syn08 said:
Sorry, I guess you meant the 33ohm for the parallel regulators, I was looking at the feedback 39ohm. Yes, they would dissipate quite some power (I'm using 5W resistors right now), but 33ohm is a grossly overestimate in terms of current. The whole input stage won't use more than 100mA DC. Double that, and the resistors can be 68ohm/2W.
syn08,
Just a short OT interruption.
Finally found the time to fiddle with my browser settings and your website and postings here with attachments are visible when using a proxy-server as PM suggested earlier.
Edmond hope this also solves your connection problems, if not already solved.
Regards,
Franklin
Just a short OT interruption.
Finally found the time to fiddle with my browser settings and your website and postings here with attachments are visible when using a proxy-server as PM suggested earlier.
Edmond hope this also solves your connection problems, if not already solved.
Regards,
Franklin
Franklin, you would have saved time if you had read
Edmond,
Thanks for your response. My solution is just the same, but using my provider's proxy server. Works like charm though.
Franklin
Thanks for your response. My solution is just the same, but using my provider's proxy server. Works like charm though.
Franklin
syn08 said:
Well, even 68 Ohm 2 watt resistors will be beyond their dissipation limits. At least the dissipation limits usually considered for good reliability.
Obviously, you could go with multiple resistors, higher wattage resistors, or as somebody else suggested, current sources. Or some combination. Even an LM317 connected as a current source will offer additional raw DC supply ripple rejection up through a few KHz if you want that.
BTW, since much of this circuitry is run in Class A, I'm not sure you need to double the calculated source current into the shunt regulator. But, I suppose it does get especially cold in Canada...
CG said:
It is cold here (and I dearly miss California) but that's not the main reason.
As discussed above, the max current at the THS4031 output is 100mA (essentially in the feedback resistors). Add this to the 80mA for the current mirrors and you get 180mA. With the opamp quiescent (10mA) and some margin, here's 200mA peak to take care of.
Othwerwise, a current source won't help with the power dissipation. If I have to dissipate power, I'd rather do it in resistors. But then I'm right now taking a second look at the parallel regulator, I may improve it a little.
If nothing else, this exercise has highlighted why some designers have chosen to use open loop transconductance stages for the input sections of their phono preamps.
syn08 said:
Ahh. I was just using the 100 mA value you quoted earlier.
Of course - watts is watts. Resistors do well there. Just not 2 watt resistors...
Good point CG, and that is why I didn't try to use a feedback loop more than 25 years ago. Too hard to drive properly, and still get low noise performance.
That was more of a generic response to Jan. Everything that is engineered involves compromises of some sort.
I'm not sure about the TI device, but have you checked on the thermal performance of the part? You probably already have and are familiar with this, but for people who aren't, this might be helpful:
http://waltjung.org/PDFs/WTnT_Op_Amp_Audio_1.pdf
For the record, the dual feedback loop can be made into a single feedback loop with a 'bridge connection between the 1 ohm resistors consisting of 2 low value resistors and then the 'feedback resistor' can be made to almost 40 ohms. Then the loading will be approximately 40 ohms, rather than 20 ohms which pushes the IC to its thermal limit, at the very least.
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