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http://education.lenardaudio.com/en/14_valve_amps_6.html Choke input filter
The primary limitations of early valve rectifier power supplies was the high forward resistance in the rectifier valves and the small storage capacity of the Electro's (approx 20uF). 470uF Electro's are often used in high quality valve amps today.
Most valve amps above 40 Watt are configured in Class AB (Negative bias). The quiescent current through each output valve is approx 50mA. When the amp is driven to full power the current can increase to approx 150mA. The B+ Voltage from a traditional Capacitor input supply will readily drop to a lower level. The B+ Voltage will be modulated by the varying current through the output valves. As the B+ Voltage reduces, so does the gain of the output valves. A modulating B+ Voltage causes massive inter-modulation distortion.
Choke input filter power supplies, also referred to as Swinging choke power supplies, are often seen in industrial electronics where a large amount of DC current is required with high regulation. Choke input filter supplies were often used in early valve rectifier power supplies for 100+ Watt valve amps. The Swinging choke is nearly as large as the power tranny. Two extra large valve rectifiers were often used. Each valve rectifier has the Anodes in parallel to reduce its internal resistance, enabling greater current.
Swinging choke power suppy The pulsating DC from the rectifier valve goes through a large Choke (non saturable Inductor). The Swinging choke converts the pulsating DC to its RMS value without loosing energy. The internal losses in the 2 valve rectifiers is minimal. AC to DC conversion is close to the academic formula.
AC Volts x 0.9 = DC Volts
640V AC x 0.9 = 576V DC
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https://el34world.com/Forum/index.php?PHPSESSID=7itm19jj08hjksv28ebkb1ujs7&topic=622.msg1936#msg1936
I went to a reliable source (Morgan Jone's book Valve Amplifiers) and looked up some choke-input supply info. He notes that in the «old days» the solution was to use a swinging choke. You have a very tough time finding those these days, so that's probably out of the question.
He gives a detailed formula for the minimum current draw needed, but then goes on to note an approximation for 50/60Hz power supplies. It's Currentmin. (in milliamps) = VIn (RMS) / L (in henries). So for your 500vac transformer (which is RMS volts), you'd divide that 500v by the inductane rating of the choke you intend to use to see what the minimum current draw must be. As an example, for a 20H choke, 500v/20H = 25 milliamps of minimum current draw. The implication is you want a very high inductance rating for your choke.
He also notes that it would seem the choke only needs to be rated for the maximum d.c. load current, but that the choke really needs to be rated in excess of this. Why? The choke generates a magnetic flux in the core proportionate to the size of the current passing through it, and if that flux is too great, it saturates the core and the inductance of the choke falls to zero. Bam! You're back at 700v.
The current that the choke will support is the max d.c. load current and the instantaneous a.c. charging current (meaning the current used to charge the filter caps). So that's IDC + IAC = Itotal peak current.
He derives a formula for the a.c. peak current. For a 60Hz wall outlet, it's
IAC (positive peak) = VIn (RMS) / [1386 * L)
Vin is the RMS transformer voltage again, and L is choke inductance in Henries.
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Обсуждение
BRIDGE LC Vac = Vdc x 1.11 Iac = Idc x 1.06 Pac = Pdc x 1.18
FULL WAVE LC Vac = Vdc x 2.22 Iac = Idc x 0.65 Pac = Pdc x 1.44
BRIDGE CLC Vac = Vdc x 0.71 Iac = Idc x 1.61 Pac = Pdc x 1.14
FULL WAVE CLC Vac = Vdc x 1.41 Iac = Idc Pac = Pdc x 1.41
https://www.sowter.co.uk/rectifier-transformer-calculation.php