Аудио-википедия

Схема:

Формула рассчета:

Программа для расчета БП:
http://www.duncanamps.com/psud2/

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

A Choke input power supply is approx 90% efficient, therefore the B+ will remain relatively stable when a Class AB amp is driven at high power. In the above circuit, 16V loss is caused by the forward resistance of the 2 large valve rectifiers. 16V valve rectifier loss is small compared to B+ 560V.

The advantage of Choke input supply is its excellent regulation and minimal pulsating DC ripple on the B+ Voltage. In the earlier times when only small Electro's were available (20uF) increasing the Inductance of the choke could directly compensate for the small value Electro's. A Choke input filter using modern lager 470uF Electro's can achieve a close to perfect ripple free B+ supply.

CIP disadvantage. Choke input filter supply must have a load on it at all times for it to stabilise. Without a load, a Choke input filter supply will behave as a capacitor input supply and the B+ Voltage will raise to the peak of the pulsating DC Voltage. In the example power supply, 640V AC x 1.414 = 904V DC. It is essential to have an external resistive load on the B+ output, at all times, or have a circuit that switches a resistive load onto the B+ if the amp is turned on without the output valves.

Swinging choke The term «Swinging choke» was often used to represent Choke input filter supplies in valve amps. In industrial electronics the term Swinging choke is used to represent the smallest physical size a choke can be to stabilise the DC current (requires further explanation). A large linear choke is normally used for valve amp power supplies, not a small swinging choke.

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.

Here's an example he gave in the book, but with a slightly different formula for 50Hz operation:

«A Class A power amplifier using a pair of push-pull 845 valves requires a raw HT of 1100v at 218mA, and a 10H 350mA choke is available, but is this adequate? The transformer supplying the choke has an output voltage of 1224VRMS, and using the equation for a 50Hz mains: IAC = VIn / 1155 * L = 1224 / (1155 * 10) = 106mA Itotal peak current = IDC + IAC = 218mA + 106mA = 324mA

The total peak current is 324mA, so the 350mA is just barely sufficient … »

That shows that the choke needs to be rated in excess of the peak d.c. load current, which is the current the amp draws from the power supply at max power, not at idle. Once again, a choke with more inductance tends to ease your required ratings.

But in all, you need to flesh out the design a bit more to properly rate the choke. Looking at this info, it's a very bad idea to just «slap something in» when you're thinking about a choke-input power supply.

Jones also notes that regarding the power transformer, it simply needs to be rated for the d.c. load current. He gives an explanation for why, but it's enough to know that's the case.

He does also note that the input choke creates voltage spikes seen by the power transformer. The solution is to have a snubber network, and he notes that the apparently best way of going about this is to fit 2 caps across the choke, with their «center-tap» connected to 0v. What this means is to use two 0.22uF 1kV caps connected in series, then take the outer leads and connect those to the choke to put this pair of caps in parallel to the choke, then connect the junction of the 2 caps to ground.