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Adding a feedback voltage divider according to datasheet specifications.I also have LM2576S-ADJ P+. How can i get +5V output from it?
Already answered, I believe. The reference circuit shown in post #1 is expected to work.Can anyone give me solution about how can i get +5V from LM2576-5.0 P+?
The output can go to 6V if you don't have any load. Try connecting 5-50 ohm resistor to output.Sorry, But i connected 330uH inductor between pin 2 and 4.
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Can anyone give me solution about how can i get +5V from LM2576-5.0 P+?
It baffles me when a company buys (supposedly USA) devices from a Far East supplier for a likely cheap price and is surprised that they are counterfeit. :-?............
The only problem we have ever experienced, related to this IC, is when our contract manufacturer purchased them from a Far East broker, and started failing.
Failure analysis by National Semi revealed counterfeit devices.
Hi,
good point.
Klaus
The leakage current specification doesn't exactly answer the minimum load current problem. The -5 version has already an internal 4k feedback divider load. All datasheet output voltage specifications are for > 0.5 A load current.
I'm not sure if I ever used a LM2576 without load, but I remember many other buck converter chips, e.g. newer LM26xx simple switchers that work well without it.
Even counterfeit in the West is expensive. Prices in the West are very high because in the West 95% of the money goes to the salesman who has to have a private airplane and bribe the government to allow him to steal. PRICE INDICATES NOTHING.It baffles me when a company buys (supposedly USA) devices from a Far East supplier for a likely cheap price and is surprised that they are counterfeit. :-?
How do they think the prices got so cheap??