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-   -   Way to reduce rail voltage ? (wise type ?) (http://www.diyaudio.com/forums/power-supplies/207811-way-reduce-rail-voltage-wise-type.html)

guitar89 29th February 2012 12:53 PM

Way to reduce rail voltage ? (wise type ?)
 
1 Attachment(s)
I'm currently planning an amplifier, and it seems that I couldn't find a good transistor that have high enough Vce and other specs I wished, so I only have my BC550C/560C to use, which is lower Vce. As my rail voltage from my transformer after filtered and rectified, it is roughly 27V-28V, I need to lower it to something from 20V-22V to cope for BCs.

I'm current finding some method of reduce rail voltage (regulator ?) with minimum dissipation, without insane complexity & cost.
And method of protection against or able to withstood small overvoltage. (spike as attached description)

I'm also thinking incorporate some inductor, which will improve the PSRR and decrease some rail voltage ? what is the drawback of using inductor in power supply ?

Osvaldo de Banfield 29th February 2012 01:13 PM

A simple resistor and a zener?

guitar89 29th February 2012 01:40 PM

Would you give me typical value of resistor ?
well... zener just get one of correct voltage should be jsut nice.
According to my simulation, it degrades PSRR alot.

Osvaldo de Banfield 29th February 2012 04:06 PM

Quote:

Originally Posted by guitar89 (Post 2927771)
Would you give me typical value of resistor ?
.

Search in any datasheet, the recommended minimum current in the zener under question. Add the load current to it. So, then knowing voltage drop in the resistor, calculate the resistor value and power loss in it. Finally, choose a smaller normal value resistor, and the power of such slowly greater than it. Example: supuose diode current 10mA, load current 50 ma and voltage drop 15V. Then, Ir = 10mA plus 50mA = 60mA.

R = 15V / .06A = 250ohms. Choose 220R.
P = 15V * .06A = .9W, choose 1W.

If the power is not so big, you can also think in a TL431 voltage regulator, that has lower output resistance. But calculus is a bit more complex.

Good luck.

guitar89 29th February 2012 10:52 PM

just to ask, it is wise to only provide regulated supply to non- output stage (output stage using raw unregulated supply), like this I reduce the power (& dissipation) needed to pass through regulation unit. (output use tons of currents)
Quote:

Through simulation, other than output, the amplfieronly consume 20mA maximum, giving extra 10mA = 30mA.
While output is using about 6A (OMG), I think last time because of that output it got worst PSRR
So here I have attached the link for the choice i would take, so i would like your suggestion on resistor I need :
http://malaysia.rs-online.com/web/p/zener/0812538/
its datasheet : http://docs-asia.electrocomponents.c...6b800297e6.pdf
Just to be sure, in my input stage, I have another pair zener is 15V from ground, I suppose it won't have problems ?

PRR 29th February 2012 11:35 PM

> I only have my BC550C/560C to use, which is lower Vce. As my rail voltage ... is roughly 27V-28V, I need to lower it to something from 20V-22V to cope for BCs.

BC550 is rated 50V.

http://www.nxp.com/documents/data_sheet/BC549_550.pdf page 2

> reduce rail voltage (regulator ?)

Yes, 7818 will drop 27V to 18V. One chip one capacitor and maybe a little scrap metal.

> inductor, which will improve the PSRR and decrease some rail voltage ? what is the drawback of using inductor in power supply ?

Cost; and if you _want_ voltage drop an inductor gives "small" voltage drop.

> spike as attached description

The raw power line does have spikes. However a well-filtered audio power supply has some transformer resistance and large capacitance. Spikes on the filtered DC are typically VERY small.

nigelwright7557 29th February 2012 11:58 PM

Just buy the right transformer.

Or buy a 5VAC transformer and wire it in series anti-phase with your transformer.

guitar89 1st March 2012 04:36 AM

Quote:

Originally Posted by PRR (Post 2928477)
BC550 is rated 50V.

Rated 45V for Vce. The VAS is resposible for full swing, from + to - (mine is dual supply amplifier), so it is exposed to over voltage if maximum swing is over +-22.5V

Quote:

Originally Posted by PRR (Post 2928477)
Yes, 7818 will drop 27V to 18V. One chip one capacitor and maybe a little scrap metal.

well.... 18V is a bit out of my limit. Intended for lower limit of 20V. anyway, its more expensive than zener. However, is it better performance than simple zener regulator ? Another question is that, are regulator of same unit able to provide negative voltage without complex ? (For my information, need another regulator that design for minus voltage ? zener just reverse direction then become minus)

Quote:

Originally Posted by nigelwright7557 (Post 2928490)
Or buy a 5VAC transformer and wire it in series anti-phase with your transformer.

Co-incidentally, there are no good transformer at the rate with approx same price, for slightly lower voltage I need. So I just brought the current one. (last time didn't think of anything, then brought a 40V, worst ! ) Also though that slight regulation will increase the clipping and PSRR.

guitar89 5th March 2012 01:36 AM

Quote:

Originally Posted by Osvaldo de Banfield (Post 2927943)
Search in any datasheet, the recommended minimum current in the zener under question. Add the load current to it. So, then knowing voltage drop in the resistor, calculate the resistor value and power loss in it. Finally, choose a smaller normal value resistor, and the power of such slowly greater than it. Example: supuose diode current 10mA, load current 50 ma and voltage drop 15V. Then, Ir = 10mA plus 50mA = 60mA.

R = 15V / .06A = 250ohms. Choose 220R.
P = 15V * .06A = .9W, choose 1W.

If the power is not so big, you can also think in a TL431 voltage regulator, that has lower output resistance. But calculus is a bit more complex.

Good luck.

Thanks for your advise, i think it is good for my solution. But I want to consult you about the whole setup. For my case, I'm dropping voltage from 27.4V-28V to 20V using Zener diode. And I get from simulation, the value of usage is 20+mA so for safe assume is 25mA.

so my setup in kinda involve filter/reservoir capacitor(I have some 2.2mF Caps or 1mF is sufficient ?) for 20V rail (regulated by zener) installed after resistor. What should be the value of resistor ? 0.5W 200ohm would doing good job ?

This are the following item :

ChrisA 6th March 2012 12:43 AM

Quote:

Originally Posted by guitar89 (Post 2927713)
I'm currently planning an amplifier, and it seems that I couldn't find a good transistor that have high enough Vce and other specs I wished, so I only have my BC550C/560C to use, which is lower Vce. As my rail voltage from my transformer after filtered and rectified, it is roughly 27V-28V, I need to lower it to something from 20V-22V to cope for BCs.


A regulator could work. But don't use anLM7818. Use the 317 adjustable reg and then adjust it to exactly what you want. Use one of those 15 turn trimmer pots to set the voltage. There is a negative version of the 317 also. But these chips will only pass about 1 amp. Is that enough. If not you will need a large size pass transistor and a large heat sink in addition to the regulator. So if this amp needs more then about an amp then I'd just replace the transformer.

Look at the data sheet for the LM317 it will have some good examples of how to do the above.

Also, you could simply use a resistor but I'd take advantage of it and put it to work rather then simply burning off power build a CRCRC" filter, This will not only drop volts but will dramaticaly cut the ripple voltage in the raw power. Not as much as a regulater would but it can handle more amps at lower cost. I'd use the CRCRC for larger currents and the LM317 for a smaller supply.

In the above "CRCRC" the lat "C" is the filter caps you have now, the two "Rs" are resistors to drop voltage use two of them to spead out the heat. the the other two "Cs" can be anything, say 100uF


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