Inverting Buck-Boost Step by Step Design Guide

If you are planning to make your own inverting buck-boost but don’t know where to start, therefore this inverting buck-boost step by step design guide is right for you. A buck-boost converter is a DCDC switching converter that combines the function of a buck and boost converter. Inverting buck-boost is a variant wherein the output is negative with respect to the ground.

Detailed Inverting Buck-Boost Step by Step Design Guide

1. Supply the Known Parameters

Start by defining the basics like input and output values. They must be given during the design stage.

Example:

Vin = 12V, Vout = -5V, Iout = 11A

Where;

  • Vin – is the input voltage of the buck-boost converter
  • Vout – is the output voltage
  • Iout – is the load current

2. Compute the Ideal Duty Cycle

Buck-boost is a duty cycle controlled DCDC converter. Once you able to derive the duty cycle, you can compute the rests of the important parameters.

Duty = – Vout /  (Vin – Vout )

Using above given;

Duty = – Vout /  (Vin – Vout ) = – (-5V) / [12V – (-5V)] = 29.412%

3. Define the Switching Frequency

You need to define what switching frequency level you will set the converter. The defining factor in selecting the switching frequency is power density, controller capability and EMI noise.

Example: Fsw = 250 kHz

4. Determine the Inductance Value

L1 = Ton X ( Vin – VQ1 ) / ( %di X Iout )

  • Ton = Duty / Fsw = 0.29412 / 250kHz = 1.18 usec
  • VQ1 – this is the voltage drop of the switch. Example: VQ1 = 0.2V
  • %di –This is the set level of the inductor ripple current. This must be provided during design stage. A good rule of thumb is 20%-40% of the load current. Example: %di = 25%
  • Iout – this is the output current declared above

Thus;

L1 = Ton X ( Vin – VQ1 ) / ( %di X Iout ) = 1.18 usec X ( 12V – 0.2V ) / ( 0.25 X 11A ) = 5.06 uH

Consider a standard Inductor, say

L1_selected = 5uH

5. Select the Inductor

Inductor Peak Current

Inductor DC Current

Idc_L1 = Imax – di_actual + di_actual / 2 = 16.972A – 2.78A + 2.78A / 2 = 15.582A

Inductor RMS Current

Irms_L1 = [ di_actual / sqrt ( 3 ) ] + Imax  – di_actual = [ 2.78A / sqrt ( 3 ) ] + 16.972A – 2.78A = 15.78A

The selected inductor must have current rating that is higher to all the computed values above.

6. Select a MOSFET Switch

Peak Current

Imax = same inductor Imax above

DC Current

RMS Current

The selected switch must have a current rating higher than all the computed values above.

Voltage Stress

VQ1_max = Vin + VD1 – Vout

VD1 – this is the voltage drop of the diode. Example: VD1 = 0.7V

Thus,

VQ1_max = 12V + 0.7V – ( -5V ) = 17.7V

The selected MOSFET must have a voltage rating higher than this value with ample of margin.

Power Dissipation

Pdiss_Q1 = Ploss_conduction + Ploss_switching

  • Ploss_conduction = Irms_Q1 X Irms-Q1 X RDSon_Q1
  • RDSon_Q1 = on state resistance, Example: RDSon_Q1 = 0.01 ohm
  • Thus,
  • Ploss_conduction = Irms_Q1 X Irms-Q1 X RDSon_Q1 = 8.568A X 8.568A X 0.01ohm = 0.734W

  • Ploss_switching = Ploss_gatecharge + Ploss_COSS + Ploss_risefall
  • Plos_gatecharge = ½ X Qgtotal X Vdrive X Fsw
  • Qgtotal – this is the total gate charge indicated in the MOSFET datasheet. Example: Qgtotal = 1nC
  • Vdrive – this is the voltage applied to the gate to source of the MOSFET. Example: Vdrive = 12V
  • Thus,
  • Ploss_gatecharge = ½ X Qgtotal X Vdrive X Fsw = 0.5 X 1nC X 12V X 250kHz = 0.0015W
  • Ploss_COSS = ½ X COSS X ( Vdrain_max )2 X Fsw
  • COSS – this is the output capacitance of the MOSFET. Example: COSS = 1nF
  • Vdrain_max – this is the peak drain voltage. This is equal to the VQ1_max above.
  • Thus,
  • Ploss_COSS = ½ X COSS X ( Vdrain_max )2 X Fsw = 0.5 X 1nF X ( 17.7V )2 X 250kHz = 0.039W
  • Ploss_risefall = 0.5 X ( trise + tfall) X Irms_Q1 X Vdrive X Fsw
  • trise – this is the rise time of the MOSFET. See datasheet. Example: trise = 1nsec
  • tfall – this is the fall time of the MOSFET. See datasheet. Example: tfall = 1nsec
  • Thus,
  • Ploss_risefall = 0.5 X ( trise + tfall) X Irms_Q1 X Vdrive X Fsw = 0.5 X ( 1nsec + 1nsec ) X 8.568A X 12V X 250kHz = 0.025W

Finally,

Pdiss_Q1 = Ploss_conduction + Ploss_switching = 0.734W + 0.0015W + 0.039W + 0.025W = 0.8W

The selected MOSFET must have a power dissipation rating higher than this value with ample of margin.

7. Select the Diode

Peak Current

The diode peak current is the same to the peak inductor and MOSFET current.

Imax = same inductor Imax above

DC Current

The DC current of the diode is just equal to the output current

Idc_diode = Iout

RMS Current

The selected diode must able to handle all the computed currents above with ample of margin.

Peak Reverse Voltage

PRV_D1 = Vin – VQ1 – Vout

Using all the values declared above,

PRV_D1 = Vin – VQ1 – Vout = 12V – 0.2V – ( -5V ) = 16.8V

The selected diode voltage must be higher than this with more margin.

Power Dissipation

Ploss_diode = VF_diode X Irms_diode

VF_diode – this is the forward voltage of the diode. See the datasheet. Example: VF_diode = 0.7V

Thus,

Ploss_diode = VF_diode X Irms_diode = 0.7V X 13.273A = 9.29W

The selected diode must have a power rating higher than this with a good margin.

8. Select the Output Capacitor

Ripple Current

The selected capacitor must have a ripple current higher than this with ample of margin.

Voltage

The selected output capacitor must have a voltage rating higher than the output voltage by a enough margin.

Minimum Capacitance

Where;

Capacitor ESR

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