Ameritron soft start: How do I stop the arcing?

[brandon7861]

Ok, hold on a minute. I just looked at the schematic. I don't think messing with the capacitor charge time is the way to do this.

The schematic shows two resistors of interest. There is one that feeds the HV capacitor at the plate (the 22Ω resistor charging the .001 cap) and there is the 10Ω resistor (R3) across the soft start relay contacts.

Let me see if I understand how this works.... The 10Ω resistor (R3) limits the current to the transformers primary which makes the secondary charge the cap a bit slower because the current draw puts a voltage across R3 which reduces the voltage across the primary. Once the capacitor charges some, the current through the transformer drops and that raises the voltage across the primary. Once there is enough voltage across that one half of the primary, the relay engages and disables the current limiting at the primary.

If this is the case, it would make more sense to alter R3, the 10Ω resistor and leave the capacitor RC alone. If the capacitor RC is increased, the current will decrease, and I think that would prematurely activate the soft start relay (essentially doing the opposite of what you want).

Take a look at it drawn a little different.

 

[brandon7861]

So lets say you could scope that resistor (you should use two probes and use the difference math because hooking the ground lead to something hot is not something the scope will like), but assuming that was done, it might look like this:

Example:
If the voltage cuts to zero from, say, 36v, then we would know that the relay activates at (240v-36v)/2=102v. The current through the resistor when the contacts close is 36v/10Ω=3.6A, way too much. Next we measure the steady state current through the primary after the relay is closed, call it 500mA in this example, then add 50% for cushion, so 750mA. What resistor value will pass 750mA when dropping 36 volts? 36/.75A=47Ω With that resistace, the relay should close when there is only a quarter ampere difference and since it was based on more than steady state current, it will not be so high the relay never closes.

Explained a different way, if we know the primary will have X current in it with the relay closed and everything charged, then while the current is slightly above that, we want the relay to engage. In the above example, we saw how a 10Ω resistor at 3.6A dropped 36v. Consider what happens when our 47Ω resistor is pulling 3.6A. That resistor has 176v across it and the primary does not have enough voltage to activate the relay.

Hey, this was fun figuring out!!! But, I'm dead tired and half asleep so this is probably all screwed up lol. Gotta be up in 3 hours. Long day tomorrow.

Edit: couldn't sleep. I overcomplicated that. Just pop out the relay, put it on the variac and see where she clicks, then measure the steady state current and bingo. no scope needed. OK going back to bed.

 

[nomadradio]

I discovered that the power drain from the HV bleeder resistors is part of the picture. Make the start resistor too high and the voltage stops rising well short of full bore. The start resistor will become a 50/50 voltage divider when the wattage absorbed by the bleeders more or less equals the wattage being thrown by the start resistor. If you set the full-bore turn-on at 2/3 charge you can't use a start resistor with a 50/50 ratio. The HV will never rise to the 2/3 turn-on point.

We used the rectified HV DC voltage to control the step start years ago. A voltage divider is tapped between ground and the 'bottom' bleeder resistor. Borrowing three Volts from a bleeder with 400 Volts on it has no adverse affect. A solid-state relay that turns on at 3 Volts DC gets fed from a pot, the SSR turns on the 120-Volt AC coil of the "RUN" relay. Putting a pot between the divider resistor and the SSR input allowed you to tweak the percentage of full voltage the caps must charge to energize the run relay. Settled on a 50 percent of full HV with this setup.

One neon pilot lamp showed the power switch was on, another one labeled "start" would light while the HV filters were charging. It would go dark when the run relay and the third light marked "run" came on.

As it happens, the clients who used these power supplies tended to be rowdy with their high-power RF toys. Shorted tubes were a regular occurrence. If there is a dead-short load on the high voltage, the "start" resistor will have to safely dissipate the full 240 Volts coming out of the wall. Remembering that 240 squared is 57,600 a 1 ohm resistor would dissipate 57.6 kilowatts. A 1000 ohm resistor would dissipate 57.6 Watts. We found that 500 ohms or so would always charge the filter caps beyond the halfway point. It would dissipate 115.2 Watts under the worst case, a dead-shorted HV circuit. A 240-Watt resistor would tolerate this far longer than the operator, wondering why it's stuck on the "start" light, and the "run" light is still dark. And when he smells the dust layer cooking on the 240-Watt start resistor he'll have an incentive to switch it off.

Alas, 240-Watt resistors were surprisingly cheap 25 years ago, but not so much now.

So long as the start resistor can safely throw all 240 Volts without damage, it can serve as an overload-protection feature. A sideways one, though.

One operator reported that a gassy 3-500Z in his six-tuber would flash purple, followed by a "dunk" sound as the overload would cause the run relay to drop out when the HV fell below the step-start level. When the run relay disengaged it put the start resistor in line with the HV transformer. This would cause the 'start' light to come on as long as the HV stayed below the 'run' level. The "run" light would stay dark until he unkeyed, removing the tubes' load current.

It wasn't installed with that feature in mind, but this fella was used to the fireworks this had caused him in the past. Said he liked it this way much better. He went through a lot of tubes, just the same.

73

 

[brandon7861]

I discovered that the power drain from the HV bleeder resistors is part of the picture. Make the start resistor too high and the voltage stops rising well short of full bore. The start resistor will become a 50/50 voltage divider when the wattage absorbed by the bleeders more or less equals the wattage being thrown by the start resistor. If you set the full-bore turn-on at 2/3 charge you can't use a start resistor with a 50/50 ratio. The HV will never rise to the 2/3 turn-on point.

We used the rectified HV DC voltage to control the step start years ago. A voltage divider is tapped between ground and the 'bottom' bleeder resistor. Borrowing three Volts from a bleeder with 400 Volts on it has no adverse affect. A solid-state relay that turns on at 3 Volts DC gets fed from a pot, the SSR turns on the 120-Volt AC coil of the "RUN" relay. Putting a pot between the divider resistor and the SSR input allowed you to tweak the percentage of full voltage the caps must charge to energize the run relay. Settled on a 50 percent of full HV with this setup.

One neon pilot lamp showed the power switch was on, another one labeled "start" would light while the HV filters were charging. It would go dark when the run relay and the third light marked "run" came on.

As it happens, the clients who used these power supplies tended to be rowdy with their high-power RF toys. Shorted tubes were a regular occurrence. If there is a dead-short load on the high voltage, the "start" resistor will have to safely dissipate the full 240 Volts coming out of the wall. Remembering that 240 squared is 57,600 a 1 ohm resistor would dissipate 57.6 kilowatts. A 1000 ohm resistor would dissipate 57.6 Watts. We found that 500 ohms or so would always charge the filter caps beyond the halfway point. It would dissipate 115.2 Watts under the worst case, a dead-shorted HV circuit. A 240-Watt resistor would tolerate this far longer than the operator, wondering why it's stuck on the "start" light, and the "run" light is still dark. And when he smells the dust layer cooking on the 240-Watt start resistor he'll have an incentive to switch it off.

Alas, 240-Watt resistors were surprisingly cheap 25 years ago, but not so much now.

So long as the start resistor can safely throw all 240 Volts without damage, it can serve as an overload-protection feature. A sideways one, though.

One operator reported that a gassy 3-500Z in his six-tuber would flash purple, followed by a "dunk" sound as the overload would cause the run relay to drop out when the HV fell below the step-start level. When the run relay disengaged it put the start resistor in line with the HV transformer. This would cause the 'start' light to come on as long as the HV stayed below the 'run' level. The "run" light would stay dark until he unkeyed, removing the tubes' load current.

It wasn't installed with that feature in mind, but this fella was used to the fireworks this had caused him in the past. Said he liked it this way much better. He went through a lot of tubes, just the same.

73

In measuring the steady-state primary-side current with relay closed, that current should also include that of the bleeder resistor in the secondary-side circuit. So, thats already accounted for in my analysis.

If the maximum current with the relay closed is known, and the voltage at which the relay clicks is known, the resistor should be easy to calculate. 240v-2*Relay voltage=resistor voltage at click over. That voltage divided by anything higher than the steady-state current will provide a resistance value that will still activate the relay.

The relay coil is acoss half the primary, so whenever the primary has 2X the relay click voltage, the relay activates. If the relay clicks at 104v, that means the transformer has 208v when it clicks. That means there is 240-208=32v on resistor. Anything higher on the resistor and the relay wont click. As long as that resistor has less than 32v across it at the current you want it to switch at, it will switch. This means (in this example) that 32v/Imax(primary) = highest acceptable resistor value.

The only issue with using close to the highest acceptable value of resistance is that as the capacitor ages and losses increase, startup might get unreliable. But I bet if you calculate it to switch at 50% higher than steady state current, it will be reliable and not arc at all.

Edit:cleaned up my multiple posts, sorry

 

[357]

Well the amp is still chugging along. A small flash on start-up is just how it goes. Other than thats this thing is a beast 1650 with 65 in all day every day. I'd say short of a henry 3K, this is the best amp for the money, at least 20 years ago it was.
I'm not fond of the non Peter Dahl iron or the Z7 variant.

I can live with a new relay every 10 years.

The A7 tube is tough enough to last me out I'd say.