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Old 14th September 2010, 05:53 PM   #1
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Default Simple Xformer+R (oh, mysterious R!) question

I'm duplicating the basic PSU and regulator designs from Roman Black's website Roman Black's gainclone amp for a LM3886 chip-amp.

He provides a lot of information on the post-rectifier design (and I understand the regulator design - very simple and sensible design commonly used) but doesn't comment on the 3R3 resistor before the primary transformer.

Why would you put a small value (though not that small - 3 ohms is nothing to be ignored) resistor in series in the live (in this case 240v 50Hz) part of the circuit?

I understand how much energy dissipation would occur if a resistor was the only load on the circuit but am less than familiar with how much such a resistor would be expected to dissipate in series with this, a rather theoretical load... or what it's doing there to start with. The amp chips can draw, in theory, (60+60)*2and-a-bit watts (as a general worst case guide) but it's very unlikely to be more than 20+20W (efficient speakers, small room - my Warfedale bookshelf speakers will blow at more than 40W+40W anyway and I can't begin to imagine how loud that would be!)...

I'm using a 30V toroidal transformer rather than Roman Black's 42V. Thus there's a bit more current in the circuit.

Any comments appreciated. I'm planning on just leaving the resistor out: but if it's useful, I'd like to know the theory behind the design. The design and web page are so old I'd prefer to ask here first rather than send the guy an email that'll never get answered. *sigh* I hate not knowing answers to questions even my basic knowledge of theory should be able to answer (but can't, obviously).
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Old 14th September 2010, 07:01 PM   #2
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Probably for turn-on surge current limiting. There's better ways to do that (e.g., thermistor).
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Old 14th September 2010, 07:53 PM   #3
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As Sy says, it's a poor effort to reduce the turn on transient that is always present in high end amplifiers.

What the author is trying to achieve is a low start up current to protect the rectifier diodes. Any transformer that is directly connected to a large set of reservioir capacitors will initially see a very high inrush or initial charging current to these capacitors. If the transformer is large enough and the diodes are large enough then this will be largely irrelevant in a Class A/B Gain Clone.

In a true High Power Class A design this becomes a different issue. A typical 100W Class A amplifier may use a 500VA transformer and (mine uses) 500000uF (yes 0.5F) of capacitors. The inrush current is sufficient to trip the mains circuit breaker for the whole house.

In this instance it is usual to limit the turn-on surge current.

There are two distinct methods of achieving this. One is to use high power thermistors which have a cold resistance of about 150Ohms, they heat up within milliseconds and then become almost short circuit. OR. You can use a relay and a timing circuit to apply about 150Ohms of series resistor to the primary circuit for about 10mS then use the relay to short out the resistor. Once the capacitors are charged you really want the full mains potential across the transformer.

Last edited by Andy5112405; 14th September 2010 at 07:56 PM. Reason: Spelling mistakes
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Old 15th September 2010, 12:08 PM   #4
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Thanks for all that. Appreciated. Am only using 38,000uF + dual rectifiers rated 35A each (they're very cheap, so why not) so no foreseeable issues with inrush current.
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