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Old 1st October 2006, 11:40 PM   #11
john65b is offline john65b  United States
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JesseG -->

How can you get +/-34 at your chip with a 38.5V 0V 38.5V trannie? After your 12,000uF caps you should be at approx +/-54VDC, right?

Now what am I missing here? Do I need to go back to EE101 again?

And the LM4780 max voltage was 84V (+/-42V), which is a tad more than the LM3886...not that 2V matter much here...


I have thought of taking the 40V 0 40V Tranformer connections right after the rectifier (should be a bit under 40VDC here - no smoothing caps) direct to the two 2200uf caps right on the chip (Cs - just like how the real gaincard has it - no smoothing caps)...this supply voltage would be in compliance with the LM3886 datasheet...

But I really don't wanna blow the chip, tho...kinda gettin attached to it...its my first foray into soldering all components right on the chip (exact component values of the gaincard - except no electrolitic input cap - what were they thinking...I have a 2.2uF polypropolyne cap here)

Anyone?
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Old 2nd October 2006, 07:15 AM   #12
AndrewT is offline AndrewT  Scotland
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Hi,
the AC voltage from your transformer is rectified by the diode bridge.
The capacitors after the rectifier TRY to charge up to the peak voltage of the AC waveform. That's where the square root of 2 comes from.

40Vac has a peak voltage just about 58V. The bridge rectifier drops about 1.4v leaving about 56Vdc across the smoothing capacitors. As mains voltage fluctuates the transformer output will also fluctuate (in the UK our tolerance is -10% and +6%)
The transformer will also produce more voltage when on a light load. A small transformer is worse here, compare the regulation, it varies from 4% for 1000VA to about 30% for 10VA.

Now you need a transformer and rectifier and smoothing capacitor to make a PSU so you are lumbered with the root 2 times Vac rule.

The maximum voltage of the chipamps is relatively low but you would expect them to perform when the incoming voltage is normal and at the two extremes I quoted. You have to ensure that after taking regulation and maximum mains that you do not destroy your chipamp.

There is one other design decision to take account of. Load impedance.
The chipamp datasheet will specify the maximum voltage for each load and you will see that as load impedance falls that the chipamp maximum voltage also falls.
Many speakers are 8ohm but some are 6ohm and a few are 4 to 8ohm due to using a low sensitivity 4ohm bass unit in conjunction with an 8ohm treble unit. Be carefull to use a PSU supply voltage that suits the load YOU will be hanging on the end.

There is a very expensive alternative, the choke regulated PSU, but it has a very major drawback, namely, if the current draw drops below a design limit the output voltage rises up towards the waveform peak. It works very well with ClassA designs that have a very high quiescent current and tends to work better when the maximum current demand is very low, ideal for Tube designs. Forget it for solid state.
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Old 2nd October 2006, 10:00 AM   #13
Nordic is offline Nordic  South Africa
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Andrew any idea how to bring the dropout voltage of the actual chip into the equation too?

It seems to specify 2 and 3V, but I can't figure whicc applies when, or how to calculate mamimum supply ( not that I'd ever want to drive one at max volts).

the datasheet note says:
Note 12: The output dropout voltage is defined as the supply voltage minus the clipping voltage. Refer to the Clipping Voltage vs. Supply Voltage graph in the
Typical Performance Characteristics section.
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Old 2nd October 2006, 12:30 PM   #14
AndrewT is offline AndrewT  Scotland
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Hi,
if one were designing an amplifier for a customer and had to meet a particular power spec. then a first guess would be based on the various voltage drops that occur in the system from transformer to output.
Based on that one builds the prototype and tests it.

For your build the loss through the chip is unimportant. It simply makes power predictions more difficult. The other voltage drops will be much more significant.

Now, if you were to produce a model of maximum power vs all the other variables then it would be instructive, but is it any more useful when most of the design decisions are out of your hands and made by the chip designer?
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Old 2nd October 2006, 12:35 PM   #15
john65b is offline john65b  United States
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Thank you Andrew.

You seem most knowledgable in power supplies, as this may be your bread and butter. I work with many EE's, but they do not seem to be able to remember and be able to help much of what I ask - they are mostly involved with control systems, COGEN Units, and high voltage power transformers.

Anyway, if your read the beggining of my post, you will see the first question I raised about hooking the 40V 0V to the rectifier to get the same result as the 20V 20V. Can you comment on if this is possible?

Also, as JesseG mentioned, is there truely a way to get a 38.5V 0 38.5V trannie rectified/smoothed with a 12,000uF caps to be able to get +/-34VDC with nothing fancy inbetween (LC filter or choke)? From your post, we are in agreement and should not be possible.

And the size of the caps on the downstream side of the rectifier does not matter as far as far as outputting peak voltage - in other words, post rectifier to 12000uF or 1000uf, they both will act as smoothing caps and should output peak voltage (x 1.414)...not less with smaller caps.

Sorry if these questions are too basic, but I am sure there are a few of us out here that could use the education.

Thanks
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Old 2nd October 2006, 01:39 PM   #16
Nordic is offline Nordic  South Africa
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The caps are charged at 100 or 120 times per second depending on whether you have 50 or 60Hz power...

The size and quality of the cap will influence how quickly they will charge. All that said, the moment the amp tries to draw more current than what the PSU is able to store and supply the voltage will drop...
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Old 2nd October 2006, 04:12 PM   #17
AndrewT is offline AndrewT  Scotland
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Hi John,
I might be a techy teacher but my bread and butter was Civil Engineering.

This simple answer assumes no regulators.
+-34Vdc from 2*38.5Vac is not possible. Ignore that poster until he can prove his assertion.

Have you considered rewinding the secondary to get your lower voltage?

Nordic,
your simplistic view on voltage available from the PSU is misleading.

Think about the emf available from the source (transformer) and then the string of resistances from source to load and all the way back to the source.
Now apply ohm's law and calculate the voltage drops when no current is flowing. Increment the current and again calculate the voltage drops. Subtract the voltage drops from the emf and the remainder is the available voltage to drive the amplifier.
You could get very sophisicated and draw a graph of current vs voltage drops and (emf-voltage drops) to see how the available voltage varies with current draw.
You will see that voltage available starts to fall as soon as even a tiny current is drawn by the load (the load includes the amplifier quiescent current). There is no point at which the voltage holds steady and then suddenly starts to fall when a threshold is reached, unless your threshold is set at zero.

I just lost my addendum on ripple voltage and the charging pulse current voltage loss. But you can work those ones out.
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Old 2nd October 2006, 05:57 PM   #18
john65b is offline john65b  United States
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Andrew,

That's what I thought about the previous post. Looks like I don't need to re-review my EE classes after all (I am an ME working 25 years in the Petroleum refining industry). Great hobby here BTW...

I should just get the correct transformer instead of try to get this one to work...

I thought about the Unwinding of the toroid, but I would need to remove the nice plastic wrapping...and I would also assume that unwinding the wraps on the toroid to reduce voltage would also reduce VA rating? If I had to derate (unwrap) the windings from 40V to 25V (is that almost 1/2 the winding?), that would mean I would have about 250VA? Not that it matters that much, since 400 is way more than I need, and 250 is fine...

As you can see from my questions, I have never unwound a torroid before, so a bit leary. Is it that cumbersome? Will I need to re-wrap it? If so, can I re-wrap with packaging tape? Common sense says just unwrap the 40V windings and leave the centertap windings alone?? Do I just unwind, cut wire, scrape off the coating and test the leads to the centertap until I get to 25V?

Thanks in advance for your help
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Old 2nd October 2006, 06:07 PM   #19
poobah is offline poobah  United States
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John,

Unwinding is not that big of a deal.

Your VA rating will remain intact (in fact, it will increase slightly).

You do not need to replace the tape... if you do just buy some polyester (mylar) or kapton.

Ideally the remaining winding should be redistrbuted evenly. Forget about it... you will get a slight increase in leakage inductance and flux... neither of which will matter for a PSU.

Both halves of the secondary are generally wound at the same time... side by side... "bifilar". It is the connections that creates the center tap... you'll see this as you remove the tape. You will have to recreate the center tap when you're through... very simple... make a sketch BEFORE you take things apart. The center tap is created by joining the START of one winding with END of the other... pretty hard to screw up, but it is entirely possible to SHORT the a winding by connecting its own start to its own end. After you break connections, ohm through and paint or label one of the windings.



P.S. Don't unwind all the way to your desired voltage you should be about 5% high to allow for voltage drops that will occur when the trans is loaded.
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Old 2nd October 2006, 06:53 PM   #20
Nordic is offline Nordic  South Africa
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Andrew, I stand corrected.
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