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?



Quote Originally Posted by ecat View Post
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!!!