3
\$\begingroup\$

When designing a boost converter to take a low input voltage to a high output voltage, there are many knowns that you control, such as the choice of inductor and the duty cycle of the switching that is at the heart of the boosting voltage. The below circuit is a demonstration of what I'm talking about - no values since its not a question about specifics.

A basic example of the inductor circuit being described.

If you're running the boost converter in discontinuous mode, with a known duty cycle and period (\$D\$ and \$T\$ respectively), input voltage (\$V_i\$) and inductance (\$L\$), you can calculate various important values such as peak current through the inductor. But the sticking point that I keep arriving at is how exactly do you calculate the output voltage over a theoretical load, R? Or, as I'm being led to believe, does the output voltage/current actively depend on the value of the load R?

Much of the literature I read assumes that you know \$V_o\$ and \$I_o\$ already (which when you're designing this, it does make sense to know what your desired output is), but I'm approaching this as if I don't, yet not getting anywhere, since the equation regarding the voltage gain:

\$ \dfrac{V_o}{V_i} = \dfrac{V_i D^2T}{2LI_o} + 1\$

includes both output voltage and output current, which seem to depend on each other from other equations:

\$I_o = \dfrac{I_{max} \cdot \delta T}{2}\$

which if you sub in the equation for \$\delta T\$ and \$I_{max}\$ just results in the first equation. I can't find a way of isolating one variable. Trying to equate the energy across the capacitor similarly isn't leading me in the right direction.

What I don't understand is that for the given four variables at the top of my post, I'm left with the output voltage and current as free variables, which leads me to posit that these then must depend on the load? Is my thinking correct, or do I misunderstand the concept of what is going on? I know I'm going about this the 'wrong' way, it's just a curious thought experiment as I'm trying to make sense of other circuits, where the PWM/duty cycle switching and other components are already set.

\$\endgroup\$
0
\$\begingroup\$

To determine the output voltage of the DCM-operated converter, you have the choice between several approaches such as the PWM switch model, volt-seconds balance etc. A simple one is the volt-seconds balance, assuming the average voltage across the inductor is always 0 V in DCM. To apply this, we can follow the path described here and draw simple waveforms first:

enter image description here

The upper curve is the ideal voltage across the low-side switch while the second one is the instantaneous current in the inductor. Considering a 0-V average value across \$L\$, we can write \$<v_L(t)>=<V_{in}>-<v_{sw}(t)>\$. The average voltage across the switch is obtained calculating the area of the SW curve and stretching it across the switch period: \$<v_{sw}(t)>=D_2V_{out}+D_3V_{in}\$. We know that \$1=D_1+D_2+D_3\$ so extracting \$D_3\$ and plugging into the previous equation leads to \$\frac{V_{out}}{V_{in}}=\frac{D_1}{D_2}+1\$ if you consider \$<V_{in}>=<v_{sw}(t)>\$. If you now consider a 100%-efficient converter and determine \$D_2\$ via the average inductor current, you should find \$M=0.5\times(1+\sqrt{1+\frac{2T_{sw}RD_1^2}{L}})\$. As you can see, yes, if you change the loading resistance or the switching frequency, you will change \$V_{out}\$. The load point at which the converter operates at the border between CCM and DCM is called the critical resistance value: below it you operate in CCM, at the exact value you operate in BCM for boundary-conduction mode and above it, you are in DCM.

\$\endgroup\$
  • \$\begingroup\$ Cheers for this, never considered calculating by taking the area of the top graph, despite having it in my notes. Its good to confirm that the load resistance will change the output voltage, whilst also encapsulating the switching frequency. Cleared up a headache for sure. \$\endgroup\$ – Simon M Dec 5 '17 at 21:53
  • \$\begingroup\$ My pleasure if I could modestly help. The other way is via the PWM switch model derived in DCM. It will quickly give you the dc transfer function including losses if needed. \$\endgroup\$ – Verbal Kint Dec 5 '17 at 22:09
1
\$\begingroup\$

That's the problem — a discontinuous-mode boost converter with no feedback does not have a well-defined output voltage.

Depending on the on time of the switch during each cycle, the value of the inductance and the input voltage, a certain amount of energy is stored in the coil. When the switch opens, this energy is dumped to the output. If the output is only a capacitor, the capacitor voltage will rise indefinitely until something breaks down.

If there's a resistance, the output voltage and current will rise until the average power dissipated matches the input power (energy per switching cycle × switching frequency).

With this type of converter, if the load is variable, then you have to use feedback to vary the input power to match, usually by varying either the duty cycle or the frequency of the switching (or sometimes both).

\$\endgroup\$
0
\$\begingroup\$

Let's take things a pulse at a time, and do away with the output resistor for a start.

Charge the inductor up with current to a certain energy \$E=\frac{1}{2}LI^2\$. When the switch opens, that same energy will be dumped into the capacitor, to give a voltage given by \$E=\frac{1}{2}CV^2\$.

After the next pulse, the energy will be doubled. If you keep on like that, the capacitor voltage will increase with each pulse until something breaks.

However, we have the output load resistor, bleeding voltage off the capacitor.

A converter like this is never used open loop. If the output voltage goes above a threshhold, the pulses are stopped. If it goes below a threshhold, they are restarted. So feedback ensures that enough pulses happen, on average, to keep the output voltage between two limits, whatever load current the load is drawing (up to some maximum of course).

The output voltage ripple is given by the energy of the inductor pulse, the output capacitor size and the output voltage.

\$\endgroup\$
0
\$\begingroup\$

I referred to this book Switching Power Supply Design 3ed. by Pressman, Keith Billings and Morey, because of my experience with K.B. in the mid-80's, who did great work to supply our company with good PSU products for mass production.

Simple DCM Boost Equation (ideal parts)

\$\frac{V_o}{V_i}= T_{on}\sqrt {\frac{R}{2L~T}}\$ ...1

Since the diode current return time ,\$T_r\$ may be longer than the switch on time, \$T_{on}\$ as the diode ESR may be greater than the switch for a switching cycle T = Ton+Tr+Toff and load R with switched inductor L.

During \$T_{on}\$, current ramps to peak, \$I_p\$ and energy per cycle, \$T\$ stored is as follows;

\$P_L=\dfrac{½L I_p^2}{T}\$

After gate off the current ramps down thru the diode as follows;

\$P_{dc}=½V_{i}I_p~T_r/T~~\$ then with \$L*I_p/T_{on}=V_{i}\$ we get; Discontinuous mode (DCM) has a fraction <1 of cycle T when it continuous = k.

\$\frac{V_o}{V_i}=\sqrt{\frac{kRT_{on}}{2L}}\$ ... 2

the negative-feedback loop keeps the output constant against input voltage changes and output load Ro changes in accordance with Eq.(above) As Vin and R (the load current) go down or up, the loop will increase or decrease T on so as to keep Vo constant.

DCM Boost adds some interesting instabilities due to the high impedance Toff period that must be examined.

\$\endgroup\$
  • \$\begingroup\$ And so if Vin or R decreases, through feedback to your controller of choice, T increases, and if you drop either sufficiently you'll increase T to the continuous regime (?) - which could happen if you're running a SMPS off a battery where its voltage drops sufficiently during discharge, I guess (but would require a non-trivial drop in voltage over its charge life...). I should look into some of the literature suggested here, including the book you mentioned. Fascinating stuff! \$\endgroup\$ – Simon M Dec 5 '17 at 22:23
  • \$\begingroup\$ Yes Vi sensitivity in terms of incremental impedance looks like a negative impedance load. As such unless input Cap ESR is really low I once had a buck reg. choke feeding a boost for low current LCD bias and the boost choke went into Chaos Mode and a little SMT part generated enough acoustic noise to sound like running water in a noisy lab like a piezo speaker. \$\endgroup\$ – Sunnyskyguy EE75 Dec 5 '17 at 22:38
  • \$\begingroup\$ The incremental resistance is negative only when the converter is operated in closed-loop control where the input voltage rejection is infinite. As such, the phase of the input impedance is truly 180° in the low-frequency range, when the open-loop gain is high, up to a few tens of Hz. As you approach crossover, the phase quickly becomes that of the open-loop impedance. \$\endgroup\$ – Verbal Kint Dec 6 '17 at 6:37
  • \$\begingroup\$ Yes when regulating, the input current drops when input voltage rises averaged over say 10 cycles, so the step input voltage negative incremental impedance actually is much wider bandwidth. \$\endgroup\$ – Sunnyskyguy EE75 Dec 6 '17 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.