Help me debug the transformer sizing for my lab psu

[pitagoras]

Hello,
I design a LAB PSU.
Let's say it will be capable out providing regulated 24V@2A.
It will be linear regulated.
It will have a pre-regulator to lower power dissipation in the linear stage.
The problem is choosing the power transformer and capacitor for the standard 4 diodes unregulated supply.
Let's say the linear regulator will have a drop of 2V.
Let's say the pre-regulator needs at least 2V drop.
So, 24V at the PSU output, 26V at the output of the pre-regulator, 28V as the minimum in the unregulated power supply.
Let's say I use a 5000uF capacitor. At 50Hz fully rectified, will be recharged every 10mS. At 2A it gives me
2A * 10ms / 5000uF = 4V.
So the peak has to be 4V above the 28V "minimum". That is 32V peak.
Plus the 1.5V drop of the rectifier bridge, 33.5V
33.5V peak are a bit less than 24V RMS, right?
So I buy a transformed rated at 24V 2A and everything will be fine. Hm.
I guess not. Because 24V@2A is 48VA, which is hopefully 33W. And my supply will provide 48W.
What is wrong with my reasoning?
What other considerations I am missing?









 

[orolo]

You have 24V@2A (rms) at the secondary of the transformer. And you want 24V@2A dc at regulated end. That's not possible from the energetic point of view.

Imagine your transformer provides V_i@I_i (peak) ac current, and that you rectify it without losses. The energy carried by a cycle of the rectified wave is \$\int_0^{\pi}V_iI_i\sin^2t\,\mathrm{d}t = \frac{\pi}{2}V_iI_i\$. In the same period, the energy carried by a dc voltage Vo at current Io is \$ \pi V_o I_o \$. If you equal them, you get: \$ V_o \le \frac{I_i}{2I_o}V_i \$.

If we apply this to your supply: 24V@2A (rms) = 34V@2.8A (peak). You want 2A of output dc current, so the maximum attainable dc voltage will be:  Vo = (2.8/4)*34 = 23.8V.  About 24V, which was expected: disregarding losses at rectification and stepping down it would be a perfect transfer of energy.

With your linear regulator scheme we can estimate what you can get out including the losses. First, the diode bridge removes 1.5V from the input, so you are left with  32.5V@2.8A (peak). Then the linear regulation, working at 2A dc and dopping about 4V will remove 8W power from the wave, so, adding all the energies in a cycle: you have 32.5*2.8*3.1416/2 = 142.94J/cycle incoming. Remove 8*3.1416 = 25.13 from losses, so 117.81 remain. Divide by pi times the dc current: 117.81/(2*3.1416) = 18.75V maximum at 2A. You don't have enough input energy per cycle to output 24V at 2 Amps. You need either more input current, or more input voltage.

[David Hess]

Another way to think of this is the power factor.  The power factor for a bridge rectifier and capacitor input circuit is about 0.7 so only 70% of the transformer power is available and the current rating for the transformer needs to be 40% higher.

In theory the switching regulator could be configured for active power factor correction after the transformer to deliver 100% of the available power allowing a smaller transformer but I have never seen this done.

[Ian.M]

Assuming you don't use a buck switching preregulator, its even worse.  The current waveform in the secondary has a high crest factor due to the bridge only passing current near the peaks of the voltage waveform.  A simple LTSPICE simulation shows that anything under 24V RMS 80VA is going to be unsatisfactory.  Also that makes no allowance for low line conditions - if you need the PSU to be able to produce its rated output with the mains input 10% low, you'd better use a 26V RMS 100VA transformer with double the reservoir capacitance.

This simplistic model does assume that the transformer regulation can be modelled as a resistive loss transferred to the secondary, but that should be good enough to size the transformer appropriately.

Code: [Select]

.model 1 D(Ron=1u Roff=1G Vfwd=0) ;Ideal diode
.model 2 ako:MURS320 ;Si 3A diode

;.step param D list 1 2
.param D 2defaults the diode model used to MURS320, but if you remove the ; commenting out the .step, it runs the sim twice, first with an ideal diode and then with the MURS320.so you can see the typical diode drop for the bridge.   Unfortunately, LTSPICE cant average or calculate the RMS (ctrl-click a waveform legend) for .stepped simulations, even if only a single step is selected, so if you want to do stuff like check the RMS current in the secondary V1, you need the .step commented out.

Comment out the .ic and change the .tran simulation command to .tran 0.2 to see the initial inrush current to get an idea if you are going to need a soft-start circuit.  If you need better resolution on the current peaks, set the max timestep to 20us., at the expense of sim running time, but that's pretty much gilding a turd as the transformer model isn't good enough for accurate numerical modelling, which will also be highly dependent on the actual diode used.  Worst case, with a 10000uF cap and a 26V 100VA transformer with the line 10% high, its going to be over 50A, so you are either going to need a very beefy bridge or a soft start circuit.

[Kleinstein]

Due to the pulsed current into the filter capacitor the rated transformer current needs to be higher.

The larger the filter caps and the lower the transformer impedance (e.g. larger toroid type), the higher the peak charging current and thus lower the power factor. As a conservative estimate you need an AC current rating nearly twice as high as the DC current you get.

With a switched mode pre-regulator one use a relatively small filter cap and allow the switched mode stage to compensate for the ripple. So instead of something like 26 V and 5000-10000 µF of filter cap one could alternatively choose a 30 V transformer and maybe 3300 µF or maybe even 2200 µF filter cap. This would ease a little on the current pulses and allow for a slightly lower transformer rating.