Skip to main content
10 events
when toggle format what by license comment
May 15, 2022 at 4:11 comment added pipe @Codebeat Few components does a better job at attenuating a signal than a voltage divider if you can deal with the impedance.
May 15, 2022 at 3:00 history tweeted twitter.com/StackElectronix/status/1525672316079943681
May 15, 2022 at 1:42 history became hot network question
May 14, 2022 at 20:27 history edited JRE CC BY-SA 4.0
"Thanks a lot" is a sarcastic remark that means "your assistance actually made the problem worse."
May 14, 2022 at 18:18 vote accept Libegria
May 14, 2022 at 18:03 answer added Spehro 'speff' Pefhany timeline score: 9
May 14, 2022 at 17:51 comment added Codebeat Voltage divider(s network), the center (2.5V) of value is 0. However it is never that precise because you squeeze a scale of -15V to +15V = 30V into a range of 0v to 5v. In fact useless. So you need an instrument /component to do this.
May 14, 2022 at 17:47 comment added DKNguyen You need to bias your signal at 2.5V after stepping it down with a divider (and you may want a capacitive divider in parallel with it as well for frequency compensation so the resistor divider doesn't form an RC lowpass filter that slows down higher frequencies). Using a differential amp on the signal after is better too. That makes a differential probe as seen here here which you could actually just outright do if you wished. Just this one doesn't have a biased output: circuitcellar.com/research-design-hub/…
S May 14, 2022 at 17:42 review First questions
May 14, 2022 at 18:20
S May 14, 2022 at 17:42 history asked Libegria CC BY-SA 4.0