A buck converter delivers a lower DC voltage than the supplied voltage.

Circuit Diagram Edit

Picture 15

Analysis Edit

Picture 16

Input output voltage relationship Edit

For continous conduction mode. Remember energy is conserved! Energy coming into inductor leaves the inductor each cycle.

This is best expressed in using

$ v = L\frac{di}{dt} $ and conservation of energy $ \int_0^{T_s} v_Ldt = 0 $

$ \therefore (V_d-V_0)T_sD = V_o(1-D)T_s $

$ \therefore \frac{V_o}{V_d} = D $

Boundary between dis and cts Conduction Edit

Discontinuous conduction is more likely the lower the duty cycle, as the inductor has less time to store energy and more time dispensing it.

find $ I_L $ when the minimum current is zero.

$ i_{min} = I_L - \frac{\Delta i_L}{2} $

$ \Delta i_L = \frac{V_d -V_o}{L} T_sD $

from the inductor equations

$ \therefore I_L = \frac{V_d -DV_d}{2L} DT_s $

$ \therefore I_L = \frac{1-D}{2L}V_dDT_s $ the load current for which we will enter into discontinuous mode.

Note: maximum current supplied when D=0.5, so the maximum boundary current is when D=0.5

It is useful to find the minimum conditions for the inductor and frequency to avoid discts conduction mode

$ \frac{V_o}{R}= \frac{1-D}{2L}V_dDT_s $

$ \therefore Lf_s = \frac{(1-D)R}{2} $


$ Lf_s > \frac{(1-D)R}{2} $

DCM Edit

Picture 17

Start anaylsis similar method as to ccm.

$ (V_d -V-_o)T_sD = V_o(T_s(1-D)-\Delta_1) $ But now we need to find an expression for $ \Delta_1 $

Output Voltage Ripple Edit

Picture 14

If our output voltage is constant the output capacitors average current is zero.

Hence $ I_L = I_o $ and $ i_{ripple} = i_c $ So capacitor essentially smooths the output current wave form, filling in the supply gaps when the inductor is out of energy.

Ok so we know

$ C = \frac{dq}{dv} $ so

$ \Delta V_o = \frac{\Delta Q}{C} $

We then have $ \Delta V_o = \frac{1}{C} \frac{1}{2} \frac{\Delta I_L}{2} \frac{T_s}{2} = \frac{\Delta I_L T_s}{8C} $ essentially the area under capacitor current shown in figure above(triangle).

$ \Delta I_L = \frac{V_o}{L}(1-D)T_s $ during Toff the voltage across the inductor is Vo.

$ \therefore \frac{\Delta V_o}{V_o}=\frac{T_s V_o}{8CL} (1-D)T_s $


$ f_c = \frac{1}{2\pi\sqrt{LC}} $ is the cut off frequency of the LC filter, hence it is very desirable for $ f_s>>f_c $

In practice, the voltage drop across the equivalent series resistance (ESR) of the capacitor may contribute significantly to the output voltage ripple. This drop has been neglected.

Practical Considerations and Ratings Edit

Transformer Considerations Edit

Switch Ratings Edit

Two Switch Flyback Converter Edit

$ Insert formula here $

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