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Help designing a choke coil to limit inrush current?

fyrstormer

New Member
Hi, first time poster but long-time tinkerer. Here's my problem:

I have an incandescent flashlight running on 2x RCR123 lithium-ion batteries, for a peak voltage of 8.4V. The batteries have protection circuits on them that limit their output to ~1.5A, and also act as low-voltage cutoffs if the batteries drop below 2.7V (the minimum voltage for lithium-ion batteries to avoid premature damage). They are powering a bulb rated for a max of 9V, unknown peak amperage but I'm guessing somewhere around 1.75A @ 9V. When I turn on the light, the batteries' protection circuits instantly engage. If I "blink" the switch several times in a row so the filament gets hit with several split-second bursts of current, it will warm-up enough (and its resistance will increase enough) that I can latch the switch and the batteries will power the bulb without their protection circuits engaging.

I would like to build a circuit, as compact and simple as possible, that can limit the inrush current enough to prevent the batteries' protection circuits from engaging even when the bulb filament is cold and its resistance is low. A resistor is the easiest solution, but also a poor solution, because it would provide constant resistance when all I need is extra resistance on startup. I've seen simple schematics for resistor+transistor circuits that claim to do the job, but I think they have some parasitic drain even when the switch is off. I think a choke coil is the right solution for this application, but correct me if I'm wrong. Are there any other options? If a choke coil is indeed the best approach, how can I calculate the physical specs for a choke coil that will do the job?

- - -

Numbers!

Supply voltage: 8.4VDC
Load resistance: ~0.3ohm (cold, measured) to ~6.75ohm (hot, calculated)
Sustained current: ~1.25A (measured)
Inrush current: >1.25A (and this is the problem I need to solve)
 
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rjenkinsgb

Well-Known Member
Most Helpful Member
A choke could theoretically work but one to do that would be massive.

The normal single-component solution for inrush limiting is an NTC thermistor.
They have a high resistance when cold and a low resistance when hot, so they have a kind of "soft start" effect.

eg. This one is 10 Ohms when cold and drops as it heats - down to 0.06 ohms at nine amps, steady state
https://uk.rs-online.com/web/p/thermistors/1218219/
(I'm not saying that is the ideal one, but it gives you the idea).

Another solution is a power mosfet with the gate connected to a resistor-capacitor circuit so it turns on progressively over a second or so.
As long as the whole circuit is _after_ the switch there cannot be any parasitic drain.
 

fyrstormer

New Member
I'm guessing the thermistor consumes a fair amount of the input power in order to keep itself hot while the circuit is in operation. Is that accurate?

Can you point me to a diagram of the MOSFET+resistor+capacitor circuit you mentioned? Can you give me some idea of which components would be suitable for the amount of power this application uses? Size is a very important factor, since most of the space inside the flashlight is already occupied by the bulb, reflector, batteries, and switch. I can connect components in the right order and cram them into the available space, but I kinda suck at actually looking up the esoteric part numbers that electronic components are labeled with.
 

alec_t

Well-Known Member
Most Helpful Member
Another solution is a power mosfet with the gate connected to a resistor-capacitor circuit so it turns on progressively over a second or so.
That should do the trick if you can find a MOSFET which doesn't have a sharp turn-on knee. Another option would be to use the MOSFET to bypass a resistor of, say, 10\( \begin{array}{l}\Omega\\\end{array} \) after a brief delay
 

fyrstormer

New Member
I suspect the second scenario you describe is the one I've seen diagrams for, except they seem to favor resistors in the multiple-megohm range for reasons that are not apparent to me. Anyway, how would the brief delay you mention be achieved?
 

ronsimpson

Well-Known Member
Most Helpful Member
Try a LED bulb. There are many types for flash lights. They have a very different start up current. Much faster. They should need less current.
1549472565677.png
 

fyrstormer

New Member
I'm aware of those, but changing the bulb type isn't the stated goal of this thread. Anyway, retrofitting a front-emitting LED into a reflector designed for an omnidirectional-emitting tungsten filament doesn't work very well -- a lot more of the total light bypasses the reflector and never gets focused into the brightest part of the beam.
 

ronsimpson

Well-Known Member
Most Helpful Member
retrofitting a front-emitting LED into a reflector designed
Have you tried it? We have sold millions of these. Yes it would be best if the reflector and light were designed for each other but most manufactures don't seem to understand that.
Your running current will be less. (depending on how much power you want) and the start up problem goes away.
The batteries have protection circuits on
"In" the battery? or a separate board? If separate board, we might be able to make changes or design a new board.
 

fyrstormer

New Member
Have you tried it? We have sold millions of these. Yes it would be best if the reflector and light were designed for each other but most manufactures don't seem to understand that.
Your running current will be less. (depending on how much power you want) and the start up problem goes away.
Many times over the past decade or so.

"In" the battery? or a separate board? If separate board, we might be able to make changes or design a new board.
It's integrated into the battery's outer shell. Taking it apart is possible; putting it back together afterwards probably isn't possible. Besides, the point of the protection circuit is to prevent the battery from being overloaded, and modifying it would presumably make it less effective at doing its job.
 

ronsimpson

Well-Known Member
Most Helpful Member
1.5A current "lock out" compared to 1.25 current draw is very tight. I think component derations would get you in trouble.
If I was designing flashlights again:
I can see a PWM circuit that measures the current. It would, on power up, start at about 1V on the bulb to limit the current to 1.5A and as the bulb warms it will increase the voltage to hold the current down until the bulb is hot.
Under a short condition, it will limit the current to 1.5A for 1/2 second then give up and "lock out" or totally shut down.
 

fyrstormer

New Member
That sounds like an ideal solution, however it would almost certainly require replacing the entire switch assembly with a custom-designed one that has room for the circuit you described. Various people have made switches just like that over the years, but they were all limited-production and discontinued years ago. I'm pretty much on my own, limited to what I can do with a soldering pencil -- which I'm very good with, but making my own circuit boards is several steps beyond my fabrication capacity. Programming the microcontroller chip required for the circuit you describe is something I could definitely do (after learning yet another programming language), but I also lack the hardware required to upload my code to a microcontroller chip.
 

ronsimpson

Well-Known Member
Most Helpful Member
Can you get the next smaller bulb?
I am looking at CL-120 in-rush resistor.
Room temp = 10 ohms. At 1.5A it has a resistance of 0.35 ohms. It was designed for 120/220VAC applications. There may be a low voltage version.
When I use them on the AC line the 10 ohms limits the current. Then the part gets hot and the resistance drops way down. In a TV set the resistor heats up in a fraction of a second but takes 90 seconds to cool off. So if you turn on/off/on/off very fast it only works on the first on.
Try this: digikey
I did not find a low voltage versions. Worth a try.
 

ronsimpson

Well-Known Member
Most Helpful Member
explain it like I'm 5
Get the milk and cookies.

When the switch is closed there can be no voltage from G to S on the MOSFET transistor. When the switch opens up then current (very small amount) flows into the Gate and charges up the gate slowly. The transistor turns on slowly.

Inside the MOSFET there is a internal capacitance from G to S. I don't like using it as a timer because it is very dependent on part to part variations. So I add a cap that is external and predictable.

The time delay is R*C. The larger the R or C the longer the time delay.

47M resistors are hard to find. 1M, 2.2M maybe even 10M you can get easy.
1549506662506.png
The RC charges up slowly. The transistor turns on in the 3 to 5 volt area depending on which part is used. So at 3V G-S some small amount of current flows D-S and through the bulb. At 3.5V much more current. at 4V maybe 1A can flow. (I should look up the part and see exactly what the turn on voltage is for this part)
I don't like the idea that the switch works backwards. ON vs OFF. There is another way where the switch works like normal. Need to think about that.
 

fyrstormer

New Member
OHHHH, now I get it. It never occurred to me that the new configuration was supposed to be "off" when the switch was in the closed position. It makes so much more sense now. So when the switch is opened and the S and G lines are no longer continuously equalized, the G line will become positively charged, while the S line remains negatively charged, but the capacitor delays the voltage-change between the two lines, so the transistor becomes conductive gradually instead of all at once. The resistor appears to serve the purpose of preventing a dead-short between the + and - terminals of the battery when the switch is closed, while still allowing the G line to become positively-charged when the switch is open. Yes?

Anyway, the flashlight in question has a normally-open momentary pressure switch that can be tightened for continuous-on if needed, so a circuit that turns on when the switch is open won't work for my application. If you happen to remember how to make this circuit turn-on when the switch is closed instead, that would be very helpful. I could replace the pressure switch with a latching switch if necessary, but if possible I'd like to preserve the original functionality as much as possible.
 
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alec_t

Well-Known Member
Most Helpful Member
Here's a sim of the arrangement I mentioned :-
BulbInrushLimiter.PNG
U3 is a 10 Ohm NTC thermistor.
The yellow trace is the limited current through the bulb. The blue trace is the current without the inrush limiting.
The current v time profile is dependent on the turn-on threshold of the particular FET used, so R1 would likely need tweaking.
 

fyrstormer

New Member
Wow, this circuit keeps getting more complex. Wouldn't the thermistor take much longer to reset than the bulb filament would take to cool off?
 

alec_t

Well-Known Member
Most Helpful Member

fyrstormer

New Member
You're suggesting using alec_t's circuit, minus the thermistor, and with a diode in-parallel with R1?

Let me see if I can figure out how this one works.

When the switch closes, the diode allows the capacitor to charge instantly, but only up to the charge supported by the supply voltage minus the diode's forward voltage. R1 allows the capacitor to reach full charge, but over a longer span of time. R2 appears to serve only as a way to discharge the capacitor when the switch opens, and its extremely high resistance is to minimize wasted battery power when the switch is closed and R1+R2 form a continuous unproductive circuit.

However, having said all that, I'm still unclear how this causes M1 to come online slowly. When I look at the circuit I see voltage between M1's + terminal and gate terminal dropping to nearly-zero instantly as the diode allows C1 to charge-up most of the way; without a significant voltage between those two terminals, no power will flow through to the - terminal of M1. What am I misunderstanding here?
 

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