I want to design discrete boost converter with 5 volt input and 24 volt output with load current of 1 amp. How do I choose suitable input and output capacitor and inductor values; what formulas are used to select those values.
Electrical – Boost converter component selection
boostcapacitorcomponent-selectionformula-derivationinductor
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Overly simple explanation alert!!
In very simplistic terms, a boost converter circuit looks like this: -
To the left is the power source (48 volts) and to the right is your load. In between is a switch (usually a MOSFET) that "shorts" the inductor to 0V for a short period of time then the switch goes open circuit.
The inductor will accumulate energy during the period it is grounded and, when the switch goes open circuit, that energy is released into the capacitor and load via the diode. This process cycles at many, many times per second.
My explanation is a bit of a simplistic approach but, if you look how energy is accumulated and transferred, it should help you understand how the values of L and C are calaculated. Using your numbers, you have a load that needs to have 100 volts across it. Say that load is 1kohm - this means it needs to be "continuously" fed 10 watts to maintain the 100 volts.
But, if the switch permanently remained open circuit (and the diode were perfect), the load inevitably receives a constant 48 volts (a power of 2.3 watts). So, the power needed to be generated by switching-action is not 10 watts but about 8 watts i.e. there is a constant background power of 2 watts feeding the load and propping it up. This is a tad simplistic because maybe up to 50% of the time the inductor is shorted out thus not delivering that "background" power to the load. You could make an argument for assuming the background power is more like 1 watt on a 50:50 switching cycle. Moving on...
Let's say the switch operates 10,000 times per second - that tells us that the energy transferred per switch operation is: -
Energy per cycle is \$\dfrac{8W}{10,000}\$ = 800 \$\mu\$J.
Let's say the inductor is 100 \$\mu\$H and see how the numbers stack-up. The energy contained in an inductor is: -
Energy = \$\dfrac{L I^2}{2}\$ so, to get 800 \$\mu\$J, a current of 4 amps is needed.
How long does the switch need to be closed for to get 4 amps through the inductor? The switch closes and the current ramps up at a rate determined by the power source voltage and the inductor's value: -
V = \$L\dfrac{di}{dt}\$ is the formula to use.
Power voltage is 48 volts and this divided by 100 \$\mu\$H = 0.48 amps per \$\mu\$s. We need 4 amps so this means the switch needs to be "on" for about 8.3 \$\mu\$s. This informs us that the duty cycle is about 8.3% (10 kHz switching frequency).
Remember, this was just me throwing numbers together to help you follow the process and I think that a more appropriate duty cycle would be closer to 50% - this is just a simplistic look at a simple boost converter.
If a 1mH inductor was chosen, a peak current of 1.265 amps is needed to create a stored energy of 800 \$\mu\$J. V/L implies a current rate of 48mA per \$\mu\$s and therefore the on period of the switch needs to be about 26 \$\mu\$s.
How big should the output capacitor be? This is primarily a question of controlling the ripple voltage across the load. Let's say that ripple should be 1V p-p - a simple approximation is to assume the load always takes a constant 0.1 amps i.e. there is 100 volts across a 1kohm resistor. The voltage droops in the 26 \$\mu\$s that the inductor is being charged so knowing that....
I = C \$\dfrac{dV}{dt}\$ we can see how much capacitance is needed.
I = 0.1 amps and dv/dt is 1 volt per 26 \$\mu\$s. Hence C = 2.6 \$\mu\$F
Remember, this is a very simplistic look at how a theoretical boost converter would work.
So here is my take, since this application is microcontroller based, couldn't one simply just stop the discharge of the inductor when the current nears zero?
Think about this - the only way to stop the "discharge current" is to prematurely ground the transistor in the boost device to start "charging" current thru the inductor - you have no option - to keep the inductor open circuit is to enter DCM and that is what you are trying to avoid.
The cycle for a boost converter (or flyback converter) can be: -
- Ground the inductor thus current ramps up and inductor stores energy (charge)
- Un-ground the inductor - energy gets released to the output cap and load (discharge)
- If inductor can't sustain current into load via diode you enter DCM and basically the inductor becomes open circuit except for the parasitic capacitance of the MOSFET switcher i.e. you get a damped oscillation until...
- The cycle begins again.
OK, so then what happens (in that cycle) is a tiny little too much energy becomes transferred to the output capacitor and load (during inductor discharge). This causes the output voltage to rise fractionally higher than what it would and, over a period of a few milli seconds, you might exceed the output voltage that is safe for the load. A few seconds later and you have a dead load and a few seconds later you have a dead boost regulator.
OK before you get to this, the control loop would probably have implemented cycle skipping but cycle skipping is noisier than ordinary DCM so why bother?
BTW, going into DCM isn't that bad - the line regulation of the output isn't as good but it's still quite controllable. There isn't much on the web about this but, due to the self-resonance of the inductor and MOSFET drain capacitance, once DCM is entered there is a small oscillating current in the inductor that is asynchronous to the PWM and when the inductor restarts charging it does so at sometimes a slightly positive or negative current. This can cause the noise on a simple controller such as a fixed-on-period controller.
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Best Answer
You are overly enthusiastic.
This is pushing the envelope for a boost. In fact, for a boost to do that, it would require an injection of steroids which would put the olympic athletes from the pinnacle of Soviet era to shame.
You'd only do that if you had no choice. So, the solution to your problem is most likely to put the 5V power supply on the shelf for another project, replace it with a 24V supply, and if you still need 5V, use a buck converter for that.
The reason is that inductor current in a boost converter is proportional to Iout*(Vout/Vin) and all losses (like I2R losses) depend on that.