Question

A buck-boost converter has an input voltage V_in= 10 V, switching frequency fs= 100 kHz, duty ratio d= 30%, load resistance R_L= 20Ω, inductance =100 μH, and

capacitance= 680 μF.

1) Calculate the following values, and include units in your answers:

a) Peak-to-peak current ripple through the inductor, Δ_iL

b) Average inductor current that will cause the converter to enter DCM

b) Load resistance that will cause the converter to enter DCM (R_crit)

d) Average voltage across the inductor insteady-state

Answer 1

For the discontinuous mode we can fairly easily equate the input energy to the output power and come up with a value for the output voltage or the duty cycle, but it will be different than for the continuous mode.
For example, the input current peak is:
di=Vin*dt/L

where dt is the 'on' time. We also know that the energy in an inductor is:
Uin=i^2*L/2

Substituting di in for i and computing over the full period we get:
Uin=F*(Vin*dt/L)^2*(L/2)

Replacing F with 1/T where T is the total period we get:
Uin=(dt^2*Vin^2)/(2*L*T)

The output power is:
Pout=Vout^2/R

Equating the input energy to the output power we have:
Uin=Pout
(dt^2*Vin^2)/(2*L*T)=Vout^2/R

Solving for Vout^2 we get:
Vout^2=(dt^2*Vin^2*R)/(2*L*T)

and of course then:
Vout=dt*Vin*sqrt(R/(2*L*T))

Starting again from:
Vout^2=(dt^2*Vin^2*R)/(2*L*T)

and since dt^2/T^2 is D^2 we can write:
Vout^2=(D^2*Vin^2*R*T)/(2*L)

and solving for D^2 we get:
D^2=(2*Vout^2*L)/(Vin^2*R*T)

and for D:
D=sqrt((2*Vout^2*L)/(Vin^2*R*T))

Here we see that the duty cycle calculation is more involved than with the continuous mode converter. We have dependence on both L and R and also the total period as well as before when we had dependence on only Vout and Vin.

An example would be for Vin=12v and Vout=6.23v (or -6.23v), and L=100uH and R=60 and T=10us.
The value for D then comes out to:
D=0.29974

and here we see with almost the same duty cycle as before we are now getting a voltage output of over 6v when before we had something around 5v. That is typically what happens with R too big.

I think what we can do next is substitute the old duty cycle into the new formula and solve for R, which will then be the maximum value of R to maintain continuous mode operation.

Keep in mind these calculations, as are many of the basic calculations for converters, based on ideal elements.

The MOSFET peak current is based on:
v=L*di/dt

again, and rearranging we have again:
di=Vin*dt/L

so the peak MOSFET current is Vin*dt/L, but that is only if the converter has a slow start mechanism. If not, the initial current peak could go much higher.

The MOSFET voltage could be estimated from the specs of the MOSFET, that is, the Ron spec. If you know the peak current (as above) then you know the peak voltage, at least as an estimate. The power dissipation however depends on both of those plus the switching losses. The switching losses are often ignored in the basic theory calculations with ideal elements, but are usually very important in all but the lowest frequency real life practical converters.