# Current and voltage value before inductor pin in buck converter

In buck converter we know that pure DC current will be after inductor pin. But what will be the current/voltage value before it. Let’s take an example, the buck converter circuit with 12 V input to 3.3 V output and current of 2 A, operating at 1 MHz. Can some one explain it with example? It seems like simple question, but I have confusing it for long time.

• Pure DC current will not exist after Inductor and currents in inductor are same in both ends.. Commented Mar 17, 2023 at 6:38
• Wave you tried to simulate it? Commented Mar 17, 2023 at 7:33

simulate this circuit – Schematic created using CircuitLab

Hi Selva . You can see above picture general voltage mode buck converter. The Iinput peak current level while the M1(N-MOSFET or P-MOSFET) is on time(DxTperiod) and Ioutput peak current level while M1 is off time((1-D)xTperiod) are same peak current level. I attach it as a picture. If we want to see the voltage change on the L1, we need to analyze while the M1 is open and the M1 is closed.When M1 is closed(it is on time), there is a 12V supply voltage on the left side of L1 and 3.3V load voltage on the right side of L1. After that when M1 is opened (it is off time) there is only -3.3V load voltage.In this way, two areas are formed when M1 is open and closed. You can see below the picture. These areas must be equal so that L1 is not saturated. If we multiply these areas and analyze, we find the general gain formula of Buck Converter.(D=Vo/Vin)

I prepared the formulas and charts for the ideal situation. You should not forget the voltage drops M1 and D1 when calculating. I've also added a resource on this topic.

https://www.microchip.com/stellent/groups/sitecomm_sg/documents/devicedoc/en542810.pdf

Buck converter waveforms are readily available if you search for it. It might also be helpful to run your own simulation e.g. in LTSpice or something similar to get a better understanding.

Below is an example to answer your specific question. The waveforms show the voltage on the left inductor pin (switch node) V_A, the inductor current I_L, the voltage across the inductor V_L and the current through the transistor (or input current) Iin.

I also put three current arrows on the schematic. The current on both pins of the inductor (orange arrows) is exactly the same (I_L), where else could the current go?

The current through the green arrow (to the load) totally depends on the load. It could be constant, but it could also be changing. For example if the buck converter supplies a processor, the current will change depending on the processor activities.

The difference between inductor current and load current will be supplied by the capacitors as in Kirchhoff's current law (KCL).

Source: industrial-electronics

Note the example is for continuous conduction mode (CCM), the waveforms for discontinuous conduction mode (DCM) will look slightly different.

As you made the simplification that inductor current is considered as pure DC without ripple, i.e. the inductances and capacitances are so large they have so little ripple that it can be ignored, the answer is easy.

You have 3.3V output at 2.0A which is 6.6W.

On the 12V side, that means the average DC input current is 6.6W/12V = 0.55A.

And because the switching duty cycle is 3.3V/12V = 0.275, for only 27.5% of the time input current flows and for the rest of the time the current is zero.

So when input current flows, it must be 0.55A / 0.275 = 2A, so it was just mathematically proven that when input current flows during the ON cycle, it must match the output current, and thus input ripple current is equal to output DC current.