The hold-up time of a Switched-Mode Power Supply (SMPS) refers to the amount of time the power supply can continue to provide the required output power after the input power is interrupted. It's primarily determined by the energy stored in the bulk capacitor(s) on the input side of the power supply.
### Steps to Calculate Hold-Up Time:
1. **Find the total energy stored in the capacitor:**
The energy stored in a capacitor is given by the formula:
\[
E = \frac{1}{2} C V^2
\]
Where:
- \( E \) is the energy in joules (J),
- \( C \) is the capacitance in farads (F),
- \( V \) is the voltage across the capacitor in volts (V).
2. **Determine the minimum operational input voltage:**
The minimum voltage at which the power supply can still operate, denoted as \( V_{\text{min}} \), is crucial in this calculation.
3. **Calculate the available energy:**
The usable energy stored in the capacitor is the difference in energy between the full voltage and the minimum operating voltage. This is given by:
\[
E_{\text{usable}} = \frac{1}{2} C (V_{\text{input}}^2 - V_{\text{min}}^2)
\]
Where:
- \( V_{\text{input}} \) is the normal input voltage (before the interruption),
- \( V_{\text{min}} \) is the minimum input voltage at which the SMPS can operate.
4. **Determine the output power:**
Let \( P_{\text{out}} \) be the power consumed by the load at the output of the SMPS. This should be in watts (W).
5. **Calculate hold-up time:**
Finally, the hold-up time \( t_{\text{hold-up}} \) is calculated by dividing the available energy by the output power:
\[
t_{\text{hold-up}} = \frac{E_{\text{usable}}}{P_{\text{out}}}
\]
This gives the time in seconds (s) that the SMPS can continue to supply power after input power is lost.
### Example Calculation:
Suppose we have:
- Input capacitor \( C = 470 \, \mu F \),
- Input voltage \( V_{\text{input}} = 400 \, V \),
- Minimum operational voltage \( V_{\text{min}} = 300 \, V \),
- Output power \( P_{\text{out}} = 100 \, W \).
Using the formula:
1. Calculate the usable energy:
\[
E_{\text{usable}} = \frac{1}{2} \times 470 \times 10^{-6} \times (400^2 - 300^2)
\]
2. Divide the usable energy by the output power to get the hold-up time.
By plugging in the values, you can get the exact hold-up time in seconds. Would you like me to calculate this example for you?