Switching regulator peak current

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If a boost converter IC with a builtin switch rated for 5amps peak is powering a 2.8W load from a 4V input power supply at 70% efficiency, is the switching current going to be I = (2.8 / 4) * 1.7 = 1.19amps or is it always going to peak at 5amps but with a very small duty cycle?
 

Hi,

If a switch is rated for 5A, that value has nothing to do with the actual current passing through the switch. The actual current going through the switch is dependent on the inductance and series resistances along the path of the current flow including that of the switch and that of the inductor.


A good approximation can be made if the in converter is designed to operate in continuous conduction mode and the inductor ripple current is small enough to be approximated as negligible.
 
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Disregard the original post and their values. Having issue with equation 3 for determing total system switching current for the attached datasheet IC using application note https://www.ti.com/lit/an/slva372c/slva372c.pdf

Parameters are
IC = mt3608 Switch Limit = 4a
Vin min&max =8v Vout = 25v
Iout=25mA L= 6.8uH η = 70%

Eq 1 Duty cycle = 22.4
Eq 2 ΔIL = 22
Eq 3 Imax-out = (4 - 22/2) * (1-22.4) = 149.8amps

View attachment MT3608.pdf
 

0.025A * 25V = 625mW, / 0.7 (eff) = 892mW, / 8vin = 111mA average, the peak current depends on the L value and the freq ( which you haven't disclosed )
 
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Easy peasy said:
0.025A * 25V = 625mW, / 0.7 (eff) = 892mW, / 8vin = 111mA average, the peak current depends on the L value and the freq ( which you haven't disclosed )
Click to expand...

The inductor was listed, it is 6.8uH but the frequency wasn't, the IC switches at fixed 1.2Mhz and employs pulse frequency modulation for light loads.

Parameters are
IC = mt3608 Switch Limit = 4a
Vin min&max =8v Vout = 25v
Iout=25mA L= 6.8uH η = 70% Frequency 1.2Mhz

Therefore with an average of 111mA is nearly any 1amp inductor good enough or is the peak current really the determinant for the inductor selection?

The application note equation 3 yields large current.
 

Hi,

For the inductor selection you need to take care of "worst case" inductor current.
Worst case is not the steady state. I assume the highest peak current you can expect at powering up the device and tge switching regulator charges the output capacitor.

As I currently see nothing else than the regulator that prevents from overcurrent ... I'd expect a peak inductor current of the given 4A.
--> I'd use a 4A inductor.

For sure the inductor will see much less current all the time...

If you choose an inductor with lower current rating you may expect saturation ... and thus decreased inductance and thus higher dI/dt.
In worst case the current rises faster than the regulator can activate it's overcurrent limiter .. maybe causing damage of the regulator.

The other option is to use a more suitable regulator IC for low output current or adjustable current limit.

Klaus
 
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25 mA output and 6.8 uH will result in discontinuous mode, peak current about 0.5 A. In so far an 1A rated inductor can work (if the switcher IC is performing proper current or voltage mode regulation), for 1.2 MHz operation frequency, core losses are however an important parameter
 
Eq 3 Imax-out = (4 - 22/2) * (1-22.4) = 149.8amps
Click to expand...

the above is clearly wrong in some way - perhaps a formula copied wrong ...?

As FvM states, for this boost chip & ckt you will be in DCM, and 0.5A pk, you may find it hard to get a boost choke off the shelf that can handle 1.2MHz and 500mA pk without getting very hot - you will need a very low loss ferrite, ideally a toroid with 5-10 turns ....
 
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Don't use % values.

Eq 1 D = 1 - (8 * 0.7 / 25) = 0.776
Eq 2 ΔIL = (8 * 0.776) / (1.2M * 6.8u) = 0.76A
Eq 3 Imax-out = (4 - 0.76/2) * (1 - 0.776) = 0.81A
 
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    Zak28

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Is there a definitive method to measure peak current of the inductor on a fabbed board?

Peak current at the inductor likely cannot be measured by using a DMM, perhaps a low resistance current shunt is required series to the inductor and switch node of the IC but Im not sure on proper method.

Eq 1 Duty Cycle
D = 1 - (8 * 0.7 / 25) = 0.776

Eq 2
ΔIL = (8 * 0.776) / (1.2M * 6.8u) = 0.76A

Eq 3 IC Peak Current Requirement
Imax-out = (4 - 0.76/2) * (1 - 0.776) = 0.81A

Eq 4 Inductor Peak Current
0.76/2 + 25e-3 / 1 - 0.776 = -0.491A

Is the peak maximum current in the inductor 0.491A?
 
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Theory of energy conversion in the boost converter:

X amount of Amperes going through...

Y Henries inductor value ...

yields X times Y Webers of energy.

At switch-Off this translates to Z amount of volts increase on the output capacitor. The value of Z depends on the value of output capacitor.
The smaller the output capacitor, the greater is Z.
The equation for this process has to do with time constants for RL and RC. Also the formula that equates Amperes in a capacitor with its voltage change.

Whatever Z is, the supply voltage is added to that.

The load drains a certain amount of voltage from the output capacitor.

The inductor always generates sufficient voltage to overcome volt level on the capacitor. Thus if there is no load, then the output capacitor theoretically rises to infinite voltage.
 
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for a boost converter Vo/Vin = 1 / (1-D) so for 8V to 25V, D will be >= 0.68, say 0.7 (losses)

at 1.2MHz the max avail on time is 833nS, x 0.7 = 581nS

from V/L = di/dt, we see that the current in 6u8 inductor will ramp to 0.683 amps in 581nS

the off time is 833 - 581 = 252nS, the current will ramp down with net Vopposing = 25-8V, giving 0.63A ramp down in 252nS ( V/L = di/dt, again )

thus the converter is just in CCM, and the true peak current is closer to 700mA ...

- - - Updated - - -

the above applies for a load on the 25V out, at light loads the D will shrink to regulate the Vo, at no load the D will shrink to near zero.

My DVM measures duty cycle ( D ) but not at 1.2MHz ...

energy is not measured in Webers, Webers per square metre is a measure of flux density, B in Tesla.
 
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It was already posted the peak current is ~0.5A which isn't much since there are already many high frequency shielded ferrite inductors available.

Considering ~90% of the 25mA is constantly consumed does the peak current happen only during startup? The soft start feature should eliminate most of the peak inrush current during initial power of the device to charge the filter capacitors.
 

for 80mA in ave ( approx ) & about 0.32Apk, you won't make it into CCM, DCM only. You say there are many shielded inductors available. From experience I can say that you might well be surprised at how hot some get even at this modest current at 1200 kHz ...
 
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Peak currents cause more heating to inductors than a steady average current?

Already fabbed a board with this IC and a 6.8uH shielded high frequency 5.5A inductor with output current only ~8mA 27.5V it makes a very small buzzing noise likely due to the peak current.
 
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Zak28 said:
Peak currents cause more heating to inductors than a steady average current?
Click to expand...
They can, especially with very high switching frequencies, but usually if this is the case the design hasn't been properly optimized. A good engineer will estimate both DC and AC losses for the inductor. Many manufacturers make this easy with online tools.
 
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Does this imply CCM - a state were there is only an average current in the inductor is the least wasteful state of the boost converter?
 

The buzzing noise may be instability in the control - going from DCM to CCM and back again - or you may have a pulsing load?
 

It infact was a pulsed load ~4.5mA and with a 6.8uH inductor it must have been pulsing large currents regularly.

Made an openoffice Calc spreadsheet which rapidly computes the 4 equations based on their input.

View attachment boost-converter.zip
 

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