Help me design a 13.8v 5A low noise linear cheap power supply

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Cathodes together .... to the bulk capacitor "+".

Klaus

Here is you power supply in the free download LTspice...just hit run and watch it...use it to size your capacitors etc

give me a shout if you want it re-done with a split coil output rectifier...or you could just put it in yourself.

neazoi said:

I am not sure if I really need a pi L/C filter before the LT1083, I think this is too overkill as most commercial receivers should cope with a little rip. I do not know.

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You are correct that a pi filter can increase the source impedance of the power supply. In fact, a small inductor in series with the mains (this may reduce the surge current of the transformer) and a MOV across the primary of the transformer (reduce the voltage spikes) are often considered a good safety feature. Because this is a relatively low power device, they can be safely omitted.

Some one mentioned a centre-tapped Tx sec as a lower rms solution - unfortunately not so, the rms is 70.7% of the full current in each wdg for a centre-tapped output - so instead one one winding with 100% you have two widings carrying 70.7% - you do win on the diode conduction though - except if you have to swap from schottkies ( <=45V rated ) to Si diodes - generally you can get 0.8V warm for Si, good schottikies can get to 0.55 volt ( but only go to 45V for true schottkies ) - if you have limited winding space on a Tx - you will get more power out for a single output wdg than for centre tap

Easy peasy said:

Some one mentioned a centre-tapped Tx sec as a lower rms solution - unfortunately not so, the rms is 70.7% of the full current in each wdg for a centre-tapped output -

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Once you put the filter capacitor after the rectifier, the picture changes drastically. Every half cycle, the transformer charges the capacitor to the peak voltage (approx) and the capacitor supplies current to the load (approx). The transformer supplies current only for a short time if the load is low. If the mains freq is 50Hz, each half cycle is 10 ms and the transformer supplies current to the capacitor for less than 5ms (typically 2-3ms). Each winding is idle for more than 15 ms every cycle. The actual calculation is difficult to perform because of various reasons but depends on the load.

Your calculations assume that you are using the transformer output directly into a resistive load. The real life is different.

the calcs are pretty standard and 0.6 - 0.65 power factor is surprisingly standard for cap input filters

In any event if the rms in a single sec = 10A, the rms in a CT sec will be 70.7% or 7.07 amps regardless of any other factors - for each wdg of the CT. Since losses = I^2R, if we assume 1 ohm for the single wdg, to make the maths easy, giving 100W losses, then the R's for the CT wdgs need to be 50 / 7.07^2 = 1 ohm each, given that the turns are the same for all 3 wdgs, you can see that 2 x wdgs thick enough to give 1 ohm each, will be double the wdg section of one wdg of the same turns to give 1 ohm.

For the CT wdgs to fit the same space, each must have half the copper area at the same turns = 2 x the R, 2 ohms in this case

so the losses for the CT wdg will be 7.07^2 x 2 ohm x 2 wdgs = 200W, i.e. double for the same wdg space on the Tx

i.e. a CT solution needs twice the sec wdg space for the same losses. This is true for any waveshape.

Alternately a CT solution in the same space can only carry 50% of that current for a singe sec solution. ( 5^2 x 2 x 2 wdgs = 100W )

Easy peasy said:

Alternately a CT solution in the same space can only carry 50% of that current for a singe sec solution. ( 5^2 x 2 x 2 wdgs = 100W )

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Right; but what is the problem?

A 15V10A 50Hz transformer and a 30V5A 50Hz transformer can use the same core because they power ratings are same. Both iron and copper losses should be very similar.

Making the 30V secondary 15-0-15 (center tapped) does not increase the winding size. Even the manufacturer will not charge extra for the center tap. Thinner wires are a pleasure for them to wind.

You will only save for two diode drops (5Ax2V) which may not appear much but it is not insignificant for a 15Vx5A supply.

It is a matter of personal preference.

Making the 30V secondary 15-0-15 (center tapped) does not increase the winding size.

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it does if you want the same VA out ...

Hi,

On a linear power supply using a center tapped transformer will not reduce overall power dissipation.
From transformer to load the voltage drop causes dissipated power, if not dissipated in the diodes it will be dissipated in the regulation circuit.
The benefit is, that the regulation has more room to regulate.

But be aware that each of the center tapped windings has higher ohmic resistance than a single winding transformer. Ideally the factor 1.4.
The pulsed current (seen on a half wave) thus will cause a higher voltage drop in the windings.
Now it depends on the true resistances if the capacitor voltage will be higher or not.

On an ideal transformer I see the benefit, curious if there still is a benefit on a real transformer.

The OP could do some simulation.

Klaus

KlausST said:

On an ideal transformer I see the benefit, curious if there still is a benefit on a real transformer.

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Right; the DC resistance of the winding can be a spoiler.

I have very little practical experience in this business and therefore I was conservative; I advised the OP to get a transformer with 4A rating (instead of 2.5 or 3A). Therefore the transformer will be up-rated to 30Vx4A rather than 15Vx5A. Up-rating the transformer will NOT add to the losses (but cost?)

It is a part of the learning process and it is a good habit up-rate individual parts too.

By the way, I believe that modern transformer technology is excellent to very good. Total transformer losses may be less than 5W at full load.

By the way, I believe that modern transformer technology is excellent to very good

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the technology has changed very little since 1945, copper has the same losses as then, as does the type of steel you find in a common transformer, for maximum power density in a Tx, single wdg output, for less power density in the Tx, centre tap o/p ...

{ foot note: for the same VA out with the same losses, the winding space has to be at least* 50% larger for the centre-tap Tx, compared to a single wdg o/p, which means the core has to be larger, to fit around the increased wdg space - and then because losses in the core are per kg - the core losses will therefore be larger - and for a larger core the mean length of turn for windings will go up* - which means even more copper is required to match the lower losses of the single o/p wdg .... }

this effect can be manged better on a toroidal core than it can be on an EI laminated core ...

Easy peasy said:

{ foot note: for the same VA out with the same losses, the winding space has to be at least* 50% larger for the centre-tap Tx,

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This has been a children's puzzle: how many circles can be packed into a square (or rectangle); a three dimensional problem also exists: how many spheres can be packed into a cube (crystal packing). For the two dimensional problem, there are two possibilities: square packing (circles on top of each other) or triangular packing (second row circles go into the valleys). The packing factor is 78% in the former case and 84% in the triangular packing).

The problem is best explained with an example: say the target is 100 VA; one case it is 10V at 10A and the other case it is 20V at 5A. 20V is split as 10-0-10 with both halves rated for 5A. To get 20V you need double the number of turns (as for 10V)

Note that the current capacity of the winding wire depends only on the wire cross section. Therefore the cross section of the 10A winding wire will be exactly 1/2 of the cross section for the 5A winding wire.

Let us say that the winding window is A; cross section of the 5A wire is a1 and the cross section of the 10A wire is 2*a1. Now many turns can be put in the same window area in each case?

I rest my case.

Hi,

So: 2 × winding_count × 0.5 × winding_cross_section
Still gives the same area.

Triangle winding density applies on both transformers, too.

I see no difference. ..except additional isolation layer, or maybe cross aligned wires...


Klaus

c_mitra said:

Therefore the cross section of the 10A winding wire will be exactly 1/2 of the cross section for the 5A winding wire.

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Um... - not really ....

Also:

one case it is 10V at 10A and the other case it is 20V at 5A. 20V is split as 10-0-10 with both halves rated for 5A.

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is only true if the 5A flows in both sec's AT THE SAME TIME, in a centre tapped use, it don't - this can be hard to comprehend for the newbie ...