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Old 9th July 2001, 09:15 PM   #1
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I have been reading the National Semiconductor datasheets for the LM317 series of regulators and saw a schematic that caught my eye. The AC from the secondary of the transformer was regulated (2 regulators required), therefore the reservoir capacitors downline of the rectification should see smaller troughs in the incoming waveform & will therefore recover charge more quickly.

I suppose it must also be possible to put a single regulator immediately after the rectifier and then smooth that.

The regulators would current limit and prevent any reservoir capacitor stressing at switch on, but with the big caps downstream of the regulator there should be oodles of current available.

Just a thought....Anybody ever tried any of this or is it just not worth it?
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Old 10th July 2001, 12:16 AM   #2
Geoff is offline Geoff  United Kingdom
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Jason

The circuit you are referring to acts as a peak limiter on the ac waveform so the end result will be a much lower maximum voltage on the smoothing capacitors, and therefore a much lower available dc voltage. It will do nothing to reduce the 'troughs' in the incoming waveform. These will still drop to 0V in line with the rectified ac waveform.

Geoff
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Old 10th July 2001, 12:08 PM   #3
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Default Post Regulator Reservoir Capacitors

I agree that the voltage between cycles must ultimately fall to zero, i also agree that the regulator will limit the peak voltage supplied to the capacitors, but say for example you require a 30 volt rail and put 40 into the regulator which loses 10 volts, the amount of power available to the capacitors must be greater than simply using 30 volt rectified by virtue of the fact that the discharging time between the (now flattened) peaks is reduced.

The post regulator waveform (without caps) would look someting like a curve-sided square wave (which i guess could bring its own problems).

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Old 11th July 2001, 09:39 PM   #4
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Default Re power?? physics explanation

Jason,

some 'simple' physics tells your sort of right. Let me explain:

First: voltage * current = power, nothing fancy about that.

Second take a pure resistor R as load.

With a normal sine-wave voltage V there is sine-wave current through the resistor.
To calculate the power during one cycle you should take the Integral-sum for all t of [(V sin t) squared] divided by R (which I don't have nearby).
In the case of the flattend wave-form this integral-sum does add up to a larger amount of power.
The maximum power is produced when the voltage is at its maximum during the whole cycle (which we normally call DC).



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Old 11th July 2001, 09:48 PM   #5
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Default Re power?? cont'd

Jason,

there are some penalties however.
1: total power supply power-consumtion rises also.
the voltage above 30V is turned into heat.

2: the squar'ish wave-form can be broken down into various sine-wave with different frequencies (fourrier analysis) [called harmonics].

You want to make sure none of the 'harmonics' bleed through the PS to the amp.

In theory increasing the frequency of the main-input voltage has the same effect.

Jos

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Old 11th July 2001, 11:00 PM   #6
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Default Reply to Dutch DIY

As previously mentioned the waveform would be similar to a sqaurewave with convex curved rising and falling edges - the mark/space ratio would be high and would as previously mentioned have higher power content than a sinewave that only peaks at the capped value.

I understand that there is potential for some of the harmonics to filter through, but sufficient reservoir capacitors would prevent that and they would have less time to discharge than if using sinusoidal waveforms.

As for the heat - if you're running single ended class A then another few watts of heat is barely a consideration!
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Old 12th July 2001, 07:24 PM   #7
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Default Re: physics

Jason,

I did a check on the calculus:

* with a plain sine-wave and pure resistor the power is 0.5 * (V squared / R) where V is the peak voltage of the sine wave.
* in the case of + & - 40V peak and + & - 30V clip the power is 0.66 * V (30v) squared / R, an improvement thus of 30% over a normal sine-wave.

good luck.

Jos
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