Power supply help

[drumsticksplinter]

Hello,

I'm looking for some assistance in building a linear power supply for a machine that uses DC servo's.

I am replacing some old drives so I can make use of the machine in a different way. The manufacturer of the new servo drives recommends that for every motor amp, 2000uf in capacitors should be used. I have 3 x motors and have measured the current for each whilst running from the existing drives. Their combined initial current is around 11A, dropping off to approx 7A after the inertia has been overcome (less that a second).

The servos are rated at 140VDC and I have a 110V transformer approx (2kva). Am I correct in thinking that I will end up with around 155VDC after smoothing? If so then this is roughly 10% over rated, which is ok right? I have 4 x 7400uf 200VDC caps and was planning on using 3 of these. I also have a 250V 25A rectifier I was hoping to use.

I suppose the questions are: are my calculations correct? Will I blow anything up if I tired this (presuming the wiring was correct)? What bleed resistor should I use and how should I incorporate it into the system?

Apologies for the questions, questions, questions, I've only been stuck on this machine power supply for 12 months now.... bout time I got it going.

Thanks

 

[(*steve*)]

According to the manufacturers recommendation, you will need 2200 * 7 uF -- that's about 16,000uF, or 3 (or 4) of the 4700uF capacitors

The main issue here will be switch on transient current. As you switch the power on with the capacitors discharged, your power supply will draw a *huge* current to charge them.

The next issue (comparatively smaller) is that your supply will draw very large current spikes at or near the top of each mains half cycle.

Your diodes and your transformer will need to be rated to cope with this. Again this is a question for the datasheets. If you assume that all the current is drawn through about 10% of the waveform the rectifier average peak current is around 110A falling to 70A. Is this within the capability of that rectifier (remember that the average current is still 7 to 11A)

You are correct about the maximum voltage for 110VAC input. This assumes something about your transformer regulation and the incoming mains voltage. In practice you may find your output voltage is somewhat higher.

I can't comment on whether 110% of rated input voltage will be a problem. This is a question for the datasheet or the manufacturer.

 

[drumsticksplinter]

Thanks for your reply Steve!

I'm a bit confused with the values you calculated though? The manufacturer of the drive recommends *2000uF* per motor amp: 2000 x 7 uF = 14,000uF, so that would equate to 2 of my *7400uF* caps. Should I be aiming to design the power supply to cope with the power the motors need to run or do I need to take into account the power they need to start?? I.e is it?:

Starting current 11A: 2000uF x 11 = 22,000uF. (3 x 7,400uF caps = 22,200uF)
Operating current 7A: 2000uF x 7 = 14,000uF. (2 x 7,400uF caps = 14,800uF)

Some one had mentioned that maybe some sort or soft start circuitry might need to be added or an inrush limiter, to restrict the rush of current to fill the caps.

My transformer was taken from a site transformer (UK) and would be used to step down 240v to 110v for power tools etc. I don't have any information on it as the casing was discarded a number of years ago. The circuit protector was a 25A circuit breaker, which I assume was correct for the rating of the transformer. Does this sound up to the job?

my bridge rectifier is this one:

http://http://uk.rs-online.com/web/...22677633D4E4F4E45267573743D3632392D3631353626

don't think that this one is suitable then?

My motors datasheet says that the rated voltage: 140VDC is +/- 10% tolerance, so hopefully they will be ok. Maybe there will be some voltage loss from the drives / cabling?

For the bleed resistors I was thinking of using a normally closed relay with power resistors shorting the caps. Once power is applied the relay takes the resistors out of circuit allowing normal operation. However, I have no idea what the size of the resistor should be? I would probably assume that if I could drop the voltage to say 10/12v over a minute of so this would be safe.

Thanks,

Adam.

 

[(*steve*)]

Thanks for your reply Steve!

I'm a bit confused with the values you calculated though? The manufacturer of the drive recommends *2000uF* per motor amp: 2000 x 7 uF = 14,000uF, so that would equate to 2 of my *7400uF* caps.

Sorry, I assumed they were 4700uF. 4700uF is a standard value, 7400uF is not.

Should I be aiming to design the power supply to cope with the power the motors need to run or do I need to take into account the power they need to start?? I.e is it?:

Starting current 11A: 2000uF x 11 = 22,000uF. (3 x 7,400uF caps = 22,200uF)
Operating current 7A: 2000uF x 7 = 14,000uF. (2 x 7,400uF caps = 14,800uF)

Ideally make it capable of a sustained 11A, but it depends on how often the motors are starting up and if all of them start up at once.

The rectifiers should be capable of the maximum currents.

Some one had mentioned that maybe some sort or soft start circuitry might need to be added or an inrush limiter, to restrict the rush of current to fill the caps.

Yeah, that could be useful.

There are several options. One is to use a resistor in series with the mains input that is shorted by a relay a couple of seconds after power is applied.

My transformer was taken from a site transformer (UK) and would be used to step down 240v to 110v for power tools etc. I don't have any information on it as the casing was discarded a number of years ago. The circuit protector was a 25A circuit breaker, which I assume was correct for the rating of the transformer. Does this sound up to the job?

If a 25A circuit breaker was used then the transformer is probably capable of 25A continuous. Sounds fine

There is a small chance it could be an autotransformer. Use your multimeter to determine if there is any connection between the primary and the secondary side.

my bridge rectifier is this one:

http://uk.rs-online.com/web/p/bridg...22677633D4E4F4E45267573743D3632392D3631353626

don't think that this one is suitable then?

That seems fine. Note that you will have to connect it to a heatsink. It will be dissipating up to 25W of power.

My motors datasheet says that the rated voltage: 140VDC is +/- 10% tolerance, so hopefully they will be ok. Maybe there will be some voltage loss from the drives / cabling?

If they're just motors then the important thing is the voltage under load. You need to place a load equivalent to a running motor onto your power supply and measure the voltage. As I said earlier it's likely to be higher off load, and also if the mains voltage is above the nominal value.

For the bleed resistors I was thinking of using a normally closed relay with power resistors shorting the caps. Once power is applied the relay takes the resistors out of circuit allowing normal operation. However, I have no idea what the size of the resistor should be? I would probably assume that if I could drop the voltage to say 10/12v over a minute of so this would be safe.

Yeah, that's fine. If you get a 25W 1k resistor it will dissipate a max of 25W, but won't get hot because the capacitors will discharge pretty fast.

something like this: http://uk.rs-online.com/web/p/panel-mount-fixed-resistors/7547828/

It will drop to 21V in 30 seconds.

 

[drumsticksplinter]

Sorry, I assumed they were 4700uF. 4700uF is a standard value, 7400uF is not.

Thats ok, I was just worried that I'd missed something.

Sorry, I assumed they were 4700uF. 4700uF is a standard value, 7400uF is not.

If I were to base the power supply on 11A, would my rectifier still be ok? Or should i up rate It?

Quote:
Some one had mentioned that maybe some sort or soft start circuitry might need to be added or an inrush limiter, to restrict the rush of current to fill the caps.
Yeah, that could be useful.

There are several options. One is to use a resistor in series with the mains input that is shorted by a relay a couple of seconds after power is applied.

I like the idea of this what rating resistor can I use? I read a bit of a paper that suggested 50w 500R, but I have no idea... I could use a timer relay that I already have!

Quote:
For the bleed resistors I was thinking of using a normally closed relay with power resistors shorting the caps. Once power is applied the relay takes the resistors out of circuit allowing normal operation. However, I have no idea what the size of the resistor should be? I would probably assume that if I could drop the voltage to say 10/12v over a minute of so this would be safe.
Yeah, that's fine. If you get a 25W 1k resistor it will dissipate a max of 25W, but won't get hot because the capacitors will discharge pretty fast.

something like this: http://uk.rs-online.com/web/p/panel-...stors/7547828/

It will drop to 21V in 30 seconds.

Thanks for this helpful information, I'm more confident that I'll actually get this working now

 

[(*steve*)]

If I were to base the power supply on 11A, would my rectifier still be ok? Or should i up rate It?

Probably, but you could always uprate the rectifier to the 35A version to be really sure.

The devices are rated for average current over a half cycle, and this takes into account the fact that you may be driving a capacitive load. You will note the huge non-repetitive single half cycle rating (300A or so).

If the 35A device is not physically much larger, and not too much more expensive you could get that and feel even more reassured.

I like the idea of this what rating resistor can I use? I read a bit of a paper that suggested 50w 500R, but I have no idea... I could use a timer relay that I already have!

500R 50W might be fine. It will be dissipating more than 50W initially, so you would have to check the datasheet to ensure you don't put it at risk of blowing like a fuse. Assume (for argument's sake) that it will have the full voltage across it for half a cycle -- so if that is 240VAC, you;re looking at a current just under an amp.

You may be able to use a relay you already have. Ensure it can take the full load current -- which will be about 6A at 240V

The resistor and the relay can also be placed on the secondary side. This may be safer, but remember that the relay will need to be able to carry twice the current.

 

[KrisBlueNZ]

This is way outside my experience, so this is just a suggestion. One, two, or even three, biiiiig chokes in pi filter configurations, with relatively low smoothing capacitance at the transformer end, increasing towards the load end.

For example, 1000 µF on the bridge output, rated for a pretty mean ripple current, then a series choke, then 4700 µF with a decent ripple current rating, then a second choke, then 2x 7400 µF across the load. Or split it into three chokes.

The chokes would have to be pretty huge - high inductance and high current. Like I said, this is outside my experience; it's just a suggestion to improve the power factor and mitigate the startup surge.

Edit: Probably the "proper" way to do it would be to feed the mains into a power factor correction stage producing around 200V DC followed by an isolated forward converter, or an input transformer with say an 80V secondary feeding a power factor correction stage that powers the load directly. Either of these options would provide a regulated output, and could include a soft start to run the motors up to speed. But they would both be pretty high power engineering. Both approaches would need a choke for the PFC section, operating around 30 kHz or so, and a transformer for isolation. The first option would be more compact because the transformer would operate around 30 kHz too, rather than mains frequency for the first option.

Edit2: It might be wiser to have a separate power supply for each motor.

 

[drumsticksplinter]

Probably, but you could always uprate the rectifier to the 35A version to be really sure.

This seems a sensible idea, they are pretty cheap I think anyway.

The resistor and the relay can also be placed on the secondary side. This may be safer, but remember that the relay will need to be able to carry twice the current.

If this this would be safer, then I'd be tempted to go with that. So I'd need something like a 12A 240V rated relay then? Also, would the resistor be the same value if placed on the 100V side?

500R 50W might be fine. It will be dissipating more than 50W initially, so you would have to check the datasheet to ensure you don't put it at risk of blowing like a fuse. Assume (for argument's sake) that it will have the full voltage across it for half a cycle -- so if that is 240VAC, you;re looking at a current just under an amp.

How do I calculate the realistic value of this resistor value? I'm going to order up all the components up asap, so if I could know which soft starting resistor would be suitable then I'll be on my way to getting this thing fired up

Thanks for jumping in on this KrisBlueNZ. Although your suggestions for alternative options might be viable, I feel that they would require much more electrical engineering, which is a skill I know nothing about....

Edit2: It might be wiser to have a separate power supply for each motor.

This might be true actually, but I like the idea of a bigger power supply like a reservoir, so should any 1 of my motors want more power then it is available, providing they all don't require it at the same time.

Thanks,

Adam.

 

[(*steve*)]

How do I calculate the realistic value of this resistor value? I'm going to order up all the components up asap, so if I could know which soft starting resistor would be suitable then I'll be on my way to getting this thing fired up

I used a fairly simple formula.

P = V^2/R -- This gives the power dissipation for a resistor of value R with a voltage V across it.

Rearranged you get R = V^2/P and we can use this to determine the resistance required for a particular power dissipation at a given voltage.

So for 110V and 50W (50W resistors are not that expensive) you get:

R = (110 * 110)/50 = 242 ohms (250 ohms would be fine).

Note that the current would fall quite rapidly as the capacitors charge, so you might be able to get away with a lower value resistor. This would require checking the datasheet for the resistor. In the previous example I halved the value of the resistor.

You are probably best off attaching the resistor to a heatsink, although it probably has sufficient thermal mass that it won't heat up a lot in the few seconds it is in circuit. If the relay fails that's another story.

I'd go for a 20A relay, but practically it won't be switching under a large load, so a lower rating may be fine.

 

[drumsticksplinter]

Thanks ever so much for this info Steve, its most helpful

I've put together a circuit diagram of the rough layout, just to see if I've got the right idea:

View attachment bpt power supply.pdf

Hopefully I'm not far off... I was thinking maybe of attaching a 10A thermostat to the soft start resistor, so in case the relay didn't fire for some reason then the primary side would go open circuit over say over 80 degrees. The thermostat's are more expensive than the actual resistor, but it would save it blowing and causing machine down time.

Something like this:

http://uk.rs-online.com/web/p/thermostats/2295878/

Thanks,

Adam.

 

[(*steve*)]

If you attach the thermal fuse to the same heatsink as the resistor and the bridge rectifier then you'll have additional protection against overload.

Send me the link for the 50W resistor you're looking at and I'll check out the specs to make sure it's OK.

edit: if your timer delayed relay has an additional NO set of contacts you could use them to only make the 155V available after the delay for capacitor charging. This would have the effect of making the output unavailable of the delay relay fails in such a way that the 270 ohm resistor was always in circuit.

edit 2: if it has SPDT output, you could use that with some judicious rearrangement of the circuit. Let me know and I'll draw the changed for you.

 

[drumsticksplinter]

Hi STeve,

I'm sure I replied to this yesterday, but for whatever reason it mustn't have posted...

Here is the link to the resistor I was looking at getting, the 1K 25w for the bleed resistor is from the same range:

http://uk.rs-online.com/web/p/products/1074197/

I've found out my timer relay and unfortunately its only rated at 8A 240V with SPDT relay. I really like the idea of making the 155v unavailable unless the relay activates and cuts out the soft start resistor. If I'm correct in thinking, would this also add protection to the resistor from overheating because once the caps were fully charged, no more current would flow until the output for the 155v "enable" was activated. Which would then switch the resistor off from the circuit?

Is there a simple way of making a delay happen for say 1 second for a normal relay? I'm having trouble finding a timer relay with contacts rated high enough. If I could use a readily available relay I could pretty much have whatever configuration I wanted.

I've tested up my transformer today. Definitely no resistance between primary and secondary. The voltage output is a true 110VAC also

Thanks,

Adam.

 

[drumsticksplinter]

Did you have a chance to look over the data sheet Steve?

Thanks,

Adam

 

[(*steve*)]

Sorry about that, and thanks for the poke.

I've actually been looking at these (or similar) resistors myself for something and I can confirm that they have good overload performance. In the case of this resistor the datasheet is pretty sparse, but they provide "technical information" which fills the void.

These resistors will handle 10x the rated load for 1 sec, 5 times for 5 secs and 2x for several minutes

If you use a 50W 500R resistor on the primary side to limit the switch on surge, it will have a max load of 240^2/500 = 115W. That's just over double the rated power, so you're fine as the capacitors will charge up pretty quickly. [edit 250R 50W on the secondary side allows half the current and the power dissipation is under 50W. A 100R 50W would have a similar limit (in terms of current to the caps) as the 500R on the primary) and would also be safe given an overload just over 2x for a short period.]

For discharging, 1k 25W... Max dissipation is under 25W.

In both cases the voltage you're supplying is well under the limiting voltage.

I think you're good to go.

Take note of the maximum dissipation with and without a heatsink. However given that your load is of relatively short duration, a smaller than recommended heatsnk (or none at all) may suffice. The best recommendation is to try it without a heatsink once and see how hot they get. Anything less than 100C should be fine.

 

[neon]

It seems to me that it depends of when the switch is turned ON for an AC input it can be zero or at the peak. At zero point obviously no big inrush but at the peak current is very large to eliminate this kind of problems SCR on the diodes input would be nice

 

[drumsticksplinter]

Thank again for the information, you have been most helpful and I really appreciate your assistance and guidance. I will order up some components and get this thing going Will try and post my findings as it might help others with similar projects.

Thanks,

Adam.