Activity for coquelicotâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #278066 |
@Olin. Thank you for answering to my last edits. The amplitude of the sine wave can be read near it (0, 1, 1000) means 0V offset, 1V amplitude, and 1000Hz in the LT spice simulator. Regarding the AC labeling, indeed I don't know why this is here but I have checked my schematic and the voltage is cons... (more) |
— | about 4 years ago |
| Comment | Post #278065 |
Following the answer of Olin, I think that the usefulness of the bootstrapping is more to provide a return path to ground without loading the AC input, than to reduce the AC input current. But I don't know how to formulate and use that quantitatively. (more) |
— | about 4 years ago |
| Edit | Post #278065 |
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— | about 4 years ago |
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— | about 4 years ago |
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— | about 4 years ago |
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— | about 4 years ago |
| Comment | Post #278066 |
@Olin No, I see no flaw in your logic, perhaps because there is no real logical deduction in you answer. Everything you say makes sense to me, but the Art of Electronics is still a widely estimated book. So, I've done a simulation and will post it as an edit to my question. You may be surprised. (more) |
— | about 4 years ago |
| Comment | Post #278066 |
Something is unclear to me in your answer: you essentially say that this circuit improves nothing, even at "high" frequencies. But this is the circuit given in the Art of Electronics, I mean it is indeed a follower. Do you mean that this book is misleading here (well, this would not be the first tim... (more) |
— | about 4 years ago |
| Edit | Post #278065 |
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— | about 4 years ago |
| Edit | Post #278065 | Initial revision | — | about 4 years ago |
| Question | — |
Design rules for opamp bootstrapping Here is the bootstrapping technique for opamps, as exposed in the Art of Electronics:bootstrap. This technique is supposed to increase considerably the input impedance of the opamp for an AC input source (usually passed through a cap). The Art of Electronics considers it somewhat outdated, as it ... (more) |
— | about 4 years ago |
| Edit | Post #277344 |
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— | over 4 years ago |
| Edit | Post #277344 |
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— | over 4 years ago |
| Edit | Post #277344 |
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— | over 4 years ago |
| Edit | Post #277344 |
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— | over 4 years ago |
| Edit | Post #277344 | Initial revision | — | over 4 years ago |
| Answer | — |
A: Constant Current Load Circuit with Op Amp let me begin with a schematic for an adjustable current source with a simple load: Image The operation is simple: The current I inside R1 gives rise a voltage drop equal to I/R1 (= I if R1 = 1). So the oamp will set the voltage to the gate of the mosfet in such a way that I/R1 = Vpot... (more) |
— | over 4 years ago |
| Edit | Post #277275 | Initial revision | — | over 4 years ago |
| Question | — |
Darlington or Sziklai configuration - why and when using complementary pairs? Why does the Darlington or Sziklai configurations need complementary transistors? For these configurations, when should I use complementary pairs and when could I use not complementary pairs ? (more) |
— | over 4 years ago |
| Comment | Post #277272 |
Thx Olin. I haven't think about the push-pull. If you can cite other configurations that need complementary pairs, I'm interested to know. (more) |
— | over 4 years ago |
| Edit | Post #277271 | Initial revision | — | over 4 years ago |
| Question | — |
Complementary transistor pairs - what is the point ? I understand that some applications require NPN-PNP pairs with identical electrical characteristics. But I would like to understand why. (more) |
— | over 4 years ago |
| Edit | Post #277090 |
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— | over 4 years ago |
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— | over 4 years ago |
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| Edit | Post #277090 | Initial revision | — | over 4 years ago |
| Answer | — |
A: Driving LED with NPN transistor from I/O pin No need to power the led with 5V. 3.3 V should suffice! Before I come to problem of regulating the current with only 3.3V, I wish to state some general simple constructions. Technically speaking, the following current sink/source is perhaps insufficiently known: DiodeRegu1 The current is se... (more) |
— | over 4 years ago |
| Comment | Post #277051 |
@Andy. Yes, I'm sure this was evident for you, and this is probably hidden inside your terms "picks up". For cranks like me, it takes some time to understand because the pickup voltage is usually measured with a probe referenced to ground (not to the pickup source). So another cap coupling to ground ... (more) |
— | over 4 years ago |
| Comment | Post #277051 |
Something here is missing to explain the L1/L2 voltage (say, a leakage from L2 to G). See the second schematic of Olin. The only difference is that you interpret his cap C1 in another way. But the logic is the same. Perhaps to complete this scenario, add another capacitive coupling to ground, so that... (more) |
— | over 4 years ago |
| Comment | Post #277048 |
Yes. The residual loads inside the house explain nicely the effect. So, it should suffice to disconnect every electrical device in the house and to measure the L1/L2 voltage again to prove that scenario. Curiously, the only difference with my scenario is how you interpret the R3 resistor (R2 in my sc... (more) |
— | over 4 years ago |
| Comment | Post #277032 |
The problem with leakage to ground is that apparently you'll always have a voltage between L1 and L2 not smaller than between L2 and G, that is 90V. So, I give up and wait for the answer of the real pro. (more) |
— | over 4 years ago |
| Comment | Post #277032 |
A small bird told me that you have an answer :-) (more) |
— | over 4 years ago |
| Edit | Post #277032 |
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— | over 4 years ago |
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| Edit | Post #277032 |
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— | over 4 years ago |
| Edit | Post #277032 | Initial revision | — | over 4 years ago |
| Answer | — |
A: How could a damaged wire in split-phase power delivery create these voltages? Well, here is was my logic tells me: The corrosion has caused a high impedance in the L2 wire, but there should be also a "short" between the L1 and L2 wire. So that the voltage between L1 and L2 falls to 30V, and the voltage between G and L1 is 90V. In other words: L1 - G = 120V = L1 - L2 + L... (more) |
— | over 4 years ago |
| Comment | Post #276885 |
Oh, and of course with one more resistor at the top, to make the schematic entirely symmetric. (more) |
— | over 4 years ago |
| Comment | Post #276885 |
Nice. Perhaps even more clever: in your second schematic, two zener diodes back-to-back in place of the single zener, and a triac in place of SCR1? (more) |
— | over 4 years ago |
| Edit | Post #276882 |
Post edited: grammar |
— | over 4 years ago |
| Comment | Post #276885 |
That's exactly what I suspected. Thank you. An almost rhetorical question: do you know some ersatz for the diac?
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— | over 4 years ago |
| Edit | Post #276882 |
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— | over 4 years ago |
| Edit | Post #276882 | Initial revision | — | over 4 years ago |
| Question | — |
Is it possible to use two zener diodes in series back to back to replace a diac? I've accidentally burnt my dimmer with a short. It is a common dimmer with a triac/diac/pot of the simplest configuration (the triac is a BTA16). After replacing the triac, this still does not work, and I'm almost sure the problem is the DB3 diac (breakdown voltage at 32 V). I don't have this beast i... (more) |
— | over 4 years ago |
| Comment | Post #276739 |
Wikipedia writes everywhere that this is a non profit, free organization. And so do many "free world" sites.
I suggest to explain in few words that "this is a non profit organization (and will always remain so),
managed by highly qualified volunteers and experts, in order to ensure a content of ... (more) |
— | over 4 years ago |