Timeline for Why not use a diode instead of a resistor for current sense circuits?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 31, 2019 at 21:34 | vote | accept | SusanW | ||
| Jul 30, 2019 at 3:01 | history | tweeted | twitter.com/StackElectronix/status/1156037101973979137 | ||
| Jul 29, 2019 at 21:19 | answer | added | hacktastical | timeline score: 0 | |
| Jul 29, 2019 at 5:56 | answer | added | Jack Creasey | timeline score: 4 | |
| Jul 28, 2019 at 0:12 | history | became hot network question | |||
| Jul 27, 2019 at 20:26 | answer | added | user136077 | timeline score: 2 | |
| Jul 27, 2019 at 17:49 | answer | added | TemeV | timeline score: 7 | |
| Jul 27, 2019 at 17:31 | comment | added | SusanW | @Andyaka Ok, I think I don't want it to be linear - I want it to snap over at around 10mA (give or take) and stay there all the way up. If I use a little \$0.2\Omega\$, it means I need to detect a 2mV gap for my 10mA trigger: might that be quite challenging with noisy power PWM and motors just next door? and it still drops \$800mW\$ at the max 2A, about same as the diode. I can make the resistor lower, but the sensitivity requirements get harder... | |
| Jul 27, 2019 at 17:17 | comment | added | glen_geek | You could reduce the bypass resistor from 25 to about 2 ohms (using a cheap comparator whose offset voltage is 10mV, sensing your 10mA decision point). Yes, the diode will clamp on big currents - its a design decision: is diode cost worth the efficiency increase? | |
| Jul 27, 2019 at 17:10 | history | edited | SusanW | CC BY-SA 4.0 |
added 208 characters in body
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| Jul 27, 2019 at 16:16 | comment | added | Andy aka | It’ll certainly waste more heat than a low impedance resistive current shunt and amplifier and, it won’t be linear. Sounds a bad idea to me. | |
| Jul 27, 2019 at 16:06 | history | asked | SusanW | CC BY-SA 4.0 |