My power supply (with 30 seconds delay and NTC inrush protection) is reaching max. voltage before the amplifier's power tubes start conducting.
In order to prevent the tubes to get too much voltage (my limit is 600 Volts), I created a shunt today using zener diodes and a high power mosfet.
This basic circuit has been tested and is working. When the voltage is just below 600 Volts (when power tubes are conducting), no current flows throught the shunt circuitry.
I'm still not sure how to share the dissipated power for voltages above 600 Volt; part is dissipated in the mosfet, part in the drain resistor. The mosfet can handle more than 200 Watts and the resistor can take 50 Watt. Fortunately the periode of power dissipation is approx. 90 seconds.
NB: I didn't choose for a series voltage regulator due to the voltage loss caused by the regulator.
Of course I will keep testing and improving. Any comments and help are welcome.
Regards, Gerrit
In order to prevent the tubes to get too much voltage (my limit is 600 Volts), I created a shunt today using zener diodes and a high power mosfet.
This basic circuit has been tested and is working. When the voltage is just below 600 Volts (when power tubes are conducting), no current flows throught the shunt circuitry.
I'm still not sure how to share the dissipated power for voltages above 600 Volt; part is dissipated in the mosfet, part in the drain resistor. The mosfet can handle more than 200 Watts and the resistor can take 50 Watt. Fortunately the periode of power dissipation is approx. 90 seconds.
NB: I didn't choose for a series voltage regulator due to the voltage loss caused by the regulator.
Of course I will keep testing and improving. Any comments and help are welcome.
Regards, Gerrit
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I wouldn't imagine the shunt components will need to dissipate much at all - they are only loaded by the difference between 600v and whatever your power supply is peaking at. I'm guessing the difference is less than 100v (if not, then your power supply has waaaaay too much sag) and for a relatively short period.
You don't get to choose which component dissipates the power - it will be the resistor since that has the larger ohmic resistance. The MOSFET is effectively a switch, not a dissipation device.
You don't get to choose which component dissipates the power - it will be the resistor since that has the larger ohmic resistance. The MOSFET is effectively a switch, not a dissipation device.
Right, in testing the max. voltage difference is somewhat less than 50 volts. I found that the mosfet gets quite hot, something I didn’t expect for a “switch”.
Perhaps the mosfet is switching very fast, too fast to be measurable with my voltmeter? Perhaps I need a capacitor between gate and source? I will continue working on this next Friday.
Regards, Gerrit
Perhaps the mosfet is switching very fast, too fast to be measurable with my voltmeter? Perhaps I need a capacitor between gate and source? I will continue working on this next Friday.
Regards, Gerrit
Hi John,
During the delay period the voltage (under no load yet, as tubes are heating) the voltage will already reach max. voltage. So the delay is good to prevent inrush currents, it is no protection against high voltage. My delay uses a relay, shorting a series resistor connected between a bridge rectifier’s minus and “ground” potential.
Regards, Gerrit
During the delay period the voltage (under no load yet, as tubes are heating) the voltage will already reach max. voltage. So the delay is good to prevent inrush currents, it is no protection against high voltage. My delay uses a relay, shorting a series resistor connected between a bridge rectifier’s minus and “ground” potential.
Regards, Gerrit
Well, that circuit is going to get REAL hot (especially the MOSFET).
You're asking it to conduct 'whatever amps' at 600 (plus a bit) volts. So, at the very least, it requires a heat-sink.
It isn't even clear that when the valves are running that it won't still be in conduction. A real watt-burner.
Myself, I would have used a bipolar emitter-follower, series regulator topology. The voltage drop is SO minimal when it is being a pass-element, that you really wouldn't have to be concerned at all about the VDROP. Maybe 1.5 volts out of 600. Nothing.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
PS — also, the string-of-zenners will be conducting quite a few watts due to the very small 1 kΩ resistor in the string.
You're asking it to conduct 'whatever amps' at 600 (plus a bit) volts. So, at the very least, it requires a heat-sink.
It isn't even clear that when the valves are running that it won't still be in conduction. A real watt-burner.
Myself, I would have used a bipolar emitter-follower, series regulator topology. The voltage drop is SO minimal when it is being a pass-element, that you really wouldn't have to be concerned at all about the VDROP. Maybe 1.5 volts out of 600. Nothing.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
PS — also, the string-of-zenners will be conducting quite a few watts due to the very small 1 kΩ resistor in the string.
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The mosfet here is not operating as a switch. It's trying to be a shunt regulator.
When the B+ voltage is under 600 volts, yes it is a switch in the off state. When the B+ is unloaded by the amplifier, the mosfet and R2 will be conducting however much current is needed to drag the B+ down to slightly over 600 volts.
Without knowing how your power supply is designed, or what kind of load it sees in normal operation, it's hard to say how much this current is, but it's easy enough to measure it. Just measure the voltage on either side of R2 and calculate it.
Beware that most mosfets designed for SMPS use are not rated for linear operation and do suffer from secondary breakdown. That 200 watt fet might only be a 50 watt fet at 600 volts.
When the B+ voltage is under 600 volts, yes it is a switch in the off state. When the B+ is unloaded by the amplifier, the mosfet and R2 will be conducting however much current is needed to drag the B+ down to slightly over 600 volts.
Without knowing how your power supply is designed, or what kind of load it sees in normal operation, it's hard to say how much this current is, but it's easy enough to measure it. Just measure the voltage on either side of R2 and calculate it.
Beware that most mosfets designed for SMPS use are not rated for linear operation and do suffer from secondary breakdown. That 200 watt fet might only be a 50 watt fet at 600 volts.
50V of surge doesn't seem particularly problematic on 600V rated parts, but 90 seconds seems kind of long.
What does your power supply look like before this device and what is going on after? Is this a class AB amp? Preamp? Class A amp?
You're like half way to a shunt regulator, maybe that's the overall way to go (or maybe not!).
What does your power supply look like before this device and what is going on after? Is this a class AB amp? Preamp? Class A amp?
You're like half way to a shunt regulator, maybe that's the overall way to go (or maybe not!).
My final tubes need approx. 2 minutes before they fully conduct. The amplifier is class A and when the tubes conduct the shunt current is down to zero. The 4 tubes will draw around 100 mA each, with 600 Volt this means 240 Watt (+ G2 power + driver stage).
I will test a series regulator as well, just to be sure what is the best solution.
Regards, Gerrit
I will test a series regulator as well, just to be sure what is the best solution.
Regards, Gerrit
I don't know anything about your power supply design, but if it is the usual FWCT circuit the solution is real easy. Stick an N channel mosfet in the CT leg. It must be rated for more than 900 volts. Control the fet with a 555 timer such that the fet IS an open switch for 2 minutes, then the timer closes the switch. Near zero dissipation in the fet, simple circuit.
For bonus points you can use a simple processor chip or board that has a PWM output pin. Program it so that the pin is low for 2 minutes, then outputs a narrow width pulse that increases in duty cycle over several seconds until it becomes 100% (fully high) turning the fet fully on. This gives you a soft start on the HV power supply.
For bonus points you can use a simple processor chip or board that has a PWM output pin. Program it so that the pin is low for 2 minutes, then outputs a narrow width pulse that increases in duty cycle over several seconds until it becomes 100% (fully high) turning the fet fully on. This gives you a soft start on the HV power supply.
If you like the shunt approach, here are 2 concrete changes that'll make things work with less over-power dissipation:
[1] Change the value of R1 to 15,000 Ω. This reduces Zener current to maybe 650 - 600 - 15 = 35 volts … = 2 mA or so. Each 200 V zenner still dissipates over 400 mW of power, though. Likely within their nominal range.
[2] Change the value of R2 to 1,800 Ω. IT will be the part 'getting hot'. Has to dissipate nearly 150 W (if your 400 mA is correct) until the tubes come on line. Takes a fair amount of heat tho' off the MOSFET.
NOTE that this is all dependent on the power supply deliver about 400 mA to the valves when they are nominally working. If the real value is substantially different, do tell … and we will recalculate the optimum values.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
[1] Change the value of R1 to 15,000 Ω. This reduces Zener current to maybe 650 - 600 - 15 = 35 volts … = 2 mA or so. Each 200 V zenner still dissipates over 400 mW of power, though. Likely within their nominal range.
[2] Change the value of R2 to 1,800 Ω. IT will be the part 'getting hot'. Has to dissipate nearly 150 W (if your 400 mA is correct) until the tubes come on line. Takes a fair amount of heat tho' off the MOSFET.
NOTE that this is all dependent on the power supply deliver about 400 mA to the valves when they are nominally working. If the real value is substantially different, do tell … and we will recalculate the optimum values.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Take a look at post #18 in this thread:
HV bench power supply design
I did a considerable amount of testing with running switching FETs in their linear region- they don't take anywhere close to what their datasheet SOA graphs would suggest.
How I would proceed:
1) What tubes are being used? Some tubes will be just fine with a bit of extra voltage for a reasonably short duration.
2) If it is a problem, have a 2-minute time delay for the plate supply. This is a better alternative to the number of power MOSFETs that have their dies blasted into a Low-Earth orbit before the circuit in post 1 is made to work.
HV bench power supply design
I did a considerable amount of testing with running switching FETs in their linear region- they don't take anywhere close to what their datasheet SOA graphs would suggest.
How I would proceed:
1) What tubes are being used? Some tubes will be just fine with a bit of extra voltage for a reasonably short duration.
2) If it is a problem, have a 2-minute time delay for the plate supply. This is a better alternative to the number of power MOSFETs that have their dies blasted into a Low-Earth orbit before the circuit in post 1 is made to work.
The DC bus in the HV power supply I designed has about 1300 uF in it. I always made sure there was a lexan shield in place when I did a power test on the regulator board. Those suckers are like small firecrackers when they go off.
Tubelab_com,
I remember a friend working with complementary Hitachi TO3 MOSFETs on a design similar Erno Borbely amplifiers and some other designs.
One of the MOSFETs ended up with a hole in the top (looked like the tip of a welder took a big chunk off the top.
They use the word 'Power' when calling certain circuits of amplifiers 'Power Supplies' for good reason.
I remember a friend working with complementary Hitachi TO3 MOSFETs on a design similar Erno Borbely amplifiers and some other designs.
One of the MOSFETs ended up with a hole in the top (looked like the tip of a welder took a big chunk off the top.
They use the word 'Power' when calling certain circuits of amplifiers 'Power Supplies' for good reason.
The old RCA 2N3055 and Westinghouse 2N3773 TO-3's had steel cases. I never blew the guts through the lid on any of those. Motorola, however used aluminum, and an experiment gone wrong in a line powered circuit could spray molten metal and "liquid fire" right through the lid of the device. it really got my attention the first time it happened.
I played with a lot of early vintage semiconductor stuff around 1970 when large surplus military 60 Hz to 400 Hz power converters were dumped on the surplus market for scrap metal value because the technology was changing almost monthly. I made the biggest audio power amp of my life at age 18 when I found such a beast in a scrap yard. It had a huge tubular heat sink with big boxer fans on each end and 4 rows of 6 each Westinghouse 2N3773 transistors. It used a real simple power supply that can not be discussed here and made somewhere over 1000 watts into 2 ohm load. The "test" load was a long piece of 1/4 inch steel wire!
Yep… all those possibilities, or replacing the resistors as I indicated. Make sure the R₂ of 1800 is at least a 25 W resistor. More likely 4 ea of 470 Ω resistors in series, each 10 W. About the cheapest solution that's rock-solid. Conveniently, you can 'try different ohmages' by just jumpering out any of the resistors in the series. Crude, but effective.
$1.09 ea, QTY 1, Mouser: 280-CR25–470-RC, 25 W, 470 Ω, use 4 ea.
$0.80 ea, QTY 1, Mouser: 280-CR10-330-RC, 10 W, 330 Ω, use 5 ea.
,
Just saying… cheap simple solutions.
⋅-=≡ GoatGuy ✓ ≡=-⋅
$1.09 ea, QTY 1, Mouser: 280-CR25–470-RC, 25 W, 470 Ω, use 4 ea.
$0.80 ea, QTY 1, Mouser: 280-CR10-330-RC, 10 W, 330 Ω, use 5 ea.
,
Just saying… cheap simple solutions.
⋅-=≡ GoatGuy ✓ ≡=-⋅
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