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My power supply design, comment please
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Voodoochile
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I can't really answer your question reliably- I'd suggest trying it both ways.
What does concern me is the lack of protection diodes for the regs. You don't want the post-reg bank draining back through your regulator. The data sheet has some good info on this. You should be able to air-wire it easily even though your baord is already in fab.
What does concern me is the lack of protection diodes for the regs. You don't want the post-reg bank draining back through your regulator. The data sheet has some good info on this. You should be able to air-wire it easily even though your baord is already in fab.
Garbz
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Voodoochille can you please elaborate on this? I've seen designs with the diode in place. The original Gilmore PSU schematic doesn't have the diode in, and mine appears to work now. So what does the diode actually prevent, and how do such cases occure?
Voodoochile
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The output caps can discharge into the output of the regulator if there is a short (not that that would happen, but sometimes it does), and ruin it. The reg can handle 15 to 20a if the spike is short enough... sounds like a lot, but even a 10uF cap can hit 20A briefly if the ESR is low enough.
Page eight of this PDF at Nat Semi talks about the details. Not mandatory, but there isn't any performance hit that I am aware of for having them installed, either.
Page eight of this PDF at Nat Semi talks about the details. Not mandatory, but there isn't any performance hit that I am aware of for having them installed, either.
Tomo
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Hi,
Agree with Voodoochile. When turning off, the charge accumulated in the POST-regulator stage may go back to the regulator. (I am think the POST-regulator stage will draw a lot of current while turning on, but I am not sure if it is fatal.)
It is really easy to do. You just connect the diode so that returning current will be bypassed to the PRE-regulator stage. Note here that even then you should not stack a lot of capacitance in the POST-regulator stage. Your diodes and tranny are going to take the discharge.
I would amass the capacitance at the PRE-regulator stage and use modest amount at the post-regulator stage. I want at bit over 4700uF there. You should get around 0.5V ripple before the regulator.
6,000uF in POST-regulator stage is a bit overkill. But then ... WE LIKE OVERKILL DON'T WE? YEAH, SIR/MA'AM! WE LIKE OVERKILL, SIR/MA'AM!
Tomo
P.S. Just use tough rectifiers and trannies.
Agree with Voodoochile. When turning off, the charge accumulated in the POST-regulator stage may go back to the regulator. (I am think the POST-regulator stage will draw a lot of current while turning on, but I am not sure if it is fatal.)
It is really easy to do. You just connect the diode so that returning current will be bypassed to the PRE-regulator stage. Note here that even then you should not stack a lot of capacitance in the POST-regulator stage. Your diodes and tranny are going to take the discharge.
I would amass the capacitance at the PRE-regulator stage and use modest amount at the post-regulator stage. I want at bit over 4700uF there. You should get around 0.5V ripple before the regulator.
6,000uF in POST-regulator stage is a bit overkill. But then ... WE LIKE OVERKILL DON'T WE? YEAH, SIR/MA'AM! WE LIKE OVERKILL, SIR/MA'AM!
Tomo
P.S. Just use tough rectifiers and trannies.
hmm.... i have received the pcb's today - 3A peak 1.9 cont rated bridge rectifier.... but it also says maxmum load capacitance 5000uF
I dont know whether to try it or not... new BR's on the way now anyway..
Also the transformer is 50VA, delivering 1.67A
now if the transformer shouldn't deliver more current than that....what happens? the BR should be ok, and the charge times should be longer reducing the demand.. however the voltage regulator is only rated to 1.5A anyway..
In kevin gilmores design, he has 4700uF after the regs, so i presume that shouldn't be too much of a problem.
I dont know whether to try it or not... new BR's on the way now anyway..
Also the transformer is 50VA, delivering 1.67A
now if the transformer shouldn't deliver more current than that....what happens? the BR should be ok, and the charge times should be longer reducing the demand.. however the voltage regulator is only rated to 1.5A anyway..
In kevin gilmores design, he has 4700uF after the regs, so i presume that shouldn't be too much of a problem.
tangent
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Quote:
They're not kidding. The designers know the effective impedance of the diode bridge, and you can work out the current that will be required to charge a cap through that resistance. The limit on capacitance is there because the higher the value, the longer the charging takes, hence the more heat you generate in the bridge. The more heat, the more likely you will damage the bridge.
That's one reason I use big TO-220 discrete rectifiers instead of monolithic bridges on the STEPS, and also why I recommend only about 2000-4000uF of capacitance.
Quote:
The more current you pull from a transformer, the higher the magnetic flux, and thus the higher the transformer's impedance. This source of inefficiency is why the voltage at the secondary of a transformer drops as current increases. The rated voltage on a transformer is where the voltage is supposed to be when the rated current is drawn from it. So, if you draw more current, the voltage drops below the rated voltage. If you draw enough current, you will generate enough heat in the windings to damage them. If you short the windings, your life can get momentarily exciting.
| it also says maxmum load capacitance 5000uF |
They're not kidding. The designers know the effective impedance of the diode bridge, and you can work out the current that will be required to charge a cap through that resistance. The limit on capacitance is there because the higher the value, the longer the charging takes, hence the more heat you generate in the bridge. The more heat, the more likely you will damage the bridge.
That's one reason I use big TO-220 discrete rectifiers instead of monolithic bridges on the STEPS, and also why I recommend only about 2000-4000uF of capacitance.
Quote:
| now if the transformer shouldn't deliver more current than that....what happens? |
The more current you pull from a transformer, the higher the magnetic flux, and thus the higher the transformer's impedance. This source of inefficiency is why the voltage at the secondary of a transformer drops as current increases. The rated voltage on a transformer is where the voltage is supposed to be when the rated current is drawn from it. So, if you draw more current, the voltage drops below the rated voltage. If you draw enough current, you will generate enough heat in the windings to damage them. If you short the windings, your life can get momentarily exciting.
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Dreamslacker
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I might be wrong but I do believe ever seeing Mr Gilmore post in the forums that the person who builds it would know what to do. ie. Bypass the opamps with ceramic caps, diodes across the regulator etc.
| Originally Posted by Garbz Voodoochille can you please elaborate on this? I've seen designs with the diode in place. The original Gilmore PSU schematic doesn't have the diode in, and mine appears to work now. So what does the diode actually prevent, and how do such cases occure? |
I might be wrong but I do believe ever seeing Mr Gilmore post in the forums that the person who builds it would know what to do. ie. Bypass the opamps with ceramic caps, diodes across the regulator etc.
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