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Arouse1973
Adam
Your 1K will be dissipating 24.7 mA^2 * 1000 = 0.6 W if the resistor has a thermal impedance of 120 deg C (Check your data sheet) per Watt, then the temperature will be 70 deg C + ambient.
Adam
Adam
KrisBlueNZ
Sadly passed away in 2015
Why are you using such low values for the voltage feedback resistors?
As Adam said, the power dissipation in R2 will be significant.
P = V2 / R
= 262 / 820
= 0.82W
The wattage rating of the resistor doesn't affect how much power it is forced to dissipate. A 2W resistor in that position still has to dissipate 0.82W. The reason a 2W resistor will run cooler than a 1W or 0.5W resistor is just that it has more surface area from which to radiate the heat. So you can expect your 2W resistor to get pretty warm.
I don't know why the regulation is poor. Make sure your inductor isn't saturating. You need to look at the current waveform in the inductor. If you have a current probe, great. Otherwise, put a small resistor e.g. 0.1Ω in series with the inductor (preferably on the left side) and put your scope across that. You'll need your scope and/or the regulator power source to be isolated from ground, otherwise you'll short out the power source when you connect the scope.
The scope should show a nice linear ramp. If the ramp suddenly starts climbing sharply, the inductor core is saturating.
As Adam said, the power dissipation in R2 will be significant.
P = V2 / R
= 262 / 820
= 0.82W
The wattage rating of the resistor doesn't affect how much power it is forced to dissipate. A 2W resistor in that position still has to dissipate 0.82W. The reason a 2W resistor will run cooler than a 1W or 0.5W resistor is just that it has more surface area from which to radiate the heat. So you can expect your 2W resistor to get pretty warm.
I don't know why the regulation is poor. Make sure your inductor isn't saturating. You need to look at the current waveform in the inductor. If you have a current probe, great. Otherwise, put a small resistor e.g. 0.1Ω in series with the inductor (preferably on the left side) and put your scope across that. You'll need your scope and/or the regulator power source to be isolated from ground, otherwise you'll short out the power source when you connect the scope.
The scope should show a nice linear ramp. If the ramp suddenly starts climbing sharply, the inductor core is saturating.
The ripple is large because the load is small and the amount of energy supplied with each cycle is large. This leads to a relatively low frequency and larger ripple.
From a look at the trace, the frequency is only 300Hz. You should be aiming for something much higher (20kHz to 150kHz perhaps).
Try a smaller inductor, i.e. lower inductance, but same or higher current rating.
From a look at the trace, the frequency is only 300Hz. You should be aiming for something much higher (20kHz to 150kHz perhaps).
Try a smaller inductor, i.e. lower inductance, but same or higher current rating.
Arouse1973
Adam
The reference design in LT Spice uses significantly higher feedback resistors. Why have you chosen such low values? You are wasting approx. 30 mA in the feedback network, why?
Adam
Adam
It was merely because of my silly mistake. I had made an Excel worksheet to find out the suitable standard values for R1 and R2 in (k) but due to tired mind I overlooked "k" and just used 33/820 instead of 33k/820k. The power dissipation problem is resolved.
Working on improper regulation.
Arouse1973
Adam
Cool 
KrisBlueNZ
Sadly passed away in 2015
Ah, OK. But re Steve's comment in post #4, what is the switching frequency? The data sheet says it's fixed at about 23 kHz. But there's ripple at a much lower frequency on the output.
Show us a photo of your construction. Did you build it up on a breadboard? (Hint: the right answer is NO, definitely not!)
Show us a photo of your construction. Did you build it up on a breadboard? (Hint: the right answer is NO, definitely not!)
I changed R1 and R2 to 18k and 470k (didn't had 820k), now I am getting 34V (instead of 33.75V).
But with 1k load, it drops to about 28V.
I checked the frequency at pin #8, it is only 400 - 425 Hz.
And yes, I have construct the circuit on breadboard (for trial only) and I do understand that the results are not reliable.
But with 1k load, it drops to about 28V.
I checked the frequency at pin #8, it is only 400 - 425 Hz.
And yes, I have construct the circuit on breadboard (for trial only) and I do understand that the results are not reliable.
Arouse1973
Adam
Lol It is now cool
Arouse1973
Adam
But smps do have more ripple. You dont think your expecting too much. Try this put a 100R and a 100uF cap on the output. Connect the resistor to the output and the cap from the other side of the resistor and then to 0V. To form a lowpass filter.
Adam
Adam
KrisBlueNZ
Sadly passed away in 2015
OK, the reason for the low switching frequency is that this converter is a gated oscillator type. It doesn't switch while the output voltage is above the target voltage. You need to put a significant load on it before you will see the actual oscillator frequency at the SW1 pin.
If you've built the circuit on breadboard, and you know the results aren't going to be reliable, why are you asking us why it doesn't seem to be working properly?
If you've built the circuit on breadboard, and you know the results aren't going to be reliable, why are you asking us why it doesn't seem to be working properly?
I assume that breadboard results are at least 80% close to the actual so for trial, it is ok.
R1=18k and R2=470k is supposed to give 33.75V as per formula mentioned in the datasheet. And 34V is quite close to the target.
I am constructing on a PCB to see the actual results and behavior against the actual load.
KrisBlueNZ
Sadly passed away in 2015
Bzzzzzt! LOL wrong. Even if "80%" was a meaningful way to quantify how well the circuit works, it would be wrong.
Good. You don't need a PCB; prototyping board is enough.
Sorry, I was quite busy; could not report earlier.
While constructing on veroboard, I noticed that the value of current limiting resistor which was connected to pin #1 wrong 750 instead of 150. Now both circuits are working properly.
I apologize all of you guys for my mistake. Lastly, thanks for sparing your time to help me.
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