Solid State power amp supply
I want to build a new HT amp with 3 amp modules each putting out around 200 watts at 8 ohms or so. I want to design a power supply for it, and I'm trying to use what I have. I could probably get away with a lot less but I designed a supply which uses 30,000uf per rail, but then has a series resistor followed by a 4700uf then another series resistor then a paralleled 4700uf cap again. I found this drasticly reduced ripple, and was wondering what you guys thought. Am I better going to a smaller third cap for this CRCRC network, say a 47uf, maybe even bypassing it with something like a .47uf film cap?
Excuse the crudeness of that schematic, but that should basicly explain what I mean. The last capacitor could change to something small like 4.7uf instead, or 47mf, whatever. It could also be bypassed by a very small value low esr film cap. However as is I think it drasticly reduces ripple over just a 30,000uf suppls, or even a 60,000uf. I know about the voltage drop, but the design calls for around 65-70 volts suppl, and if I use 56-0-56 tranny instead of the 50-0-50, I can get away with this supply. Oh and is the below correct for a dual supply.
Why not use LC filtering instead or RC? You would get better filering and less resistive loss (waste heat). It is easy to wind your own inductors, or you can salvage some out of an old PC power supply. Blown PC power supplies should be free for the asking at so many places, and the inductors will probably be fine.
The simplest inductors are nothing more than a length of wire wound into a coil, with maybe a bit of string tied in a few places to hold it together. You do need to use enameled wire with a large enough diameter for the current.
You can get much fancier by looking at a few web sites that have tutorials on inductor design. They have all the equations you need to predict the inductance of your creations, info about different core materials, etc.
Its easy. Its fun. Its cheap. What more could a DIYer ask for?
This college has a nice intro website:
Besides, I think you have a little design problem:
w = I^2 * R
So 3 Amps into 8 Ohms can only give you 72 Watts with a 100% efficient amplifier. You would need 5 Amps to get to 200 Watts.
If your 3 modules need 600 Watts total at 60 volts, then
60w / 60 volts = 10 Amps
But 10 Amps through 500 Ohms of series resistors would require:
10A * 500Ohm = 5000 Volts!!!!
If you are checking the ripple on a real circuit or a simulation, then you have no load on the output of the filter. With a load current down in the milliamps, RC filtering is reasonable. With an 8 Ohm load, you need a resistive loss in the power supply of less than one ohm. That's why you should use inductors instead of resisters in the filter.
|All times are GMT. The time now is 11:05 AM.|
vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2016 DragonByte Technologies Ltd.
Copyright ©1999-2016 diyAudio