# why Boost converter output voltage decreases at high power?

I am implementing non-isolated Boost Converter, capable of handling a power of 1Kw and output voltage is to be maintained at 400V at an input voltage of 120V . At smaller load current (around 2.5 Amp input, 0.8Amp output current) the output voltage across the capacitor stabilizes and it attains a stable value with a small ripple. But as I increase the load current (4.5 Amp input, 1.43Amp output current) the output voltage across the capacitor starts discharging fastly and never attains a stable voltage level. I am using two 3300uF capacitors rated at 350V each, connected in series at the output. I don't understand why the capacitor discharges as load current is increased?

the schematic for boost converter is attached here .

• Where's your sense point for the feedback loop? When you say it doesn't stabilize what do you mean? Is there an oscillation on the output or does it just droop? Can you post a schematic? – John D Apr 7 '16 at 18:08
• Output voltage goes down continuously at high power. and it is just a simple boost converter – Misal313 Apr 7 '16 at 18:10
• i attached the schematic of boost converter in the question. @JohnD – Misal313 Apr 7 '16 at 18:13
• By "just a simple boost converter" do you mean that there's no feedback, it's just running open loop? How are you setting or controlling the duty cycle? – John D Apr 7 '16 at 18:15
• What is the value of the inductor, and what is the switching frequency? It sounds like you're not pumping enough energy into the inductor on each switching cycle. – Dave Tweed Apr 7 '16 at 18:26

If good regulation of a DC-DC converter is necessary, the typical method to achieve it is to use feedback. This effectively reduces the output impedance of the supply so that it doesn't vary with load current.

An open-loop converter operating at a fixed duty cycle has a much higher output impedance due to the resistance of the switches, the DCR of the inductor, the resistance of the traces, the source resistance of the input supply, and the drop across the diode.

So droop of the output voltage with load current is expected and normal.

• ok.. thank you @johnD for response . can you tell us about the feedback .. how can we design feedback resistor . ? – Misal313 Apr 7 '16 at 18:25
• will be any limit of duty cycle variation fro duty cycle – Misal313 Apr 7 '16 at 18:26
• how we limit the duty cycle during feedback ? will it go above the 70% when output voltage will increase.? @johnD – Misal313 Apr 7 '16 at 19:29
• @misal313 This site does not operate like a conventional forum. Comments are not the place for a multitude of ancillary questions. Either edit your original question or pose new questions, but bear in mind that questions like "how do I design a regulated boost converter" tend to be closed as being too broad. – Adam Lawrence Apr 7 '16 at 19:49

At a switching frequency of 15 kHz, you need to put

$$\frac{1000 W}{15 kHz} = 66.7 mJ$$

of energy into the inductor on every switching cycle. With a 4 mH inductor, this would be a peak current of

$$\sqrt{\frac{2 \cdot 66.7 mJ}{4 mH}} = 5.77 A$$

However, when you apply 120V to that inductor, the current ramps at a rate of

$$\frac{120 V}{4 mH} = 30 A/ms$$

but you're only giving it

$$\frac{0.68}{15 kHz} = 45.3 \mu s$$

to charge, which only lets it get to

$$45.3 \mu s * 30 A/ms = 1.36 A$$

Therefore, with or without feedback, this design is incapable of converting 1000W.

BTW, at these power levels, I would be looking at building a multi-phase converter, with 3 or 4 sets of coils and switches. They all operate at the same duty cycle, but with staggered timing. This makes the input and output ripple a lot easier to deal with.

EDIT: The above analysis is for DCM (discontinuous conduction mode) only. The converter can be made to work with those parameters in CCM (continuous conduction mode), as shown by the simulation below. (If you run it, note that it takes a few hundred milliseconds for the startup transient to die out. I added R2 to help tame that.)

simulate this circuit – Schematic created using CircuitLab

The key thing to note here is that the peak current in the coil is almost 13 A, so its saturation current rating must be comfortably above that. The switch and the diode need to be able to handle this current, as well.

• now , what we have to do. ? – Misal313 Apr 7 '16 at 18:50
• You have to pick a different coil and/or a different switching frequency. Or, as I just added to my answer, turn it into a multi-phase converter. – Dave Tweed Apr 7 '16 at 18:52
• Can you please explain how did you come up with the conclusion that the converter is incapable of converting 1000W based on your calculation? What's the current value of 1.36Amp telling us? – Misal313 Apr 7 '16 at 18:58
• Increasing the frequency alone is not enough. This inductor simply will not work with the input voltage, output voltage and power level you've specified. How did you come up with it in the first place? – Dave Tweed Apr 7 '16 at 19:05
• I have extended my analysis to include CCM, which should actually work. But note the caveats about the current levels in the various components. Are your components beefy enough? – Dave Tweed Apr 7 '16 at 22:11