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mengghee

Full Member level 3
In the theoretical design of the boost converter, if i am not wrong, changing the load current wouldn't affect the output voltage. but in practical design, it does ... can anybody tell me what causes the variation of output voltage ? thank you.

regards,
mengghee

as for all converters variation in output voltage
causes :
For DC :
= duty cycle
= input voltage
For AC :
= all above
= switching frequency

Because real elements contains conduction loss as minimum (serial resistors connected with ideal element) and paracitic inductance and capacitance.

For example your mosfet contain drain to source
resistance in open state. This resistance create voltage drop that impact on output voltage.
the same story with diode voltage drop,
inductance dc resistance.

mengghee

Points: 2
thank you,

my question was ... what causes the vout to vary if i change my load resistor ? in the equation vout = vin / 1- D . and through the equation. it has nothing to do with RL. thank you

regards,
mengghee

yes for ideal case vout= vin/1-D , but for real case
when your inductance has DC resistance RL connected in series with it formula will be :
vout = [vin/1-D]*[1/(1+RL/(R*(1-D)^2))]

mengghee

Points: 2
thanks rts,

do you have any reference that i can see ? i am just wondering how that is derived

regards,
mengghee

take a look at page 30 (output voltage) and page 31 (efficiency) of this document

The expressions show how the output voltage $V$ and the efficiency $\eta$ are influenced by the diode forward drop $V_D$, diode on resistance $R_D$, inductor series resistance $R_L$, MOSFET on resistance $R_{on}$ and load resistance $R$. Note that $V_g$ is the input voltage, $V$ is the output voltage, $D$ is the duty ratio and $D\prime=1-D$.

By the way, if you take the expressions for the output voltage and efficiency and set $V_D$, $R_D$, $R_L$, $R_{on}$ to zero, you will arrive at the theoretical values of $V=V_g/D\prime=V_g/(1-D)$ and $\eta=1$=100%.

Best regards,
v_c

Points: 2