Low Frequency AC->DC for 5V USB Buck

[ACharnley]

Hi,

I've an efficient buck circuit which works well. It has a pin which is grounded disabled the chip, however the voltage coming from this varies (seems to be a % of the input voltage).

My problem is that at the beginning of power on the frequency is low and this causes the circuit to burst 5v, which sometimes locks USB devices or causes them to on/off. Due to tight space I have limited capacitance, but then the amount required to overcome this would be substantial.

I've also played with using two resistors to lessen the voltage difference on the disable pin and a capacitor, but this varies and has not proven reliable.

Are there any other ideas? I'm thinking I could have a small linear regulator with a resistor/capacitor/transistor to switch the pin, but how to discharge the capacitor if the power becomes intermittent?

Cheers!

Andrew

 

[alec_t]

Things would be clearer for us if you post a schematic.

 

[ACharnley]

Ok let me explain it better. I have a MP1584 buck, which on the datasheet is pretty much using diagram below.

https://www.monolithicpower.com/Pro...on/SwitchingRegulators/Step-down(Buck)/MP1584

The voltage divider to EN controls the min voltage before turn on. I've added a capacitor onto the 24.8k so that initially there is a delay.

However if power is interrupted I need to restart the delay, so short the capacitor.

I've tried a PNP and MOSFET-P between the capacitor and + rail for base/gate, which if interrupted should cause power to flow and short the capacitor, but it's not working. If I use a diode to test, it does.

Strangely, if I connect base on the transistor to ground it works!

Any idea?

 

[Les Jones]

If you can't be bothered to draw the EXACT schematic then you at least need to describe waht you have done accuratelly. for example you say "I've added a capacitor onto the 24.8k" You should have said in parallel with or in series with. You have not said HOW you have connected the transistor or mosfet.

Les.

 

[ACharnley]

I don't have schematic software installed right now.

Let me put it another way. How do I discharge a capacitor rapidly when the power to a circuit it has charged from is turned off?

 

[alec_t]

Have you tried this?

 

[ACharnley]

Oh god, that makes perfect sense!

 

[ACharnley]

Not working as I'd expected. I thought the chipset would consume the power from the capacitor, but it doesn't probably because it switches off. That leaves it to go through the resistors, which brings the speed issue again unless I make them small?

 

[Mikebits]

My problem is that at the beginning of power on the frequency is low and this causes the circuit to burst 5v, which sometimes locks USB devices or causes them to on/off.

What frequency are you talking about? This does not make sense to me? Your Vin should be DC > Vout. Once Vin ramps up, En pin should start the internal regulator of the chip, and lastly output a stepped down Vout.
What are you using to input Vin? This may be the source of some problems. Your attempt to delay the startup may only be masking the true problem which could very well be your Vin.

 

[ACharnley]

It's a bike dynamo, so the frequency can go very low.

I thought I'd cracked it, it looked good on the bench, but on testing the discharge time of the delay capacitor is still too great for the low frequency. I stuck a 1k res across the input to try and drain it, and used a skhotty but it looks like I need to short the thing. Perhaps back to the PNP in parallel with it?

 

[alec_t]

Ah, now that we know the supply is a bike dynamo I think you should remove the capacitor altogether and instead adjust R5/R6 (as per the 1584 datasheet) so that the EN input gets >3V (its typical threshold voltage) only when the dynamo is running at a high enough frequency (whatever that is?).

 

[ACharnley]

Some USB devices lock if the voltage going in is < 5v, so the DC-DC must give out only 5V. That's where the EN input works.

When the dynamo speed is low it gives out 6V still but at very low frequency. This causes the USB to cycle which again causes some devices to lock (especially smartphones). This is where the delay capacitor comes in.

However for hysteresis, if the rider dips below the threshold, the fact that the frequency is too low has to be 'detected' and the capacitor discharged (quickly) to redo the delay. The easiest way I could see to do this is with a PNP over the capacitor and base wired to +v. When the frequency drops and the +v starts pulsing the PNP would enable and short the capacitor.

For some reason the PNP isn't doing it's job though! I tried a MOSFET -P as well but with the same result; no discharge of the capacitor.

 

[ACharnley]

Any idea on how to get this capacitor discharged rapidly?

 

[alec_t]

Perhaps you could use a relay, with NC contacts wired cross the cap?

 

[ACharnley]

I considered it, but the circuit won't open ... to charge the relay.

 

[alec_t]

I'm surprised that if you've got 6V from the dynamo at the input to the converter it can't directly drive, say, a reed relay?

 

[ACharnley]

Well yes, I could put another LDO before the relay, but there's no hysterisis. I'm really trying to avoid using a relay if poss. I'm surprised there's not some obvious way to reset the capacitor. Perhaps I need to look at it another way, some type of digital counter?

 

[ACharnley]

Looking at how a 555 does it, it has an internal NPN to discharge the capacitor so my idea of using a PNP seems to be along the right lines. My understanding is current flows from emitter to collector if there's no ve at base. If a voltage is applied at base it greater than the emitter voltage it turns off.

In theory then this should work, but in my tests it did not! I could only get current to flow when I applied ground to the base.

 

[alec_t]

I'm surprised there's not some obvious way to reset the capacitor.

There is: use a relay . One difficulty is that, when the regulator output voltage drops, you lose the power supply for any active circuit/transistor which might discharge the cap. That's why I suggested the relay; since loss of voltage would cause it to drop out and hence close the NC contacts across the cap.

 

[ronsimpson]

I am a little lost by now.
So the circuit in #6 almost works? Does not work?
The diode that is added usually works but:
If Vin has no load there is no reason for Vin to go to zero volts so the 100k resistor + diode can not discharge the delay cap because Vin might not go to zero.
So try a 1k resistor (or some value) across Vin for a test. That will make a path back to ground for the delay cap discharge.
This is based on the assumption that when Vin is too low there is no load on the the dynamo.