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Reverse engineering a cheap eBay ultrasonic power supply

A project log for Improve the Haber process

See if ultrasonic cavitation can be used to fixate atmospheric Nitrogen less expensively than the Haber process.

Peter WalshPeter Walsh 04/10/2015 at 20:2219 Comments

TL;DR

I purchased and reverse engineered a cheap eBay ultrasonic power supply, to see if it held any value for the hobbyist.

This first post explains the process of reverse engineering the (or any) board.

A subsequent post will discuss the specifics of this board.

NB: Can anyone tell me what R4 (1 ohm, 5 watt) does in the schematic below?


How to reverse engineer a simple PCB

Dave Jones posted a great video (from his blog) showing how to reverse engineer a PCB. He prints and overlays transparencies of the board top and bottom, so that he can visually align the traces with the components.

I did essentially the same thing, but within a paint program - eliminating the need for physical copies. This has some advantages over his method (zoom, annotation, and contrast adjustment), but some disadvantages as well (small visual aperture).

Step 1: Take a picture of the top and bottom of the PCB

Step 2 Flip and flood

Flip the trace image left-for-right, and flood-fill the copper traces with a high-contrast color. Use the pen/brush tool to touch up areas that didn't fill properly (ie - the shiny, reflective bit in the image above).

Step 3: Make a sandwich image

Compose a 3-layer image with the components image on top, the traces in the middle, and a white bottom layer. Set the bottom (white) layer opacity to 100%, and adjust the opacity of the trace and component layers to give a nice X-ray view of the board.

(This will be specific to your eye sensitivity and the characteristics of your display, so adjust to taste. For my setup, the best settings seem to be White: 100%, Trace 100%, Components: 78%.)

Rotate and adjust the trace layer so that the holes in the trace layer line up with the components.

PCB X-ray view

Step 4: Zoom and annotate

The image can be zoomed and annotated as needed. Also, sections and components can be erased once deciphered. (I found this particularly useful - certain sections are "distracting", and blocking them helps me concentrate on other sections.)



The results: a shiny, new annotated schematic, ready for the surgeon's table:

A high-resolution version of this (pdf and svg) will be available on GitHub presently, as well as the KiCad schematic.

Discussion of this specific circuit will be in a subsequent log.

Discussions

Mjolinor wrote 04/15/2015 at 12:09 point

Is R7 right. I think it should be  base collector.

I have been banging my head on this for a few weeks: http://forums.hackaday.com/viewtopic.php?f=4&t=5345

After reading your article it seems I am right in my hypothesis that the Chinese thought they could make it run from 230 volts by removing one of the diodes in the bridge, it is indeed a 120 volt circuit.

  Are you sure? yes | no

Peter Walsh wrote 04/15/2015 at 23:05 point

After reading your comment I took a close look at the PCB - I think R7 is correct as shown.

I took a section of my annotated image and flood-filled the traces to highlight. The transistor reads BCE left to right, and the resistor connects B and E, both of which go to the negative side through R4.

On startup Q2 is biased "on" through R2 and Q1 is biased "off" through R7. When current flows through the transducer, the corresponding output voltages turn Q2 off and Q1 on. Then current flows the other direction and so on.

A link to the image is below.

https://hackaday.io/project/5283-potpourri/log/16563-r7-looks-correct

Also, the instructions that came with my power supply have an image that shows the difference between 120V and 220V boards, which is the selection of transformer winding. I've included a copy of that image in the link above.

Also also, the board indicates that it can be used for a 40kHz transducer. I assume this is winding selection on the other transformer: smaller winding => less inductance => higher filter center frequency 

  Are you sure? yes | no

Mjolinor wrote 04/16/2015 at 07:18 point

Well you are right about the resistors, it is easier to see in green and it makes a lot more sense. My board however does have both resistors base-collector. I also have a 500 watt Ultrawave USC and that too has the resistors base-collector on all four transistors. This Chinese one does not have any links or extra windings on the transformers so either they ignore the fact that it is a higher voltage or they change the transformers when built.

I think that the Chinese boards are copies of the Ultrawave boards ( or vice versa :) ) but the Ultrawave board has R6 replaced with two diodes in series, same for R8.

I am still having either puny power out or failing components no matter what I change. The original transistor is max 400 volts and rectified mains is 325. Given the number of inductors and the oscillations in there that is way to little headroom and changing the transistors seems to change the circuit responses as you would expect.

  Are you sure? yes | no

Peter Walsh wrote 04/16/2015 at 07:51 point

It sounds like you have a different board.

Another potential problem is shoot-through (look up "shoot through failures" in Google). That's when both transistors conduct at the same time and therefore conduct too much power from the high rail to the low. It can happen for obscure reasons, such as the capacitance of the B-E junction slowing down the switching, so that both transistors overlap "on" for a brief period each half cycle.

I'm going to make a mod to my board where the feedback is amplified and goes into a Schmitt Trigger, then through a half-H-bridge HI/LO driver to control the transistors. The drivers have a programmable dead time so there's no chance of both transistors turning on at the same time. And the Schmitt trigger should ensure feedback even when the transducer is run at low power.

The circuit in this link shows both resistors gobase to collector,maybe some insight there:

http://jumperone.com/2012/02/ultrasonic-cleaner-re/

Good luck with the board - I know those transistors don't come cheap.

  Are you sure? yes | no

Mjolinor wrote 04/16/2015 at 07:58 point

Preventing the crowbar of the supplies is what I think the 1 ohm resistor is for on your board.

I am thinking of just binning the feedback and sticking a PIC in there driving the transistors directly.

  Are you sure? yes | no

Mjolinor wrote 04/16/2015 at 08:53 point

My Chinese board. Transformer top left, choke top right. I am thinking that thsi is wrong. The labelling is not really clear but from what you have written and what I have read this transformer is taking a 110 input and making it 220 for the transducer. As the 230 volt I am supplying is already 230 then this transformer is not needed. I may try just removing it and see what happens.The problem then becomes one of isolation and safety I think. Maybe I need to rewind it as 1:1.

  Are you sure? yes | no

Mjolinor wrote 04/19/2015 at 12:02 point

Cores from the inductor (left) and the transformer (right). Spot the air gap on the transformer core. This would be why mine didn't work when I used an inductor core for it.

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mime wrote 04/16/2015 at 14:51 point

Yeah I realised that as soon as I sent it, that's why I deleted my comment.

  Are you sure? yes | no

Peter Walsh wrote 04/15/2015 at 23:50 point

I've just now read your linked post in detail. Your circuit is very close to mine, check my images above and see if they are the same.

The printed instructions that came with my supply say to never connect the supply to the transducer alone; ie - always connect it to the cleaner bath. Otherwise it'll get very hot.

This indicates to me that these supplies are built to power transducers off-resonance. If you use a tuned horn or the transducer by itself, the supply will draw too much power and overheat.

Okay, some background. The transducer will have an impedance based on the frequency it is being run at. At the transducer's native resonant frequency, the impedance is less than 25 ohms, and if it's anywhere away from this the impedance is much higher - perhaps 1000 ohms.

When you attach a horn or water bath to a transducer, the attached item will have its own resonant frequency and the overall resonant frequency will be a mix of the two.

An untuned horn connected to a transducer will resonate in a frequency that's not the transducer native frequency, so the transducer looks electrically like a 1000 ohm resistor. The power taken from the board is V^2/R, so an output voltage of 220 volts will result in about 50 watts of power, and all is well.

If you connect the transducer with no horn, or if you use a tuned horn, that same 200 volts will push almost 2000 watts into the transducer. The transducer will get hot and so will the supply board.

This may also explain why the boards have blown transistors - if these are cleaner systems and the water evaporated out or something, it would put the transducer close to resonance, draw too much power, and blow the board.

I predict that if you use a tuned horn with one of these boards, it will do the same thing; ie, heat up and blow out something.

Also also, you can use a variac to adjust the power, but only up to a point. At lower power levels the feedback to the transistors is also lower. Eventually you'll reach a point where the feedback is too feeble to control the transistors, and I *suspect* that when you reach this point one of the transistors will blow up.

(The power through the transistors is only 50% of the system power, because each transistor is only on half the time. Disconnecting the transistor keeps Q1 on all the time, and this may put it outside it's safe range.)

Maybe. Keep an eye out for these problems.

I'm planning to publish a modification to the eBay supplies which allow the operator to use lower power w/o the feedback problems. Check back in a month or two - if I can get it to work.

  Are you sure? yes | no

mime wrote 04/14/2015 at 08:22 point

Ideally, the circuit should indicate with a dot the directions of the windings of L1. (Also true for the transformer wires btw).

http://electronics.stackexchange.com/questions/66708/dots-in-a-transformer-symbol

I suspect that L1-2 and L1-3 are wound in opposite directions. In that way only, would Q1 and Q2 conduct in alternate phase of the sine wave. Otherwise, they would conduct simultaneously and short out the power.

I think R4 is there just for current limiting. The FJP13007 can widthstand 8A of DC current and 16A of pulse current. The AC current and the average AC current have a maximum somewhere between those numbers.

If you measure the inductance of T1+T2 at 50/60 Hz, and also their DC resistance, then you could work out the current on the primary side. 

Or, as The Lightning Stalkr suggests, replace R4 by a wire. Directly measure AC current before and after. But at your own risk, and the meter could mess up your measurements.

  Are you sure? yes | no

Peter Walsh wrote 04/16/2015 at 01:37 point

I've updated the build log image w/dots. Good call - thanks for pointing that out.

I should be able to measure the voltages on both sides of the transformer directly. I've been putting off doing that until I have an off-resonance transducer/horn combination. I can burn out the supply by using only the transducer, even on a variac.

  Are you sure? yes | no

mime wrote 04/16/2015 at 14:30 point

Well, I'm not sure what your load is supposed to be (how many watts), but how about adding a rectifier (with schottky diodes because they're faster), a filter cap, and adding a 'normal' load such as an incandescent bulb?

Should be fairly easy, and the voltages are in the same order of magnitude as what a bulb would expect.

I'm not sure how much a bulb would enjoy 40 KHz without a rectifier.

  Are you sure? yes | no

The Lightning Stalkr wrote 04/14/2015 at 00:46 point

Waste power?  I'm not exactly sure myself, but it might "cushion" the low side a little against resonances and form part of a snubber along with the bridge caps.  It could also protect against the oscillator going into a hard lockup, or getting "weird" for some reason.  They probably found an instability and putting a resistor there gets rid of it and no one really knows why.  You could try removing it from the circuit and just replacing it with a piece of wire and seeing what happens.  The transistors are a matched pair, so if one of them goes out, you need to find one with the same gain (β).

  Are you sure? yes | no

Peter Walsh wrote 04/14/2015 at 00:32 point

The connections to T2 are correct, it's not a typo, and I double verified them to be sure.

They're using it (the second winding) as an inductor, so shorting and grounding the primary makes sense.

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AKA wrote 04/14/2015 at 00:14 point

Nice work, thanks for sharing! Good to see this process made "official" with documentation ;-)

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Thomas Shaddack wrote 04/13/2015 at 23:38 point

My guess is an inrush current limiter. These are fairly common in smaller-current power supplies.

A sometimes-encountered variant of inrush current limiters employs NTC thermistors that start as higher-resistance and get more conductive as they warm up. But for a low-current application where the heat losses on the resistor won't be excessive, a simple resistor will do the job too.

  Are you sure? yes | no

pastcompute wrote 04/12/2015 at 11:25 point

I'd naively guess it is either some kind of current limit protection - being connected to the negative side of the full bridge rectifier - or it somehow ensures that one of the transistors turns on after the other to 'kickstart' the oscillation, by adding a small bias. But its been a very long time since I've done any of that kind of analogue circuit analysis...

Are pins 1/2 of T1 supposed to be shorted?

  Are you sure? yes | no

Peter Walsh wrote 04/12/2015 at 15:02 point

I verified the T1 connections - I think they're using it as an inductor. The board has the capability to choose between taps on the other winding, I think they use one winding for 28kHz and the other for the 40kHz version. To an effect, they're grounding the unused side of the transformer.

  Are you sure? yes | no

Peter Walsh wrote 04/10/2015 at 22:44 point

NB: Can anyone tell me what R4 (1 ohm, 5 watt) does in the schematic above?

It's a mystery to me.

  Are you sure? yes | no