I’m wondering if it’s possible to use the PG pin on the D24V22F12 to drain the regulator after shutdown? I have a circuit which requires that the capacitors in the regulator drain as quickly as possible after shutdown. Otherwise, the remaining charge (voltage level) will cause problems when trying to power back up.
My experience from another Pololu regulator is that it will bleed off the remaining charge veeery slowly (several seconds before it drops below 2V). Adding an old-fashioned bleeder resistor is not an option. For the circuit to drain fast enough, this resistor wiould have to continuously waste 3 times more current than the entire main circuit is using.
My idea is to connect Vout to PG through a small bleeder resistor. When the regulator operates normally, the two voltages will balance each other. When power to the regulator is cut off, PG will drain the capacitors from Vout through the resistor.
Could this work? The regulator specs don’t state any absolute maximum current/voltage rating for the PG pin and I don’t particularly want to risk frying a $10 regulator by attempting to do this without checking first…
I’m of course assuming that PG has a “high” level similar to Vout, although I see that 6V is mentioned in a post (is this true?).
The PG pin on the D24V22F12 regulator is only able to handle up to 6V, so it should not be pulled up to the 12V output. You might be able to make a circuit with a MOSFET to connect a bleed resistor when the input voltage or power good goes low. What load are you powering from the regulator?
By the way, the PG pin on the regulator is an open drain output, so it is open or floating when VOUT is within a good range and pulled low when the output is not in the good range.
Hi Claire, and thanks for the reply,
OK, using the PG pin to drain caps is definitely not an option
My circuit is powering a LED lamp (ca. 300mA @12V) controlled by a NE555 chip. The chip is there to provide low voltage disconnect for the battery.
A normal resistive load would of course drain the caps in a flash, but the threshold voltage of the LEDs prevent the caps from further discharge once their voltage reaches about 8V. The trigger of the NE555 doesn’t reset until the voltage falls below around 4V. So if I try to restart the circuit before that, it will not respond.
I’m really new to MOSFETS or transistors in general, but I have done something similar to what you suggest (I think).
A J175 P-channel JFET is connected with source/drain between regulator ouptut and ground. The gate is connected to the 18V regulator input, providing about 6V of bias (VGS) wit a fresh battery. Once the 18 supply is cut off, the JFET will open and drain the caps.
The solution works, but only just. As the battery discharges, this bias shrinks to just a couple of volts and the JFET starts to “leak” considerably.
I’m also worried that the JFET may cause unexpected behaviour during start-up as it will conduct intermittently until the input voltage stabilizes to approximately 18V (there is a considerable amount of switch bounce at startup).
So, I was looking for a more elegant solution. But with my limited understanding of MOSFETs, I cannot see a way to use the PG pin to properly bias a MOSFET when the VDS voltage varies all the way from 12V down to 0V during the discharge process…
If your JFET is working, it seems like that should be fine. I thought a little more about using a MOSFET to do something similar and it actually seems pretty impractical to connect a circuit like that to the PG pin, since the PG pin needs a pull up to 6V or less, and that is not something you have in your system. You could connect a p-channel MOSFET with the gate to VIN, source to VOUT, and drain to ground through a bleed resistor. That way it would turn on and allow discharge when VIN is below VOUT, which would only occur after the input is turned off. That configuration might still have issues as VOUT discharges and gets close to the threshold of the MOSFET though.
Thanks for your help. It seems a MOSFET would definitely be an alternative, I’ll have to look into this more closely to find the pros and cons.
Another question regarding the Power Good pin on the D24V22F12. Could I simply connect the EN and PG pins to make sure PG would pull down the EN pin once power is below drop-out and thus shut down the regulator?
This would help a great deal in tidying up start-up transients in my circuit.
It sounds like you expect the power good pin to change based on whether the input voltage is in a valid range, but that is not how that pin (or most power good outputs I have encountered) works. The power good pin is pulled low when the output of the regulator is not within the valid range. So if PG was connected to EN (which needs to be high to enable the board), the regulator would probably never be able to turn on.
Ok, I’ll try to explain a little better.
My understanding of “open drain” and what I read about the regulator is that the PG pin will open towards GND when Vin is out of range. When Vin is good, the PG pin will be “disconnected” i.e. high impedance.
The EN pin on the other hand, is pulled up through a 270k resistor. If the EN pin is grounded (pulled below 1V) the regulator output will be disabled.
By connecting the two, the PG pin would pull down the EN pin towards ground when Vin is out of range. Conversely -if Vin is good - the PG-pin will be high-impedance (much higher than 270k) and allow the EN pin to go high again.
See attached figure.
The only showstoppers I can think of are:
- PG pin changes behaviour when the board is disabled i.e. becomes “disconnected”.
- Voltage drop across PG pin (Vds) when open is close to or above 1 volt and will therefore not be able to pull EN below the required 1V
Does this make sense? I’m quite a rookie when it comes to these things, so I might have misunderstood completely
As I mentioned in my last post, the power good pin changes state based on the state of the regulator’s output, not the input.
Wov, how did I miss that despite reading the spec several times. That certainly makes the PG and EN pins pretty useless for my application. Sorry for wasting your time.