I would like some more explaination as to why  the voltage drop across a resistor based on current draw is going to pose a problem. I have a little better  understanding of the difference between 'switching' and 'unregulated' now, and my understanding is that 'unregulated' supplies do not compensate and the voltage drops when the draw goes up. Gecko recommends 'unregulated' supplies, so I presume they must be designed to handle the voltage drop.

A couple of problems  with using a resistor to drop voltage is that not only is it wastefull (heat producer) it is of no practical use unless the current remains a constant value, any variation in current with affect the output voltage.
Switching supplies are regulated, linear supplies may be regulated or un-regulated, Linear unregulated are sufficient.
Although unregulated, if designed for the maximum current, there should be minimum volt drop when highest current is drawn.
The aim is for the reservoir capacitance to maintain the peak DC.
Incidentally, a little known fact is when building your own linear supply, if you over design the sizing of the smoothing capacitor, you will increase the VA rating required for the transformer.

AC is not part of the question. The thread started with me wanting to power a 72v DC motor with TWO 36V EC supplies.

I decided to use a 36V motor instead of the 72V, so the question became acedemic. I  threw things off by saying my PD was unregulated when it is actually switching and is regulated. I did not know the difference at first. So I learned something very useful about PS, but I don't think there was agreement on an answer.

The situation has changed now and I have kiled the 36v PS and replaced it with a 72V unregulated PS and now seek to run a 36V motor off the new 72V power.

I know it is possible, I just don't know exatly how to go about it and I also know it would not be 'regulated', I just don't know if that will be a problem, or why. I am mostly interested in the why.

At least one of the power supplies listed at Keling have an unregulated 83V output and also a 5V regulated output, so obviously it is doable.

If indeed 'unregulated' means the voltage fluctuates with load, and these are the type of power supplies recommended for my CNC setup, why then would a simlar fluctuating voltage downstream of a resistor be unacceptable?

This seems on the surface to be contradictory, and I would like to understand the issue better. If anyone can shed some light on these question, it would be much appreciated.
 

" . . . any variation in current with affect the output voltage."

This is what I don't get. Isn't this exactly what an 'unregulated' power supply does anyway? If so, why then is this an argument against knocking down the voltage with a resistor?

I'm not concerned with efficienecy or the resistor getting hot. The ones I bought are contained in finned alumuninum cases. I only spent 9 buck on the pair of them, so it's no biggie if I don't use them, but I would like to come away with a better understanding of this issue.

"Although unregulated, if designed for the maximum current, there should be minimum volt drop when highest current is drawn."

The little servo motor runs at 4A and draws a max of 20A at 36V. An onbvious solution is to buy another 36V 9A power supply and set the output current limit on the Gecko drive to 9A. But that doesn't make for a very interesting discussion.

How would one design an unregulated supply for this if they had availble 72V DC to start with instead of 110V AC? Is a transformer the answer?

" . . . any variation in current with affect the output voltage."

This is what I don't get. Isn't this exactly what an 'unregulated' power supply does anyway? If so, why then is this an argument against knocking down the voltage with a resistor?

I'm not concerned with efficienecy or the resistor getting hot. The ones I bought are contained in finned alumuninum cases. I only spent 9 buck on the pair of them, so it's no biggie if I don't use them, but I would like to come away with a better understanding of this issue.

"Although unregulated, if designed for the maximum current, there should be minimum volt drop when highest current is drawn."

The little servo motor runs at 4A and draws a max of 20A at 36V. An onbvious solution is to buy another 36V 9A power supply and set the output current limit on the Gecko drive to 9A. But that doesn't make for a very interesting discussion.

How would one design an unregulated supply for this if they had availble 72V DC to start with instead of 110V AC? Is a transformer the answer?

The problem is the resistor will only give you the desired motor voltage at one current setting.  If the current is higher, the power wasted in the resistor increases, and the voltage at the motor decreases.  If the current is lower, the power wasted in the resistor decreases, and the motor voltage increases.  If you use a 9 ohm resistor you will get 36V on the motor *only* when it's drawing 4A, and it'll work fine under those conditions.  Under and conditions where the motor is drawing less than 4A, the motor will get higher voltage, because the voltage drop across the resistor will be proportionately less.  At 1A, the motor will see 63V.  The motor does *not* always draw 4A.  It draws 4A only at maximum load.

It's difficult to explain to someone without an electrical background why this is a bad idea, but it is.  The high voltage the motor *will* see at times will greatly increase the eddy currents in the iron of the motor.  Eddy currents cause heating, and the amount of heating is proportional to the square of the applied voltage, so doubling the voltage will quadruple the amount of heat generated.  This is why when choosing stepper power supplies you select the voltage based on the rated voltage and inductance of the motor, and don't simply go for the highest violtage you can put your hands on.

Designing an unregulated supply to deliver a specific voltage and current requires having a transformer with the proper voltage and current rating.  The output voltage will be the transformer rated voltage multiplied by the sqaure root of two.  So, a 24V transformer would be used to build a ~34V power supply.

Regards,
Ray L.

This is what I don't get. Isn't this exactly what an 'unregulated' power supply does anyway? If so, why then is this an argument against knocking down the voltage with a resistor?


The little servo motor runs at 4A and draws a max of 20A at 36V. An onbvious solution is to buy another 36V 9A power supply and set the output current limit on the Gecko drive to 9A. But that doesn't make for a very interesting discussion.

How would one design an unregulated supply for this if they had availble 72V DC to start with instead of 110V AC? Is a transformer the answer?
I decided to use a 36V motor instead of the 72V, so the question became acedemic. I  threw things off by saying my PD was unregulated when it is actually switching and is regulated. I did not know the difference at first. So I learned something very useful about PS, but I don't think there was agreement on an answer.

The situation has changed now and I have kiled the 36v PS and replaced it with a 72V unregulated PS and now seek to run a 36V motor off the new 72V power.


If indeed 'unregulated' means the voltage fluctuates with load, and these are the type of power supplies recommended for my CNC setup, why then would a simlar fluctuating voltage downstream of a resistor be unacceptable?

This seems on the surface to be contradictory, and I would like to understand the issue better. If anyone can shed some light on these question, it would be much appreciated.

OK, If I understand correctly, now you have a 72VDC supply and you really need 36V?
Is this unregulated supply you have a linear supply or switching type? Does it have either a large Laminated or Toroidal transformer or does it appear to contain a fair amount of electronics?
This will pretty much decide as to whether you have viable options.
The difference between a dropping resistor and an unregulated DC supply is the resistor is going to drop voltage in a linear fashion directly related to current.
The direct output of a properly rated DC supply, whether unregulated or not will tend to maintain the voltage either due to Capacitive storage as in linear or by electronic means as in switching supply.
The slight change in voltage due to change in current will be no where as drastic as it will be by use of a dropping resistor.

Thanks very much for the info. I know it takes patience to explain things to persons uneducated in your field. I have simlar problems on the mechanical end.

I think I understand why the resistor is perhaps not a good choice. If I understand it correctly, the higher voltage available to the motor when it is not drawing 4A might cause it to overheat. I can get my brain around that idea. The motor actually draws 20A max, presumably when stalled . . .  which is how I cooked the little 9A PS . . I had the lock on the indexer and started the program running. I was wondering for a few seconds why the indxer was not spinning until after the flash of light on the shelf over my head . . . and everything went dead  


The power supply I just bought is listed as unregulated.     http://www.kelinginc.net/SwitchingPowerSupply.html   It is Model KL7220

Can't say if it is 'linear' because I do not know what that means relative to a power supply. What is 'toroidal' by the way? I know what the word means, but as with 'linear' I can't relate that to a power supply in any way that seems logical.   

So there is still the option of a step down transformer. If I found a transformer that halves voltage and is rated to carry 20A @ 36V on the output, (or 10A 72V input, I suppose), would that be suitable for running the little servo off the big 72V supply?

OR . . .  venturing farthur into the swamp . .  to tap the transfomer where the voltage is 36V? I have no idea if this is  a reasonable idea, but it makes sense to pose the question as long as you guys remain willing to tollerate me pestering you.

Quote from: yosefi83 on April 24, 2009, 07:00:56 AM

Ray,

What determines the current capacity is the capacitors. And as you know, a capacitor can deliver enormous currents. As to the capacitance, you're right - you need the same capacitance twice - one for each half of the rectified signal.

Daniel

Daniel,

Sorry, but that's not correct.  Take a 24V/5A, or 170W, transformer, connect  it to a full wave rectifier and filter, and you'll have a 34V/5A, or 170W,  power supply.  Connect that same transformer to a voltage doubler and filter, and the power the transformer can deliver remains 170W, but you've douibled the voltage, so the current must be halved.  There's no free lunch here.

Regards,
Ray L.

Ray,

Got your point, you were meaning the bulk current capacity the transformer can deliver. But you never walk on the edge of the transformer's current capacity, and what is more important for a motor drive is the peak current - and this one can be quite large with any circuit using that big a capacitance.

With steppers for example, the drive would hardly draw 1/3 the motor's rated current. I have mounted an ammeter on the lathe where I used the voltage doubler circuit and the current never goes above 0.5~0.7A. The steppers are rated 2.1A.

However, as you pointed, a voltage multiplier circuit might not be good in systems that use servos, due to their tendency to draw large currents under load.

Daniel

A transformer, static type that is, can only process AC, and you cannot drop the AC input to your switching  supply as the supply will put out (or attempt to) the rated voltage as the voltage drops.
The reason I was asking is if you had a linear supply that had a Toroidal transformer, this looks like a Large donut BTW, you can quite easily take off windings to tailor it to the voltage you need.
If I were you I would look into building your own Linear P.S. There are only basically three components, not counting fusing etc.
If you a fairly capable with a soldering Iron, it should be no sweat.
Between previous postings and forums, together with web articles , there must be hundreds  of how-to's out there.
If you do find a transformer closer to your required voltage/current, you are half-way there.
The formula for the transformer AC  output is DC output required x .7      DC x.7 = AC output.

Quote from: HimyKabibble on April 24, 2009, 07:52:04 AM

Quote from: yosefi83 on April 24, 2009, 07:00:56 AM

Ray,

What determines the current capacity is the capacitors. And as you know, a capacitor can deliver enormous currents. As to the capacitance, you're right - you need the same capacitance twice - one for each half of the rectified signal.

Daniel

Daniel,

Sorry, but that's not correct.  Take a 24V/5A, or 170W, transformer, connect  it to a full wave rectifier and filter, and you'll have a 34V/5A, or 170W,  power supply.  Connect that same transformer to a voltage doubler and filter, and the power the transformer can deliver remains 170W, but you've douibled the voltage, so the current must be halved.  There's no free lunch here.

Regards,
Ray L.

Ray,

Got your point, you were meaning the bulk current capacity the transformer can deliver. But you never walk on the edge of the transformer's current capacity, and what is more important for a motor drive is the peak current - and this one can be quite large with any circuit using that big a capacitance.

With steppers for example, the drive would hardly draw 1/3 the motor's rated current. I have mounted an ammeter on the lathe where I used the voltage doubler circuit and the current never goes above 0.5~0.7A. The steppers are rated 2.1A.

However, as you pointed, a voltage multiplier circuit might not be good in systems that use servos, due to their tendency to draw large currents under load.

Daniel

You still size the power supply to the maximum continuous load it needs to support.  If the 36V supply is rated to handle the maximum load, and you then put a voltage doubler on it, you can no longer support the maximum load.  If your 36V supply has lots of "extra" capacity, then a doubler will probably work, but that would not normally be the case.  And a stepper motor *will* draw its maximum rated current when it is under maximum load.  That may not occur often, but when it does, if your supply drops out, the stepper *will* lose steps, if not stall.  In any case, the supply must be sized to support the maximum load, always.  My point still holds - whatever the continuous rated current of the 36V supply (and that is the way power supplies are spec'd and rated - not by instantaneous peak current), adding a voltage doubler *will* reduce it by half.

Regards,
Ray L.

http://www.mouser.com/Search/ProductDetail.aspx?qs=sGAEpiMZZMtwaiKVUtQsNVph2yGGEGyO0kpwY0%2fK3NA%3d

This looks like it would do the trick, but the cost is more than a new 36v power supply, so it would not be a practical solution.