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Boost converter - how it works.

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skyhawk
I've been thinking a lot about your post, and wanted to ask you please,
What guaranty you that the system will reach the point where the amount of current decrease when the switch is open is the same as the increase when the switch was closed - meaning that the system will reach the periodic state?

indulis
Thanks for the article.
I read it and it only explains about steady state, not how you reach steady state which is what i'm trying to realize.
 
skyhawk
I've been thinking a lot about your post, and wanted to ask you please,
What guaranty you that the system will reach the point where the amount of current decrease when the switch is open is the same as the increase when the switch was closed - meaning that the system will reach the periodic state?

indulis
Thanks for the article.
I read it and it only explains about steady state, not how you reach steady state which is what i'm trying to realize.

First we prove by construction that such a periodic state exists. I've discussed it for a simplified situation where the resistance of the inductor and the voltage drop across the switching element are neglected. The article linked by indulis covers the more realistic case as well as continuous vs discontinuous modes.

Then we examine whether this state is stable or not, i.e. when you are away from this state does the system want to return. The key here is the equation:

Ldi/dt = V1 - Vd - Vc

Vc is also the output voltage and controls current delivered to the load, iout = Vc/R. The integral of iout over one cycle plus the change in the charge of the capacitor over the cycle must equal the integral of the current through the inductor while the switch is open. In the periodic state the charge on the capacitor does not change over a cycle.

Now suppose Vc is higher than that for a periodic state, then the integral over the cycle of the current delivered to the load will be higher that the case where Vc corresponds to the periodic state. On the other hand the current through the inductor will decrease more rapidly than in the periodic state with the result that the integrated current through the inductor while the switch is open will be less. Thus the output from the capacitor increases while the input to the capacitor decreases and the capacitor discharges and Vc decreases toward its periodic value.

By a similar argument if Vc is lower than that for a periodic state it will tend to incease toward its periodic value.

editted to add correction in bold
 
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In a switching regulator there is feedback, of course, that detects the output voltage and adjusts the "on" period of the switch to give the desired output voltage, increasing the on-time to increase the voltage and reducing the on-time to reduce the voltage.
 
Look at this:
 

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Vc is also the output voltage and controls current delivered to the load, iout = Vc/R. The integral of iout over one cycle plus the change in the charge of the capacitor over the cycle must equal the integral of the current through the inductor. In the periodic state the charge on the capacitor does not change over a cycle.

First, Thank you alot! I understood this transient stage thanks to you :)
I just wanted to correct you (I think), the integral is over Toff, and not over the whole cycle, since only during Toff: iL(t)=iC(t)+iR(t), isnt it?

In a switching regulator there is feedback, of course, that detects the output voltage and adjusts the "on" period of the switch to give the desired output voltage, increasing the on-time to increase the voltage and reducing the on-time to reduce the voltage.
Is it a must in order to reach a steady output voltage or is it just for decreasing output ripple?
Since there's no feedback in the boost converter.
Morover, according to MikeMl simulation, the output is pretty steady.


Look at this:
Thank you!
excellent:)
 
It surely doesnt come from the capacitor, since the direction of the capacitor current is towards the capacitor, since during the off period, the capacitor gets charged (otherwise the capacitor will never get charged).

That leaves us with the either the inductor or the voltage source as the current source for the load.
How can you know that if Isource = IL?
 
For a stable boost converter, Iinductor = Iload. Isource is going to be higher than Iload as the extra current is used for the voltage boost.
 
First, Thank you alot! I understood this transient stage thanks to you :)
I just wanted to correct you (I think), the integral is over Toff, and not over the whole cycle, since only during Toff: iL(t)=iC(t)+iR(t), isnt it?


Is it a must in order to reach a steady output voltage or is it just for decreasing output ripple?
Since there's no feedback in the boost converter.
Morover, according to MikeMl simulation, the output is pretty steady.



Thank you!
excellent:)

Good catch. The integral of the inductor current is only over the time the switch is open since that is the only time the diode conducts. The integral of the load current is over the full cycle.

Your solution to the differential equation for current is incorrect, but that's not relevant for this discussion.

The simple equation for Vout as a function of Vin and duty cycle is only valid for continuous mode operation. When the load current drops below a critical value the mode of operation becomes discontinuous and the formula for Vout becomes more complex. This is one reason for using feedback. If the "on time" is not reduced for low load currents the Vout will not obey the simple formula and will rise too high. Also the simple formula is only an approximation for the continuous mode, and feedback will permit the controller to adjust the duty cycle so that exactly the desired Vout is produced. The converter will reach a periodic state without feedback, but the output voltage may not be what is desired.

Ripple is reduced by using a larger capacitor and/or a higher frequency. Higher frequencies also permit the use of small value inductors, while maintaining continuous mode operation.
 
Is it a must in order to reach a steady output voltage or is it just for decreasing output ripple?
Since there's no feedback in the boost converter.
Morover, according to MikeMl simulation, the output is pretty steady.
Feedback has no significant effect on ripple. It's to maintain a constant output voltage independent of a varying input voltage or output load.

MikeMl's simulation output is steady because the input voltage and output load are constant. It has reached an equilibrium point for the particular input voltage, output load, and duty-cycle he selected.
 
It surely doesnt come from the capacitor, since the direction of the capacitor current is towards the capacitor, since during the off period, the capacitor gets charged (otherwise the capacitor will never get charged).

That leaves us with the either the inductor or the voltage source as the current source for the load.
How can you know that if Isource = IL?

The voltage source is in series with the inductor so the current through the inductor is the same as the current through the voltage source at all times. When the switch is open this same current flows though the capacitorand load with it splitting and part going through the capacitor and part through the load.

Neglecting losses due to the resistance of the inductor and the voltage drops across the diode and switch, the power supplied by the voltage source must equal the power delivered to the load. Therefore

Vin*Isource = Vout*Iload

or

Iload = Isource*(Vin/Vout)

where Iload is the average current supplied by the voltage source and Iload is the average current supplied to the load.
 
What keeps the cap from overcharging is the feedback loop. If the duty cycle was not controlled based on the output voltage, a boost converter can output voltages much greater than the input.

The output voltage is sampled and fed back, compared to a fixed stable reference, and a pwm signal is generated. The pwm drives the power switch at the right duty cycle to keep the output voltage regulated. Does this help?
 
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Claude, only at low loads. Even modest loads will limit the peak voltage.
 
run that by me again please

Claude, only at low loads. Even modest loads will limit the peak voltage.

Sure it will *limit* the peak voltage, but will it be at a safe value? If a boost converter inputs 12V, and outputs 24V, its duty cycle is 50% steady state. Without a servo loop, if the pwm maxes out at, let's say 90%, then under heavy load in continuous conduction mode, ccm, the output is 120V! At light loads it exceeds that value.

Of course loading heavier tends to reduce the peak output voltage. But it's already high enough to be dangerous to personnel and components.

At all loads up to 100%, the servo loop regulates the voltage. To verify, take a boost converter and short out the resistor in the lower leg of the feedback divider. The pwm will max out to it's upper limit. The output will rise to a value larger than the input.

Vout = Vin / (1-D). If D = 0.90, then VOut = 10* Vin, for continuous conduction mode, ccm. For discontinuous mode, dcm, a lower duty cycle is needed for a given output voltage. In other words, a given duty cycle produces a larger output in dcm.

Trust me, I've been doing SMPS for a long time. I'm not making this stuff up.
 
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