Lately, I have been learning how to design 3.7V to 5V boost converter for powering esp32-cam. Then I thought how one can get exactly 5V or other desired output voltage from a boost converter. I know there is equation relating the input, output voltage and duty cycle of the pulse that is fed into the gate of the mosfet. However I did not understand how this equation was derived. And so I wanted to write a note on how this boost converter voltage equation is derived.
To understand how the output voltage of a boost converter is derived, using basic energy conservation and circuit behavior during the switching cycle.
🔁 Boost Converter Switching Operation
A boost converter works in two phases:
1. Switch ON (transistor conducts)
- Inductor is connected directly across input voltage.
- It stores energy as a magnetic field.
- Diode is reverse biased → no current flows to output.
2. Switch OFF (transistor open)
- Inductor releases stored energy.
- Polarity of inductor reverses, adds to input voltage.
- Current flows through diode to the output → boosted voltage.
🧮 Step-by-Step Derivation (Ideal Case)
Assume:
- : input voltage
- : output voltage (greater than )
- : duty cycle (fraction of time switch is ON, between 0 and 1)
- : total switching period
- : inductance
🔋 ON Period (duration = ):
- Switch is closed.
- Voltage across inductor:
V_L = V_{in}
\Delta I_L = \frac{V_{in} \cdot D \cdot T}{L}
🔌 OFF Period (duration = ):
- Switch is open.
- Inductor discharges its energy into the load.
- Voltage across inductor:
V_L = V_{in} - V_{out} < 0
\Delta I_L = \frac{(V_{in} - V_{out}) \cdot (1 - D) \cdot T}{L}
📘 Key Idea: In steady-state, current change per cycle is zero
\text{Total } \Delta I_L = 0
So:
\Delta I_{\text{ON}} + \Delta I_{\text{OFF}} = 0
Substitute expressions:
\frac{V_{in} \cdot D \cdot T}{L} + \frac{(V_{in} - V_{out}) \cdot (1 - D) \cdot T}{L} = 0
Cancel from all terms:
V_{in} \cdot D + (V_{in} - V_{out}) \cdot (1 - D) = 0
Expand:
V_{in} \cdot D + V_{in}(1 - D) - V_{out}(1 - D) = 0
Combine terms:
V_{in}(D + 1 - D) - V_{out}(1 - D) = 0
V_{in} - V_{out}(1 - D) = 0
Now solve for :
V_{out} = \frac{V_{in}}{1 - D}
✅ Final Result:
\boxed{V_{out} = \frac{V_{in}}{1 - D}}
This is the ideal boost converter equation.
⚠️ Real-World Effects
The real output voltage is slightly lower due to:
- Diode forward voltage ()
- MOSFET voltage drop
- ESR losses in inductor and capacitor
So, a more realistic version:
V_{out} \approx \frac{V_{in}}{1 - D} - V_{\text{losses}}
Let me know in the comment if you want a diagram to illustrate this switching cycle!