Is Inductor peak current same as the Inductor Inrush current

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
I am using a DC-DC converter which is having an inductor rating of 68uH and 2.05A.
Output Voltage of dc-dc is 23V and output current max is 300mA
Schematic :
enter image description here
I am using a current probe and measuring the current through the inductor.
When I turn ON the DC-DC converter, the Inductor draws 10.75A. Should I consider this as Inductor Inrush current or peak current? If I consider this as Inrush current, as I do this during start-up of the converter, then how to measure the peak current of the inductor?
And since the inductor is rated for only 2.05A, won't this high current damage the component even though it is for a short duration?
Current probe is set as 0.1V/A in the probe and 0.1V/A in the scope as well.
Inductor Peak current (below is the waveform captured during the start-up of the DC-DC converter):
enter image description here
Inductor RMS Current (below is the waveform captured during normal running condition of the DC-DC Converter) :
enter image description here
 
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ronsimpson

Joined Oct 7, 2019
1,007
Is this a 12V to 23V DC to DC power supply? (non isolated)?

The inrush current is for a short time. There is not time for the wire in the inductor to heat up. I worry about the diode.
What is the load? on the 23V? It appears to have capacitance.
What is the source of power? 12V battery?
 

Thread Starter

Electronic_Maniac

Joined Oct 26, 2017
253
Is this a 12V to 23V DC to DC power supply? (non isolated)?

The inrush current is for a short time. There is not time for the wire in the inductor to heat up. I worry about the diode.
What is the load? on the 23V? It appears to have capacitance.
What is the source of power? 12V battery?
Provide the details and the schematic. Please check and help
 

ronsimpson

Joined Oct 7, 2019
1,007
the details
Thanks.
The inductor current is measured after things are up and running.

This inrush current, will be there even if you removed the IC from the board. It is power from the battery going through the inductor through the big diode to charge up the output capacitors to 12 volts. If you look at the current into the board you will see this current and the current to charge up your input capacitors.

What IC are you using? I do not understand the "gate drive" two transistor circuit. What is the part number of the inductor. The resolution on the schematic is low and I can not see it. What diode?
 

MrAl

Joined Jun 17, 2014
7,849
I am using a DC-DC converter which is having an inductor rating of 68uH and 2.05A.
Output Voltage of dc-dc is 23V and output current max is 300mA
Schematic :
enter image description here
I am using a current probe and measuring the current through the inductor.
When I turn ON the DC-DC converter, the Inductor draws 10.75A. Should I consider this as Inductor Inrush current or peak current? If I consider this as Inrush current, as I do this during start-up of the converter, then how to measure the peak current of the inductor?
And since the inductor is rated for only 2.05A, won't this high current damage the component even though it is for a short duration?
Current probe is set as 0.1V/A in the probe and 0.1V/A in the scope as well.
Inductor Peak current (below is the waveform captured during the start-up of the DC-DC converter):
enter image description here
Inductor RMS Current (below is the waveform captured during normal running condition of the DC-DC Converter) :
enter image description here
If your current probe is not fast enough to see the inductor current then try measuring the voltage across the two 0.3 ohms (0.15 ohms total) and use Ohm's Law to calculate the current i=V/R so i=V/0.15. There will be a lot of noise but you can still get an idea what is going on.

What is the frequency of operation?

The inrush current is almost never the same as the peak current. If there is no soft start mechanism then the inrush will usually be higher than the normal peak current of the inductor. The RMS current is nice to know too, but the peak current will tell you if the inductor is saturating or not. The frequency will help determine that too.

Most modern circuits have a soft start mechanism so that the inrush current is greatly reduce, below the normal peak current of the inductor.
 

DickCappels

Joined Aug 21, 2008
6,640
Looking at the diagram I would guess that this is some kind of mc34063 variety.
Yes, it must be an MC34063 or second source for it.

No. They are (probably) not the same thing. I have not, in my years designing power supplies as well as other circuit, heard of inductor inrush current. What I am pretty sure you are seeing is the charging of the 330 uf output capacitor on the power supply. Peak Current in an inductor is the maximum current observed during a switching cycle.
 

MrAl

Joined Jun 17, 2014
7,849
Yes, it must be an MC34063 or second source for it.

No. They are (probably) not the same thing. I have not, in my years designing power supplies as well as other circuit, heard of inductor inrush current. What I am pretty sure you are seeing is the charging of the 330 uf output capacitor on the power supply. Peak Current in an inductor is the maximum current observed during a switching cycle.
Hi,

If you do some simulations of 'regular' switchers you'll see the inductor current jump way up when the power is first applied. That's because the output capacitor is not yet charged to the normal operating voltage. I think you will see this even with a buck.
Of course this is where 'slow start' circuits where born. They provide a smaller duty cycle until the cap has time to charge up. Current limit only helps if the current limit is also monitoring the inductor current. Slow start is often implemented as a feed forward type control where there is no output monitoring just a delay with gradual ramp up of the PWM pulse
Even with no feedback the surge occurs, and feedback just makes it worse because the control circuit thinks it has to put out more current to get the output voltage to come up to normal.
This can be simulated quite easily. Just provide a set duty cycle that would normally provide a normal output voltage level. When the circuit starts up, the current starts out higher than normal. So in a buck with 50 percent duty cycle we would normally have about 1/2 the output as the input is and the current would be whatever the output load required. During startup though the output voltage is zero so the cap draws a lot more current with a 50 percent duty cycle and of course it has to get its current through the inductor.
The input cap surge can also be a problem of course but that happens faster while the inductor current is still high.
 
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MrAl

Joined Jun 17, 2014
7,849
This is a boost. The inrush problem will happen (with this supply) before the PWM starts to work. Soft start will not help.
Hi,

That's the first time i ever heard that :)

Switching converters have gotten more sophisticated and TI knows how to handle boost circuits for one with a three mode startup regimen. But soft start is not new i knew about it back in the early 1980's. We did synthesized sine converters with low THD even back then, and soft start was in every one of them.

But the inrush also happens with a buck but i think that is only in the case where there is output voltage feedback and no current limiting. The control tells the switcher that there is low output voltage so to increase the duty cycle to maximum thus the input current is high. After the output levels off the duty cycle cuts back to a more normal value and inductor current goes back to normal. Without slow start the inductor surge current could be 10 times the normal operating current.
 
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