Timeline for Current source - Achieving a defined settling time with variable load resistance
Current License: CC BY-SA 3.0
13 events
when toggle format | what | by | license | comment | |
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Dec 17, 2014 at 1:36 | answer | added | patellar-myotatic | timeline score: 1 | |
Apr 28, 2014 at 10:19 | vote | accept | Medo42 | ||
Apr 25, 2014 at 17:10 | answer | added | Andy aka | timeline score: 1 | |
Apr 25, 2014 at 15:46 | comment | added | Medo42 | @DaveTweed I'm considering switching Q1 and RLoad (and turning Q1 into an N-FET to avoid the base current error)... had tried it before, but it caused oscillation problems. However, thinking back I still had a smaller shunt with an amplifier in there back then, so I probably had a gain>1 in the feedback loop. If I see it correctly, the worst-case gain I can get if I just switch Q1 and RLoad here is 1 (when Q1 turns into a voltage follower again). Have to re-think things other parts of the circuit yet again, sigh. Being a beginner is hard. | |
Apr 25, 2014 at 14:04 | comment | added | Medo42 | Yes, it's referenced to the negative output of V1. I added a ground symbol now, hope that makes it more clear. | |
Apr 25, 2014 at 14:03 | comment | added | Dave Tweed | Simplest thing would be to put the load resistance on the collector side of the NPN, then the transistor does most of the heavy lifting with regard to compliance. | |
Apr 25, 2014 at 14:03 | history | edited | Medo42 | CC BY-SA 3.0 |
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Apr 25, 2014 at 14:00 | comment | added | Vladimir Cravero | yes, that's why I spoke of a tradeoff. you want a faster system->you raise the dominant pole | |
Apr 25, 2014 at 14:00 | comment | added | Dave Tweed | What is V_in referenced to? The negative end of V1? You should include a ground symbol to eliminate the ambiguity. | |
Apr 25, 2014 at 13:59 | comment | added | Medo42 | Wouldn't that just have the same effect as reducing the capacitor value, i.e. raising the corner frequency? | |
Apr 25, 2014 at 13:47 | answer | added | Spehro 'speff' Pefhany | timeline score: 1 | |
Apr 25, 2014 at 13:10 | comment | added | Vladimir Cravero | Can't you just get rid of \$R_2\$ or at least reduce it? The only problem I see is that the circuit becomes less stable... It's a tradeoff. | |
Apr 25, 2014 at 13:02 | history | asked | Medo42 | CC BY-SA 3.0 |