C,R//D,C? Catch diodes for the CRC? Sure. I've done it a lot.
With the forward voltage drop of the diode (none are zero), the filter effect never disappears.
Parts:
The diodes usually have to be heatsinked (especially 2 silicon diodes series for ~1.35v or similar scheme can incur the need of heatsinking). The thermal runaway needs bolted down, else your voltage drop shrinks a lot. However, very low power amplifiers (like a T amp) don't require as much voltage drop, 1 diode with 1R can do, and that doesn't need heatsinked.
Probably, you won't need high wattage resistor with such little voltage drops.
Tuning:
You need to to arrange the resistor current versus diode drops so that the harmonic differentiation from the diodes switching to the bass beat happens corresponding to the amplifier run at near maximum, in which case you've cancelled some blare in trade for a more open sound. That is the practical use of this filter. I've found this particular power filter, C,R//D,C quite helpful during MP3 playback.
Caveat:
You don't want the diodes to switch on too early (proportionate the amplifier output). Insufficient catch diode drop at the "R" of the CRC for a high power amplifier could be a problem necessitating increased resistor current for less effective CRC filter. . . exactly the same caveat as an ordinary CRC. A partial solution is to increase the diode drop and that may incur the need of heatsinking the diodes.
Case:
We could say that power supply reservoir size of a CRC power supply could be decreased by either shorting/bypassing the resistor to remove both the loss and the filter, or adding catch diodes to constrain the loss. This infers that the loss of a CRC was present to begin with. Decreasing loss works like increasing transformer VA.
Thanks:
Thank you for reminding me that if I'm going to install any sort of CRC filter (including C,R//D,C), I'll need a bigger transformer to make up for the losses of the filter. Generally, I like the catch diodes for decreasing this problem, but its no substitute for buying a big enough transformer. It seems that only a plenty big transformer avoids the need of epic capacitance.
P.S.
Series filters didn't decrease needed power supply reservoir size. Howabout parallel filters? Will they do it?
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I have seen a lot of old time designers (me included) place a diode and small resistor in series followed by some caps. My thoughts were that the diode will block current being "sucked back". The series resistor would form part of an LP filter also.
Any comment on using both as explained?
I've not tried that one, Daniel - easy enough to do, waffle or not!
What I did try a bit differently but also a bit weird was to add an extra diode between the bridge and the first cap and added a // R in a similar power sharing scheme - I first saw this in one of John Brown's 1541A dac power supplies and tried it on the F3 power amp - a definitely change but not sure if it was an actual improvement - added a further RC snubber across that and seemed a bit "quioeter" - very dependent on the diodes but no discernable effect with different size caps from 4700uF, 6800, 10mF and 15mF, apart from different types (ie Nichi FG, Elna for A, Siemens, BHC, etc)
Tried much of these variations on the supply of the F3 amp that has a Cmultiplier after the supply and quite surprised that nearly all of the variations were quite so clearly audible, altho using the headphones better test.
What I did try a bit differently but also a bit weird was to add an extra diode between the bridge and the first cap and added a // R in a similar power sharing scheme - I first saw this in one of John Brown's 1541A dac power supplies and tried it on the F3 power amp - a definitely change but not sure if it was an actual improvement - added a further RC snubber across that and seemed a bit "quioeter" - very dependent on the diodes but no discernable effect with different size caps from 4700uF, 6800, 10mF and 15mF, apart from different types (ie Nichi FG, Elna for A, Siemens, BHC, etc)
Tried much of these variations on the supply of the F3 amp that has a Cmultiplier after the supply and quite surprised that nearly all of the variations were quite so clearly audible, altho using the headphones better test.
If you want smoothing you want current to be 'sucked back' when appropriate. A diode ensures that sharp upward movements in the first cap get transfered to the second, so no smoothing. I have seen so-called 'diode isolation' in circuits, but I have never been convinced that it works. I suspect it is a popular myth.
There is one place where a diode might do some good: isolating a separate supply rail for the earlier part of the circuit if you want to guarantee that it doesn't droop as much as the main supply rail during LF peaks. The aim is not smoothing.
There is one place where a diode might do some good: isolating a separate supply rail for the earlier part of the circuit if you want to guarantee that it doesn't droop as much as the main supply rail during LF peaks. The aim is not smoothing.
The resistor paralleled to it stops that problem--the diode is off except when the amplifier is near full blast.
Edit: For a larger amplifier this can take more than one diode in series to get appropriate voltage drop.
Tha's a different topic entirely. Left, right channels schottky outputs could be added to the power board for broadening a stereo amplifier. Right, The aim is not smoothing. Apples versus oranges. That's a different circuit.
Perhaps it is useful that the amplifier board bypass caps don't need to fight with all of the other power caps?
P.S.
If I did All of that to a dual secondaries supply, it would have 16 diodes (8 for dual rectifiers, 2 for 1.4v V+ catch diode, 2 for 1.4v V- catch diode, and 4 schottky for outputs). How many audiophile sound effects does one power supply need and who pays that much attention to power supplies anyway? I think a lot of that enhancement is possibly noise, up until trying to replay an mp3, whereupon the enhancements could be of service.
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I think this parallel R and D is a bit of a tangent to the thread, but I must be missing something.
This approach simply allows one to use a larger resistor value for filtering and not incur the IR voltage drop during peak currents.
At lower power there is just an RC filter, for the higher peak currents the 2 capacitors act as one reservoir, the resistor also acts to dampen potential resonances between the two capacitors.
Dont know of the overall benefit, as stated it adds some R to the filter during low power times.
Thanks
-Antonio
This approach simply allows one to use a larger resistor value for filtering and not incur the IR voltage drop during peak currents.
At lower power there is just an RC filter, for the higher peak currents the 2 capacitors act as one reservoir, the resistor also acts to dampen potential resonances between the two capacitors.
Dont know of the overall benefit, as stated it adds some R to the filter during low power times.
Thanks
-Antonio
That's about it.
The resistor can be set same as for an ordinary CRC. In this case the catch diodes (usually a series pair unless the amp is tiny) just make the CRC's voltage drop more predictable despite the reactive speaker load. Reducing loss reduces needed power supply reservoir size.
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Terry,
I guess my question is how to set the value of a component, given the per-unitized expressions.
Using Rs as an example:
Rs_u = Rs/Zsb = Rs x (Ssb/Vsb^2) = (0.18)Rs = Rs/5.551
Since I need to have 100% of Rs when nothing has changed yet, I could either multiply by a fixed Zsb0 that stays constant even if the bases change:
Rs = Rs_u x Zsb0 = Rs0 Zsb0 / Zsb ,
or, I could add a fixed constant:
Rs = Rs_u + c = Rs0/Zsb + Rs0(1-(1/Zsb0)) .
Wait, it looks like L12 (secondary leakage inductance) might be a better example:
L12_u = L12/Lsb = L12*2*Pi*f*Ssb/(Vsb*Vsb) = 67.91 * L12
Using a factor would give
1a) L12f = L12_u x Lsb0 = L120 Lsb0 / Lsb
which is L120 when Lsb = Lsb0,
or adding a constant instead would give
1b) L12c = L120/Lsb + L120(1-(1/Lsb0))
which is also L120 when Lsb = Lsb0.
My brain is just not working yet today because it's not immediately clear which approach might be valid. Or maybe I'm barking up the wrong tree altogether. But 1a seems like it has to be correct, since we're talking normalization/denormalization, which implies factors.
I guess I could see what happens with each approach, when changing a base. I'll try changing VA from 120 to 240:
We have original values of Zsb0 = 5.551, Lsb0 = 0.01473, and L120 = 0.01488.
New Ssb = 240, giving Zsb = 2.776, giving Lsb = 0.007363.
1a above would give L12f = L120 Lsb0 / Lsb = 0.02977 H. (Reasonable-looking)
1b above would give L12c = L120/Lsb + L120(1-(1/Lsb0)) = 2.021 - 0.9953 = 1.026 H. (Whacky-looking)
I'll try 1a's approach.
Cheers,
Tom
Terry,
OK, here's what I got and it appears to almost work!
I might need to add corrections for the voltage drops across the primary leakage inductance, and across the series combo of the secondary resistance and leakage inductance.
I have it set up so that the input voltage is a fixed constant (unless it's manually changed), since in reality it shouldn't change.
; "PER-UNITIZED" PARAMETER SCHEME,
; FOR PSU SCALING:
.param fb0=60
.param Vsb0=25.81
.param Ssb0=120
.param Isb0={Ssb0/Vsb0}
.param Zsb0={Vsb0*Vsb0/Ssb0}
.param Lsb0={Zsb0/(2*pi*fb0)}
.param Vpb0={N*Vsb0}
.param Spb0={Ssb0}
.param Zpb0={Vpb0*Vpb0/Spb0}
.param Lpb0={Zpb0/(2*pi*fb0)}
; SET NEW VALUES HERE:
; Frequency:
.param fb=60
; Ouput Volts RMS:
.param Vsb=36
; Volt-Amp rating:
.param Ssb=120
; Input Volts RMS:
.param Vpb0=115.9
.param Vpb={Vpb0}
.param N1={Vpb0/Vsb}
.param Isb={Ssb/Vsb}
.param Zsb={Vsb*Vsb/Ssb}
.param Lsb={Zsb/(2*pi*fb)}
.param Rs_1u={Rs*Zsb/Zsb0}
.param L12_1u={L12*Lsb/Lsb0}
.param Spb={Ssb}
.param Zpb={Vpb*Vpb/Spb}
.param Lpb={Zpb/(2*pi*fb)}
.param L11_1u={L11*Lpb/Lpb0}
.param Lm_1u={Lm*Lpb/Lpb0}
.param Rp_1u={Rp*Zpb/Zpb0}
Cheers,
Tom
OK, here's what I got and it appears to almost work!
I might need to add corrections for the voltage drops across the primary leakage inductance, and across the series combo of the secondary resistance and leakage inductance.
I have it set up so that the input voltage is a fixed constant (unless it's manually changed), since in reality it shouldn't change.
; "PER-UNITIZED" PARAMETER SCHEME,
; FOR PSU SCALING:
.param fb0=60
.param Vsb0=25.81
.param Ssb0=120
.param Isb0={Ssb0/Vsb0}
.param Zsb0={Vsb0*Vsb0/Ssb0}
.param Lsb0={Zsb0/(2*pi*fb0)}
.param Vpb0={N*Vsb0}
.param Spb0={Ssb0}
.param Zpb0={Vpb0*Vpb0/Spb0}
.param Lpb0={Zpb0/(2*pi*fb0)}
; SET NEW VALUES HERE:
; Frequency:
.param fb=60
; Ouput Volts RMS:
.param Vsb=36
; Volt-Amp rating:
.param Ssb=120
; Input Volts RMS:
.param Vpb0=115.9
.param Vpb={Vpb0}
.param N1={Vpb0/Vsb}
.param Isb={Ssb/Vsb}
.param Zsb={Vsb*Vsb/Ssb}
.param Lsb={Zsb/(2*pi*fb)}
.param Rs_1u={Rs*Zsb/Zsb0}
.param L12_1u={L12*Lsb/Lsb0}
.param Spb={Ssb}
.param Zpb={Vpb*Vpb/Spb}
.param Lpb={Zpb/(2*pi*fb)}
.param L11_1u={L11*Lpb/Lpb0}
.param Lm_1u={Lm*Lpb/Lpb0}
.param Rp_1u={Rp*Zpb/Zpb0}
Cheers,
Tom
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Tom @ #951
Ah, I see where you are going awry, and its conceptual. The "base impedance" is analagous to the load - its the ideal impedance that sucks "base power" from the ideal "base voltage" - we completely ignore all parasitics. In a perfect world that means our actual load is 100% = 1PU. Of course back in the real world everything is lossy, so when you per-unitise the load, it'll be a bit less than 1PU = 100% - say 95%. that pretty much represents the efficiency - and the missing bit (5%) will be made up by the parasitics.
So the transformer impedances (Rs/p, Ls/p_leak) should be very small. I calculated Rs_pu = 5.2% and Ls_leak_pu = 9.5%.
You've gone wrong in assuming Rs_pu should be 100% of Rs when nothing has changed yet. it shouldnt, it should be very small (ideally 0%). you are conflating the normalising and de-normalising processes. If you de-normalise a per-unit value without changing any of the base parameters, you indeed get 100% of whatever you started with (the same is true for the leakage inductances).
Also, if you normalise the primary and secondary separately (Vp_base & Vs_base, same Pbase) then you dont need any complex calculations whatsoever to de-normalise any of the Ls or Rs:
- define Pbase_new, Vp_base_new, Vs_base_new
- derive Zp_base_new, Zs_base_new, Lp_base_new, Ls_base_new
- derive the new turns ratio: N_new = Vp_base_new/Vs_base_new
- directly calculate the new primary impedances:
Lp_leak_new = Lp_leak_pu*Lp_base_new
Lp_mag_new = Lp_mag_pu*Lp_base_new
Rp_new = Rp_pu*Zp_base_new
- directly calculate the new secondary impedances:
Ls_leak_new = Ls_leak_pu*Ls_base_new
Rs_new = Rs_pu*Zs_base_new
HTH
Ah, I see where you are going awry, and its conceptual. The "base impedance" is analagous to the load - its the ideal impedance that sucks "base power" from the ideal "base voltage" - we completely ignore all parasitics. In a perfect world that means our actual load is 100% = 1PU. Of course back in the real world everything is lossy, so when you per-unitise the load, it'll be a bit less than 1PU = 100% - say 95%. that pretty much represents the efficiency - and the missing bit (5%) will be made up by the parasitics.
So the transformer impedances (Rs/p, Ls/p_leak) should be very small. I calculated Rs_pu = 5.2% and Ls_leak_pu = 9.5%.
You've gone wrong in assuming Rs_pu should be 100% of Rs when nothing has changed yet. it shouldnt, it should be very small (ideally 0%). you are conflating the normalising and de-normalising processes. If you de-normalise a per-unit value without changing any of the base parameters, you indeed get 100% of whatever you started with (the same is true for the leakage inductances).
Also, if you normalise the primary and secondary separately (Vp_base & Vs_base, same Pbase) then you dont need any complex calculations whatsoever to de-normalise any of the Ls or Rs:
- define Pbase_new, Vp_base_new, Vs_base_new
- derive Zp_base_new, Zs_base_new, Lp_base_new, Ls_base_new
- derive the new turns ratio: N_new = Vp_base_new/Vs_base_new
- directly calculate the new primary impedances:
Lp_leak_new = Lp_leak_pu*Lp_base_new
Lp_mag_new = Lp_mag_pu*Lp_base_new
Rp_new = Rp_pu*Zp_base_new
- directly calculate the new secondary impedances:
Ls_leak_new = Ls_leak_pu*Ls_base_new
Rs_new = Rs_pu*Zs_base_new
HTH
[EDIT: You posted while I ws posting this. I'll digest your post and come back with changes.]
This seems to work. But I don't really know how to test the resulting transformer to see if it's reasonable, because the output voltage depends on the load. I guess I can try running the standard tests on it, to check its regulation etc.
I'm still not sure that the de-normalization is correct. I used factors like Zsb/Zsb0. Should any of those be inverted?
Also, I had to add an additional change to the turns ratio (in addition to the one for a change in Vsb with an unchanging Vpb), based on accounting for the voltage drop across L12, the secondary's leakage inductance, calculating backward to the ideal transformer's secondary voltage, and then using that (instead of Vsb) and the unchanging Vpb to calculate the turns ratio. But that required specifying a load resistance, to be able to calculate the inductor current. So it seems rather arbitrary.
[My notation is not really correct, for PU, I guess. The parameters that I named like Rs_1u are set to the component values that are used in the simulation. I don't actually calulate the PU value, just the bases and the de-normalized values.]
; FOR PSU SCALING:
.param fb0=60
.param Vsb0=25.81
.param Ssb0=120
.param Isb0={Ssb0/Vsb0}
.param Zsb0={Vsb0*Vsb0/Ssb0}
.param Lsb0={Zsb0/(2*pi*fb0)}
.param Vpb0={N*Vsb0}
.param Spb0={Ssb0}
.param Zpb0={Vpb0*Vpb0/Spb0}
.param Lpb0={Zpb0/(2*pi*fb0)}
; SET NEW VALUES HERE:
; Frequency:
.param fb=60
; Ouput Volts RMS:
.param Vsb=36
; Volt-Amp rating:
.param Ssb=240
; Input Volts RMS:
.param Vpb0=115.9
.param Vpb={Vpb0}
.param N1={Vpb0/Vsb}
; Calculate new secondary bases:
.param Isb={Ssb/Vsb}
.param Zsb={Vsb*Vsb/Ssb}
.param Lsb={Zsb/(2*pi*fb)}
; Calculate new secondary component values for model:
.param Rs_1u={Rs*Zsb/Zsb0}
.param L12_1u={L12*Lsb/Lsb0}
; Calculate new primary bases:
.param Spb={Ssb}
.param Zpb={Vpb*Vpb/Spb}
.param Lpb={Zpb/(2*pi*fb)}
; Calculate new primary component values for model:
.param L11_1u={L11*Lpb/Lpb0}
.param Lm_1u={Lm*Lpb/Lpb0}
.param Rp_1u={Rp*Zpb/Zpb0}
; NEW TURNS RATIO, with adjustment for Vdrop across L12:
;.param Nt3={Vpb0/Vsb}
.param loadavg=2
.param Vsecideal={Vsb*(1+((Rs_1u)*0+((L12_1u*2*pi*fb))/loadavg))}
.param Nt3={Vpb0/Vsecideal}
; Adjust for Vdrop across L11:
;.param XL11={2*pi*fb*L11_1u}
;.param VL11={0.02*XL11 + (XL11*Vsb/loadavg/Nt3)}
;.param Nt3={(Vpb0-(VL11))/Vsecideal}
This seems to work. But I don't really know how to test the resulting transformer to see if it's reasonable, because the output voltage depends on the load. I guess I can try running the standard tests on it, to check its regulation etc.
I'm still not sure that the de-normalization is correct. I used factors like Zsb/Zsb0. Should any of those be inverted?
Also, I had to add an additional change to the turns ratio (in addition to the one for a change in Vsb with an unchanging Vpb), based on accounting for the voltage drop across L12, the secondary's leakage inductance, calculating backward to the ideal transformer's secondary voltage, and then using that (instead of Vsb) and the unchanging Vpb to calculate the turns ratio. But that required specifying a load resistance, to be able to calculate the inductor current. So it seems rather arbitrary.
[My notation is not really correct, for PU, I guess. The parameters that I named like Rs_1u are set to the component values that are used in the simulation. I don't actually calulate the PU value, just the bases and the de-normalized values.]
; FOR PSU SCALING:
.param fb0=60
.param Vsb0=25.81
.param Ssb0=120
.param Isb0={Ssb0/Vsb0}
.param Zsb0={Vsb0*Vsb0/Ssb0}
.param Lsb0={Zsb0/(2*pi*fb0)}
.param Vpb0={N*Vsb0}
.param Spb0={Ssb0}
.param Zpb0={Vpb0*Vpb0/Spb0}
.param Lpb0={Zpb0/(2*pi*fb0)}
; SET NEW VALUES HERE:
; Frequency:
.param fb=60
; Ouput Volts RMS:
.param Vsb=36
; Volt-Amp rating:
.param Ssb=240
; Input Volts RMS:
.param Vpb0=115.9
.param Vpb={Vpb0}
.param N1={Vpb0/Vsb}
; Calculate new secondary bases:
.param Isb={Ssb/Vsb}
.param Zsb={Vsb*Vsb/Ssb}
.param Lsb={Zsb/(2*pi*fb)}
; Calculate new secondary component values for model:
.param Rs_1u={Rs*Zsb/Zsb0}
.param L12_1u={L12*Lsb/Lsb0}
; Calculate new primary bases:
.param Spb={Ssb}
.param Zpb={Vpb*Vpb/Spb}
.param Lpb={Zpb/(2*pi*fb)}
; Calculate new primary component values for model:
.param L11_1u={L11*Lpb/Lpb0}
.param Lm_1u={Lm*Lpb/Lpb0}
.param Rp_1u={Rp*Zpb/Zpb0}
; NEW TURNS RATIO, with adjustment for Vdrop across L12:
;.param Nt3={Vpb0/Vsb}
.param loadavg=2
.param Vsecideal={Vsb*(1+((Rs_1u)*0+((L12_1u*2*pi*fb))/loadavg))}
.param Nt3={Vpb0/Vsecideal}
; Adjust for Vdrop across L11:
;.param XL11={2*pi*fb*L11_1u}
;.param VL11={0.02*XL11 + (XL11*Vsb/loadavg/Nt3)}
;.param Nt3={(Vpb0-(VL11))/Vsecideal}
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jh: correct. Tom & I are discussing the transformer alone.
The capacitors you refer to via post #928 are the DC bus caps.
But you are also correct in that the transformer model does not include stray capacitance - it is a low frequency model. Even including the first 20 harmonics, the AC line current is low frequency, and the stray capacitances can be ignored without affecting the behaviour.
If you were interested in the propagation of rectifier-related ringing into the AC line, you would need to include the stray capacitance - 1.4mH of secondary leakage ringing with 10nF of diode capacitance is about 42kHz (I just made up the 10nF number). At those sorts of frequencies the transformer inter-winding capacitance will be important (if not dominant) and best be considered.
Otherwise it just complicates things unnecessarily
The capacitors you refer to via post #928 are the DC bus caps.
But you are also correct in that the transformer model does not include stray capacitance - it is a low frequency model. Even including the first 20 harmonics, the AC line current is low frequency, and the stray capacitances can be ignored without affecting the behaviour.
If you were interested in the propagation of rectifier-related ringing into the AC line, you would need to include the stray capacitance - 1.4mH of secondary leakage ringing with 10nF of diode capacitance is about 42kHz (I just made up the 10nF number). At those sorts of frequencies the transformer inter-winding capacitance will be important (if not dominant) and best be considered.
Otherwise it just complicates things unnecessarily
,
For everyone's amusement ...
If you really want to frighten the horses, your amplifier, your speakers, and your bank manager, consider this "perfect"?? cap for your next amp project:
* metallized polypropylene film
* 17,000uF
* 700 VDC
* ESR: 0.0002 ohms
* ESL: <30nH
* RMS Current: 400 amps
* Peak Current: 132,000 amps
* Life Expectancy: >100,000 hrs
Downsides? How about dimensions of 145mm x 330mm x 470mm ... and cost? If you have to ask, then you can't afford it ...
And what's the beastie? A UL9_BE1718K from Electronic Concepts, http://www.ecicaps.com; used for DC links in power distribution. These people make some pretty OTT devices, some almost affordable ...
Frank
If you really want to frighten the horses, your amplifier, your speakers, and your bank manager, consider this "perfect"?? cap for your next amp project:
* metallized polypropylene film
* 17,000uF
* 700 VDC
* ESR: 0.0002 ohms
* ESL: <30nH
* RMS Current: 400 amps
* Peak Current: 132,000 amps
* Life Expectancy: >100,000 hrs
Downsides? How about dimensions of 145mm x 330mm x 470mm ... and cost? If you have to ask, then you can't afford it ...
And what's the beastie? A UL9_BE1718K from Electronic Concepts, http://www.ecicaps.com; used for DC links in power distribution. These people make some pretty OTT devices, some almost affordable ...
Frank
Thanks Frank,
the various transformer capacitances and the effects of the current harmonics generated in both the sec and pri windings plus the effects of the load/filter capacitances reflected thru the diode bridges - perhaps the SIMs just gets too complicated to include these other mechanisms?
the various transformer capacitances and the effects of the current harmonics generated in both the sec and pri windings plus the effects of the load/filter capacitances reflected thru the diode bridges - perhaps the SIMs just gets too complicated to include these other mechanisms?
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