Dear all,

I've done a boost converter with the help of allaboutcircuit.

Things go well. I got my desired output.

But, the problem is, the output is increasing.

I connected it onto my load and it increases from 45V to 48.5V in 50minutes.

Could anyone tell me the reason??

Please refer to the link below to know about my circuit:
http://forum.allaboutcircuits.com/showthread.php?t=47918

Thanks a lot.

 

There are fourty(40) posts in that topic. Please put a schematic of your current circuit HERE.

 

Are you still using this circuit? http://forum.allaboutcircuits.com/attachment.php?attachmentid=26749&d=1295646374

Your output sensing arrangement seems to use a simple common-emitter transistors. That is going to have a very substantial temperature dependency, so the output level will not be very stable.

Of course, your load may also be changing in some way, either because it is heating up, or for some other reason. A battery on charge would be expected to show a rising voltage if it initially drew enough current to activate the current limit facility.

 

Are you still using this circuit? http://forum.allaboutcircuits.com/attachment.php?attachmentid=26749&d=1295646374

Your output sensing arrangement seems to use a simple common-emitter transistors. That is going to have a very substantial temperature dependency, so the output level will not be very stable.

Of course, your load may also be changing in some way, either because it is heating up, or for some other reason. A battery on charge would be expected to show a rising voltage if it initially drew enough current to activate the current limit facility.

YUP, you're right. The BJT i use inside my circuit is 2N2222. Do I need to change it ?

 

You have never mentioned what your load is.

Trying to charge a battery would give results like that.

The boost converter circuit wasn't supposed to be a precision supply; just a simple way to build a current-limited roughly-regulated output.

If you want better regulation, then look at using "real" regulator ICs that were designed to do this type of thing in the first place. I simply don't have the time nor inclination to improve upon it.

 

You have never mentioned what your load is.

Trying to charge a battery would give results like that.

The boost converter circuit wasn't supposed to be a precision supply; just a simple way to build a current-limited roughly-regulated output.

If you want better regulation, then look at using "real" regulator ICs that were designed to do this type of thing in the first place. I simply don't have the time nor inclination to improve upon it.

I got a question, that is, if the resistance of load changes, will the output voltage change also in my circuit?

In addition, when I connect to my intended load, the voltage is decreased (compare to resistor as load), i know this is normal. But, how should I explain it? What theory is involved, is it "loading effect"?

Thanks in advanced.

 

is it loading effect?

 

I have no idea what your load is.
You may be exceeding the capability of the circuit to maintain the set voltage.
If it is a battery, that would be a good reason why.
Perhaps your load changes resistance over temperature.
I can't really help any more, as you don't provide any information.

 

I have no idea what your load is.
You may be exceeding the capability of the circuit to maintain the set voltage.
If it is a battery, that would be a good reason why.
Perhaps your load changes resistance over temperature.
I can't really help any more, as you don't provide any information.

Oh ya, the resistance changes as time goes,because it is a gel-like material, and it melts when i connect output voltage onto it.

thanks. I just want to make sure with it~~

 

I have no idea what your load is.
You may be exceeding the capability of the circuit to maintain the set voltage.
If it is a battery, that would be a good reason why.
Perhaps your load changes resistance over temperature.
I can't really help any more, as you don't provide any information.

These are information I got.

Beginning: Input Voltage 11V;Input Current:37.5mA;Output Current 5mA...Vout = 45V;
after 1 hour: Input voltage 11V;Input Current:30mA;Output Current 3mA...Vout = 46V;


It is so strange.... I don't understand the result.

 

My input is Battery, but why it would be a good reason???

 

These are information I got.

Beginning: Input Voltage 11V;Input Current:37.5mA;Output Current 5mA...Vout = 45V;

Input 0.4125w, Output 0.225W, .225/.4125*100 = 54.5% efficient

after 1 hour:Input voltage 11V;Input Current:30mA;Output Current 3mA...Vout = 46V;

Input 0.33W, Output .138W, .138/.33*100 = 41.8% eff

It is so strange.... I don't understand the result.

actually, that's only a 2.2% increase in output voltage.

However, keep in mind that you're probably operating with the minimum amount of load the boost converter can operate and still regulate the output voltage. I didn't test the circuit with that light of a load; you'd said it measures around 6k Ohms or so. That's around 7mA current. The voltage adjustment network of R3/VR1/R4 is about 0.9mA current at 47v.

 

My input is Battery, but why it would be a good reason???

No, it's your load increasing in resistance.

 

Input 0.4125w, Output 0.225W, .225/.4125*100 = 54.5% efficient


Input 0.33W, Output .138W, .138/.33*100 = 41.8% eff


actually, that's only a 2.2% increase in output voltage.

However, keep in mind that you're probably operating with the minimum amount of load the boost converter can operate and still regulate the output voltage. I didn't test the circuit with that light of a load; you'd said it measures around 6k Ohms or so. That's around 7mA current. The voltage adjustment network of R3/VR1/R4 is about 0.9mA current at 47v.

I think you're right(the red words), but why would this happen?

I have tested the different resistor (10k and 47kOhms), then both of them give me different output voltage. Once the load resistance increase,the output increase also.
10k resistor gives me 45V;47k resistor gives me 50++V instead( I didn't measure it properly)
Anyway, I thought larger load resistance will reduce the output voltage( I am really having a wrong concept)

Since my initial load would be 45V/5mA = 9k.
So, I think, if put a parallel resistor of 10k, it would manage to maintain its output at 45V for 1 hour.

 

You might be seeing increased ripple on the output.

Try increasing the size of the capacitor (C5) on the output; perhaps starting with 100uF, then 500uF to 1000uF.