# Bi-directional buck converter

Discussion in 'Homework Help' started by AlexMcDuffMiller, Sep 29, 2012.

1. ### AlexMcDuffMiller Thread Starter New Member

Sep 16, 2012
5
0
Hi everyone,

I'm looking to calculate the instantaneous current in the output inductor of a bi-directional buck converter.

I believe I am right in modeling the average current as a transformer (i_vp/D=i_cp). However, I am not sure how to get the instantaneous ripple current that rides on top of that average current, any ideas?

Thanks!

2. ### Ron H AAC Fanatic!

Apr 14, 2005
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You need to post a schematic.

3. ### AlexMcDuffMiller Thread Starter New Member

Sep 16, 2012
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Hopefully that image somewhat makes sense....The switches are actually transistors and are given complementary inputs to each other. Assume them to be ideal.

4. ### crutschow Expert

Mar 14, 2008
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Where's the inductor?

5. ### AlexMcDuffMiller Thread Starter New Member

Sep 16, 2012
5
0
The little bumpy thing on the right side of the image.

Also, the line going into the conductor is connected to the diode/switch pairs where it crosses over.

6. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
675
The general equation for current through an inductor is i=(1/L)∫v*dt, where v is the voltage across the inductor, and L is the inductance.
For a step of voltage, this simplifies to i=v*t/L.
When the voltage across the inductor is positive, the current will be a rising ramp. When the voltage is negative, the current will be a falling ramp. For steady-state conditions (constant average current),the amplitude of the two ramps will obviously have to be equal.

See the attached simulation. The input voltage is a 0 to 25V pulse waveform with a duty cycle of 20% (2μS on, 8μS off). This will give us an average of 5V across the filtered output. The average current will be 5V/5Ω=1A, but the ripple current in the inductor will be independent of that value, as long as the duty cycle remains constant.
During the 2uS ON time (t1), the voltage across the inductor is (25-5)=20V, so the current will rise by i=v*t1/L=20v*2uS/200uH=200mA. During the 8uS OFF time, the current will fall by -5v*8uS/200uH=-200mA. Therefore, the ripple current is 200mA peak-to-peak.

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