Electronic – Can someone please help me with these questions on DC Chopper Circuits

The continuous mode boost and buck-boost converters exhibit control-to-output transfer functions \$G_{vd}(s)=\hat{v}(s)/\hat{d}(s)\$ containing two poles and one RHS (right half-plane) zero, called zero of nonminimum phase.

Your original transfer funtion is:

$$G_{vd}(s)=\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\times \frac{1/\underline{L}C\times\left ( 1-s\left ( \underline{L}/R \right ) \right )}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$

Starting from a simpler function (without the RHP zero), named \$ G_{vd}'\$:

$$G_{vd}'=\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\times \frac{1/\underline{L}C}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$

Or, placed as a standard second order t.f:

$$G_{vd}'=K_{DC}\times \frac{1/\underline{L}C}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$

where the DC gain is \$K_{DC} =\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\$.

The equation can be rewritten as:

$$G_{vd}'=K_{DC}\times \frac{1}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }$$

With \$\omega_0=1/\sqrt{\underline{L} C}\$ and \$\omega_0Q=R/\underline{L}\$

Similarly, the original transfer function can be expressed as (RHP zero included):

$$G_{vd}=K_{DC}\times \frac{\left (1-\frac{s}{\omega_{RHP}} \right )}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }$$

The response will be:

$$\hat{v}(s)= \frac{K_{DC}}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1}\times\hat{d}(s) - \frac{K_{DC}/\omega_{RHP}}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }\times\hat{d}(s)\times s$$

The response of the original system is the sum of two components: The first is equivalent to the modified system response (without the zero) and the second is the derivative (scaled) of that one. For the case of a stable system with a step input in \$t = 0\$, this last component will have substantial influence at the beginning and then will vanishes when \$t\rightarrow\infty \$. Note the negative sign leads to a momentary opposite effect in output (nonminimum phase).

UPDATE:

The presence of zero RHP in the model is explained as follows: For the output voltage to increase, the duty cycle must be increased in such a way that the inductor will be disconnected from the load for a long time, causing the output voltage to drop (i.e. in the opposite direction to the one desired). The controller must be designed to meet the project requirements and avoid oscillations while maintaining duty cycle below an undesirable 100% - being limited by the PWM integrated circuit itself.

With a boost converter you need to transfer energy to the load by charging an inductor with energy and then releasing that stored energy. In CCM, inductor current never falls to zero so the energy transferred is dependant on the peak current and the min current.

What the current peaks at during charging represents the maximum inductor stored energy and, what the current drops to (during load replenishment) represents the energy left in the inductor just as the cycle repeats.

The difference in energy is what is "given" to the load.

For a fixed value inductor and input power supply voltage, the rate at which current linearly rises (di/dt) is constant and depends entirely on V = L di/dt, that well-known formula. I'm assuming perfect lossless components of course!

So, Imax, for a given operating frequency will always end-up at some fixed-value above Iaverage and, Imin will be the same fixed-value below Iaverage. We could call that fixed value Ipeak.

So, energy given to the load, W is: -

\$\dfrac{L}{2}[I_{MAX}^2-I_{MIN}^2]\$

Re-arranging using \$I_A\$ and \$I_P\$

W = \$\dfrac{L}{2}[(I_A+I_P)^2-(I_A-I_P)^2]\$

where \$I_A\$ is average current and \$I_P\$ is the peak above (or below) \$I_A\$. You also might be able to get to this formula (hopefully): -

W = \$2L\cdot I_A\cdot I_P\$

This means that both the average current and the peak current determines the energy transfered to the load. But, for a given input supply voltage, operating frequency, duty cycle and inductor value you cannot control \$I_P\$.

So, if the load resistance increases in value BUT you wanted to keep the average output voltage the same, the only option (other than D) is to increase operating frequency to reduce the peak current attained by the inductor during charging. This of course reduces average current gradually over a few cycles and what you find, in some controllers, is that the average current becomes lower and the frequency slews back to the original value.

More load current means lower frequency, less load current means higher frequency.

At the end of the day, I believe you still need a "clever" controller (to alter frequency) so I'm not sure that this question and the optional answers have much bearing on the practical world.

That's the way I see it anyway. Good (but flawed) question!!