Capacitor too big?

[Ed.]

Hi Spec,
1) Mains would be preferable depending on cost, budget is almost non existant, just bought a newish car.
2) 36V at 30A+

Hope this helps.
Ed.

 

[Ed.]

Did as TCR suggested and did a quick check, removed the capacitor, the measured output of the bridge rectifier is 39V DC connected it to the 3 x batteries connected in series and it measured 47.5V+ DC at the end of the battery terminals. What gives? How can the voltage get higher than the sum of the batteries or the sum of the rectifier output charging them. Is this because of the RMS factor?

 

[spec]

Hi Ed

Did as TCR suggested and did a quick check, removed the capacitor, the measured output of the bridge rectifier is 39V DC connected it to the 3 x batteries connected in series and it measured 47.5V+ DC at the end of the battery terminals.

I thought TCM was saying to just try the whole system, inverter and all, and hope for the best.

It is handy that you have done that test. I was about to ask you to do a similar test except with a resistor in series with the bridge rectifier output.

Can you do the test again but set your multimeter to AC volts and connect a 100nF, or over, non-polarized capacitor in series with your multimeter and measure the ripple voltage across the three batteries. The ripple voltage should only be a volt or so, because there is no load on the batteries.

What is the maximum DC current range on your multimeter because it would be handy to know what current is being pumped into the batteries. I would guess 10A average. The battery charge current will be in gulps at the peaks of the rectified AC.

If it were not for the absolute maximum input voltage of 40V for the induction heater, all you would need to do would be to connect up the transformer, bridge rectifier, a resistor, and the induction heater and that would do your job. But it would not be advisable to leave the transformer connected to the batteries or they will be over charged and damaged.

The question now is, with a 30A load on the three serially connected batteries, what would the battery voltage then be. I would guess around 38V, which would be OK for the induction heater, depending on the ripple voltage.

What I was going to suggest when you get a working induction heater is to connect four power diodes in series between the battery and induction heater (to drop the voltage seem by the induction heater by 4V or so) and then adjust the series resistor between the bridge rectifier and batteries for an average voltage of 42.3V, which is a good charge voltage for the three batteries in series.

What gives? How can the voltage get higher than the sum of the batteries or the sum of the rectifier output charging them. Is this because of the RMS factor?

Yes, The output from the bridge rectifier will be a rectified sine wave. The peaks of the sine wave will force gulps of current into the battery and the battery will act like a huge voltage clamp at a voltage related to the charging current. The peaks will occur at 100 Hz, which is double your mains supply frequency of 50Hz for Australia.

One question, when the bridge rectifier was connected directly to the three batteries in series, was the transformer humming loudly and did it get hot?

spec

 

[spec]

Hi again Ed,

The non-battery approach I had in mind was to use an off-the-shelf, 36V, 30A mains-powered switch-mode power supply in place of the three serially connected batteries. This would be the best system, but the power supply would cost around $90US from what I could tell after a quick search on the net.

As a result of the cost, I have abandoned the switch mode power supply approach for the time being.

spec

 

[Ed.]

Hi Spec, when I had it connected to the batteries I could not really hear any humming or noise apart from the split second when I threw the switch on and it made a very very slight noise but that was it, as for heat, I didn't leave it working long enough (30 seconds only) as I was watching the voltage slowly increase from 47V to 47.5V and it would have kept going, so switched it off to not damage the batteries.
I couldn't detect any heat increase in the transformer in that time.

"Can you do the test again but set your meter to AC volts and fit a 100nF, or over, non-polarized capacitor in series with your multimeter and measure the ripple voltage across the three batteries."

I don't have one of those capacitors that I know of, unless there is one on an old circuit board that I can use and identify, The multimeter I have is a Fluke 177 which has a 10A max although I suspect that it has a fault as is not working when measuring current so I can't rely on it.

I checked out the price of large transformers a while ago which is why I didn't consider them as they were in the hundreds of dollars, and that is why I contemplated at one stage of using the MIG welder.

There are two other options I considering.

1) Adding another battery to the group, connecting the DC output to the ends of the 4 batteries and just connecting the heater to only three batteries, so the idea being that the 47-55V charging voltage being shared between the 4 batteries but still supplying the correct voltage to the heater, although I suspect that it won't be good for the extra battery! Having said that, due to the short time the batteries would be supplying power it may not be an issue.

2) Using 2 pairs of 18A x 2 batteries, so 4 batteries in total running at 24V @ 36A, it would give me greater reserve but wont be able to charge them till I finished heating.

 

[Les Jones]

Hi Ed,
would it not be possible to remove some turns from the secondary of the transformer so that it gives the voltage that you require ?

Also I suspect that you could have blown a fuse in your meter. They very often have a separate socket for the high current range which is protected by it's own fuse.

Les.

 

[Ed.]

Hi Les, I would most likely stuff it up if if I tried to reduce the windings, it looks like it may have been dipped and if I start pulling the windings off I may damage the insulation on the wire or possible short circuit it. I certainly don't trust myself to do that sort of work.

I hadn't thought about the fuse being blown, might have to take the back off it to have a look. Been meaning to get a digital amp meter for ages, will have to get one next year.

 

[spec]

Hy Ed

...when I had it connected to the batteries I could not real hear any humming or noise apart from the split second when I threw the switch on and it made a very very slight noise but that was it, as for heat, I didn't leave it working long enough (30 seconds only) as I was watching the voltage slowly increase from 47V to 47.5V and it would have kept going, so switched it off to not damage the batteries.
I couldn't detect any heat increase in the transformer in that time.

Hmm, sounds like a meaty transformer.

"Can you do the test again but set your meter to AC volts and fit a 100nF, or over, non-polarized capacitor in series with your multimeter and measure the ripple voltage across the three batteries."

I don't have one of those capacitors that I know of, unless there is one on an old circuit board that I can use and identify, The multimeter I have is a Fluke 177 which has a 10A max although I suspect that it has a fault as is not working when measuring current so I can't rely on it.

OK - no sweat

There are two other options I am considering.

1) Adding another battery to the group, connecting the DC output to the ends of the 4 batteries and just connecting the heater to only three batteries, so the idea being that the 47-55V charging voltage being shared between the 4 batteries but still supplying the correct voltage to the heater, although I suspect that it won't be good for the extra battery! Having said that, due to the short time the batteries would be supplying power it may not be an issue.

Interesting approach, but bulky and expensive. By the way can I ask how much each battery costs?

There are two other options I am considering.
2) Using 2 pairs of 18A x 2 batteries, so 4 batteries in total running at 24V @ 36A, it would give me greater reserve but wont be able to charge them till I finished heating.

This is similar to my suggested 24V system, except with four batteries.
Just to clarify a few things:
(1) An 18 Ah lead acid battery will source more than 18A, more like 100A
(2) The induction heater current is dependent on its input voltage, not the Ah rating of the batteries. With 24V input the induction heater current will still be 20A, even if you had a pair of 100Ah lorry batteries in place of your 18 AH batteries. The amount of induction heating power with a 24v input would thus be 24V *20A= 480 Watts (to a very rough approximation)
(3) With a 36V input the induction heater current would be 30A so the induction heating power would be 36V * 30A = 1080 Watts (once again to a very rough approximation)

I think you ruled out a 24V system before because you wanted as much induction heating power as possible- is that correct, or would you in fact be happy with the heating power from 24V.

Don't get too hung up about transformer output voltages- there are always ways of sorting things in electronics.

spec

 

[Ed.]

Hi Spec, I am not sure about true battery costs as I bought a heap of them from a scrap metal dealer at about $2/kg, these ones were only a year old at the time and were just replaced during routine maintenance. I got 2 x 60A ones which are now connected to a 1Kw UPS, 6 x 36A connected to two more UPS's and a heap of the smaller ones which are spare. 2 are connected to a drum sander which I built and drive the conveyor belt, and I still have 4 spare to play with, which I can use for this project. I suspect that the 18A ones are about $60-110 each new. The reason for so many UPS's is that we have a lot of summer electrical storms and blackouts are common, so all the TV's, my computer network and sound system are all protected, so I scored well when I got them.

Ideally I would like to use a 36V system if possible as it will generate more heat quicker and I can use it for other projects as well so that would be my first choice, but if not then 24V will have to do but more work to charge the batteries afterwards and heating the metal will take longer.

With the battery amp hours the more reserve I have the less chance of the batteries being drained too much and quicker recharge time.

 

[spec]

I am not sure about true battery costs as I bought a heap of them from a scrap metal dealer at about $2/kg, these ones were only a year old at the time and were just replaced during routine maintenance. I got 2 x 60A ones which are now connected to a 1Kw UPS, 6 x 36A connected to two more UPS's and a heap of the smaller ones which are spare. 2 are connected to a drum sander which I built and drive the conveyor belt, and I still have 4 spare to play with, which I can use for this project. I suspect that the 18A ones are about $60-110 each new. The reason for so many UPS's is that we have a lot of summer electrical storms and blackouts are common, so all the TV's, my computer network and sound system are all protected, so I scored well when I got them.

That is an excellent purchase.

Ideally I would like to use a 36V system if possible as it will generate more heat quicker and I can use it for other projects as well so that would be my first choice.

Thanks for clearing that- your requirements come first- implementation and compromises come later, but I do not think there will be any compromises in this case.

With the battery amp hours the more reserve I have the less chance of the batteries being drained too much and quicker recharge time.

You can put as many batteries in parallel as you like. You can even have different numbers of batteries in parallel in the three battery stack.

spec

 

[spec]

Hi again, again Ed,

Here is another approach for you to consider:

(1) Three battery stack
(2) Mains switch mode power supply to charge the battery to a precise voltage of 42.3V with a current of 10A.

A suitable power supply would probably cost around £30UK from eBay.

How does this approach grab you (don't worry about the implementation details for the time being)?

And here is another question relating to the transformer charging approach. I take it that it would not be practical to add any more turns over the existing turns on the transformer, but do you have another high current but low voltage mains transformer knocking around in your work-shop?

And another general question, how confident are you about assembling a simple electronic circuit, if all the details were explained? Also, do you have the necessary tools: soldering iron etc?

spec

 

[Ed.]

Hi Spec,

Unfortunately I don't have another transformer with the same amperage output. I do have one that is the same physical size but with twin outputs, a 127V and a 57V, however the wire in the 57V coil is about half the thickness of the one I am currently using, it is also wrapped in a different manner. The first transformer is in two parts with the 240V winding on the bottom half of the steel former and the 55V is above it on the former with considerably thicker wire.

This one looks like the 240V primary is in the center wrapped around the former and full height, and the 2 other windings are wrapped around the primary coil also full height. The wire for the 57V and the 127V coil is about the same thickness as the primary coil. They are also separate coils and not just tapped with a common lead. So I don't know how many amps the 57V coil will output but that is all I have apart from a similarly large transformer out of a microwave but that is probably a very high voltage output so not suitable.

That 30 pounds cost may not translate to being able to source it in Australia in equivalent AU$ , as shipping would be a killer if it was only available from overseas, even interstate freight can be quite prohibitively costly here in Oz. Lastly these transformers even if available locally could be vastly overpriced here as well. Yes I have the necessary tools, and should be able to follow simple circuits.

However it sounds interesting, what do you have in mind? If there is a transformer you have in mind , could you send me a link and I will see what is available locally to match it.

Ed.

 

[spec]

Hi Ed

Unfortunately I don't have another transformer with the same amperage output. I do have one that is the same physical size but with twin outputs, a 127V and a 57V, however the wire in the 57V coil is about half the thickness of the one I am currently using, it is also wrapped in a different manner. The first transformer is in two parts with the 240V winding on the bottom half of the steel former and the 55V is above it on the former with considerably thicker wire.

This one looks like the 240V primary is in the center wrapped around the former and full height, and the 2 other windings are wrapped around the primary coil also full height. The wire for the 57V and the 127V coil is about the same thickness as the primary coil. They are also separate coils and not just tapped with a common lead. So I don't know how many amps the 57V coil will output but that is all I have apart from a similarly large transformer out of a microwave but that is probably a very high voltage output so not suitable.

OK, no probs about transformer.

That 30 pounds cost may not translate to being able to source it in Australia in equivalent AU$ , as shipping would be a killer if it was only available from overseas, even interstate freight can be quite prohibitively costly here in Oz.

If you could get a PSU for around £30, UK including delivery, would that approach be acceptable?

Yes I have the necessary tools, and should be able to follow simple circuits.

However it sounds interesting, what do you have in mind?

Below is a rough schematic that I knocked out in a few minutes. It is very much notional but it gives an idea of the sort of arrangement I have in mind. Although the circuit may look complicated it is not. Would you be happy building something like that?

spec

 

[spec]

Hi Ed,

here are a few simpler mains driven induction heater solutions with no batteries:

(1) Just stack three of these power supplies for a 36V, 30V system: **broken link removed** at $24.53Au each supply, including delivery, post, and packing. Total gross cost for all three supplies = $73.59Au or £42.68UK

(2) Or just one of these power supplies for a 27.6 (24v adjusted up to +15%) 30A system: **broken link removed** at $52.67Au, or £30.55UK, each supply, including delivery, post, and packing.

Other similar vendors will do a custom output voltage, so maybe this one will, so you could probably get a 36V, 30A or close power supply.

Alibaba are also a source of similar good-value power supplies, but I can't find out their delivery charges for Australia.

spec

 

[shortbus=]

spec, not meaning to hijack the tread, but have a question. The schematic in post #33, you show four diodes in series to drop voltage. I other threads I was going to do that in a circuit and was told it was a bad idea by a few people. Is it a good practice to do this? It was the only way I could/can figure out how to give a specific voltage drop in my circuit, that is stalled because of people saying it's not good practice. Thanks for an answer.

 

[Ed.]

Hi Spec, I should be able to handle that circuit build of yours, but before I do, would my option which I mentioned in an earlier post about the 4 batteries work? 'cause if it would work that would be the simplest and quickest option as I already have the batteries.

1) Adding another battery to the group, connecting the DC output to the ends of the 4 batteries and just connecting the heater to only three batteries, so the idea being that the 47-55V charging voltage being shared between the 4 batteries but still supplying the correct voltage of the 3 batteries to the heater, although I suspect that it won't be good for the extra battery! Having said that, due to the short time the batteries would be supplying power it may not be an issue.

Incidently I took my transformer to Jaycar which is an electronics supplier here in Australia and showed it to them, and the two guys who work there both agreed that it is probably about a 40-50A transformer.

 

[tcmtech]

An easy cheat to get your transformer output voltage down would be to reconfigure a common smaller unit with a 10:1 reduction ratio as a autotransformer and use it to knock off ~10% of your input voltage to the main unit.

Most automotive battery chargers use a center tapped secondary so between that and the multiple taps already in place on their primary you can make a autotransformer with a pretty wide selection of buck/boost capabilities.

 

[spec]

spec, not meaning to hijack the tread, but have a question. The schematic in post #33, you show four diodes in series to drop voltage. I other threads I was going to do that in a circuit and was told it was a bad idea by a few people. Is it a good practice to do this? It was the only way I could/can figure out how to give a specific voltage drop in my circuit, that is stalled because of people saying it's not good practice. Thanks for an answer.

Hi SB,

I can quite understand your dilemma about using diodes to generate/drop a voltage, but to say that it is a bad idea full stop is too wider statement- it all depends on the application and the implementation.

Diode and BJT VBE drops are ubiquitous in electronics and are used in pretty much all integrated circuits, one way or another, with great success.

One of the things about using diode drops, that puts people off, is the temperature dependence of roughly -2mV deg C. Sometimes that voltage change with temperature is a nuisance and other times it is essential, for example in temperature sensors or VBE multipliers in typical class AB audio power amplifiers.

To put the voltage/temperature variation in perspective though:
(1) The forward voltage drop of a standard silicon diode, at relatively low currents, is around 600mV at a junction temperature of 25 Deg C. This is based on the 'band-gap' for a silicon semiconductor junction which is a physical characteristic.
(2) If you take an equipment local temperature as ranging from 10 Deg C to 50 Deg C, that gives a temperature change of 50-10= 40 Deg C.
(3) With a 40 Deg C temperature change the diode forward drop is going to change by 2 * 40 = 80mV
(4) So the total percentage voltage change is, (80mV/600mV) * 100 = 13.33% or +- 6.67%, which is not an umanageble voltage change.

Of course, all this is very simplified and is only a rough guide; as soon as you start taking significant current, compared to the diode size, the diodes resistance come into play and that resistance is temperature sensitive too.

In the circuit of post #33, the requirement is extremely loose: just make sure that the supply voltage to the induction heater never exceeds 40V and, ideally never drops below 36V so, in that application, the string of four diodes will do the job, especially as it is guaranteed that the induction heater will be taking around 30A, which should give a forward voltage of around 1V per diode.

It is not a pretty circuit though I must admit, and for a production item would not be used (the whole approach would be different), but it is cheap, flexible, and effective; that is the aim anyway.

I hope all that answers you question SB; if you would like to discuss the circuit where it was thought that using diode drops was not ideal, just post the circuit on a new a thread and we can all evaluate it- I bet there will be a wide range of views.

spec

 

[spec]

Hi Ed,

I should be able to handle that circuit build of yours.

That is good. I hope you didn't mind me asking.

...would my option which I mentioned in an earlier post about the 4 batteries work? 'cause if it would work that would be the simplest and quickest option as I already have the batteries.

1) Adding another battery to the group, connecting the DC output to the ends of the 4 batteries and just connecting the heater to only three batteries, so the idea being that the 47-55V charging voltage being shared between the 4 batteries but still supplying the correct voltage of the 3 batteries to the heater, although I suspect that it won't be good for the extra battery! Having said that, due to the short time the batteries would be supplying power it may not be an issue.

Yes, take your point- I will have a look at this approach in more detail.

Incidentally, I took my transformer to Jaycar which is an electronics supplier here in Australia and showed it to them, and the two guys who work there both agreed that it is probably about a 40-50A transformer.

Hmm- tasty transformer.

spec

 

[spec]

Hi again Ed,

About the four battery approach: you have got the opposite problem. That is not enough volts from the bridge rectifier to charge the four battery stack to the ideal charge voltage of 2.35V per cell (24 cells in a four battery stack = 56.4V total required).

But the real worry is that the top battery will be badly overcharged (in view of the current capability of the transformer) while the lower three batteries will be under charged.

Also, you would still need to limit the induction heater voltage to 40V, so a few diodes would again be required.

Sorry for the negative view.

As an aside, I have just realized that it would be wise to put a 50A or so fuse in series with the output of the bridge rectifier as a precaution.

spec