Hi Eric,hi John,
Found a Video that should help you with your SEPIC SMPS design
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Hi Eric,hi J,
Is that 33Vout with any load resistor.???
Seems very high.
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Hi Eric,hi J,
So to summarise.
You are using a 12Vdc wall wart.?
What Vout voltage range do you get.?, with the 47k pot.
What typical load currents.?
Details of the caps used.?
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hi,How do i obtain the load current? I'm currently using 1K resistor as load.
The output capacitor i've replaced is the same 35V 100uF electrolytic.
I have added the capacitor in parallel, my Vout is now reaching 42V when pot is set at 47K at 5V voltage in. Could it be that when my old output capacitor blew up, it messed up other components as well?hi,
To calculate the Load current= Vout/ Rload eg: 30Vout/1000R = 0.03Amps
If you increase the capacity of the Vout cap it will improve the smoothing, add another 100uF 35v in parallel.
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Hi Ebp,Input power supply: The current limiting behavior of most switchmode wallwarts is likely to be quite soft and unpredictable. The best thing to do is just check the voltage any time you see inexplicable behavior.
Excessive output voltage: This is most likely due to a problem in the feedback resistor. Any time you get odd behavior like that it is best to test with a fixed resistor instead of the pot. Pots are somewhat delicate and can wear out. High resistance between the element and the wiper can arise. When using a pot as a variable resistor, as opposed to using it as a voltage divider, always connect the unused terminal to the wiper. On a breadboard, try to do it by connecting with a separate wire to the source or destination. In both cases, the objective is to try to make sure that the end-to-end resistance of the pot remains in-circuit even if something goes wrong with the wiper or its connection. If the feedback resistors goes open, the switcher runs open-loop (unregulated). With a boost converter this can lead to grossly excessive output voltage, frequently resulting in something failing catastrophically.
If the feedback circuit goes open, the FB pin of the IC will usually drop to 0V. This can be confirmed by measurement. Probing the feedback pin is perilous in a working switcher. Noise injection can make a well-behaved switcher go berserk and of course shorting it to anything is bad news. Volt meters are usually unsuitable due to the long unshielded leads. An oscilloscope is the best instrument to use.
Hi Eric,hi J,
Look at this clip from the d/s it shows the resistive divider chain for the Vout.
Vref is 1.245Volts
Note the recommended R2 value of 6k2
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View attachment 148178
Ebp,I'm sorry, from earlier sim schematics I thought you were using the pot as a variable resistor making up part of the total resistance between the output and the FB pin. It is OK with this controller to use it as you are, except I would put a fixed resistor between the bottom and ground to limit the maximum output voltage. Without such a resistor you could set the output voltage to something extremely high.
It is interesting that the circuit max'es out at 38 volts. This may be due to internal circuitry to protect the switch from excessive voltage (rated at 35 V). I haven't read the data sheet carefully - it may be mentioned.
Hi Eric,hi John
This circuit should give the Vout range required.
Note 47k pot set for full 47K and 0k.
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Hi Eric,hi John,
Post 56 circuit looks OK.
What current loads are expecting across the full Vout range.?.
If you are considering using a USB voltage supply source, take care not to exceed the USB rating.
That Amazon variable power resistor seems an 'over kill' to me.
Be aware the 25W power rating of the pot, may only apply to the maximum resistance of the 1k track.
Do a few calculations at different pot settings, Vout and load currents, post your results
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