[time-nuts] Re: 5061A HV Supply, et al.

[ed breya]

I doubt that the laser supply hurt the A18. If it did, it would have 
likely broken the rectifiers in reverse, and you wouldn't be getting the 
normal HV and 4 V monitor signal and almost-working behavior afterward.

I assume this laser supply is the kind for small HeNe tubes. These 
things have provisions for striking the tube with higher voltage to get 
it going. It's not unusual for the unloaded output to get up to 15 kV or 
so, either continuous, or as ignition spikes superimposed on a medium HV 
output. Once the tube lights, it loads it down to the normal operating 
power of the tube plus the ballast resistor, and it stops trying to 
light it up.

This could be a handy unit to serve as the external HV source to 
possibly restore the Cs tube. You just have to add a few things around 
it to make it do approximately the right thing. In principle, all you 
need is to load the supply down just right to get the right voltage and 
fool it into thinking it's running a stable HeNe tube, drawing the 
normal current or power range, and not trying to strike or re-strike.

These things are not very smart or precise, so you have lots of room to 
tweak things around without too much grief. I'm going to make some 
assumptions about the power supply and the HV tools and parts you may 
have at your disposal.

First, if you actually had 7 kV output continuous for a few minutes, I 
assume the supply is a constant-power flyback type, so the output likely 
can reach 10-15 kV or so to strike the lamp, and does not use an 
ignition pulse circuit. This is best for the application, since you 
won't have it decide to make nuisance pulses. You can't tell yet for 
sure if this is the case, until you scope it out.

The main HV tool you seem to have is the "172 meg divider," which I 
assume is a string of megohm range resistors. Do you believe it can take 
10 to 15 kV for a short time without breaking down or overheating? 
You've already proofed it to 7 kV, no sweat. If it can take the higher 
voltage, it makes it more convenient, but it's not absolutely necessary.

Now thinking about the supply rating and characteristics, it normally 
would be running a small (like 0.5 mW) HeNe tube plasma load, plus a 
ballast around maybe 75 k ohm or more, at the nameplate 1250 V by 4 mA, 
so 5 W going in. The ballast swamps out the tube's negative resistance, 
and drops say 300 V, leaving about 950 V on the tube.

Since we're assuming approximately constant output power, and we know it 
can reach 7 kV at some load, then it can easily reach the desired 3,500 
V ion pump supply level. This is 2.8 times the normal laser voltage, so 
the available current should be around 4 mA/2.8 = 1.43 mA, which is a 
very nice current level in the scheme of things here. So, that 5 W gives 
a good current at the desired higher voltage, just by loading it down 
right. A load of 3,500 V / 1.43 mA is about 2.5 megs, which should be 
readily realizable with a decent resistor inventory. It would be an 
assembly of lots of whatever resistors it takes to get the right value 
and handle over 3,500 V, and over 5 W total power dissipation.

So, if you start with something around 2.5 megs, 10 W, 5 kV max rating, 
and hook up your 172 meg divider to monitor the situation, you can see 
what it's doing, and tweak the system accordingly. This all is without 
any hookup to the 5061A, just all by itself until it's ready.  At this 
point you can also scope it - hook it to the HV divider output and look 
for any signs of ignition pulses that may cause trouble. Note that the 
frequency response will be quite low. I'm guessing you'll just see some 
high frequency ripple from the laser supply. If it is re-striking, then 
certain changes may be in order to suppress it, or it may not even 
matter for this purpose.

You can also form a divider into the load resistor assembly itself, so 
you can monitor it at any time, at some some convenient scaling factor 
like 1000x. For instance, in a very simple form, whatever the load R 
needs to be for the right voltage, you can put 99.9 percent of it in the 
power section, and 0.1 percent in series at the bottom. Or, the divider 
can be a small portion of all the resistors that make the whole thing.

Once you proof it, then it can be attached to the Cs tube, and you can 
see what it's doing voltage-wise. With no tube load, your PS circuit 
should be around the right 3,500 V nominal, and can only go down from 
there as it pulls ion current.

BTW these little constant-power supplies can be dangerous, since they 
can put out quite a bit of current as the voltage drops. A 125 V shock 
can pump 40-50 mA through your arm or whatever, with good contact. 
Always be careful.

Ed