Help designing a choke coil to limit inrush current?

[fyrstormer]

While I'm at it, is there any way to design this circuit to use a PTC thermistor instead of a NTC thermistor? Because the bulb is already a PTC thermistor, and it would be super-handy if it could perform double-duty as the load and part of the regulator.

 

[ronsimpson]

When the switch closes, the diode allows the capacitor to charge instantly,

No. No current flow at start-up. The diode has current flow at turn off to help discharge the cap.
At turn on the cap charges up through the resistor and the diode does nothing. Slow charge up.

 

[ronsimpson]

Turn on

At turn off there is two ways that C1 discharges.
C1R2 discharge.
and
C1, diode, bulb, transistor. This only happens when C1 has enough voltage to keep M1 on, even partly on.

 

[rjenkinsgb]

You could add a moderate value fixed resistor across the FET to give a fast discharge path - eg. 1K.

If the supply is less than 15V or so, you probably don't then need R2, so it stays quite simple.
That's the exact circuit I was imagining in my earlier answer.

 

[fyrstormer]

No. No current flow at start-up. The diode has current flow at turn off to help discharge the cap.
At turn on the cap charges up through the resistor and the diode does nothing. Slow charge up.
View attachment 116536

Isn't the diode pointing the wrong way for current to flow from the capacitor to the + power lead? My understanding of the diode symbol is, the arrow with the line blocking it indicates which direction classical current (i.e. positively charged, before people discovered electrons were negatively charged) is blocked. In this case you have the diode blocking classical current from flowing towards the + power lead.

Also, wouldn't R1 serve the same purpose as the diode, allowing the capacitor to discharge to the + power lead? Or is the point of the diode to make the discharge happen faster than it would if there were only a resistor connecting C1 to the + power lead?

 

[fyrstormer]

You could add a moderate value fixed resistor across the FET to give a fast discharge path - eg. 1K.

If the supply is less than 15V or so, you probably don't then need R2, so it stays quite simple.
That's the exact circuit I was imagining in my earlier answer.

The supply will never be higher than 9V in this application.

What would a resistor across the MOSFET do, that other components of this circuit don't do? Does it simplify the circuit in some way?

- - -

Sorry for all the noob questions; I have an engineering degree and I've been working in my field for 12 years, but every time I dip my toes into circuit design I'm reminded just how different electrical engineering is from other forms of engineering. There's so much memorization of what components do in various combinations, and that was not part of my curriculum when I was in college. I was required to take a year of physics with calculus, but no circuit design. Maybe I'll suggest to my university that they add this, if they haven't already. But my curriculum was already a master's degree crammed into 4 years with only two elective slots, so there might simply be no room for it.

 

[rjenkinsgb]

When the power switch is opened, the lamp will act as a discharge path to ground so the diode conducts then.
An added resistor across the FET ensures the "ground" is still present when the FET gate voltage is too low for it to conduct.

Without a separate discharge path, the "turn-off" time would be teh same or longer than the turn-on ramp time, so rapidly switching off & on could cause current surges.

 

[fyrstormer]

Ahh, I wasn't aware that MOSFETS have a forward voltage like diodes do. So the resistor across the MOSFET lets the last little bit of charge from the capacitor bleed-off after the MOSFET has become non-conductive.

 

[rjenkinsgb]

It's not so much a diode-drop effect as the inverse of the soft-start; as the cap discharges and gate voltage drops, the FET will gradually stop conducting so the cap discharge will slow down & eventually stop, unless there is an alternate path.

The resistor across the cap would work with a low on-off rate, but a diode and the resistor across the FET provide a fast discharge path so R2 is then not needed for discharge.

 

[fyrstormer]

Aha! I was just about to ask what would be the relative advantages and disadvantages of having R2 bridging the input and output terminals of the MOSFET, vs. bridging the two terminals of C1.

So it sounds like bridging the MOSFET with R2 allows for faster cycling of the circuit, but the tradeoff is there's more wasted power as R2 is constantly conducting straight to ground when the circuit is on. The original circuit also wastes some power since R1 + R2 are also continuously conducting straight to ground when the circuit is on, but because R2 has a much higher resistance in the original circuit design, the amount of wasted power is greatly reduced. Is that about right?

 

[rjenkinsgb]

There is no wasted power via the resistor across the FET - it provides a tiny current to the lamp before the FET starts to conduct & is totally shorted once the FET is fully conducting and the "soft start" phase is complete.
All the power is used by the lamp, other than tiny circuit losses.

R1+R2 would take a tiny amount of power, 10 - 15 microamps or so depending on voltage.
Without R2, current through R1 stops when the cap is fully charged and the only waste is any leakage via components.

 

[fyrstormer]

*facepalm*

Right...a resistor bridging the MOSFET would only be able to conduct power that was also going through the bulb and doing useful work.

It's amazing how I can write multi-threaded recursive algorithms with ease, and even calculus finally makes sense to me after a decade of mulling it over, but if you put a circuit diagram in front of me, my IQ drops by 50 points.

I guess where I really get hung-up is, I can trace the possible paths that current will take through a circuit diagram, but I can't look at the components involved and "just know" which effects will be significant and which will be insignificant. I guess that's the part that comes from years of practice.

 

[fyrstormer]

So it seems like the modified version of the circuit shown above, with R2 bridging the MOSFET instead of bridging the capacitor, is the best solution for my application. But just for the sake of thoroughness, what is the difference in operation between the original circuit including R2, vs. the original circuit excluding R2? I can see that R2 in its original location would help to discharge the capacitor a little faster when the circuit shuts off, but what other effects would R2 have in that position, since it's also connected to the gate terminal of the MOSFET?

 

[ronsimpson]

R1 can be much larger. 1M or 10M and C1 then can be much smaller.

 

[fyrstormer]

Am I correct in surmising the advantage of having a higher-resistance R1 and a lower-capacitance C1 is, C1 would be physically smaller and thus easier to fit into a confined space? Or is there an electronic advantage to that arrangement as well?

Also, just to verify: Does C1 need to be a directional electrolytic capacitor? Or can it be a solid capacitor? I only ask because the symbol in the circuit diagram indicates a directional electrolytic capacitor.

 

[ronsimpson]

I was trying to get C1 down in size so a solid cap can be used. The ele caps die with age, are not good with wide temperatures, large in physical size. ….

 

[fyrstormer]

That's what I figured. Thanks.

Last question: When I'm prototyping this circuit, which components should I adjust to make the power-on ramp-up longer or shorter as needed? I assume the answer is "C1 and/or R2", but I'm not entirely clear on how to adjust them to get the needed effect. Will increasing the resistance of R1 be sufficient to increase the ramp-up time, or do I need to increase the capacitance of C1 as well? Or should I adjust it some other way? And can you recommend a specific MOSFET that is suitable for the voltage and wattage range I'm dealing with? (6V-9V, <=12W)