New Measurements
Right Channel
Positive:
Output=24
R101 Vdrop =0.32
V Ground to Source=25
Q101 VGS=9.8
Negative:
Output=24
R201 Vdrop=0.26
V Ground to Source=26
Q101 VGS=5
Left Channel
Positive:
Output=24
R101 Vdrop=0.31
V Ground to Source=25
Q101 VGS=7.9
Negative:
Output=24
R101 Vdrop=0.27
V Ground to Source=25
Q201 VGS=5
The weird measurements had nothing to do with the reg. and everything to do with the way I was measuring.
The only questionable measurement now is the VGS on both positive regs. I'll get some new 9610s and see what happens.
Would it be OK to use it the way it is now?
Right Channel
Positive:
Output=24
R101 Vdrop =0.32
V Ground to Source=25
Q101 VGS=9.8
Negative:
Output=24
R201 Vdrop=0.26
V Ground to Source=26
Q101 VGS=5
Left Channel
Positive:
Output=24
R101 Vdrop=0.31
V Ground to Source=25
Q101 VGS=7.9
Negative:
Output=24
R101 Vdrop=0.27
V Ground to Source=25
Q201 VGS=5
The weird measurements had nothing to do with the reg. and everything to do with the way I was measuring.
The only questionable measurement now is the VGS on both positive regs. I'll get some new 9610s and see what happens.
Would it be OK to use it the way it is now?
Its the power. 9610 will blow if you run say 2A for a T-Amp constantly in a CCSed shunt reg. You will need BIG sinks also. 9640 is heavy duty. Slower than 9610 of course. Use it to replace both 9610 and 9530. You can find it easily in the open market.
The way they are now there is very low CCS current indicated. Trim them for 20V output and see if the 10R VR101 drop goes 2V and over.
It can be derived from what peak power it can deliver to your speakers impedance plus those few mA MAX it uses for itself because so efficient. The reg must cater for the load and the amp. And the load is speakers i.e. heavy. Can't escape it.
Also from the data sheet:
PO Output Power (Continuous Average/Channel)
THD+N = 0.1% R/L = 4Ω 13W, R/ L = 8Ω 8W
THD+N = 10% R/L = 4Ω 22W, R/ L = 8Ω 12W
My speaker is 92db (Ariel ME2); and I am planning to use one T-2040 per channel, as in passive bi-amping.
From the above, can we have an idea what the peak power to the speakers?? (then plus 70-75mA (for the T amp), and plus 100mA for the reg.....is that correct??)
Cheers,
King
Considering this one channel.
Vin = {V Ground to Source} + {R101 Vdrop} = 25Vdc + 0.32Vdc = 25.32Vdc
Vout = 24Vdc
Vin-Vout = 1.32Vdrop
Am I reading the data correctly?
When building a bipolar supply to feed a preamp, where should I unite the psu output to make it +-32?
Or Can I maintain a dual mono arrangement all the way to the preamp psu input?
I'm not quite sure on how to configure the wiring and where to establish the central ground
I'll have the transformer and pre-regulator (diode bridge and CRC) in one case and the sslv boards near the preamp without the diodes, just the big reservoir cap.
Or Can I maintain a dual mono arrangement all the way to the preamp psu input?
I'm not quite sure on how to configure the wiring and where to establish the central ground
I'll have the transformer and pre-regulator (diode bridge and CRC) in one case and the sslv boards near the preamp without the diodes, just the big reservoir cap.
That is why I asked him to test with near 20V output so to see if the CCS wakes up. He could trim down to 21.5V output and he reported no significant change in R101/201 Vdrop. That is weird.
The bib pcbs are made as independent sections so the straightforward way is to use four wire secondaries (dual type transformer), each wires pair to each bib, and two bridges (available bridge diodes places are on each bib already) then unite at zero at the preamp's input. That's best.
If you must use a three wire secondaries (centre tap type transformer) with a single bridge because that is all you got, you should wire in series the (-) and (+) of the positive and negative section's reservoir capacitors respectively. Then feed the raw (+) (-) from the single bridge across them reservoirs free pins and the (0) on their common wired middle.
Your ME2 has two 8 Ohm woofer-mid drivers in parallel? If yes then its a roughly 4 Ohm nominal speaker. Depends on the crossover for how low and where in frequency band it dips. That's 44W at 10% THD for two channels from your amp's data. Mr. Ohm said I=sqrt(P/R). Thus I=sqrt(44/4)=3.31A RMS. Such a shunt reg would need to be set at 3.5A RMS CCS to cover such THD peaks having a bit of spare to maintain the efficient amp and itself. Or at 1.75A RMS when using two amps and two regs. RMS is 1.41 times less than peak. If you set them at 1.75A CCS per channel it will current limit and hard clip at the now 10% THD wattage figure. So the setting must be even richer. When the medium power delivered while playing music could be 10% to 20% of peaks that means serious heat waste and sinking. Better avoid it. Inefficient and maybe problematic to build and run flawlessly. Expensive at least and most probably you will end up with fan cooling. You are reversing the efficient chip amp's reason of existence by making its PSU max idling "Class A"
