Boost converter transfer function with non-ideal components

Thread Starter

J_Rod

Joined Nov 4, 2014
109
Hi, I am stuck trying the derivation for a boost converter transfer function in continuous conduction, open-loop operation mode to find Vout/Vin with non-ideal components, that is, a Rsw, the MOSFET switch on-state resistance, ESR, an equivalent series resistance with the capacitor, RD, the on-state resistance of the diode, and RL, the inductor's resistance.

When the switch is ON:
\( L \frac{di_L}{dt} = V_{in} -i_L (R_L +R_{DS,ON}) \)

\( C \frac{dV_C}{dt} = \frac{-V_{out}}{R + ESR} \)

\( C \frac{dV_C}{dt} = \frac{V_C - V_{out}}{ESR} \)

When the switch is off:
\( L \frac{di_L}{dt} = V_{in} -V_{out} -i_L(R_L + R_D) \)

\( C \frac{dV_C}{dt} = i_L -\frac{V_{out}}{R +ESR} \)

\( V_C = V_{out} -i_CESR \)

Then
\( \Delta i_{Lon} + \Delta i_{Loff} = 0 \)

But to get \( \Delta i_L \) terms I would have to integrate the equations with the inductor voltage, which both contain a term with \( i_L \). How can that be done? Thanks.
 

wayneh

Joined Sep 9, 2010
16,155
Had diffy Q yet? I believe that's a first order differential equation. I'm pretty rusty on the terminology. It's been a long time but I believe almost any text on the subject will cover the solution.
 

Thread Starter

J_Rod

Joined Nov 4, 2014
109
Had diffy Q yet? I believe that's a first order differential equation. I'm pretty rusty on the terminology. It's been a long time but I believe almost any text on the subject will cover the solution.
That's how to solve for \( i_L \), but I want to find the expression for \( \Delta i_{Lon} \) to add to the expression for \( \Delta i_{Loff} \) to sum to 0, and then rearrange to find \( \frac {V_{out}}{V_{in}} \). Would it be incorrect to set \( i_L = I_L \), the average value, on the right side of the equations, if I assume a small ripple?
 

wayneh

Joined Sep 9, 2010
16,155
Well if you have i as a function of time, can't you integrate to get ∆i ? (I'm just hoping that asking questions helps you see a solution. I haven't worked this out.)
 

anhnha

Joined Apr 19, 2012
880
Hi, I am stuck trying the derivation for a boost converter transfer function in continuous conduction, open-loop operation mode to find Vout/Vin with non-ideal components, that is, a Rsw, the MOSFET switch on-state resistance, ESR, an equivalent series resistance with the capacitor, RD, the on-state resistance of the diode, and RL, the inductor's resistance.

When the switch is ON:
\( L \frac{di_L}{dt} = V_{in} -i_L (R_L +R_{DS,ON}) \)

\( C \frac{dV_C}{dt} = \frac{-V_{out}}{R + ESR} \)

\( C \frac{dV_C}{dt} = \frac{V_C - V_{out}}{ESR} \)

When the switch is off:
\( L \frac{di_L}{dt} = V_{in} -V_{out} -i_L(R_L + R_D) \)

\( C \frac{dV_C}{dt} = i_L -\frac{V_{out}}{R +ESR} \)

\( V_C = V_{out} -i_CESR \)

Then
\( \Delta i_{Lon} + \Delta i_{Loff} = 0 \)

But to get \( \Delta i_L \) terms I would have to integrate the equations with the inductor voltage, which both contain a term with \( i_L \). How can that be done? Thanks.
There are some principles you should use here to solve for the result:
1. Inductor volt-second balance
2. Capacitor charge balance
3. Small ripple approximation
Refer to this for detail.
With small ripple approximation, you can assume that Vout and iL are constant (and let's call it IL).
With #1 and #2, you have set of two equations and you can now solve for Vout and IL.
But to get
terms I would have to integrate the equations with the inductor voltage, which both contain a term with
. How can that be done?
With small approximation, iL in the right hand side can be considered as a constant.
However, if you want an exact calculation, then I think it is much easier to use Laplace transform.
 
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