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I am looking for feed back on the following power supply design.

This power supply is intended to have 2 outputs: 24V @ 500mA and 32V @ 850mA. When the power supply is in standby, J1 pin 1 is @ 0V, the outputs change to 7.48V and 9.75V respectively. When the power supply is active J1 pin 1 is @ 3V.

  1. Is my feedback to the LT1370HVCT7 going to work with my mosfet
    switches?
  2. What kind of power requirements am I going to need for the 12V input? I haven't been able to figure this out.
  3. What, if anything, am I missing?

Higher Res Schematic Dual Rail PSU

UPDATE:

  1. Changed input voltage to 5V
  2. Tried to follow theamk's advice and use series R2 resistors for the voltage divider so I could short out one of one of them thus changing the voltage divider output
  3. Using a p-channel mosfet to drive the 2 n-channel mosfets
  4. Added logic level shifter 3V to 5V for the standby input

Higher Res Schematic Dual Rail PSU Rev B

How is it looking now?

UPDATE 2:

  1. Used gsills standby divider switch
  2. Removed level translation

Higher Res Schematic Dual Rail PSU Rev C

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  1. No, your feedback will not work.
    • Your schematics has N-channel mosfets. They require gate to be more positive than source. In your design, when the system is on, you will have +24/+30v on source, and only 0/+12V on gate.
    • Even if you change mosfets to P-channels, you will still have at least 12v across gate-source when system is on is open. You would not be able to move to standby once you turn on your supply.

My recommendation is to place mosfets between ground and feedback pin (split R3/R7 into two series resistors, and use mosfet to short it)

  1. Power requirements are simple to calculate:

    I_input = ( I_output * V_output ) / V_input / efficiency

So assuming 85% efficiency, you will need 1.2 + 2.7 = 3.8A @ 12 volts.

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  • \$\begingroup\$ If I use 5V input @ 85% efficiency I will need (0.5/24)*(5/0.85) = ~0.12A, (0.85/32)*(5/0.85) = ~0.16A ... 0.12 + 0.16 = 0.28A @ 5V? That doesn't look right. \$\endgroup\$ – Hair_of_the_Dog May 7 '15 at 23:50
  • \$\begingroup\$ Yes, thanks. The values were correct, the formula was not - now fixed. \$\endgroup\$ – theamk May 8 '15 at 1:05
  • \$\begingroup\$ Ah... :) So (0.5*25)/5/0.85 + (0.85*32)/5/0.85 = ~9.4A @ 5V. I am trying to figure out your recommendation for the feedback right now. I'm not sure I get it, but I'm going to try. \$\endgroup\$ – Hair_of_the_Dog May 8 '15 at 1:08
  • \$\begingroup\$ I think I understand. I need to recalculate my resistor divider, though. \$\endgroup\$ – Hair_of_the_Dog May 8 '15 at 1:51
  • \$\begingroup\$ I think this is what you were talking about. Can you take another look? \$\endgroup\$ – Hair_of_the_Dog May 8 '15 at 4:14
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"What, if anything, am I missing?"

The main thing would be that you are using non-isolated boost regulators. With a non-isolated boost, the minimum output voltage will be Vin-Vd or in this case ~11.6V, since Vin is 12V and D1 and D2 are Schottky diodes. A transformer coupled Flyback or Sepic would be needed to have standby output voltages less than Vin.

Also, as already pointed out by theamk, Q1 and Q4 don't function as switches here, but as voltage followers. So, the voltage divider pull-up from them would be 12V-Vth. But that won't really matter until you change topologies.

Edit for change of Vin:


Using 5V for Vin will allow a boost to be used while in the lower voltage standby mode. But, the standby switch is very complicated, and doesn't work. If standby control voltage range is really just zero to 3 volts, there would be no need to level shift using Q3. Q2 doesn't switch, it voltage follows, so never really turns on, which means that Q1 and Q4 may never turn off.

Why not try something like this:

enter image description here

When Vstandby is 0V, Q1 is off and you get an output voltage less than 24V. When Vstandby is 3V, Q1 is on paralleling R3 and R4, allowing the higher output voltage of 24V.

A similar arrangement could be used for the 32V output.

You will need to work out the divider ratios, but should be possible.

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  • \$\begingroup\$ What if I use a 5V input instead of 12V? I should be able to get the lower standby output voltages, yes? \$\endgroup\$ – Hair_of_the_Dog May 7 '15 at 23:38
  • \$\begingroup\$ @Hair If 5V input was possible a boost could still be used for the standby voltages you have in your question. It would require ~8A out of the 5V. Then you would just need to change the function of Q1 and Q4. \$\endgroup\$ – gsills May 8 '15 at 3:16
  • \$\begingroup\$ It is possible and I have updated the schematic to reflect this. I have also incorporated what I think @theamk was trying to tell me. Can you give is another look? \$\endgroup\$ – Hair_of_the_Dog May 8 '15 at 4:12
  • \$\begingroup\$ Nice! That is much easier. I will update the schematic and update. \$\endgroup\$ – Hair_of_the_Dog May 22 '15 at 16:36
  • \$\begingroup\$ Schematic updated. I think this is going to work. What do you think? \$\endgroup\$ – Hair_of_the_Dog May 24 '15 at 1:05

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