Boost converting not working correctly in LTspice.

Thread Starter

ANewYorker

Joined Feb 12, 2024
7
I am not sure what is wrong with my boost converter. Normally, at 0.5 duty cycle, Vout should be twice Vin. But somehow it is much higher. For my switch, Vto=4.99V (Is Vto equal to Vthreshold? As in switch turned on when Vgs>Vthreshold). With gate voltage varying between 0 and 24V, Vout would be 700V. When gate voltage varied between 0 and 9V, Vout would still go to 700V but much more slowly.

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crutschow

Joined Mar 14, 2008
34,455
A buck converter has a theoretical output proportional to the duty-cycle and independent of load, but not a boost converter.
It's output is determined by the duty-cycle and the load.

And where did you get the idea to vary gate voltage?
That Vgs voltage should always be high enough to fully turn on (well above the Vgs threshold volage, as given in the data sheer where the Rds on-resistance is stated).
You vary the duty-cycle of the pulse, not the gate voltage.

You need to study more about boost converters, and using MOSFETs as switches, as you knowledge is clearly deficient.
 

Thread Starter

ANewYorker

Joined Feb 12, 2024
7
A buck converter has a theoretical output proportional to the duty-cycle and independent of load, but not a boost converter.
It's output is determined by the duty-cycle and the load.

And where did you get the idea to vary gate voltage?
That Vgs voltage should always be high enough to fully turn on (well above the Vgs threshold volage, as given in the data sheer where the Rds on-resistance is stated).
You vary the duty-cycle of the pulse, not the gate voltage.

You need to study more about boost converters, and using MOSFETs as switches, as you knowledge is clearly deficient.
Actually buck converter output is also affected by the load current.
 

Papabravo

Joined Feb 24, 2006
21,225
The 10K load means a very small load current and the default diode model is definitely sub-optimal. Use a real diode instead, preferably a Schottky diode. I agree that Vgs should be limited to 2 to 2.5 times Vgs(th) and should not exceed 20V per the datasheet.
 

Ian0

Joined Aug 7, 2020
9,826
For both buck and boost converters, output will be proportional to input and duty cycle whilst it is in continuous conduction mode, and the value of the inductance only affect the ripple current.
In discontinuous mode the energy transferred (not the voltage) is proportional to duty cycle (and to inductance). Therefore, with no load, the output voltage increases indefinitely (or until it enters continuous current mode)
 

Papabravo

Joined Feb 24, 2006
21,225
Is this closer to what you had in mind?

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The design procedure is given in Hart, Daniel W., Power Electronics, McGraw Hill, 2011, §6.5, The Boost Converter pp. 211-220
 
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Thread Starter

ANewYorker

Joined Feb 12, 2024
7
The 10K load means a very small load current and the default diode model is definitely sub-optimal. Use a real diode instead, preferably a Schottky diode. I agree that Vgs should be limited to 2 to 2.5 times Vgs(th) and should not exceed 20V per the datasheet.
I changed diode to Schottky. I think no matter how I change Vgs, Vout is always around 700V. So The formula that Vout=Vin/(1-D) doesn't really work.
 

Papabravo

Joined Feb 24, 2006
21,225
I have added Vg, which is equivalent to Vgs, because the source of the MOSFET is at GND potential.

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You can clearly see the sharp edges that switch the gate between +12V and GND. This is what you do in all SMPS implementations. Look at the relative values of my inductor, capacitor, and resistor and ask yourself why my design is close to the original goal and yours is not. If your answer is that I designed it that way and I know what I am doing, you would be correct.
 
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Papabravo

Joined Feb 24, 2006
21,225
I don't have any idea if 250 mA is anywhere near the current you had in mind. If you tell me what you had in mind, I can refine the design and consider other things like closing the loop and soft startup.
 

pwrtrnx

Joined Feb 1, 2024
20
Just to answer the original question - your ckt is ringing at turn off - the top of the ring waveform is peak rectified by the output diode - causing the high Vout ( look at the waveform across the mosfet ) - if the load were heavier the Vout would drop. Also, the high duty cycle and light load will allow the volts to trend up - as you are seeing. The energy built up in the inductor every cycle has to go somewhere ! - if the load is too light the Vout will go up until the power in the load = the power processed by the booster !

Also, tie the gate to the source and start the sim, you will see a ringing up of Vin to Vout, this can be as high as 2 x Vin on Vout - for low loss components - some thing to be careful about when applying starting power to any booster.

There are some great examples above on how to implement a booster properly - fixed D on the gate is not that great - as the volts go up with load R ( and vice versa ) - a proper control loop is needed - or at least a volt limit to cut the gate drive when Vout target is reached.
 

pwrtrnx

Joined Feb 1, 2024
20
Also for the booster, delta I in the L = dV / dt, you have 100 volts applied to 0.1uH for 2uS - this is 2kA built up in the inductor over the 2uS ( V/L = di/ dt ) - so it is no surprise the Vout is rising to absorb this energy.
 

Thread Starter

ANewYorker

Joined Feb 12, 2024
7
Is this closer to what you had in mind?

View attachment 315095

The design procedure is given in Hart, Daniel W., Power Electronics, McGraw Hill, 2011, §6.5, The Boost Converter pp. 211-220
I don't have any idea if 250 mA is anywhere near the current you had in mind. If you tell me what you had in mind, I can refine the design and consider other things like closing the loop and soft startup.
I had no idea a booster converter could regulate current. Thanks for showing the example and the textbook.
 

Papabravo

Joined Feb 24, 2006
21,225
I had no idea a booster converter could regulate current. Thanks for showing the example and the textbook.
I think you misunderstand. This basic design example is operating in an open loop fashion. There is no feedback and no actual regulation. If you look at the current waveforms you will see that there is significant current ripple in the inductor around the operating point with a FIXED load. It is true that the output current ripple in the fixed load is much less than the inductor ripple. To make a practical power supply you would need to monitor the output voltage and have a mechanism to vary the duty cycle which will vary the output voltage. I have updated the example with some measurement instructions. Don't take this too seriously because we are using ideal components. A more realistic example would show higher losses and the actual efficiency would likely be between 85% to 94%.

It may also be worth mentioning that the components may or may not survive the actual circuit conditions. The simulator will not tell you that a component is being stressed.

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BobaMosfet

Joined Jul 1, 2009
2,113
Just adding this- No buck/boost converter will operate correctly without a load it was designed for. If you haven't, add an appropriate load on it and see.
 

Thread Starter

ANewYorker

Joined Feb 12, 2024
7
@Papabravo just wondering, for continuous mode operation, does inductor have to be always positive? The textbook says the minimum inductance for continuous mode is \[ \frac{D(1-D)^2R}{2f} \]. For D=0.5, R=787 and f=250e3, that is equal to 0.2mH. But when I put 0.25mH for the inductor, I get a negative current. Do you know what went wrong? Vout is around 100V, which is correct.

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