Choke loaded power supply

[TUBEDUDE]

Anyone built a 15-20 watt single ended amp using a choke loaded power supply?
I'm thinking of a low impedance (1.6k) , low voltage, high current design using a choke loaded supply.

[R.G.]

Choke loaded, or choke input filtered?

Those are different.

[Guy77]

Do you mean something like this:

Rectifier --- 40uF ----choke --- 40uF (B+ Output Trans) --- 1K ---- 30uF (B+ screen) --- 10K --- 30uF (B+ V2) ---- 10K ---- 30uF (B+ V1)

So that your output transformer connects after the choke as opposed to before it?

Cheers
Guy

[Stevem]

Is using a rectifier tube part of this plan?

A single ended OT that can pump out 20 watts is a big puppy!

[didit]

Only a "HiFi". And only prototype, that's then sat awaiting some else to provide "Lightning-connected" DSP+D/A front-end intended to mate it as system. Pair of Hammond 1627 or 1630 plus KT88s. Had a single shared choke input followed by moderate capacitor and then a per channel choke then large capacitor, and then a smaller higher Henry choke isolating each driver triode (1/2 a 6SN7). Detailed recall poor as it was several years ago. Required large signal input AC volts to make power swing. We tried guitar but only with a too weak preamp. Sounded HiFi.

Best .. Ian

[Phil_S]

If you mean choke input, there is a long thread I started some time ago when I built a SE EL34 amp. The project worked out well, except that its still on the bench waiting for me to finish the reverb circuit. What can I say? Life got in the way. It is on my list for this winter. https://ampgarage.com/forum/viewtopic.php?f=6&t=28888

I tried several chokes and I believe the data are reported in the thread. If not, there, another tread. Good luck.

[TUBEDUDE]

I was going to try 5AR4 》10H 200mA 》 200uF 》 5H 20mA 》20uF 》 etc.
With the plate connected after the choke.

[R.G.]

I was going to try 5AR4 》10H 200mA 》 200uF 》 5H 20mA 》20uF 》 etc.
With the plate connected after the choke.

OK, makes sense. That qualifies as what I called a choke input filtered power supply.

Choke input power filters are complicated in other ways from capacitor input filters. Capacitor input filters produce an output DC of close to the peak of the incoming AC voltage. Choke input filters produce an output voltage of about 0.637 times the peak of the incoming AC voltage UNDER CERTAIN CONDITIONS.

Capacitor-input filters work by having the capacitor charge up to the peak of the incoming rectified AC half-cycle, and storing energy to release during the time until the next charging pulse. It's kind of "capture and release" for electricity. Capacitor input filters start at Vpeak and have increasing ripple voltage under higher load current. In extreme cases, they "sag" until the capacitor isn't providing any significant filtering, and you're just getting full wave rectified half-sine voltage. The rectifiers in capacitor-input filters are always working intermittently, and there is always a portion of the incoming AC half-wave cycle where the rectifiers are not conducting.

Inductor input filters have an odd (to people used to capacitor-input filters) changeover in output voltage and ripple. At quite low output currents, the inductor has nearly no energy stored per half cycle of AC voltage in, and the capacitors dominate the output voltage - it looks and acts like a capacitor-output filter, and the output voltage is close to the peak of the incoming AC half cycle voltage. But as output current increases, the current per half-cycle stored and released in the inductor increases. This results in the inductor's stored energy lengthening the time per half cycle that the diodes conduct. Effectively, the inductor's stored current during charging pulses is used to add voltage to keep the diodes conducting.

So as load current increases, the rectifier diodes' conduction time per half cycle lengthens much more than it would in a capacitor-input filter. At some point, the stored energy in the inductor (per half-cycle) is able to lengthen the conduction time of the diodes so that the one diode or the other is always conducting. The full half-cycle of sine wave out of the diodes flows through the inductor and the inductor current never goes to zero.

Output voltage from an inductor input filter starts at close to Vpeak of the incoming AC waveform, and drops as output current increases until the inductor is conducting continuously. The output voltage at this point is a bit less than 0.637 times the peak of the input AC waveform, but it remains at 0.637 with increasing load current until things start burning up. From very light load to this "critical" loading, the output voltage from the filter drops by 36%, but then remains constant. The whole trick with inductive input filters is then to make sure that the load stays high enough (or the inductor/capacitor are sized correctly) to keep the output load current above the critical loading current for that inductor and cap. Many power supplies from back in the day of inductor input filters had load resistors that simply burned up the minimum load to force the inductor current above the critical value.

A Class A amp where the output current from the power supply is a good place for an inductor input filtered power supply. There are some gotchas, of course. When the AC power is turned on, the tubes that make the load are not conducting. If you use solid state rectifiers, the output voltage goes semi-instantly to full Vpk, so your filter caps and everything else on the B+ lines needs to be rated for what amounts to a 36% overvoltage for a while. When the tubes heat up and current rises, the output voltage drops to the intended voltage. Another gotcha is that the inductor and first filter cap have to be calculated. It takes the right inductor/capacitor combination and the right minimum current to make this work out so the inductor goes into continuous conduction. I once learned how to do that calculation, but I've forgotten and would have to go look it up. There were designs for inductors that would have a portion of their iron saturate at low currents, providing a high inductance at low current, which got into continuous conduction early, then changed over to the lower inductance but higher current capability as current rose.

It will take some calculation to ensure that the choke you have is run under the correct conditions for a choke input filter and give the desired output voltage and ripple reduction.

[TUBEDUDE]

I am currently using a 6DQ5 single ended In enhanced mode with an Ep of <300V @ 150mA with a CLC filtered supply. I want to try the choke input to get a lower Ep, as the available transformers at 200mA have high voltages. I need that 64% final voltage.
I'm using a seperate xfmr for a higher preamp voltage. With the 5AR4 warmup time, skyrocketing B+ shouldn't be an issue, and at 150+ mA, i think the current draw wil keep the choke out of oscillation.

I might have tried higher Ep/ lower current but am determined to use this high quality O.T. that has a 1800 Ohm Z.

[R.G.]

That's a reasonable scenario. If your rectifiers, caps and tubes can stand the temporary overvoltage during warmup and your voltage drops to 64% of the no-tube-installed case, you're good.

[Tony Bones]

I recall a rule of thumb for the minimum inductance required for a choke input filter:

L >= V/I

In this case I is in milliamps while V is in volts and L is in henries.

It's a complete coincidence that the calculation turns out to be so simple, and looks so much like Ohm's law, but it works.

[TUBEDUDE]

That sounds low to me. 300V÷ 150 mA = 2H?

[TUBEDUDE]

That's a reasonable scenario. If your rectifiers, caps and tubes can stand the temporary overvoltage during warmup and your voltage drops to 64% of the no-tube-installed case, you're good.

With the 5AR4 tube rectifier warmup at about 18 seconds, the 6DQ5 heater should have electrons ready to meet the rising voltage immediately, avoiding any real overvoltage.

[Tony Bones]

That sounds low to me. 300V÷ 150 mA = 2H?

That's just the minimum (critical) inductance. I was looking through a textbook last night that went into great detail on the calculation. The more accurate calculation would give something like 1.7H. But, the author (Landee) suggested that the optimum inductance is twice the critical inductance. Something like 3.5H. Her didn't offer much technical justification...

Anyway, 10H is ok and will result in lower ripple. 5H would probably work fine.

[R.G.]

With the 5AR4 tube rectifier warmup at about 18 seconds, the 6DQ5 heater should have electrons ready to meet the rising voltage immediately, avoiding any real overvoltage.

That's probably OK.

What's really going on is that the tube rectifier and output tubes are in a warm-up horse race. To the extent that the output tube(s) are either faster or the same speed as the rectifier, no overvoltage occurs.

The issue with this is that sometimes output tubes die or are removed. In that instance, the overvoltage happens and is a long-term thing. Realistically, in any amplifier that can have an overvoltage situation on warm up, it will have that condition sometime.

In my mind, I make a distinction between two eras of engineering. The first era was engineering controlled by inherent qualities, where the simple nature of the devices controlled the operation. Using tube rectifier warm up time to prevent overvoltage because the other tubes warmed up as fast or faster is an example of the first era. I have a very cautious approach to over-voltage/-power/-current conditions and I was expressing that in my comment - use the inherent properties of the capacitors et. al. to withstand the inevitable overvoltage.

The second era is what we're in now. I don't have a good label for this era yet, although I could probably make one up. In today's era, the fundamental idea is to sense conditions, then use logic, either hardwired or programmed, to make the desired things happen. As an example of the second-era form of solving the possibility of overvoltage, one would sense the voltage in some way, then add a switched-n fix for the overvoltage, perhaps also telling the owner/user that there is a problem.

As an example, one could add a MOSFET and power resistor set up so a voltage ov er X would cause the MOSFET to switch in the resistor as a load on the power supply to wait till the power tubes warmed up, if they ever did. That's probably a clumsy way to do it. Another is to add a series MOSFET in the B+ line that was run saturated to an ohm or so in normal conditions, but which changed over to not letting voltages over X get through, an adaptation of a series regulator. Another is to add some CMOS logic that just sensed thing and turned on/off some switch that prevented the overvoltage from getting out.

Lots of ways.

The era I shudder about is what I'd call the zero-th era: "... well, it mostly works. I don't know what could go wrong, and so I'm not going to worry about any of that".