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Roseljr
New member
Username: Roseljr
Post Number: 2
Registered: 10-2021
| | Posted on Tuesday, December 21, 2021 - 8:51 pm: | 
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When designing a switching power supply with UC3843, how is the current sampling circuit determined?
When designing a switching power supply with UC3843, some of the current sampling circuits are taken through a coil, and some are taken directly from the source stage of the switching tube. How is it determined? How is that coil calculated?
https://www.utmel.com/components/uc3843-pwm-controller-pinout-datasheet-and-uses?id=388 |
Ke0koy
Junior Member
Username: Ke0koy
Post Number: 45
Registered: 12-2018
| | Posted on Monday, December 27, 2021 - 2:32 pm: |
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There is a lot going on in that chip, I'll try to explain it. This video might be a better one to watch as it shows the inner workings of the chip.
https://www.youtube.com/watch?v=RnHjBvKqpBY
Long story short, the chip creates a reference voltage (5v), which is cut in half by an internal divider, and that goes into one side of an op amp (internally). Pin two takes in another voltage from an external divider that goes into the other side of that op amp, and its output goes through the comp resistor at pin one which creates feedback from the op amps output changing its overall output value to something lower. The output voltage from there (still internal to the chip) is fed into one side of a comparator, and the voltage generated across the current sense resistor (R9 in the video you linked to) goes into the other side of that comparator. The switching output from that comparator is what switches the mosfet on and off. The short version is that if you leave everything else in that schematic the same, R4 will change the PWM width (so would changing R9) which changes the final voltage and using a pot there with the wiper to pin 2 would allow for adjustment.
Calling it a current sense pin is a bit confusing as pin 3 is really taking the voltage generated across a low value resistor. That resistor needs to be very low ohm (so it can handle the current the transformer needs). When I said changing R9 also changes the PWM width, do not take that to think a pot can go there. The entirety of the transformer primary current must go through it. You could MacGyver a voltage divider leaving that point, but why, that's what pin 2 is for 
Take a look at the other video I linked to, the circuit is better and he makes it easier to understand (in clean english too). |
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