which diode for 55 Khz switching supply??

[albert]

You missed something, but it's probably my fault for not being more
specific:>:

The discussion above is c-o-m-p-a-r-i-n-g the CW mult to the 'other'
multiplier (since the previous post had suggesting looking at the
'other' multiplier as an alternative). In the CW, each stage needs
about 100 volts PIV per diode and all of the diodes can be the same
rating.

In the 'other' multiplier, caps and diodes need to withstand the
higher voltages as the number of stages is increased. So, a PM Tube
supply based on the 'other' would need 500v and 1000v diodes and caps
in the upper stages even though each stage produces only 100v.

My point (above) was that higher voltage rated diodes get very poor in
terms of electrical losses and high frequency switching losses-which
is what you need to use if you put together the 'other' multiplier in
a PM Tube supply.

The 'other' multiplier is well suited for higher power, but lower
voltage aps for this reason. I think the perfomance of the 'other'
multiplier in a PM tube supply would be poor:>:

Regards,

A

 

[John Woodgate]

<[email protected]>) about 'which diode for 55
Khz switching supply??', on Sat, 13 Dec 2003:

Yes, I have looked atthat circuit also. It is commonly referred to as a
CW also, but it isn't!

We may not be talking about the same circuit. Do you have a URL for the
one you looked at?

[millisnip]

The OTHER multiplier needs different component values with respect to
voltage ratings whereas the CW doesn't have this restriction.

The capacitor voltage ratings do have to increase, but I suppose your
capacitors are not of very large value anyway.

Honestly, I haven't modeled this alternative yet, but the upper stages
would require greater than 1 KV PIV, and these types of diodes are very
lossy. Even 500 volt PIV dioded have dismal switiching losses and the
middle stages of the circiut you sugeest would need to have those
diodes.

I don't see at present why the diode ratings need to go up. Each diode
is between an Nth-stage and an (N+1)th-stage. so it seems that PIVs
would be all the same.

 

[Jim Adney]

If you think this is OK, well, I suppose we have different
standards::> But, typical values for the divider resistor are 330K
(each), in series this is (about) 3.3 meg ohms (assuming 10 resistors
are needed for the divider string). This 3.3 meg total resistance
appears across the 1KV supply whether the tube is conducting or not.
In fact, you can take the tube away completely and those resistors
still load the power supply and dissipate the same amount of heat into
the air.

The 3.3 MOhm across 1000 volts disipates ~1/3 Watt. If you think that
your losses with a 55 kHz, 10 stage C-W will be less, then you must
never have tried it before.

Resistive dividers dissipate big amounts of heat and are cheap. In
some aps, they are clearly the way to go.

In general I agree with you. I just don't think your instincts were in
the right direction this time.


-

 

[John Larkin]

The 3.3 MOhm across 1000 volts disipates ~1/3 Watt. If you think that
your losses with a 55 kHz, 10 stage C-W will be less, then you must
never have tried it before.

If done right, the C-W losses will be much less than 1/3 w, and
provide lower dynode impedances too. Night-vision gear uses HV
supplies that have amazingly low idle currents. WWII-vintage
sniperscope supplies would run for weeks on one zinc-carbon flashlight
battery.

A sot-23 dual diode and a couple of 2 nF surface-mount caps don't cost
much, and dissipate nil. Several companies make commercial C-W PMT
sockets, but they tend to be expensive.

John

 

[Roger Gt]

Good Day All.

I am building a cockcroft-walton voltage multiplier. It will have 10
stages, with each stage having a 100 volt output, so the total output
is ~1000 volts. It will operate at 55 Khz and only needs to produce
small currents (microamps). I want to have minimal switching diode
losses since I will have lots of diodes:>:

I need a fast diode with small input capacitance. I need to have it
withstand 200 volts at 100 microamps.

Which diode should I use??????

All suggestions are appreciated.

http://usa.hamamatsu.com/hcpdf/parts_HC/HC120-01.pdf

It provides 1100VDC from +/-18VDC in. Maybe it would be as easy to just
buy one?

 

[Albert]

If they made it as a building block assy, I'd think about it. The way
they currently package thier cw supplies is a severe limitation
because you can't wire it up to other tubes. I've spoken to them about
this. they told me to pound sand.

I think you should check the price before suggesting Hamamatsu::>

I've chosen to 'just say no' to the big H. Your mileage will vary, I
shall pray for you.

 

[Roger Gt]

If they made it as a building block assy, I'd think about it. The way
they currently package their cw supplies is a severe limitation
because you can't wire it up to other tubes. I've spoken to them about
this. they told me to pound sand.

I think you should check the price before suggesting Hamamatsu::>

Considering the problems you seem to have, I didn't feel price was the
critical parameter. Besides all High voltage units are relatively pricey!

I've chosen to 'just say no' to the big H. Your mileage will vary, I
shall pray for you.

Okay, your call. Sorry I made any sugestion.
I was unaware this project had religious significance, I do not need (or
Want) your prayers. I was trying to help. Only a suggestion. It has been
sometime since I have done any really high voltage power supplies, I had no
where near the problems but the criteria may have been different. Took
three attempts to get the circuit to work, but my problems were insulation
leakage due to a confined space.

Recent designs were only about 200 volts and at 330Khz. No problems there
either.

 

[Roger Gt]

Good Day All.

I am building a cockcroft-walton voltage multiplier. It will have 10
stages, with each stage having a 100 volt output, so the total output
is ~1000 volts. It will operate at 55 Khz and only needs to produce
small currents (microamps). I want to have minimal switching diode
losses since I will have lots of diodes:>:

I need a fast diode with small input capacitance. I need to have it
withstand 200 volts at 100 microamps.

Which diode should I use??????

All suggestions are appreciated.

OKAY:

Possible
1N4936-T DO41 Case, 1 amp rating, 15pf 400PIV (@1A 1.2V forward, since your
operating at less current, this will be somewhat less.)
DL4936-13 MELF all ratings the same.
Both will operate at 55Khz. I use them at 330Khz!

The other problems I can not address.

 

[Roger Gt]

<albert> wrote in message
OKAY:

Possible
1N4936-T DO41 Case, 1 amp rating, 15pf 400PIV (@1A 1.2V forward, since your
operating at less current, this will be somewhat less.)
DL4936-13 MELF all ratings the same.
Both will operate at 55Khz. I use them at 330Khz!

The other problems I can not address.

Modeling the circuit:
Using the 1N4936 with all caps equal to .33uF and a 55Khz input.
I show a rise time of 83ns and a simulated voltage of 1.154kv with a load of
10Meg ohm in ten segments.
I did not model the Capacitor losses, but at this frequency it will not be
major.
The ripple with this kind of multiplier is significant, and may require post
generation filtering.

 

[John Woodgate]

Using the 1N4936 with all caps equal to .33uF and a 55Khz input.

That seems to be rather a large capacitor for 55 kHz and hardly any
current. I once used 68 nF at 50 Hz.

 

[Albert]

That seems to be rather a large capacitor for 55 kHz and hardly any
current. I once used 68 nF at 50 Hz.

Hi John,

OK, I know the Big H uses .15 uF in their 120 Khz sine based products.
I modeled a 5 stage cw in my old cripple ware spice, and found .1 uF
worked well at 100 Khz square wave input model.

When I had the cripple ware spice running, I had a 5 stage cw modeled
with appropriate load resistors and discovered that the I got the same
results with a 2 stage divider. I think this is because the load is
very small past the second stage, the leakage current of the diodes in
the upper stage exceeds the load current of the output current of the
upper stages and because the caps in the upper stages are not
completely discharged and recharged (as the are in the first stage). I
concluded that one could use a properly loaded 2 stage cw and get
nearly the same results as the full 10 stage circuit (in a pm tube
model only).

Regards,

A

 

[Albert]

Modeling the circuit:
Using the 1N4936 with all caps equal to .33uF and a 55Khz input.
I show a rise time of 83ns and a simulated voltage of 1.154kv with a load of
10Meg ohm in ten segments.
I did not model the Capacitor losses, but at this frequency it will not be
major.

If you really want to model it, set your load resistors so they
approximate an actual tube.

Set up a 10 stage multiplier and signal generator with the output set
so that the average dc output of the first stage = 100v. Assume each
dynode has a gain of 4 and set the load resistor on the first stage so
that the current through the load on the first dynode is 75ua. Set the
resistor across the second stage so that the load resistor draws 25
ua. Set the load r on the third stage so that the load current is 25/4
ua and so...all the way to the top stage. In actuality, you don't need
to put resistors above the 5th stage as the r gets so small it doesn't
impact the end result. This models a PM tube at it's worst case output
current of 100 ua.

To model it at a more typical operating point, use about 1 ua output
current. This is easy to do, change the load resistors so that each
one is 100 times larger than the worst case model above.

You can then measure the power dissipation of each load r and add them
up. This is the total power delivered to the load.

Measure the current from the 55 Khz source (average) and multiply that
towards the actual output voltage of the generator and multiply them
together to get the power delivered to the input of the cw.

With the power in and the power out values, you can determine
approximate efficiency. My guess is that it will be fairly efficient.

I will be doing this procedure in the near future as I become more
familiar with my spice program, I'm definately a newbee with respect
to spice:>:

Regards,

A

 

[Albert]

I was unaware this project had religious significance, I do not need (or
Want) your prayers.

Not a problem, and I understand your statement. I could have been
praying to Murphy, asking him to avoid visiting your workshop.

Please accept my apology.

A

 

[John Larkin]

That seems to be rather a large capacitor for 55 kHz and hardly any
current. I once used 68 nF at 50 Hz.

My kilovolt C-W uses 2 nf 0603-size caps, except the first stage used
5 nf, which seems to help (didn't model it; this just works.)
Frequency is about 20 KHz, which is efficient at my light load. If you
seem to need big caps, maybe your diodes are too big.

Incidentally, I tested 0603 resistors for max voltage capability,
hoping to use two in series as the feedback divider. High-value 0603s
typically break down at 1500 volts or so.

John

 

[Roger Gt]

I read in sci.electronics.design that Roger Gt wrote
about 'which diode for 55 KHz switching supply??', on Mon, 15 Dec 2003:


That seems to be rather a large capacitor for 55 kHz and hardly any
current. I once used 68 nF at 50 Hz.
--

And I used 100nF at 330Khz for two stage. But I was looking at the ripple
which was 8VPP and it was only an attempt to see if the diodes work (also I
HAVE .33nF-400V caps) Other values Will do as well!

 

[Roger Gt]

If you really want to model it, set your load resistors so they
approximate an actual tube.

Set up a 10 stage multiplier and signal generator with the output set
so that the average dc output of the first stage = 100v. Assume each
dynode has a gain of 4 and set the load resistor on the first stage so
that the current through the load on the first dynode is 75ua. Set the
resistor across the second stage so that the load resistor draws 25
ua. Set the load r on the third stage so that the load current is 25/4
ua and so...all the way to the top stage. In actuality, you don't need
to put resistors above the 5th stage as the r gets so small it doesn't
impact the end result. This models a PM tube at it's worst case output
current of 100 ua.

To model it at a more typical operating point, use about 1 ua output
current. This is easy to do, change the load resistors so that each
one is 100 times larger than the worst case model above.

You can then measure the power dissipation of each load r and add them
up. This is the total power delivered to the load.

Measure the current from the 55 Khz source (average) and multiply that
towards the actual output voltage of the generator and multiply them
together to get the power delivered to the input of the cw.

With the power in and the power out values, you can determine
approximate efficiency. My guess is that it will be fairly efficient.

I will be doing this procedure in the near future as I become more
familiar with my spice program, I'm definitely a newbee with respect
to spice:>:

Later today perhaps, I need to get some work done too! My last serious use
of spice was modeling all 14 boards in a weather satellite! Found many
interesting things in the circuits that no one expected!

 

[Roger Gt]

Not a problem, and I understand your statement. I could have been
praying to Murphy, asking him to avoid visiting your workshop.

Please accept my apology.

A

If we must invoke help, let us beseech Chernabog! After all HE is a party
animal!

 

[albert]

We may not be talking about the same circuit. Do you have a URL for the
one you looked at?

[millisnip]

Honestly, I haven't modeled this alternative yet, but the upper stages
would require greater than 1 KV PIV, and these types of diodes are very
lossy. Even 500 volt PIV dioded have dismal switiching losses and the
middle stages of the circiut you sugeest would need to have those
diodes.

I don't see at present why the diode ratings need to go up. Each diode
is between an Nth-stage and an (N+1)th-stage. so it seems that PIVs
would be all the same.

I have a pdf, but can't post it here. Now that I think about it, it
might have been the caps that need higher ratings-when I wrote the
original comment about needing higher PIV, I could have err'd.

Regards,

A

 

[albert]

I don't see at present why the diode ratings need to go up. Each diode
is between an Nth-stage and an (N+1)th-stage. so it seems that PIVs
would be all the same.

Yes John, you are correct!

Clearly the caps are in series, not the diodes!

Thanks for pointing this out, I think I should model that circuit
before deciding which one to use in the hardware version:>:

The diodes only need to be the same rating!

Regards,

A

 

[ctsbillc]

WW2 vintage sniperscopes used a Zamboni pile, a series arrangement of
copper-zinc primary cells.

as I recall....

Bill