I'm designing a power supply that has an input of 100VAC to 300VAC. On the primary side on the main transformer, I have a 15V rail to power the control chip (a FAN6920) and a gate drive chip (a FAN7382). On this 15V rail, I have roughly 400uF of capacitance. Because of this amount of capacitance, the hiccup time of the supply is about 10 sec. That's the time it take for the 15V rail to decay down to about 5V, at which point the control chip will attempt to start back up. This is acceptable during an actual fault (e.g. an over voltage condition on the output or the output shorted), but this hiccup time is also present when turning the line off and then back on within that 10 seconds. The supply is used to power lights, so as you can imagine, no one wants a potential 10 second delay when they hit the light switch.

To that end, I'm trying to figure out a way to design a circuit that senses whether the AC line has gone away, and if so, to pull down the 15V rail quickly, while at the same time not impacting steady state operation.


simulate this circuit – Schematic created using CircuitLab

My first attempt at this is above. This idea is that C1 will charge to 10V when the line voltage is applied, holding M2 on. Holding M2 on will hold M1 off. At the same time, C2 is charging to 10V. When the line voltage is disconnected, C1 discharges much quicker than C2, allowing M2 to open up, and allowing C2 to change the gate of M1, turning it on and rapidly discharging the 15V rail through a 10 ohm resistor. However, in practice, C1 doesn't discharge nearly quick enough. It maintains enough voltage to keep M2 on for several seconds, so it doesn't really buy me anything (and also, the charge of C2 decays enough that it can't turn M1 on at that point, anyway). I can't change the time constant of C1 and R2 by much because the voltage across C1 will start drooping significantly when the line crosses 0V.

I am not tied to this topology by any means; this was just my first pass at a solution. And while I've tried a few other things, nothing has really worked, so any suggestions would be appreciated. Thanks.

  • \$\begingroup\$ Are D1 and D2 connected to a transformer secondary with a grounded center tap? The way it's drawn it won't work at all. \$\endgroup\$ – Spehro Pefhany Oct 23 '14 at 18:57
  • \$\begingroup\$ No, D1 and D2 are coming right of the AC line. V1 is the 120VAC input. The voltage at the node of R1 and R3 is a rectified sine wave when input power is connected. \$\endgroup\$ – flettz Oct 23 '14 at 19:11
  • \$\begingroup\$ Can you post enough more of the circuit to see where else the 120VAC goes? D1 and D2 are in addition to the main input rectifier (bridge?). \$\endgroup\$ – Spehro Pefhany Oct 23 '14 at 19:29
  • \$\begingroup\$ D1 and D2 are in addition to the main bridge rectifier. The tap off the line right before the bridge. The output of the bridge rectifier becomes the input to a boost stage, which takes the line voltage and jumps it up to 450VDC. That voltage is put through a 2-switch flyback to generate the output of the supply. I didn't include more of the schematic because it didn't seem relevant. I can't use anything downstream of the AC input, as I still need the long hiccup time during actual fault conditions. The only time I want to discharge the 15V rail is when the AC power goes away. \$\endgroup\$ – flettz Oct 23 '14 at 20:00
  • \$\begingroup\$ I will edit the schematic to show a little more detail to help, though. \$\endgroup\$ – flettz Oct 23 '14 at 20:00

The circuit you've shown (assuming there is 120VAC connected to an unseen bridge rectifier with the '-' connected to circuit common) 'looks' like it should work. Obviously those diodes cannot actually be 1N4148s since they're rated at only about 75V.

The time constant of R2/C1 is less than 10msec so there is no way it should take seconds to discharge (even if you allow for the gate capacitance on the MOSFET- presumably you're not actually using that part number, but something much smaller anyway). If there was enough leakage from the main supply capacitor through the main bridge rectifier that could explain it (so the problem would be 'off the page').

Perhaps if you split the (off the page) bleeder resistor for the input filter into two resistors, each to the common, rather than one resistor across the line..

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  • \$\begingroup\$ You are correct in that the part numbers are not accurate - I was just sketching the topology of the circuit. The actual components have the necessary ratings. You are also correct that C1 should dissipate much faster, but it does not for reasons I have not figured out. It will drop from 10V to about 5V is quick order, and then takes forever to go the rest of the way down. I am using a logic level fet, here - the irll014npbf. It's Vgs is 1.5V. It's possible that using a FET with a higher Vgs - in the 4V range would perform better. I will try that. \$\endgroup\$ – flettz Oct 23 '14 at 20:17
  • \$\begingroup\$ Caution that if it is leakage as I suspect then the effect may be highly temperature sensitive (as well as unit-to-unit variations). \$\endgroup\$ – Spehro Pefhany Oct 23 '14 at 23:06

You didn't mention whether you would need to pass UL. If so then you need to be careful about isolation. An easy implementation is to use a relay with a 120V coil voltage. Connect the normally closed (NC) side of the relay to your 15V rail and a bleed resistor. When the 120VAC is turned off then the relay is switched to route the 15VDC through the bleed resistor. These type of safety circuits are great for high voltage rails too.

Another UL acceptable solution would be to use a bridge rectifier to an opto-isolator. The output of the opto would be connected to a FET to pull down the rail. Again, easy and straight forward.

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