Choke Input Filter—Watts up with that?

[KD1SH]

  Okay, haven't seen any dumb questions here for a while; can't let that continue, can we?
Looking at Hammond's rectifier guide, a very handy thing:
https://www.hammfg.com/files/products/700/hammond-recitifier-guide.pdf?v=1697662070
Note in the attached image that:
IDC, the current into the load, is shown to be 1.54 times the secondary AC current.
  The average DC output is .45 times the primary AC voltage, since we're using the center tap.
Assuming a transformer secondary with 2000 volts between each end—1000 volts from each end to the center tap—we'll have a rectified and filtered DC output of 900 volts across the load. No problem there.
  But here's the problem: assuming we pull 250ma through the load, that would mean a total energy output of 900 X .250, or 225 watts (ignoring series resistances and such for simplicity). Going by Hammond's numbers, the actual secondary AC current must be 250ma divided by 1.54, for 162.3ma, giving us a total of 162.3 watts on the secondary side. So, 162.3 watts available at the secondary, and 225 watts available at the output of the rectifier and filter. Watts up with the extra Watts?
  Okay, a just-before-hitting-the-post-button epiphany: secondary voltage is given in RMS, but the filter—both choke and cap—are responding to peak voltage. I think I've answered my own question. Must be the heat; it's as good an excuse as any.

Edited at 17:05 to correct an embarrassing math error.

[W7TFO]

Yep, .707 vs. 1.414 

73DG

[K8DI]

No free lunch…  The ideas are fine except the Hammond pic says Sec. V, not half of Sec V….2000 not 1000 volts…162.3mA times 2000 volts, 325 watts.  That times .707, as it is peak secondary current, not rms…

Ed

[KD1SH]

  Indeed, no free lunch. In a purely DC circuit, more current leaving a circuit element than what went into it would have Mr. Kirchhoff turning in his grave. The same prohibition of free ham sandwiches must apply to AC circuits as well, but we've got to pay attention to the measurement methods—RMS vs. peak vs. average—or we get hoodwinked.
  Hammond's drawing looks okay to me; it accurately depicts the secondary VAC as appearing between both ends of the winding, and then declares that the DC output of the filter is 45% of that. Referenced to the center tap, of course, the secondary VAC would be half of the full value, but Hammond is more concerned with the output of the filter.

No free lunch…  The ideas are fine except the Hammond pic says Sec. V, not half of Sec V….2000 not 1000 volts…162.3mA times 2000 volts, 325 watts.  That times .707, as it is peak secondary current, not rms…

Ed

[KD1SH]

  Exactly! And don't forget about 0.636, the average of the peak value of the sine wave. That was the giveaway. Why is the DC output 45% of the full secondary RMS voltage, rather than 50%? Because the 45% number gives us the average of one half of the full secondary voltage.
Full secondary = 2000 Vrms.
Half of that (because of the center tap) is 1000 Vrms.
1000 Vrms is 1,414 Vpk.
The average of 1,414 Vpk is 1,414 X 0.636, or 900 volts.
Hammond's 45% of full secondary takes us on a little shortcut, and we wind up in the same place.
Basic theory, but I hope I never cease to be fascinated by the magic of it all!

Yep, .707 vs. 1.414  

73DG

[Tom WA3KLR]

This is a simulation of a lower voltage receiver supply, but note the current waveform from the 1/2 secondary. It is essentially a square wave with a 120 Hz ripple imposed on it. The current snaps to the average value of the output current right away, due to the action of the choke input filter. The true power of each half cycle is not intuitive.
But we all know of the law of conservation of energy. If the diodes and the choke are lossless, the power delivered to the load resistor has to equal the power coming from the secondary windings.

* I did another simulation using the Hammond values and show the instantaneous power curve for a half cycle operation, whose average value is the true power value. What's actually going on is somewhat complex and you can't just multiply independent voltage and current values.

I didn't dig out the Hammond transformer catalog to see if it is like transformer catalogs I used to see years ago with all the various rectifier configurations and resulting secondary currents versus load current. They do this because the secondary current is not intuitive and they want you to choose the correct power transformer rating for your application.

By the way, in my simulation with ideal components I came up with 1.37 X for secondary current versus load current.

* Trace lines are thickened. The dark blue line is the instantaneous power curve (secondary acv X diode current).
1/2 secondary voltage is pink.
Diode current is light blue.
DC Load voltage is olive drab.

If you take the average value of the instantaneous power curve for this 1/2 cycle, it is 1/2 of the load power.

[Tom WA3KLR]

Changes to post above.

[KD1SH]

  Very nice work, Tom, with the graphing. Yes indeed, once we change from DC to sine, intuition goes out the window. I've found, over the years, that whenever I go from the calculator to the breadboard, the results rarely mirror the calculations. The tolerance stack rules.
  Interesting about the difference between Hammond's 1.54 vs your 1.37 numbers for load vs secondary current. Ballpark, though; as you say, Hammond's intent was to guide us in selecting the right transformer, and since they don't offer thousands of transformers with secondary outputs in 1 volt increments, it's down to hand-grenades and horseshoes—close enough is fine.

[K8DI]

These simulation graphs show an error in the simulation. They do not take into account properly the inductance of the transformer, and show a nearly instantaneous change in the secondary current as the diode begins to conduct. In the same way that the choke takes time to change its rate of current flow (why it works as a filter) so does the transformer's inductance make the secondary current flow take time to change. Whenever you see that super fast rise time and peak, you have to ask yourself whether it would really look like that...think of the harmonics it would have to make to have that kind of rise time and knife-edge waveform...


Ed

[Tom WA3KLR]

Good catch Ed, I didn’t model the transformer I just used voltage sources for the secondary windings so the effective transformer impedance is zero. I was interested in showing the basic diode current waveform.

I was thinking last night to at least add a realistic winding resistance to the simulation and a realistic choke resistance. The diode loss may be close to zero – I did not look at the model for the diode. I can switch to a 1N4007 diode type.  I knew what I put together is basically a zero-loss model – the power delivered to the load is the same as from the simulated secondary.  This results in the 1.37 factor – the lowest possible ratio.  Real additional losses will mean the power from the secondary has to be higher to supply the same output DCV and current. This will raise the current ratio to the secondary windings. The 1.54 factor by Hammond must have been based on typical real supplies. If I now add the losses I mentioned above, it will be interesting to see what current ratio I come up with.

[Opcom]

PSUD II seems to get pretty close. In that, you measure the DCR of the primary and secondary to get a better model of the transformer. But the calculations are straight out of Terman.