Boost Converter Inductor Current and Switching Current

boost converter inductor current

I have attached a simulation result of the inductor current and switching current of a boost converter. can anybody please tell me why the inductor current rises so high until 7A and decreases .... so does the switching current. thank you very much. the boost converter i am designing requires a 500mA current. from 5v to 10v. thank you very much again.

Regards,
Meng Ghee

boost converter inrush current

Before the boost converter can operate in the boost mode, the output voltage must rise immediate to the supply(5V). There will be in-rush current from supply to charge up the cap . The inductor current will be governed by V = Ldi/dt with 5V = 100uH*di/140us => di = 7A.

Added after 9 minutes:

By the way, the turn on @periodic 20us of power NMOS also will sink in alot of current. You should have a delay of ramp gen to wait for the output to rise up to VBAT before start the ramp gen.

boost converter ripple current

Even though the steady-state output current is only 500mA, you can get a higher inductor current, especially in a simulation where you do not control the duty-cycle dynamically and you do not have any feedback, nor any elements to prevent the inrush current..

Plus, you need to charge up the output capacitor from zero. While the output voltage is lower than the input voltage, the output capacitor charges through the inductor and the diode and there is nothing in your circuit to limit that current. The inductor current keeps increasing, even though the transistor is off, because the current flows from Vin through the diode, out into the output capacitor.
If you were to plot the output voltage, you would see that the inductor current decreases as the output voltage increases.

inductor inrush current

thank you Hoonsk and VVV, but unfortunately... i still don't quite get it... foolish me. why must the output capacitor rise to 5v ? well, from what i know ( which is limited .. hehe ) when the switch is closed, the inductor charges up and when it is open ... it discharges the current. but from what u guys were saying, i kinda figure that the inrush current is caused by the output capacitor and load (while the switch is open). but i cannot understand how they affects the current. hope can get an answer. and also, hoonsk, how did you get dt=140us ?? thank you very much.

boost converter circuit+inrush current

I'm not sure about hoonsk's answer, but the current is definitely sinking into the capacitor on startup. I'm not really sure why you have a 5mF cap (huge) in there, but that's why the current is so high and for many milliseconds on startup. I could see the cap being several hundred uF's, but not 5000uF. I believe in Spice you can preload your caps with a voltage at t=0, but it's not really necessary or accurate to do this. It *is* helpful sometimes to do this when you don't want to have to simulate the first 40ms due to a startup transient - such as what you are seeing.

May I make a suggestion?

What you have there is a boost converter in it's simplest form, but for it to work well you need to close the loop, i.e. give it a feedback path so that it may regulate the output with variances in the output load.

Instead of using op-amps and comparitors to generate the PWM and regulate the output, why not grab one of many ICs that will do this all in a small form factor? One that I've used on many designs is the MC34063A. Last I checked it was around 25 cents in quantity, and relatively simple to calculate values for. I would suggest making an Excel spreadsheet with all of the equations built in so you can iteratively try different values quickly.

Attached is the datasheet and a very nice application note. I did not see a 5V to 10V (@ 500mA) converter in there, but there are many examples to learn from. I would suggest reading all of the data sheet, and then the app note. I would also imagine you'll need an external switch. I've also attached the Spice model.

boost converter circuit + inrush current

Thank you B.Rabbit, i have tried simulating the circuit with a reduced capacitor value. i noticed that even by reducing the capacitor value to 0.5mF. the in rush current is still there ... now. i start to think will the inrush current damage the capacitor. thank you. still wondering why the inrush.

regards,
mengghee

boost converter inductor

The inrush current will still be there, and mainly will be dependent on the voltage input, inductor value and ESR of the output cap - but these should not force you to change them to adjust the inrush current. Instead, other techniques for a "soft-start" can be implemented so as to limit the inrush current.

Just curious, did lowering the value of the capacitor shorten the duration of the 7A inrush current? It should have.

The inrush current can damage the cap, as you are concerned.. but with a soft-start circuit, it should not be a problem. Typically all you have to worry about is the peak current associated with the ON cycle (power switch closed). This same current will be present in several components, namely the input cap (which you will need), inductor, switch obviously, blocking diode, and output cap.

Again, a good resource to learn about these types of regulators is that data sheet and application note on the MC34063A...

Also, here (see attached) is a switching regulator fundamentals guide from National Semiconductor.. from the bottom right of this page:
**broken link removed**

EDIT: BTW, the capacitor acts like a low ohm resistor when t=0 (i.e. short circuit) As the capacitor initially charges, the voltage at the switch (Vd + Vc) will also rise.. and your control circuitry will not likely start PWM'ing the switch until a proper Vcc is met. These start up times are associated with pretty much any circuit that has input and power bus capacitors. They need to charge up before valid Vcc levels can be supplied to whatever circuitry you are powering.

boost-converter current-sensing

Becareful the term "soft-start" used in many appl notes, where the sense element is usually attached at the end of the power NMOS to limit the current through the power device. Due to the direct path from 5V supply->inductor->diode->cap, I'm afraid there is really nothing can stop/limit this in-rush current until the cap reaches 5v. A bigger L value would "choke" the current and slow down the charge current rate with dV/L.

switching current of converter

"nothing" ? Sure there is.. we use currently limiting devices to pre-charge thousands of uF's on various boards before connecting them directly to the power source. It just depends on how necessary it is to precharge your caps, and how much board space and money you are willing to spend.

In the case of the boost convertor, your circuit is most definitely going to have some form of a switch on it's input. That switch can either be used soley as the current limiting device for several 10's of milliseconds, after which it would be hard-on. Or you could use two paths on the input.. one for startup through a linear current limiting device, and the other for a hard switch.

There are all kinds of ways you could easily do this manually with a DPST or DPDT switch (i.e. Momentary-Off-On). The circuit could be designed to have the momentary required for precharge, then a quick flick to on gets things started.

Surely there are many many ways to achieve this that I haven't even spewed forth yet.. hee hee.

Box -> [] -------------> Thinking <

inductor current in excel

Sure there are many ways to stop/limit the current if the problem is push "externally" elsewhere and not from the boost converter itself. Certainly not the "softstart" technique from the appl notes.

boost converter with spst switches

Sure.. but I never said it was in an app note You are correct though.. if strickly constraining yourself to the boost circuit core configuration of (PWR -> L -> D -> C), there is no simple way of limiting the inrush current. One thing you might do is play with the ESR of the cap to balance your inrush current with output load requirements.

That said, my personal experience in portable battery operated circuits has never required any extra inrush current limiting in addition to the standard boost configuration. One such circuit was a 9V battery (low as 6V on input) to 15V out at 500mA. I think the output cap on that was only 120uF with a low ESR.

mengghee, what is the ESR of the output cap you are simulating? Maybe it's just TOO low, or not set at all.

soft start precharge curcuit

ESR ? i thought we are suppose to choose a capacitance with a low esr. but as what u have said, to reduce the in rush current, a high ESR is required ? i did not set any ESR on the capacitance. should i set it ? thank you.

mengghee

calculate switch current in boost

Rabbit, may I know what is the portable battery application that you mentioned that uses 9V battery and output cap of 120uF ?
It's good to have forumers like you to participate and discuss about power management. Can you mention where you're working ? I know portable power management is a booming business nowadays...

B.Rabbit said:

That said, my personal experience in portable battery operated circuits has never required any extra inrush current limiting in addition to the standard boost configuration. One such circuit was a 9V battery (low as 6V on input) to 15V out at 500mA. I think the output cap on that was only 120uF with a low ESR.
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boost inductor peak current

Limiting the inrush current can be accomplished by means of other components, not part of the boost circuit, and certinly not by increasing the ESR of the output capacitor.
As for the value of the output capacitor, it is not really that important, you select it based on the ESR and the ripple current, in order to meet the output voltage ripple spec:
ESR=Voutripple/Iripple

I will try to answer your question, though. Let's try first qualitatively, instead of mathematically.

When the switch is on, you know that the inductor is connected right across the input power and current flows through the inductor towards ground. What happens when the switch opens? You expect the voltage at the drain to rise. Well, it does, but only about 0.6V, when the diode turns on and it clamps the voltage, because the cathode of the diode is really at ground potential, because the output capacitor is discharged and it cannot charge up instantly.
So, when the transistor turns off, the inductor is still connected across the input voltage, the only difference being that there is now a diode in series to "ground". The situation is almost unchaged, so the curent continues to RISE in the inductor. As the voltage across the cap rises, things get better, but not much. Let's say the output voltage is now 1V. So when the transistor turns off, the drain voltage rises, but when it reaches 1.6V (that is, Vout+Vdiode), the diode turns on and the current still rises.
What changes is the SLOPE of the current, but it is still POSITIVE.
Mathematically:

di/dt=(Vin-1.6V)/L

When the transistor was on, the slope was dit/dt=Vin/L, also positive.

It is only when the output voltage is higher than the input, that the inductor current slope becomes negative, and so the current can actually decrease:

di/dt=(Vin-(Vd+Vout))/L this is negative when Vin<Vd+Vout

boost converter+soft switching

Someone can tell the relevant articles materials or website of digital power management??

Thanks in advance
Best regards,jinsin

inductor current rise

Hi VVV, thanks for your wonderful explaination.
could you show me som approaches abut soft-start and current sense?
thanks a lot!

A PTC ceramic resistor in series with the capacitor will restrain inrush current.. used on motor start up caps.

PTC Applications

Once the capacitor has reached operating charge and the current is reasonable, the device cools + lowers resistance
Pretty cheap and reliable..