1. The power supply has two modes of operation:

Variable, where the output is controlled by a voltage applied to pin 28 (i/Ucr) on the main connector. This control voltage is to be set somewhere between 5V and 11V.

Fixed, if voltage applied to pin 28 is lower than 5V or above 11V the power supply output is fixed to 42.8 (+/- 0.5V). It appears that leaving pin 28 unconnected is sufficient to force the supply into fixed mode.

The resistor pair (R33, R34) provide the feedback voltage used when in fixed mode. I think, but never did prove, the resistor pair (R11, R12) provides the feedback voltage when in variable mode. All comments from this point on will assume fixed mode is selected.

Without modification the R33, R34 pair act as a x10 divider so whatever the voltage at the output one tenth of that voltage appears at the junction of R33, R34. The PSU is designed to hold the voltage at this junction at 4.28 volts.

For a given output voltage the R33, R34 pair are chosen such that 4.28 volts appears at the point where they join.

For an output of 24V, R34 must drop (24 - 4.28 =) 19.72V, leaving 4.28 volts across R33. Leaving R33 at 1k this would give us a value of 4k6 for R34.

Without modification the current through R33, R34 = 42.8 / (9k1 + 1k) = 4.28mA. The power dissipated by R33 = 4.28V * 0.00428A = 0.018W and the power dissipated by R34 = 38.52V * 0.00428A = 0.165W (so why do they use such a chunky resistor for R34 ?)

For a 1k + 4k6 24V modification the current through R33, R34 = 24 / (4k6+ 1k) = 4.29mA. The power dissipated by R33 = 4.28V * 0.00429A = 0.018W and the power dissipated by R34 = 19.72V * 0.00429A = 0.085W (most resistors/multi-turns would appear ok, but keep a close eye on the temperature of the multi-turn)

Another thing to look out for is ripple, how much the output voltage varies as the switch mode switches. Indeed, the minimum duty cycle of the switching circuit may even limit the minimum voltage to something greater than 24V. Only testing will tell.

Finally. Beware of the 400 Volt lines!!!

As I type the PSU is running at 26volts, I have connected up scope, and the power output looks very clean, I have not yet put any load on the psu.

I will be connecting the PSU to my control board later on today.

3. Originally Posted by Jonathan
Would it be safe to put a few of these power supplies in parallel? Maybe put small resistors / diodes(?) in to balance the load.
I ask because it's occured to me that these are the right voltage for my big 48V brushless motor/spindle. It claims to take over 100amps but obviously I'll never need anywhere near that amount! I think 3 PSUs in parallel would be plenty.
In theory, my interpretation of the theory at least, any switch mode PSUs of similar nominal output voltage can be paralleled without the use of resistors or diodes. Since similar is not the same as exact and exact is impossible, initially one PSU will attempt to handle the full load. It will reach its limit and the OP voltage will drop, at this point another PSU will kick in and try raise the output voltage back to what it should be. Ripple may be a problem, especially when the PSU kick in and out, so test the setup and see.

N.B.
When dealing with switch modes, mains voltages or anything else dangerous I recommend wearing safety glasses, safety gloves, standing well back and getting a friend or loved one to throw the switches ;)

4. 26V with no load, if this is low enough for you then the job should be done. Lovely :)

As you have a multi-turn already set up, can you go any lower?

5. The lowest I have tried taking the PSU down to is 20volts.

I have been running the TB6560 board at 24volts fine unto my old psu gave up.

I would like to run how much voltage as possible to gain the extra speed out of the stepper motors.

I am slowly going to work my way upto 30volts.

I don't know how much, or if I will get any extra speed out of the stepper motors going from 24 to 30volts.

6. Cool.

Remember to make some allowance for the peek voltage generated by the motor when the driver stops driving, heh. There is a thread around here somewhere with more details. There's also a thread relating voltage to stepper speed.

7. I bought three of these 500W 50.5V power supplies for £12 way back in 2009.. as time has passed I mislaid the documentation.. A google search Found the missing PDF here. My Lazy way of solving this problem was to buy three 20 Amp bridge rectifiers to limit the output Voltage to Less than 50V therefore solving the slightly higher than 50v output.
My reason for this solution is messing with switch mode PSUs is quite dangerous. (a friend of mine was hit in the temple by a flying electrolytic case when delving into a switch mode psu. With such force it caused a tear in his retina.)

Your circuit hacking/reverse engineering provides a more elegant solution thank you.

Is their any reason why my "Lazy solution" of 2 x the diode volt drop was not considered as a solution in regulated mode?

Originally Posted by ecat
The power supply has two modes of operation:

Variable, where the output is controlled by a voltage applied to pin 28 (i/Ucr) on the main connector. This control voltage is to be set somewhere between 5V and 11V.

Fixed, if voltage applied to pin 28 is lower than 5V or above 11V the power supply output is fixed to 42.8 (+/- 0.5V). It appears that leaving pin 28 unconnected is sufficient to force the supply into fixed mode.

The resistor pair (R33, R34) provide the feedback voltage used when in fixed mode. I think, but never did prove, the resistor pair (R11, R12) provides the feedback voltage when in variable mode. All comments from this point on will assume fixed mode is selected.

Without modification the R33, R34 pair act as a x10 divider so whatever the voltage at the output one tenth of that voltage appears at the junction of R33, R34. The PSU is designed to hold the voltage at this junction at 4.28 volts.

For a given output voltage the R33, R34 pair are chosen such that 4.28 volts appears at the point where they join.

For an output of 24V, R34 must drop (24 - 4.28 =) 19.72V, leaving 4.28 volts across R33. Leaving R33 at 1k this would give us a value of 4k6 for R34.

Without modification the current through R33, R34 = 42.8 / (9k1 + 1k) = 4.28mA. The power dissipated by R33 = 4.28V * 0.00428A = 0.018W and the power dissipated by R34 = 38.52V * 0.00428A = 0.165W (so why do they use such a chunky resistor for R34 ?)

For a 1k + 4k6 24V modification the current through R33, R34 = 24 / (4k6+ 1k) = 4.29mA. The power dissipated by R33 = 4.28V * 0.00429A = 0.018W and the power dissipated by R34 = 19.72V * 0.00429A = 0.085W (most resistors/multi-turns would appear ok, but keep a close eye on the temperature of the multi-turn)

Another thing to look out for is ripple, how much the output voltage varies as the switch mode switches. Indeed, the minimum duty cycle of the switching circuit may even limit the minimum voltage to something greater than 24V. Only testing will tell.

Finally. Beware of the 400 Volt lines!!!

8. Originally Posted by Bodge
I bought three of these 500W 50.5V power supplies for £12 way back in 2009.. as time has passed I mislaid the documentation.. A google search Found the missing PDF here. My Lazy way of solving this problem was to buy three 20 Amp bridge rectifiers to limit the output Voltage to Less than 50V therefore solving the slightly higher than 50v output.
My reason for this solution is messing with switch mode PSUs is quite dangerous. (a friend of mine was hit in the temple by a flying electrolytic case when delving into a switch mode psu. With such force it caused a tear in his retina.)

Your circuit hacking/reverse engineering provides a more elegant solution thank you.

Is their any reason why my "Lazy solution" of 2 x the diode volt drop was not considered as a solution in regulated mode?
Good question, I think at the time my list of solutions looked like:

Plan A
Modify existing circuit.

Plan B
There should always be space for a plan B even if it lacks a concrete definition ;)

Plan C
As the required voltage drop is small use one or more semi-conductor junctions (diodes or transistors)

-------------------

Plan A
The PSUs were quite cheap so it didn't really matter if I blew one up. In fact the 'spare parts' alone would be worth more than £12.

I had two PSUs to play with, potentially losing one was not a problem. In fact I did lose one, I left it in the office and forgot to bring it home when the company closed down - Grrrr

The whole idea of switch-mode PSUs is to efficiently convert one voltage to another. Modification is the correct solution.

After all that, any modification had to be as simple as possible, as safe as possible and not involve anything near the high voltage side.

Should only be attempted by those who are aware of and acknowledge and are are happy with the risks involved in opening a SMPS

Plan B
Very simple solution.

Does not involve opening the PSU

Safe to implement but in use this sense of safety is somewhat deceptive:

1) The forward voltage (Vf) drop of a semiconductor junction (diode) is not a fixed value. Buy rule of thumb we often assume something around 0.6v, 0.7v but look at the datasheet for the device you intend to use and you may find that Vf = 0.7V at 1A but up to Vf = 1.0V at 10A and as low as 0.4V at <0.1A. So two diodes in series may give you a 1.4V drop or they may give you a 2.0V drop or a 0.8V drop - over 50% variation depending on current draw. Whether or not this matters is entirely down to what you are trying to achieve, but I'm sure some bright spark is certain to attempt to use 10 to 15 diodes in series for 7V to 10V drop - well, at 50% or more variation this is when the magic smoke can escape.

2) Heat! If each semiconductor junction drops 1V at 10A draw then each and every semiconductor junction will dissipate 10W at 10A draw, for continuous use we now need heatsinks and possibly insulation.

-------------------

That covers most of the risk / reward assesment. Two diodes are great if you've done your homework and everything checks out, in my case I was happier making the mod :)

9. Cool.. I was trying to imagine some reason I had not thought of.
The devices in question were KBPC5010, I was wrong saying they were 20A it was 50A.
With FVD variation <0.5v over the current range 0-10A. (0.5v to 1v)

They will now be put to one side for my next Green energy project.

Originally Posted by ecat
Good question, I think at the time my list of solutions looked like:

Plan A
Modify existing circuit.

Plan B
There should always be space for a plan B even if it lacks a concrete definition ;)

Plan C
As the required voltage drop is small use one or more semi-conductor junctions (diodes or transistors)

-------------------

Plan A
The PSUs were quite cheap so it didn't really matter if I blew one up. In fact the 'spare parts' alone would be worth more than £12.

I had two PSUs to play with, potentially losing one was not a problem. In fact I did lose one, I left it in the office and forgot to bring it home when the company closed down - Grrrr

The whole idea of switch-mode PSUs is to efficiently convert one voltage to another. Modification is the correct solution.

After all that, any modification had to be as simple as possible, as safe as possible and not involve anything near the high voltage side.

Should only be attempted by those who are aware of and acknowledge and are are happy with the risks involved in opening a SMPS

Plan B
Very simple solution.

Does not involve opening the PSU

Safe to implement but in use this sense of safety is somewhat deceptive:

1) The forward voltage (Vf) drop of a semiconductor junction (diode) is not a fixed value. Buy rule of thumb we often assume something around 0.6v, 0.7v but look at the datasheet for the device you intend to use and you may find that Vf = 0.7V at 1A but up to Vf = 1.0V at 10A and as low as 0.4V at <0.1A. So two diodes in series may give you a 1.4V drop or they may give you a 2.0V drop or a 0.8V drop - over 50% variation depending on current draw. Whether or not this matters is entirely down to what you are trying to achieve, but I'm sure some bright spark is certain to attempt to use 10 to 15 diodes in series for 7V to 10V drop - well, at 50% or more variation this is when the magic smoke can escape.

2) Heat! If each semiconductor junction drops 1V at 10A draw then each and every semiconductor junction will dissipate 10W at 10A draw, for continuous use we now need heatsinks and possibly insulation.

-------------------

That covers most of the risk / reward assesment. Two diodes are great if you've done your homework and everything checks out, in my case I was happier making the mod :)

10. From all the great work done a lot earlier (Many thanks guys)... we could calculate the replacement for R34 this way :-

Given that the PSU spec says "If the PSU is out of regulation then Uout=42.8+/-0.5V DC."

Taking the potential divider made from R34 and R33 is 9K1 and 1k with a +/- 1% tolerance

Applying 42.8V to this divider gives us a Junction voltage of 4.238 +/- 1%

Calculated by (R1(1k)/(R1(1K)+R2(9.1K))* 42.8V = V divider = (4.238V)

So rough calculations for 49V +/- 0.5 we need a resistor R2 of 10K561 or 10.561K (which ever standard you prefer)

The nearest we I could get with standard values (1/R1)+(1/R2)+(1/R3)=(1/Rtotal) was (1/22K)+(1/22k)+(1/270K) = 10.569K or 10K569 (which ever standard you prefer)

Thus this should give an output voltage of 49.025 +/- 0.5V eg +/- 1%

To change R34 for the 3 parallel resistors this was my process.

Warning make sure the PSU has been off for a good time as the PSU capacitors hold High voltages for a considerable time 400v DC is a skin splitting voltage and quite a Deadly!! thing to touch.

Remove the 5 outer screws, remove the cover, Unscrew the top board another 4 screws, disconnect the connector connector on the end unscrew the two screws holing the grey i/o connector disconnect the mains link connector and the other 4way connector

Remove the two PCBS.

Un screw the two screws holding the Heat sink down unscrew the screw that you can reach on Diode de solder the other diode remove the heat sink.
De-solder R34 replace it with the new resistor made from 2 x 22k 1/4W 1% and the 270k 1/4W 1% resistors connected in parallel re-solder R34 Re-assemble in reverse order...
Now to test it.. the result of this change gave me an output voltage of 48.9V very slightly out from my calculated value but well within the +/- 1% tolerance on the resistors used.

excuse dyslexic spelling

pictures to follow.
Last edited by Bodge; 29-04-2012 at 04:30 PM. Reason: added re-asembly part.

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