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Boost converter with no load

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MCG

New Member
Hi,
I got a little bit confused in this boost converter-related matter, can you help me out?

The question is can a boost converter work if the load resistor is missing from its output?

I thought the answer is no because the capacitor cannot discharge while the transistor is on, so the voltage on it just keeps building up forever. But I built the circuit on a breadboard and it worked, I could control the output voltage level by setting the switching frequency and the voltage ripple on the capacitor suggested it does discharge somehow. But how?

If infinite load resistance ( no load ) on a boost converter cannot work, how can we specify the maximum load resistance by which it can still work? ( even discontinuously )

Thank you for your time
 

smanches

New Member
edit: deleted for being stupid.
 
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crutschow

Well-Known Member
Most Helpful Member
Your converter is probably going into a discontinuous mode when the voltage increases above the control voltage under no load. The output cap then discharges through the load provided by your multimeter or scope. (Remember, it's difficult to measure something without disturbing it slightly by the measurement itself).
 

dknguyen

Well-Known Member
Most Helpful Member
Boost converters destroy themselves if there is no load (or if the load is too small) since the capacitor continues to accumulate charge from the inductor until the voltage gets high enough to cause a component to fail.

The minimum load (or maximum load resistance, as you put it) where a boost converter will run properly is the resistance that drains charge/energy from the output capacitor at the same rate as the slowest rate that boost converter can add to the capacitor (ie. when running at minimum duty cycle).

A higher resistance load drains energy more slowly from the capacitor and the boost converter cannot reduce it's charge rate to match. This causes charge to accumulate in the capacitor faster than it is drained which eventually causes the capacitor voltage to equal the inductor peak voltage. When this happens the capacitor voltage continues to build up and the converter can't do anything about it sinc eit's already running at minimum duty cycle. The voltage continues to build up until something gives- usually as a spark across the switch or capacitor membranes causing damage.
 
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MrAl

Well-Known Member
Most Helpful Member
Hi,

I have to agree that most boost converters are made to operate with at least
some load connected to the output. There are some however that do switch
modes (one for example goes into discontinuous mode) but if you dont know
that before hand you could very well blow out the output transistor when the
voltage shoots way up.
The basic theory behind these guys is that the inductor charges up with a current
when the output transistor pulls one end of the inductor to ground, and once
that transistor shuts off, the voltage shoots up high and causes current flow
in the capacitor. With cycle after cycle of on/off states the capacitor voltage
builds with no limit until something blows, and that is usually the output transistor
collector (or drain) because it's voltage rating is exceeded. Once that happens
the circuit has to be repaired before it will work again.
 
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Hero999

Banned
Boost converters destroy themselves if there is no load (or if the load is too small) since the capacitor continues to accumulate charge from the inductor until the voltage gets high enough to cause a component to fail.
It depends on the controller, that will not happen if it can work down to 0% duty cycle.
 

dknguyen

Well-Known Member
Most Helpful Member
It depends on the controller, that will not happen if it can work down to 0% duty cycle.
I then proceed to say the minimum load is dependent on the minimum duty cycle that the converter is capable of ;)
 

smanches

New Member
I think everyone is assuming a non-regulated boost converter. I have yet to see any SMPS controller that could not go down to 0% duty, only the upper end has limitations. At 0% duty on a boost, the load sees Vcc.
 

MrAl

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Hi,

I, for one, was not assuming anything like that and even stated a particular
mode of operation that would prevent destruction :)

For some though the device will blow out. Even a regulated converter can
blow out if that regulation depends on something other than output voltage.
There are a number of LED boost converter chips that will blow out because
they depend on at least some load being connected, and they dont actually
measure output voltage for feedback.
 
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dknguyen

Well-Known Member
Most Helpful Member
Hmmm that's a good point. I've never noticed the relation between minimum load requirement and regulated vs unregulated. I just always run a boost converter with a minimum load.
 
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smanches

New Member
For some though the device will blow out. Even a regulated converter can
blow out if that regulation depends on something other than output voltage.
There are a number of LED boost converter chips that will blow out because
they depend on at least some load being connected, and they dont actually
measure output voltage for feedback.
Never thought about the fact they might not regulate voltage. In that particular case it could just be looking at the current?
 

dknguyen

Well-Known Member
Most Helpful Member
Boost converters always look at inductor current as part of their control loop. If it's regulated then they look at output voltage too. I've never seen one that uses output current as part of it's control loop and that's what you would need to get around the minimum load problem.
 
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MrAl

Well-Known Member
Most Helpful Member
Never thought about the fact they might not regulate voltage. In that particular case it could just be looking at the current?
Hi again,


As dknguyen said, some boost converters look at the current through
the inductor to be able to reach two goals at the same time that
can be a problem in boost converters:
1. To protect the output transistor from too much currrent.
2. To indirectly regulate the output power.

There are numerous variations however. When the converter is to
be used for voltage applications usually the voltage is monitored.
When the converter is to be used as a current source the current is
monitored. Of course some use both, voltage regulation with current
limiting.
 

MCG

New Member
Thanks for your replies.

Your converter is probably going into a discontinuous mode when the voltage increases above the control voltage under no load. The output cap then discharges through the load provided by your multimeter or scope. (Remember, it's difficult to measure something without disturbing it slightly by the measurement itself).
I wanted to prove this theory by driving my boost converter on breadboard with no load resistor, to find out which component would be destroyed first. But what happened is that the output voltage wouldnt grow without limit, instead it settled to a final ( quite high ) DC value which pretty much depended on switching frequency. I couldnt destroy any of my components no matter how hard I tried :)
If the capacitor discharged through the oscilloscope probes, something souldve blown after I removed the probes, dont you think?

Do you have an idea what might save my circuit or what I might be doing wrong?
 

dknguyen

Well-Known Member
Most Helpful Member
We can't really tell without looking at the schematic.
 
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shasikumar

New Member
Boost Converter under no-load condition

Please help me...

Consider an ideal boost converter, what will be the output voltage under no load condition, with 50% duty cycle. What is the a mathematical expression for the voltage conversion ratio. All the devices and elements are ideal, at output side only capacitor is connected, no resistance is connected. What is the mathematical relation between input voltage and capacitor voltage.

Thank you
 

MrAl

Well-Known Member
Most Helpful Member
Hi,

You should have probably started your own thread.

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.
 

MrAl

Well-Known Member
Most Helpful Member
Hello,

I gave you the formula did you not see it ?

As D goes from 0 to close to 1 the output voltage increases, but remember there is a limit to D because as D approaches 1 there is a point that is reached where the output actually decreases instead of increasing when there are losses in the circuit, and there are always losses in the practical circuit so there is always a limit to D. We know that D can not be equal to 1, but also D can not be to close to 1 either because of the losses or else
the output no longer increases. This becomes important in the control part of the circuit where we normally want D to increase in order to maintain the output voltage with normal loading. But if D gets to a certain point the output decreases and thus we loose regulation and the control gets stuck in low gear so the output goes to a much lower value than we wanted. To prevent this, we impose a limit on D so we get normal control operation.
 
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shasikumar

New Member
It is not valid in no-load condition, because the voltage across the capacitor is keep on increasing. It don't have any path to discharge (no-load). Have you understood??
 
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