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I am working on a Nixie power supply, but I would like to improve it.

  • I have 4x9V batteries in series for a total of 36V to be switched across a multiplier.
  • A (TTL) 555 timer is running astable from only the first 9V battery to generate a 8.5-ish volt square wave, 10kHz (or any frequency you want, I guess), approx. 50% duty.
  • The 555 output drives the gate of an N-channel BS170 MOSFET.
  • The MOSFET drain is connected up to 36V through an approx 1.2kΩ resistor. This resistance needs to be as low as possible to push current into:
  • a 6-stage Cockcroft-Walton multiplier, which produces a nice ~220VDC output under no load. Unfortunately, it sags to about 155VDC when loaded by a 47kΩ resistor in series with the tube.

Schematic

IN-14 driven from 36V multiplier

Things I like about this circuit:

  • It Works™
  • It can be built by extremely common parts I'm likely to have on-hand, e.g.:
  • It requires no inductors.
  • It requires no specialized IC's such as boost converters.
  • It requires only capacitors and diodes with voltage ratings to handle each stage, not the full shebang.
  • It crashes Multisim.

Things I don't like about this circuit:

  • The output voltage sags to ~155VDC under only ~600μA load.
  • I'm too stupid to think of a better way to switch 36V across the multiplier:
  • While the 555 timer output is high, I'm wasting over 1W across the drain resistor just to drive the multiplier.
  • The multiplier input voltage is hampered by the drain resistor.

How can I:

  • make improvements that can enable ~10mA to be sourced with less than 40V drop in supply output?

I have tried:

  • Replacing the MOSFET driver section with something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

I toasted quite a few transistors trying this inverter. As shown, the gates of the inverter are pulled up to 36V by the 10kΩ resistor. Is it possible that the gate charge time is what destroyed the transistors?

EDIT: I just realized that the maximum ratings for gate-source voltage on both inverter FETs is ±20V. That would explain why they fried. Hmm, maybe instead of a single 10kΩ, I could make a voltage divider to drive each gate separately?

For these reasons, CW multipliers with large number of stages are used only where relatively low output current is required. These effects can be partially compensated by increasing the capacitance in the lower stages, by increasing the frequency of the input power and by using an AC power source with a square or triangular waveform.

  • studying other popular Nixie power supply designs, such as these.

I suspect that switching the 36V across the multiplier more efficiently would go a long way toward improving the performance.

EDIT/SUMMARY: Switching the 36V across the multiplier more efficiently went a long way toward improving the performance. As several people suggested, something called "push-pull" was a quick fix here. A CMOS inverter with separately-driven gates makes the charge pump much more effective:

555 switching 36V more effectively via push-pull

The supply now stands at ~216VDC when loaded with two tubes, a huge improvement:

Much larger loads supported

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  • \$\begingroup\$ You are probably shorting the two FETs on your push/pull config. There will be a moment at which both the devices will be on and shoot through will occur. \$\endgroup\$ – Wesley Lee Mar 17 '17 at 7:56
  • \$\begingroup\$ @WesleyLee I think it's that I'm driving the gates at voltages beyond twice their absolute maximum rating, but I think you're right even if it could handle those voltages. Like I said in the OP, the 10kΩ resistor might be charging the gates too slowly, causing too much shoot-through. \$\endgroup\$ – Yankee Mar 17 '17 at 8:43
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    \$\begingroup\$ Wesley is right. Just limiting the gate drive won't do. Let's say the Vgs (on ) is 5V for each, which is on the high side for all but the most ancient FETs. That leaves 25V of range on the first signal where both FETs are on. That's a lot of short-circuit. Driving FETs like this with a power capability requires an amount of dead-time in the drive signals and it may go into several to dozens of micro-seconds. \$\endgroup\$ – Asmyldof Mar 17 '17 at 8:51
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    \$\begingroup\$ Also be aware that above 2~5mA output the batteries may start to significantly resist the peak currents required, which will still cause droop on the output as well, because these kinds of batteries are notoriously bad at handling decent currents. \$\endgroup\$ – Asmyldof Mar 17 '17 at 8:58
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You need to ditch Rd from your first schematic, and use a low impedance push-pull output as in your second schematic. However, as you correctly say, 36v will toast the gates of 20v Vgs FETs. There are few fets with Vgsmax greater than 20v, and none to my knowledge with more than 30v.

Amongst the options are to use

a) suitable level shifters to control the FET gates, small bipolars would work well here
b) a gate drive transformer (though usually only used for higher power applications)
c) how about 18v push-pull drive from two batteries, but in push-pull, like this ...

schematic

simulate this circuit – Schematic created using CircuitLab

I've illustrated 4 stages here, the extension to more stages is obvious.

Now, I've not connected the upper capacitor. There are two options

a) Cockcroft Walton stylee, where you are limited by maximum voltage. Here, you'd connect C5 to the D1/D2 junction. This allows low voltage across each capacitor, but results in high output impedance. Also known as a Villard cascade, though invented by Greinacher.

b) Dickson charge pump stylee, which results in a much lower output impedance. C5 connects back to the driven end of C2. This means C5 needs a higher voltage rating, but if you can get caps with a suitable voltage rating cheaply, 250v or even 400v are commonly available, then this configuration has a much lower voltage droop with current.

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    \$\begingroup\$ There's a further improvement that I used in my design which I've not illustrated in the schematic. Like in your schematic, I've based the multiplier from ground, so taken my D1 to ground. If instead that point is taken to rail, then you get a whole extra rail's worth of output voltage for free. Output impedance - Cockroft Walton increases as a the number of stages squared, my 'all caps back to driver' configuration only increases linearly with the number of stages, big saving for large numbers of stages. \$\endgroup\$ – Neil_UK Mar 17 '17 at 9:19
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    \$\begingroup\$ from that page, this is the way to drive a MOSFET push-pull - talkingelectronics.com/projects/MOSFET/images/PushPull_12v.gif. Using the audio amplifier version (the one with the 'chain of diodes'?) is far too complicated and meets other requirements. \$\endgroup\$ – Neil_UK Mar 17 '17 at 9:43
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    \$\begingroup\$ From a cockroft-walton, concentrate on the chain of diodes for a moment. Now connect the bottom to ground (or rail for a free one-rail leg-up) and the top to the output. Now notice that alternate nodes are driven in anti-phase. Now disconnect the 'driven' end of each capacitor, and bring them back in two groups. Now drive the two groups in anti-phase, that antiphase could be ground and an AC signal, as in the original cockroft-walton, or it could be two anti-phase AC signals, that get you twice as much voltage swing for your rail. \$\endgroup\$ – Neil_UK Mar 17 '17 at 9:48
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    \$\begingroup\$ you don't have to use BJTs, you can use FETs. However, small FETs blow up easily, and in many ways, BJTs are easier to bias for things like level shifters. Well done for finding the name 'Dickson multiplier', that's exactly the sort of thing I had in mind. \$\endgroup\$ – Neil_UK Mar 17 '17 at 14:14
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    \$\begingroup\$ any time, glad to help. This is the sort of thing we used to make a tiny Dickson multiplier, two channels, high voltage very high current drive output, easy to drive with logic, and tiny. maximintegrated.com/en/products/power/power-switching/… Do accept the answer if you found it useful \$\endgroup\$ – Neil_UK Mar 17 '17 at 14:33
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Replace resistor \$R_D\$ by an inductor which gives you a boost converter. That way you start the voltage multiplication already with voltage \$\gg 36V\$.

But make sure that

  • the MOSFET can hable the voltage (for BS170 \$V_{DS,max}\$ is 60V) and
  • the MOSFET can handle the max. current (mainly depending on operating voltage, switching frequency and inductance); for BS170 \$I_{D,max}\$ is 0.5A.
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  • \$\begingroup\$ I appreciate solutions that use inductors, but I don't have any on hand. Thank you for your answer! \$\endgroup\$ – Yankee Mar 17 '17 at 8:48
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    \$\begingroup\$ @Yankee: tip: if you have a broken compact fluorescent lamp you can get an nice inductor that might be perfect for your purpose from the electonics that is hidden in the plastic socket (the inductor is likely not to be the broken part); make sure not to break the glass tube. The inductance value is probably in the range of a few mH. See the inductor at the right in this picture labled "3.5mH" \$\endgroup\$ – Curd Mar 17 '17 at 8:52
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    \$\begingroup\$ That's awesome. I'll definitely keep that in mind if I'm ever in a pinch. Alas, all the lights in my apartment are LEDs! \$\endgroup\$ – Yankee Mar 17 '17 at 8:57
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36 V on the gate will destroy the devices. You need to find proper MOSFET driving circuits.

Absolute maximum rating for BS170 Gate-Source voltage: \$|V_{GSS}| < 20~V\$.

The simplest way to fix the circuit will be to use a transistor push-pull stage. The pull-up resistor \$R_1\$ will only have to be a little bit smaller. Say \$3.3 ~k\Omega\$?

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  • \$\begingroup\$ Push-pull is what I went with and it worked. I'm sorry for not choosing your answer, I went with the one who helped me understand what push-pull is, since I've never seen that before today. Thank you for the answer. \$\endgroup\$ – Yankee Mar 17 '17 at 18:03
  • \$\begingroup\$ @Yankee I suggested push-pull with normal bipolar transistors, not MOSFETs. (No problem :-) \$\endgroup\$ – skvery Mar 17 '17 at 19:15
  • \$\begingroup\$ Outside of situations involving amplifiers, why is it that push-pull is normally done with BJTs, and not FETs? @Neil_UK said, paraphrasing a bit, " using BJTs is not necessary, but FETs can fail in unexpected ways, and BJTs can be easier to bias." Are there any other reasons why BJTs should be preferred to place and remove charge on the gates of FETs, vs just using other FETs? I just think it's weird that I've literally not found even one example of FETs used to drive gates of other FETs while learning about "push-pull." \$\endgroup\$ – Yankee Mar 19 '17 at 6:07
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as a voltage multiplier, its current output is inversely related to its voltage output. so to increase the current output, you have two choices, without stepping out the topology:

1) increase drive current: the 555 can deliver 200ma and your bs170 a few ma. you could try an emitter follower as a buffer; or a dedicated driver;

2) increase drive voltage: run the whole thing at as high of a voltage as you can;

if i were you, i would try driving the multiplier directly with the 555 first. if that doesn't deliver enough current, think about a different approach, like a step-up converter.

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  • \$\begingroup\$ Driving the multiplier directly from the 555 was the first thing I tried. The issue wasn't the 555's current output capability, but the maximum output voltage. I needed too many stages in the multiplier and never got to 200V. The voltage actually started going back down after about the 15th stage lol. \$\endgroup\$ – Yankee Mar 17 '17 at 18:05
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What kind of batteries do you use? The internal resistance of a 9V battery can be quite high. I think a normal alkaline can only provide around 3 Amps because of that. 36V * 3A / 220V is about 500mA at the output without considering any losses in the circuit. I think rechargeables could perform better.

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  • \$\begingroup\$ These are just run of the mill 9V batteries. For general use cases that might be a concern, but in my specific case I'm drawing much less than 20mA from the end of the multiplier. \$\endgroup\$ – Yankee Mar 17 '17 at 14:59

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