Why is my boost converter output voltage spiking when load is removed?

In summary, the conversation discusses designing a dc-dc boost converter with an 11v output from a 6v input. The converter works well on PCB but the output voltage shoots up to 30v-40v when the load is removed. The cause of this issue is believed to be due to a lack of minimum load on the converter and possible instability in the feedback loop. The conversation also suggests checking for oscillations and potential loss in components. It is recommended to investigate each component and their potential impact on the converter's performance.
  • #1
dhanavanthri
3
0
Hi need some help here.
I have designed a dc-dc boost converter to give output voltage of 11v from 6 v input.
Well it has been implimented on PCB, works fine but one thing i din't understand well.

The have adjusted the duty cycle to get my 11volts on load(LED Load), when i remove the load the output voltage is shooting to 30v to 40v.
I checked in Pspice even there the output voltage shoots to the same value.
Is the driver has bad regulation or what??
please let me know if i am missing some thing very important!
 
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  • #2
A lot of the converters have minimum load. With too little load, the converter can be unstable and get into oscillation and the voltage will change. Use the scope probe to see whether you see any oscillation at the output.

The close loop feedback is around the PWM chip, I have seen a lot of converters that is not stable, even the ones from manufacturers. That is the reason they specified the minimum load and they get "noisy" if load is too light.
 
  • #3
Do you have a feedback path to the chip to control the voltage?
Normally this reduces the output voltage to some fixed value and then the chip adjusts the duty cycle to give an output which maintains this control voltage at the fixed level.

It is possible that you should have some minimum load on the chip. This will be given in the data sheet for the chip.
 
  • #4
Thank you guys.

@Yungman :
As far the oscillation are concerned i see that output voltage has dc offset of 30v above that have oscillations of Peak to peak of 2v to 3v.

@Vk6kro :
I am not using closed loop operation so no feed back.
since the load is LED no need for voltage constant voltage arrangement.
I need to say one more thing during the load the voltage is not changing only when i remove the laod the voltage is getting shot up.

And yes , i have a very less load of 60 to 80 mA.
Any solutions to counter the over shoot or its harmless?
 
  • #5
If you have no feed back, then all bets are off. Voltage will usually jump if you remove the load. I don't even think the 2 to 3 volt signal you said is even oscillation, it might just by the frequency of the converter.

Without the schematic, it is hard to give you any solution.
 
  • #6
Ok got your point but one thing still haunts me.
The current drawn through the inductor when switch is turned ON is same when OFF , right!..
if above is true then the output voltage during load and no load should be same!(Assumming Inductor internal resistance is 0).
 
  • #7
That is the big assumption. Other components have loss also. Even your 6V supply input can have internal impedance. To find out why, you really have to chase down each component to find out where is the major loss and beef up the component.

For one, you check to make sure the 6V input change with load?

Also, is the way you boost the voltage is by hooking one end of the inductor to 6V, then use a transistor to pull the other end to ground and let the current build up in the inductor. Then you open the transistor and let the inductor fly back and kick to high voltage? If that is the case, there is no way out! That is because if you are not drawing current( when you remove the LED) then the inductor will fly to high voltage. I think the reason you only go to 30V is because some of the components zener out and start drawing current and stop the inductor from flying any higher.
 
Last edited:

1. What is a Boost converter?

A Boost converter is a type of DC-DC converter that is used to step up or increase the output voltage from a power source. It uses an inductor, a switching element (such as a transistor), and a diode to regulate the output voltage. It is commonly used in electronic devices to convert a low voltage input to a higher voltage output.

2. How does a Boost converter work?

A Boost converter works by using the inductor to store energy when the switching element is on, and then releasing that energy when the switching element is turned off. This creates a higher voltage output than the input voltage. The diode then directs this output to the load, maintaining a steady voltage level.

3. What are the main components of a Boost converter?

The main components of a Boost converter are the inductor, switching element, diode, and capacitor. The inductor stores and releases energy, the switching element controls the flow of current, the diode directs the current to the load, and the capacitor filters the output voltage to ensure a steady level.

4. What are the advantages of using a Boost converter?

Boost converters have several advantages, including high efficiency, compact size, and the ability to step up the output voltage. They also have a simpler design compared to other types of DC-DC converters, making them cost-effective and easy to implement in various electronic devices.

5. What are some common applications of Boost converters?

Boost converters are commonly used in electronic devices that require a higher voltage output than the power source can provide. Some examples include portable electronic devices, battery chargers, LED drivers, and renewable energy systems such as solar panels and wind turbines. They are also used in automotive applications, such as in hybrid and electric vehicles.

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