simulate this circuit – Schematic created using CircuitLab

I have a voltage source(V1) connected to a boost-converter, the components aren't rated(more focused into the theory than actual build). What is confusing me, is the current. The input current will be set to whatever the power-supply(V1) is inputting into the circuit, I know that input-current \$I_o\$ < \$I_i\$, I'm not sure how to rate the currents like so:

At point(A) : \$\ I_i\$

At all other points, is current simply \$\ I_o\$ = \$\ \frac{V}{R}\$ at B,C,D? What's troubling me is point A and the rest, how are they different values in series... like A and D


input current \$I_o < I_i\$

This equation refers to the average input and output currents.

During the switching cycle the input current will increase and decrease, roughly following a triangle wave shape.

When the switch M1 is closed, the current drawn from V1 increases, building up energy in the magnetic field of L1. Meanwhile the load draws current from output capacitor C2. When M1 is opened, current drawn from V1 decreases as the built up energy in L1 is delivered to C2.

At all other points, is current simply \$I_o = \frac{V}{R}\$ at B, C, D?

At B, it will be roughly \$\frac{V}{R}\$, but will still have some ripple because \$V_o\$ will have some ripple.

At C, you will have the combined return current of the load current (\$\frac{V_o}{R}\$) plus the current through C2, which will be positive in some parts of the cycle and negative in other parts.

At D you will have the return current from the load, the return current from C2, and return current from the switch, all adding up to exactly the same value as the inductor current \$I_i\$, which you could see using KCL.

  • \$\begingroup\$ Its starting to make sense, to be specific. D = Ix+Iy+Iz? Since the current is breaking up into those branches? \$\endgroup\$ – Pupil Oct 19 '15 at 2:49
  • 1
    \$\begingroup\$ You need to learn Kirchoff's Current Law. Your equation may be correct, but exactly how you write the sum depends on how you define the direction of current in each of the branches. \$\endgroup\$ – The Photon Oct 19 '15 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.