transformer power supply for audio amplifier

[devil6600]

what is BR1 in attached schematic? is it a diode bridge. how to make a diode bridge which can output +25v and -25V DC.

 

[rfredel]

Hello devil6600,

yes, it's a bridge rectifier.

Look for the circuit of a bridge rectifier. You have a center tap transformer. The ct is use as GND. The bridge is use as two-way rectifier. The cathodes of two diodes will deliver the positive voltage and the anodes will give the negative voltage against the ct-point of the transformer.

The transformer delivers 18V. After rectifying and the reservoir capacitor the voltage will be U= = sqrt2 * U~ = 1,414*18= 25V.

Regards

Rainer

 

[BradtheRad]

how to make a diode bridge which can output +25v and -25V DC.

That is the same thing as your schematic shows: a bipolar supply.

The peak that you can get from it is +24 and -24 V, however. There must be little or no load.

The more load you add, the more you reduce amplitude of the supply V.

There are ways to boost the volt level. Example, voltage doublers.

 

[godfreyl]

how to make a diode bridge...

You can make one with four diodes, but it's easier to just buy a bridge. Here are some examples: https://za.rs-online.com/web/c/semiconductors/discrete-semiconductors/bridge-rectifiers/?sort-by=P_breakPrice1&sort-order=asc&sort-option=Price&applied-dimensions=4294767855&lastAttributeSelectedBlock=4294957617

One rated at 6 amps should be fine.

 

[rfredel]

Hello BradtheRad,

There must be little or no load. The more load you add, the more you reduce amplitude of the supply V.

That is not true. The voltage declaration at a transformer output is for max. load. The open-circuit voltage is dependent from the internal resistance of the winding and from the size of the transformer. Small transformers can deliver between 0,8 and 2 times more voltage without load.

This power supply every time delivers the peak voltage of the transformer.

Real is, that the ripple voltage rise up, if the reservoir capacitor is to small. Here is the empirical formula to calculate the capacitor 1000µF / A.

There are ways to boost the volt level. Example, voltage doublers.

Why, look what I write, or make a good calculation of the PSU. Voltage doublers can not deliver so much current, only a few mA.

Regards

Rainer

 

[BradtheRad]

That is not true. The voltage declaration at a transformer output is for max. load. The open-circuit voltage is dependent from the internal resistance of the winding and from the size of the transformer. Small transformers can deliver between 0,8 and 2 times more voltage without load.

This power supply every time delivers the peak voltage of the transformer.

Real is, that the ripple voltage rise up, if the reservoir capacitor is to small. Here is the empirical formula to calculate the capacitor 1000µF / A.

Since the charging waveform is at the peak value (1.414 x nominal V) for just a brief time, this means the capacitor can only provide power to a slight load at the 1.4x level.

To be correct, we must subtract two diode drops. That is why I said the 18V transformer will produce a 24 V supply, since it must go through a full-wave diode bridge.

It can provide a little more power at the 1.3x level. And even more power at the 1.2x level. Etc. Because the charging waveform provides sufficient voltage, and for a long enough time, to be able to sustain the capacitor at that level.

That has been my experience.

Voltage doublers can not deliver so much current, only a few mA.

I believe that since one round of diodes and capacitor can carry several amps, nothing prevents a second round of diodes and capacitor from doing the same thing, in the proper configuration.

 

[rfredel]

Hello BradtheRad,

all you write is ok for someone in school or university, who has to do an examination, but it isn't good in practice. For real work you don't need exact voltage. You know, that the line voltage is not constant. Here in Europa the line voltage can vary between + and - 5%. So the output can also vary in the same value. If exact voltage is need, it is better to use a regulator or stabilization.

Since 40 years I work in developing and constuction of audio devices. Never I caculate PSUs like you do only in University. I never get problems with the PSU.

Please consider that an amplifier not every time must give full power, so that the supply voltage goes up and down. So only have a look that your reservoir capacitor has enough capacity to hold the voltage in a given tolerance.

Regards

Rainer

 

[BradtheRad]

Since 40 years I work in developing and constuction of audio devices. Never I caculate PSUs like you do only in University. I never get problems with the PSU.

Please consider that an amplifier not every time must give full power, so that the supply voltage goes up and down. So only have a look that your reservoir capacitor has enough capacity to hold the voltage in a given tolerance.

I'm delighted to come across someone who had real-life practical experience with audio equipment.

Like you, I too have observed that a sufficiently large smoothing capacitor will carry a load through a brief loud sound.

So I won't argue with you on that point.

I have also observed, on my own oscilloscope, with my own eyes, that the volt level drops if the loud sound goes on longer than a brief time.

Then when the sound is quiet, the volt level rises back up, although never higher than 1.4 x the transformer's nominal AC V, minus diode drops.

 

[rfredel]

Hello BradtheRad,

I'm delighted to come across someone who had real-life practical experience with audio equipment.

If you want, I'll share my experience with you or someone else. I know, that so much experience and knowledge is going lost and so I'm happy if I can help.

So I won't argue with you on that point.

Also I don't like to argue, because I'm very peaceful and friendly. :-D

I have also observed, on my own oscilloscope, with my own eyes, that the volt level drops if the loud sound goes on longer than a brief time.

Then when the sound is quiet, the volt level rises back up, although never higher than 1.4 x the transformer's nominal AC V, minus diode drops.

That is Ok, and their is no diskussion about it. Only I use a empirica formula to calculate the voltage. So I use only 1,414 instead of the real value of sqrt2 and I don't take care of the forward voltage of the diodes. It will be only ~1,4V less. This voltage is less then the worst case of the dropping of the line voltage.

This is what I want to tell you.

Regards

Rainer