Timeline for Startup voltage help: How can I change the input voltage to this IC when starting the circuit?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 20, 2020 at 14:28 | vote | accept | James S | ||
| Aug 15, 2020 at 9:00 | history | tweeted | twitter.com/StackElectronix/status/1294559469981052933 | ||
| Aug 14, 2020 at 2:10 | comment | added | Spehro 'speff' Pefhany | Seems to be 100K+ of UC3843B available with little trouble. | |
| Aug 14, 2020 at 2:06 | answer | added | Bruce Abbott | timeline score: 1 | |
| Aug 13, 2020 at 22:29 | comment | added | varun | @JamesS Honestly, I dint but that's a better idea .. Your divider should have enough voltage to turn on the FET initially.. then after sometimes, the resistance drops and the voltage should drop below to turn off the FET. This FET should connect 24V to U12. | |
| Aug 13, 2020 at 20:48 | comment | added | D.A.S. | Why using obsolete parts when you could be using better ones UCC38Cxx from TI | |
| Aug 13, 2020 at 17:55 | comment | added | sstobbe | Look at Fig 17 in the datasheet. You can use a weak pull-up to Vin diode-or'd to your 15V supply. Make sure to add a bypass caps near the Vcc pin. | |
| Aug 13, 2020 at 17:42 | comment | added | James S | @varun do you mean to use that divider to bias the gate of the FET? | |
| Aug 13, 2020 at 17:41 | comment | added | James S | @ThePhoton currently, this is a replacement for the 3843B due to availability | |
| Aug 13, 2020 at 17:11 | comment | added | The Photon | Why not just use UC3843B, with lower UVLO threshold? | |
| Aug 13, 2020 at 17:09 | comment | added | Voltage Spike♦ | Can you use the larger package with the split Vcc and Vc? then you could run them at seperate voltages and use a circuit to give the 5V regulator 6V and dissipate the power outside of the part, then run Vc at whatever you want. | |
| Aug 13, 2020 at 17:02 | comment | added | varun | You can use a thermistor (Negative temp coefficient) and a normal resistor and create a voltage divider that divides rectifier 24V. After certain time, the resistance of the thermistor decreases and its voltage drop decreases. So you start off with 17V then drop below 15V. | |
| Aug 13, 2020 at 16:34 | history | asked | James S | CC BY-SA 4.0 |