specific and nonvarying forward current through diodes D601, D602 for purposes of setting a stable forward voltage drop of about 725 mV (see Figure 4 of data sheet below).
But, given that Vb equals the positive rail voltage V+ minus the forward voltages drops on the diodes, wouldn't welded maintenance of a constant current through D601 and D602 mean that Vb is forced to track the power supply's ripple or other noise transients, thereby imposing the power supply voltage movements directly onto Ve and, in turn, changing the emitter current developed through resistor R603?
But, given that Vb equals the positive rail voltage V+ minus the forward voltages drops on the diodes, wouldn't welded maintenance of a constant current through D601 and D602 mean that Vb is forced to track the power supply's ripple or other noise transients, thereby imposing the power supply voltage movements directly onto Ve and, in turn, changing the emitter current developed through resistor R603?
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Your description is almost entirely correct.
To pretty good accuracy, all the diodes and the transistor's Vbe have equal voltage drops. So the the voltage at the base of Q603 is:
V+ - 2*Vbe
Voltage at emitter is Vbe higher than the voltage at the base, i.e. V+ - Vbe. So the voltage across R603 is just Vbe, and it's unchanging with variation in supply voltage--- hence, a constant current.