With regards to the simple solution of using a 12 VAC wall wart instead of 9 VAC, what is the effect of this on heat dissipation? Will the booster caps and regulators become hotter? I'm worried about additional heat that might shorten the life of these components. Why would the manufacturer use a 9 VAC wall wart when using a 12 VAC isn't going to add much to the cost? There has to be a reason why. Maybe heat dissipation is the reason? I remember that in the early releases of the Delta 1010 back in 1999, there were many complaints of failing units due to "capacitor problems", so they said, and maybe those failures were related to excessive heat causing damage to capacitors.
AndrewT said:
Thanks for the heads up! There are very small SMD caps near the reg pins but I don't know their values, let alone their voltage. There are also big electyrolytic caps near the regs and in the rectifier/doubler circuit, with voltage ranging from 35 to 63V if i recall correctly.
Reducing to +-12 VDC might not work because these regs supply power to the JRC 5532 opamps which need +-15V for optimum operation.
Hi,
have you checked the datasheet for the JRC version of the 5532?
I suspect it will operate over a much wider range of supply voltage.
What is the maximum ouput voltage of the sound card?
Most opamps works very well when outputting a signal 10db below maximum. The difference between maximum on +-15Vdc and +-12Vdc is only about 2volts (10Vac cf 8Vac).
If your maximum output is 2Vac then the overhead is 14db cf 12db.
Both these figures exceed 10db.
Does the +-15Vdc supply anything else?
have you checked the datasheet for the JRC version of the 5532?
I suspect it will operate over a much wider range of supply voltage.
What is the maximum ouput voltage of the sound card?
Most opamps works very well when outputting a signal 10db below maximum. The difference between maximum on +-15Vdc and +-12Vdc is only about 2volts (10Vac cf 8Vac).
If your maximum output is 2Vac then the overhead is 14db cf 12db.
Both these figures exceed 10db.
Does the +-15Vdc supply anything else?
I checked the datasheet and yes, it operates over a wider range, but I'm wary about deviating from +-15V since that is what's typical as shown in the graphs. I also don't want to lose headroom by going to a lower voltage.
I don't know what the maximum output voltage of the sound card is. Where do I get that info? I also don't know what other things get their supply from the +-15VDC. All I know is that the sound card has 8 analog inputs and outputs, an S/PDIF input and output, a word clock input and output, and a MIDI input and output.
BTW, the 9VAC wall wart is rated 41 VA. How do you calculate the Amps from the given VA? What is the power factor to use in this case?
I don't know what the maximum output voltage of the sound card is. Where do I get that info? I also don't know what other things get their supply from the +-15VDC. All I know is that the sound card has 8 analog inputs and outputs, an S/PDIF input and output, a word clock input and output, and a MIDI input and output.
BTW, the 9VAC wall wart is rated 41 VA. How do you calculate the Amps from the given VA? What is the power factor to use in this case?
Hi,
Iac=VA/Vac = 41/9=4.55Aac
If full wave rectified then the peak output voltage = sqrt(2)*Vac, i.e. Vdc~=1.4*Vac
If fed into a capacitor input filter then the amplifier must be de-rated to 0.7times.
The effective continuous DC current for a capacitor input, full wave rectified transformer, is Idc=Iac/2=VA/Vac/2=2.28Adc.
A half wave rectifier is different and a voltage doubler is different again.
Iac=VA/Vac = 41/9=4.55Aac
If full wave rectified then the peak output voltage = sqrt(2)*Vac, i.e. Vdc~=1.4*Vac
If fed into a capacitor input filter then the amplifier must be de-rated to 0.7times.
The effective continuous DC current for a capacitor input, full wave rectified transformer, is Idc=Iac/2=VA/Vac/2=2.28Adc.
A half wave rectifier is different and a voltage doubler is different again.
Hi, been busy with other things but now I'm back. 
I checked the output voltage of my 9 VAC wall wart. Without load it is 11 V. With load it is 10.6 V. That's a very small drop of only 0.4 V when the load is connected! Considering that the wall wart's output is actually higher (it's already close to 12 V that Richie suggested), and what you're saying that even a 9 VAC transformer can effectively produce the same DC output as an 18 VAC transformer if the doubler is operating efficiently, I'm beginning to wonder if I really have a problem at all! The key is if the doubler is operating efficiently. So, what does it take to make the doubler operate efficiently?
BTW, one strange thing I noticed is that if the unit (a.k.a. break-out box) is powered but the computer is off (the unit is connected to the PCI card inside the computer by means of an IEEE 1284 D-Sub cable), the unit's power LED does not turn on. In this state, the output voltage I get from the 9 VAC wall wart is 11 V, which is its "without load" output. After turning on the computer, that's when the unit's power LED turns on, and that's when I get 10.6 V output from the wall wart. How do you explain this?
Thanks!
AndrewT said:
I checked the output voltage of my 9 VAC wall wart. Without load it is 11 V. With load it is 10.6 V. That's a very small drop of only 0.4 V when the load is connected! Considering that the wall wart's output is actually higher (it's already close to 12 V that Richie suggested), and what you're saying that even a 9 VAC transformer can effectively produce the same DC output as an 18 VAC transformer if the doubler is operating efficiently, I'm beginning to wonder if I really have a problem at all! The key is if the doubler is operating efficiently. So, what does it take to make the doubler operate efficiently?
BTW, one strange thing I noticed is that if the unit (a.k.a. break-out box) is powered but the computer is off (the unit is connected to the PCI card inside the computer by means of an IEEE 1284 D-Sub cable), the unit's power LED does not turn on. In this state, the output voltage I get from the 9 VAC wall wart is 11 V, which is its "without load" output. After turning on the computer, that's when the unit's power LED turns on, and that's when I get 10.6 V output from the wall wart. How do you explain this?
Thanks!
Hi,
it seems the computer sends a shutdown signal to the sound card. There must be a solid state switch, or relay, in the supply to the sound card before the LED.
Voltage doublers always have a problem generating that doubled voltage. As the output current increases that loss in output voltage gets worse.
I doubt you can do anything to make the doubler work any better in the space available on the sound card.
If there is a real PSU problem in the card then bypassing the doubler with a new dual polarity supply gets rid of disease.
it seems the computer sends a shutdown signal to the sound card. There must be a solid state switch, or relay, in the supply to the sound card before the LED.
Voltage doublers always have a problem generating that doubled voltage. As the output current increases that loss in output voltage gets worse.
I doubt you can do anything to make the doubler work any better in the space available on the sound card.
If there is a real PSU problem in the card then bypassing the doubler with a new dual polarity supply gets rid of disease.
How do I reliably find out if there is really a PSU problem?
Going back to the original suggestion of using LDO regs and Schottky diodes for lower voltage drop, I think I'm beginning to see the wisdom of that. The lower voltage drop will allow the use of a lower input voltage going into the regs to make them work, maybe 1.5 volts lower. Because of this reduction in the voltage requirement, the 9 VAC wall wart in combination with the rectifier/doubler becomes adequate for the job. Considering that the wall wart is actually supplying not just 9 VAC but 10.6 VAC with load, I think the chances are now even better that this solution would work. Don't you think so? And as a bonus, this solution doesn't have the disadvantage of increasing the heat dissipation.
Going back to the original suggestion of using LDO regs and Schottky diodes for lower voltage drop, I think I'm beginning to see the wisdom of that. The lower voltage drop will allow the use of a lower input voltage going into the regs to make them work, maybe 1.5 volts lower. Because of this reduction in the voltage requirement, the 9 VAC wall wart in combination with the rectifier/doubler becomes adequate for the job. Considering that the wall wart is actually supplying not just 9 VAC but 10.6 VAC with load, I think the chances are now even better that this solution would work. Don't you think so? And as a bonus, this solution doesn't have the disadvantage of increasing the heat dissipation.
I do have a scope that I got from eBay, but it didn't come with a manual and I'm still trying to learn how to use it. Meanwhile, will a DMM suffice?
I guess that in order for the results to be meaningful, I should measure the regs' output voltage while there is a lot of activity going on in the soundcard, like 5 input/output channel pairs simultaneously performing loopback recording, right?
I guess that in order for the results to be meaningful, I should measure the regs' output voltage while there is a lot of activity going on in the soundcard, like 5 input/output channel pairs simultaneously performing loopback recording, right?
Hi,
you'll get a better idea by comparing quiescent against a little signal and then against lot of signal. By a lot I mean near the maximum that the sound card can send out which from the PSU spec is way above the input requirement of the next stage.
You will also need to add a dummy load to the sound card. Somewhere just under the lowest impedance value you are ever likely to feed.
If you have a dual beam or dual trace scope you can monitor the output signal at the same timel as the PSU voltage.
You can also monitor the voltage doubler voltage and the input to the regulators.
Why have we been guessing when you have the tools to hand?
you'll get a better idea by comparing quiescent against a little signal and then against lot of signal. By a lot I mean near the maximum that the sound card can send out which from the PSU spec is way above the input requirement of the next stage.
You will also need to add a dummy load to the sound card. Somewhere just under the lowest impedance value you are ever likely to feed.
If you have a dual beam or dual trace scope you can monitor the output signal at the same timel as the PSU voltage.
You can also monitor the voltage doubler voltage and the input to the regulators.
Why have we been guessing when you have the tools to hand?
AndrewT said:
It's because I don't know how to use my scope yet.
Been so busy and haven't had the time to figure it out. But I will soon get to that.Meanwhile, I need your advice as to which of these MBR1100 Schottky diodes is better performing or better suited for my application...
http://www.onsemi.com/pub/Collateral/MBR1100-D.PDF
-- or --
http://www.irf.com/product-info/datasheets/data/mbr1100.pdf
This will be the replacement for the 1N4001 rectifier diodes...
http://www.componentkits.com/dslibrary/1N4004.pdf
I'm not sure how to compare specs for evaluating their compatibility, performance, etc.
Thank you so much!
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