Ideal Buck, Boost and Inverting Buck-Boost Calculations

Compare ideal converter models

These equations describe ideal components in steady-state continuous conduction mode (CCM). B2 is positive input voltage in volts and B3 is switch duty as a fraction: 0.6 means 60%. This exercise uses 0<D<1, excluding switching endpoints. Losses and control-loop dynamics are omitted.

For a buck stage, Vout=D Vin. For a boost stage, Vout=Vin/(1-D). For the conventional inverting buck-boost, Vout=-Vin D/(1-D): its output polarity is negative relative to the input reference. The last equation is not a universal duty-cycle rule for every four-switch non-inverting buck-boost controller; consult its operating modes and datasheet.

The worksheet predicts ideal quantities, not component ratings or safe operating limits. In particular, a numeric result near D=1 does not establish that a real controller can attain that voltage. Maximum duty cycle, current limit, losses and load conditions constrain real operation.

Boost output voltage

E2 retains the unrounded ideal voltage. With 12 V and D=0.6 the result is 30 V. Try D=0.5 for 24 V. A duty of 0.98 produces 600 V mathematically; the example deliberately does not label this feasible or safe.

=IF(COUNT(B2:B3)<>2,"Enter numeric voltage and duty",IF(OR(B2<=0,B3<=0,B3>=1),"Require Vin>0 and 0<D<1",B2/(1-B3)))

Buck output voltage

E3 uses the ideal buck ratio, giving 7.2 V for the initial inputs. This is a separate topology comparison, not a second output of the same boost circuit.

=IF(COUNT(B2:B3)<>2,"Enter numeric voltage and duty",IF(OR(B2<=0,B3<=0,B3>=1),"Require Vin>0 and 0<D<1",B2*B3))

Inverting buck-boost signed output

E4 gives -18 V initially. Its magnitude is 18 V. At D=0.5 its output is -12 V, while the boost comparison gives +24 V. Preserve the sign when interpreting this topology.

=IF(COUNT(B2:B3)<>2,"Enter numeric voltage and duty",IF(OR(B2<=0,B3<=0,B3>=1),"Require Vin>0 and 0<D<1",-B2*B3/(1-B3)))

Boost inductor peak-to-peak ripple

B4 is switching frequency in Hz and B5 is inductance in henries. Thus 200 kHz is entered as 200000 and 45 microhenries as 0.000045. During the ideal boost switch-on interval, the inductor sees Vin; E5 calculates Vin D/(f L) in amperes peak-to-peak. The initial ripple is 0.8 A.

At fixed voltage and duty, doubling frequency or inductance halves ripple. This is the boost ripple equation; the buck on-state inductor voltage is Vin-Vout, so do not reuse this expression unchanged for a buck stage.

Ripple alone does not establish CCM, core saturation margin or loop stability. For the triangular boost-current model, CCM requires average inductor current to exceed half the peak-to-peak ripple; peak current is average current plus half the ripple. Load and current ratings are absent from this exercise, so those conditions are not verified.

=IF(COUNT(B2:B5)<>4,"Enter numeric inputs",IF(OR(B2<=0,B3<=0,B3>=1,B4<=0,B5<=0),"Require positive Vin, f, L and 0<D<1",B2*B3/(B4*B5)))

Further reading

• Texas Instruments — Basic Calculation of a Boost Converter’s Power Stage
• Texas Instruments — Inverting Buck-Boost and Cuk Converters