Hi,
admittedly I don't have any clue of what's going on exactly in a choke input filter, besides the DC voltage is about 0.9 times the AC input and regulation is comparatively good (= stiff DC voltage) at more than some minimum load. But 'till now I've never designed a choke input PSU by myself. So there are some questions where I'd need some helpful answers:
Best regards and TIA!
admittedly I don't have any clue of what's going on exactly in a choke input filter, besides the DC voltage is about 0.9 times the AC input and regulation is comparatively good (= stiff DC voltage) at more than some minimum load. But 'till now I've never designed a choke input PSU by myself. So there are some questions where I'd need some helpful answers:
- Are there any formulae that allow for designing the choke and the first filter capacitor? What are the required parameters?
- If there is a CT'ed secondary wining and I use a pair of SS diodes instead of a two way rectifier tube, there' some short time (depending on the secondary voltage and the diode forward voltage) next to the zero crossing when no diode conducts. As an inductor tries to keep the current flowing, would a third diode wired reversely between the rectifier output and the CT help the choke to do it's job? I've never seen this, anyway, but I wonder why this helper diode usually is omitted? Would regulation be compromised by such a diode?
Best regards and TIA!
I remember Patrick Turner R.I.P. giving very good insight about transformers in general but I can't get his web active nor the waybackmachine
You probably want continuous current through the inductor. The rectified voltage is then ideally on average 2/pi times the peak voltage minus some drop across the diode, so ideally 2 sqrt(2)/pi times the RMS voltage minus some drop across the diode. If the inductor had no DC resistance, the average voltage at its other end would be exactly the same. 2 sqrt(2)/pi ~= 0.9003163162, so very close to your 0.9.
To ensure continuous current, the peak ripple of the inductor current has to be smaller than the average current (the downward peaks anyway, so they stay above zero). I never tried, but it should be possible to calculate the integral of the difference between the rectifier output voltage and the voltage of 2 sqrt(2)/pi times the RMS voltage at the filter capacitor. Divide by the inductance and you know the ripple current.
I don't think there is a time where both diodes become non-conductive. The voltage at the input of the inductor will just go slightly negative, just negative enough to ensure the current keeps flowing through one of the diodes.
Thanks to all of you so far. 'Till now I've understood that the choke's inductance is determined by the minimum current from where I intend the DC voltage to be regulated?
Let's go a step further: Any iron core inductor that carrys DC current needs an airgap to prevent magnetic saturation. Apart from that, I know the inductance increases by the square of the turns count. Now, how do I chose the proper airgap width? How do I determine the core size, which, of course, seems to be dependant of the maximum DC voltage drop?
Best regards!
Let's go a step further: Any iron core inductor that carrys DC current needs an airgap to prevent magnetic saturation. Apart from that, I know the inductance increases by the square of the turns count. Now, how do I chose the proper airgap width? How do I determine the core size, which, of course, seems to be dependant of the maximum DC voltage drop?
Best regards!
I own this book, 3rd edition, and I've read it more than 10 years ago. But I'm not aware he elaborated on this?
Best regards!
MJ didn't elaborate on how do design a choke, but he definitely elaborated on how to design a choke input PS (page 310).
There is a simple rule of thumb on how to calculate the minimum DC current:
I(mA) = Uoutput(V) / L(mH)
If you have a look into tube rectifier datasheets showing output curves for different L values and voltages, this formula pretty well matches the points where the output voltages are going up as soon as the currents go belowthe critical value. For higer than this minimum DC currents, the inductor is a high impedance for all AC voltages (100Hz / 120 Hz and harmonics) and a low impedance (winding resistance) for the DC current. As soon as the minimum current is reached, there is no current gap anymore and the diodes behave as already mentioned by MarcelG. That's the reason why these types of power supplies create much less noise than capacitor loaded rectifiers which produce high current spikes when reloading the filter cap and why they are much more stiffer (only winding resistance)
The inductor itself is not much different in design compared to inductors used in CLC filter chains. The only difference is a higher AC voltage between windings and core and a good insulation is needed. To overcome this drawback, you can also put the inductor into the ground line instead on the B+ line.
I(mA) = Uoutput(V) / L(mH)
If you have a look into tube rectifier datasheets showing output curves for different L values and voltages, this formula pretty well matches the points where the output voltages are going up as soon as the currents go belowthe critical value. For higer than this minimum DC currents, the inductor is a high impedance for all AC voltages (100Hz / 120 Hz and harmonics) and a low impedance (winding resistance) for the DC current. As soon as the minimum current is reached, there is no current gap anymore and the diodes behave as already mentioned by MarcelG. That's the reason why these types of power supplies create much less noise than capacitor loaded rectifiers which produce high current spikes when reloading the filter cap and why they are much more stiffer (only winding resistance)
The inductor itself is not much different in design compared to inductors used in CLC filter chains. The only difference is a higher AC voltage between windings and core and a good insulation is needed. To overcome this drawback, you can also put the inductor into the ground line instead on the B+ line.
After years of seeing that claim.... I think it applies to high-drop (vacuum) rectifiers. Not much to crystal diodes.
Cap-input has peak currents 3X to 10X the average, and on a vacuum rect that may be many volts. Choke-input spreads the current over a whole half-cycle so the peak drop is much less.
A 0.6V 1N1007 may get as high as 2V at high peak current, but this added 1.3V is typically insignificant for >100V supplies.
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No. Not really. The inductive kick swings the dead-zone very fast.
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From a recent conversation. The formula was stolen from DIYAudio, but I forget where {EDIT: here}. The problem was very solid-state but still shows correct unit-handling.
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Narrow current peaks will also produce more voltage drop across the winding resistance and spreading inductance of the transformer, not just across the diodes. Doesn't that affect the regulation/stiffness?
if you can get hold of the ARRL handbooks, there is a good treatment of the subject on choke input filters, i have the 1060 edition, will try to retrieve it..
i have printed files of Patrick Turner saved on my drive, will try to post them if moderation agrees....
i have printed files of Patrick Turner saved on my drive, will try to post them if moderation agrees....
Psud will allow you to design an LC circuit choosing different H levels of chokes and caps. It seems to work fine.
Thank you Tony, if you have the rest of information of Patrick Turner I'll appreciate if you can send me a PM or email, thanks.
Here is the link to the copy of Patrick Turner's site :
http://naturetour.digicom.bg
Hard to find but still exists.
http://naturetour.digicom.bg
Hard to find but still exists.
I have given part of the answer here:
https://www.diyaudio.com/community/threads/ei-laminated-open-ended-inductors.378442/post-6822159
In the formula, l is twice the airgap.
Another formula you need is the number of turns required to stay below an induction B, with a maximum current I assuming an inductance L: N = L*I/B*A, A being the iron area.
Of course, even if you manipulate both formula's algeabrically, you will need some iterations, because the core can only take some much copper, the airgap needs to be reasonable (small), etc.
You should start with a core you have available, compute the maximum number of turns that fit into it taking into account the thickness to keep the DC resistance within acceptable limits and compute the B value for the target inductance. If it is lower than 1T, you can go further and compute the gap required to achieve L.
If you arrive at B significantly smaller than 1T, you can try to reduce N, and recalculate the gap.
If B>1T, you need to chose a larger core