Given :
Buck-Boost converter

Please look at the Image that I have attached also.
1.Input Voltage (Vg) : 12V
Output Voltage (V): -5V

I have obtained the Duty cycle formula for this converter as: V= -(DVg)/(1-D)

When I substitute for Input and Output voltage, I get the answers as D=0.2941 which is not correct apparently.

Please tell me where I am wrong.
Please tell me whether the formula of Duty cycle for this converter is correct.
2. And I am also supposed to calculate the current through the MOSFET and Voltage across it.
Please help me on how should I approach for the calculations of Current through the MOSFET and Voltage across it.
Thanks.

 

I have obtained the Duty cycle formula for this converter as: V= -(DVg)/(1-D).

So you think the output voltage is independent of the the size of L,C and the load? Does that make sense to you?

If you are an EE student, the first step is to derive a formula for the duty cycle of this circuit.

 

Well in theory for ideal component use in the circuit, when the inductor work in CCM mode Vout is "set" only by Vin and duty cycle.

 

Well in theory for ideal component use in the circuit, when the inductor work in CCM mode Vout is "set" only by Vin and duty cycle.

Thanks, I had forgotten that.

 

Given :
Buck-Boost converter

Please look at the Image that I have attached also.
1.Input Voltage (Vg) : 12V
Output Voltage (V): -5V

I have obtained the Duty cycle formula for this converter as: V= -(DVg)/(1-D)

When I substitute for Input and Output voltage, I get the answers as D=0.2941 which is not correct apparently.

Please tell me where I am wrong.
Please tell me whether the formula of Duty cycle for this converter is correct.
2. And I am also supposed to calculate the current through the MOSFET and Voltage across it.
Please help me on how should I approach for the calculations of Current through the MOSFET and Voltage across it.
Thanks.

Hello there,

Your wording troubles me just a little: "which is not correct apparently".

My question is, how did you determine that was not the right formula?
I ask because i am curious but also because that looks right to me.
I have to know what you are doing in order to determine what is "not correct" to you though in order to see what you are doing, which is apparently not exactly correct. We can then look into this more.
Boost converters are a little more tricky than bucks. If i had to guess, i would guess that you are running in discontinuous mode and that's not good When you jump from theory to the practical you have to take other things into consideration especially if the circuit is sensitive to certain variations.

Just to note, Wikipedia gives the same formula slightly rearranged:
Vo/Vi=-D/(1-D)

 

So, Is the answer for my Duty cycle = 0.29411 or 29.41% , correct?
And,
How do I approach for the calculations of MOSFET current and voltage?

 

So, Is the answer for my Duty cycle = 0.29411 or 29.41% , correct?
And,
How do I approach for the calculations of MOSFET current and voltage?

Hello again,

Basically yes, that duty cycle is correct, but only when the converter is running in the continuous mode.
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.

 

Hello again,

Basically yes, that duty cycle is correct, but only when the converter is running in the continuous mode.
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.

And 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.

Thank you so much. You couldnt have explained better. Very clear.