Boost converter with no load

[shasikumar]

And my converter is operating at 50% duty cycle, under no load condition. So I am not worrying about duty cycle (D), just I want to know what is the response of output voltage (=capacitor voltage), and relation between output and input (under no-load with D=0.5).

 

[MrAl]

Hello again,


Well if you have absolutely no path to discharge whatsoever then the capacitor voltage will increase forever because the inductor gets charged with a finite non zero amount of energy and that will be in the form of a current, and the current will always charge the capacitor even if it has 100,000 volts across it because the inductor acts as a current source. So what is the point, you want the time domain solution to that? For what purpose?

 

[ronsimpson]

Vout=Vin/(1-D) where D is the pulse duty cycle. But there is a limit to D when there are losses in the circuit.

We have talked about with/with-out feedback.
We have talked about with/with-out load.
How about continuous/dis-continuous.

Example: assuming a simple inductor + MOSFET + diode boost supply.

Bottom half of picture. continuous:
Green trace is current in inductor.
Blue trace is gate drive of MOSFET at 50%.
Purple is MOSFET drain voltage at 50%. Red arrow is supply voltage.
The area A must equal B area.
Area A= supply voltage for the time the MOSFET is on.
Area B= the voltage above supply for the time the diode conducts.
In this case the output voltage is 2x the supply voltage with 50%.

Top half of picture. dis-continuous:
Green trace is current in inductor. does not flow all the time.
Blue trace is gate drive of MOSFET at 50%.
Purple is MOSFET drain voltage. Red arrow is supply voltage.
The area A must equal B area.
Area A= supply voltage for the time the MOSFET is on.
Area B= the voltage above supply for the time the diode conducts.
Time C= no power is flowing.
In this case the output is 4x the supply voltage at 50% gate drive.

If you start out with a continuous supply, unload the supply, and watch the voltage go up, (error amplifier removed), (stuck at 50%), the supply will move from continuous to dis-continuous. The voltage will clime. The inductor has energy stored on it and it will unload the energy somewhere. The output capacitor will take the energy and charge up. At some point something will break. Another good reason for error amplifiers.

 

[shasikumar]

Hello MrAL..

Thanks for your reply.. I am doing a project on boost converter based modular inverter. In that the boost converter has to be operated at no-load condition. So before operating the experimental setup i want to know the time domain solution of it. Please help in this.

 

[ChrisP58]

You really don't want to be focused on when it will reach it's overvoltage point based on time. You want to design the control loop such that feedback reduces the dutycycle to a level below which the parasitic load will keep it in at the voltage you want.

If you can't get it to stay below that level with parasitic losses, then you shut it down briefly. But you make the decision based on voltage, not on how much time it has been running.

 

[shasikumar]

So it is not possible to run in open loop?

 

[simonbramble]

Firstly, my website explains how boost converters work:
http://www.simonbramble.co.uk/dc_dc_converter_design/boost_converter/boost_converter_design.htm

Many (but not all) boost converters have 2 feedback loops. One regulates the cycle by cycle current in the inductors. This protects the inductor and FET from over current, but also changes the mode of operation. This mode is called current mode control, as opposed to voltage mode control.

There is another feedback loop (a slower, 'outer' feedback loop) that regulates the voltage. The output voltage is divided down using 2 resistors and fed back to a reference point on the chip. As the voltage increases, this tells the chip to throttle back the current in the inductor, so you don't get overshoot when the output reaches regulation. When the part reaches regulation, the switching stops

This is all assuming you operate with a closed loop. If you disconnect the feedback, the voltage will ramp up and up. the maximum voltage will be determined by the maximum duty cycle the chip is able to operate at (about 90%). The chip will go into continuous conduction mode and in this mode, the duty cycle is given as:
DC = (Vout - Vin)/Vin

See my articles and have a play with LTspice. This will tell you lots about how your circuit is working

 

[MrAl]

Hello MrAL..

Thanks for your reply.. I am doing a project on boost converter based modular inverter. In that the boost converter has to be operated at no-load condition. So before operating the experimental setup i want to know the time domain solution of it. Please help in this.

Hello again,


First, because you have no load at all you can expect the voltage to go up higher than you want it to unless you have feedback, a constant duty cycle is not good.

The equations for each cycle after the switch opens are as follows...

I=(sin(t*w)*sqrt(C)*E)/sqrt(L)-(v1*sin(t*w)*sqrt(C))/sqrt(L)+i1*cos(t*w)
Vout=(i1*sin(t*w)*sqrt(L))/sqrt(C)-cos(t*w)*E+E+v1*cos(t*w)

where
I is the inductor current after the switch opens, we set I=0 and solve for t,
Vout is the capacitor voltage after the switch opens, we use t from above,
i1 is the initial inductor currrent,
v1 is the initial capacitor voltage,
w=1/sqrt(LC),
i1=dt*E/L,
dt=switch 'on' time period.

i1 always equals dt*E/L because that's the current at the end of the switch 'on' cycle.

Now we can run those equations a couple tens of thousands of times, or we can simplify based on the energy transfer between the inductor and capacitor. This leads to:

Vout=sqrt(N)*K

where
N is the number of switch on/off cycles since startup with zero capacitor voltage, and
K=dt*E/sqrt(L*C), where dt is the switch 'on' time period, and E is the battery supply voltage.

From the above simple equation for Vout, we see that because K is a constant no matter how many cycles we allow we always see an increasing voltage, but we do see a decrease in the amount of voltage increase per cycle as the voltage builds up.

For example, with E=10v and dt=1ms and L=500uH and C=5000uf after 40 cycles we see a capacitor voltage Vout of 40 volts and after one more cycle we see Vout=40.6969 which is an increase of 0.6969 volts, but after 400 cycles we see Vout=126.4911 volts and after one more cycle we see 126.6491 which is an increase of only 0.1580 volts, about 1/4 of the previous change, and after 4000 cycles we see Vout=400v and after 4001 cycles Vout=400.049996875391 which is only a change of 0.05 volts now from the previous cycle. After 400000 cycles we see Vout=4000 volts and after 400001 cycles we see Vout=4000.005 volts, a change that is 10 times less than that at around 4000 cycles. Looking at this in time, we see the following:
cycles=40, t=0.080, Vout=40v, change=0.7v
cycles=4000,t=8.0, Vout=400v, change=0.05v,
cycles=400000,t=800,Vout=4000v,change=0.005v (more then 10 minutes)
cycles=430000,t=1660,Vout=4147v,change=0.0048v (another full minute later)
cycles=40000000,t=80000,Vout=40000v, change=0.0005v (more than 22 hours operation)
cycles=40030000,t=80060,Vout=40014.9971885542v,change=0.0005v (another minute later)

So we see as time goes by it takes longer and longer to get a higher voltage of the same increment. At 10 minutes operation we get around 150v more after another minute, but after 22 hours of operation we only get another 15 volts after another minute of operation.
After three months operation being on 24 hours a day 7 days a week, we see around 400000 volts output with only a change of 5e-5 volts per cycle now. But even though that is low, if we keep it on for another three months we STILL see an increase, to 565685 volts.
So although the slope decreases given enough time the voltage always climbs. The limiting factor is how long the unit is run for before it is turned off.

Now of course that is the super ideal case which will never be seen in real life because even the tiny tiniest power loss will limit the output voltage after some number of cycles. This could be a 1 megohm resistor, but even that tiny loss of power will limit the output.

So you can see two things here:
1. You can not run the unit with no load unless it has feedback to limit the voltage, and even then the feedback might not be able to cut the pulse width back enough to limit the output anyway.
2. You need feedback and at least some load

 

[ChrisP58]

So it is not possible to run in open loop?

Possible, but why would you want to, when it is so easy to close the loop and make a much better supply.

 

[ronsimpson]

Part 1:
Using Simon's circuit but with a 555 timer in place of the controller. (set at 50%) no voltage feed back, no current feed back.
I think this is what the original question is about. Frequency = 100khz or 10uS On time=5uS off time=5us.
Supply =10V Inductor value so 10volts for 5uS will cause a current slope of 1A.
Assuming perfect parts: Inductor resistance=0, diode drop=0v, no loss in MOSFET.

At start Vin=10V, Vout=10V
T=0 MOSFET turns on. Current ramps from 0 to 1A in 5uS
MOSFET is off. Current flows into output capacitor. There is 0 voltage across the inductor so the current can not ramp down so the current remains at 1A.
T=10uS MOSFET turns on, Current starts out at 1A and ramps to 2A in 5uS.
MOSFET turns off.
Current flows into output cap. Because the out voltage is very sightly above input voltage there is 0.1V across the inductor and the current ramps down from 2A to 1.99A.
Next cycle current ramps form 1.99A up to 2.99 and back down to 2.97
Next cycle current ramps form 1.97A up to 3.97 and back down to 3.94
Next cycle current ramps form 3.94A up to 4.94 and back down to 4.89
The current builds quickly as long as Vout is less than 2x vin.

At Vout=2x vin the current ramp up=ramp down so
50A ramps to 51A and back down to 50A

At Vin=10, Vout=30 the current ramps down 2x faster then it ramps up.
10A ramps to 11A and down to 9A

At some point the supply will go dis-continuous.
Current will ramp from 0 to1A in 5uS when the MOSFET is on.
Current will ramp down from 1A to 0A in a short time (1uS when Vout = 50V) The remaining 4uS will have no current flow.
From here on a set amount of energy will enter the inductor each cycle and pump up the output like MrAl said.
I am pointing out that with out current feed back the current goes very high and the output cap will charge up faster than what MrAl said.

 

[ronsimpson]

Part 2:
Using Simon's circuit but with a UC3844 controller. (TL3844) (50% max) no voltage feed back, with current feed back.
Frequency = 100khz or 10uS On time=5uS off time=5us.
Supply =10V Inductor value so 10volts for 5uS will cause a current slope of 1A.
Assuming perfect parts: Inductor resistance=0, diode drop=0v, no loss in MOSFET.

At start Vin=10V, Vout=10V
T=0 MOSFET turns on. Current ramps from 0 to 1A in 5uS
MOSFET is off. Current flows into output capacitor. There is 0 voltage across the inductor so the current can not ramp down so the current remains at 1A.
T=10uS MOSFET turns on, Current starts out at 1A, The controller sees the current is at 1A and turns off in 100nS. As fast as it can. Power is put into the inductor for only 1% of the time. Power charges up the output capacitor for 99% of the time. One might think the diode current is a 50% duty cycle current ramp from1A to 0A. But it is 99% and 1A square wave. The output will charge 4x faster. 50% to 99% is about 2x and triangle to square wave is 2x so 4x.

When Vout=15V the current will charge from 0.5A to 1A and current will discharge form 1A to 0.5A. There is more energy in a ramp (1A to 0.5A) than in a ramp (1A to 0A). So the output capacitor will charge faster.

When Vout=20V the current will ramp from 0 to 1 and from 1 back to 0.

When Vout>20V the supply will go dis-continuous.
At Vout=50V
Current will ramp 0,1,2,3,4,5A when the MOSFET is on. Then down from 5A to 0A in 1nS and set at 0A for the next 4uS.

Current feedback keeps things from braking because of too much current.
Voltage feedback keeps things from braking because of too much voltage.