Activity for Circuit fantasist
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #279016 |
Post edited: Adding a subtitle |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: Adding a new property |
— | about 4 years ago |
| Comment | Post #279025 |
I agree with your considerations. Good find... (more) |
— | about 4 years ago |
| Comment | Post #279016 |
The difference between the potentiometer and the "single entry labeled 'light intensity signal'" is that the potentiometer is a real device... and the reader can see the path where the input current flows. For the purposes of understanding, it is very important to see where currents flow. That is why... (more) |
— | about 4 years ago |
| Comment | Post #279016 |
@coquelicot, To manually test a voltage-controlled analog circuit, you need a variable voltage source... but imagine you have only a constant one. Then, what else to use if not a simple potentiometer? I have outlined the studied circuit in yellow; the potentiometer (input voltage source) and the pow... (more) |
— | about 4 years ago |
| Comment | Post #279018 |
Good question... I have also asked myself this question... and I must admit I do not have a convincing answer. Maybe we should take a look at H&H and Student manual for H&H... (more) |
— | about 4 years ago |
| Comment | Post #279010 |
In the OP's circuit (NIC), I suppose there is a loop gain since the op-amp gain is enormous... although the total feedback ratio is a difference between the two voltage dividers ratios... (more) |
— | about 4 years ago |
| Comment | Post #279010 |
To work as a latch, we have to keep the input signal in a neutral state (in the middle of the hysteresis loop). Conversely, for it to be a Schmitt trigger, we must keep the input signal outside the hysteresis loop. (more) |
— | about 4 years ago |
| Comment | Post #279010 |
Every (op-amp) circuit with positive feedback and loop gain > 1 is a latch (with one input)... and every latch with one input has hysteresis. Depending on how we control it, it can work either as a latch or as a Schmitt trigger. (more) |
— | about 4 years ago |
| Comment | Post #279010 |
@Olin, first to move here my comment from below where I had put it wrong: If the four resistors were absolutely equal and the op-amp was ideal, the circuit would be identical as a bare op-amp without feedback (resistors). Now about your last comment above... (more) |
— | about 4 years ago |
| Comment | Post #279016 |
@tlfong01, the light-to-voltage converter is another circuit that stays before this. But it does not belong to this topic ("analog LED voltage indicator"). I have only mentioned it to make the story more human-friendly... (more) |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: About the grounding |
— | about 4 years ago |
| Comment | Post #279016 |
@tlfong01, to understand circuits you need more imagination than IQ (said Einstein)... (more) |
— | about 4 years ago |
| Comment | Post #279016 |
@Olin Lathrop, I need to think a lot before I answer you because I am quite surprised and I can not react. I just managed to ask myself the simple question, "Why are we so different?" (more) |
— | about 4 years ago |
| Comment | Post #279010 |
Olin, I understand you since the final result is an op-amp without any feedback, i.e., a "comparator without hysteresis". But actually there are two feedbacks (negative and positive) that neutralize each other and the final result is "no feedback". But this state is unstable - it is enough for the p... (more) |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: More about the unique circuit properties |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: Minor edit |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: About the unique properties of the structure |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: About the advantages of the second version |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: About the circuit input resistance |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: Added another improvement |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: Added another concept |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: Minor edit |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: Minor edit |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: Minor edit |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: Minor edit |
— | about 4 years ago |
| Edit | Post #279016 |
Post edited: Minor edit |
— | about 4 years ago |
| Edit | Post #279016 | Initial revision | — | about 4 years ago |
| Article | — |
3-LED voltage indicator (an inventor's story) Goals and objectives Motivation. Having shown how a 1-transistor circuit can be invented, now I will demonstrate how we can invent another more complex 2-transistor circuit. As before, my goals are two - specific (the very 2-transistor circuit) and general (the technology of invention). With my ... (more) |
— | about 4 years ago |
| Comment | Post #279010 |
@2kind, Thanks to your mistake you "invented" an ingenious circuit-:) This is how great inventions are sometimes made! (more) |
— | about 4 years ago |
| Comment | Post #279010 |
You have considered a very interesting situation in the OP's circuit when "left ends of R14 and R16 are tied to ground". Actually, this is a NIC. In this circuit, there are two kinds of feedback - negative (R16-R19) and positive (R14-R19). To be stable, this circuit requires the negative feedback to ... (more) |
— | about 4 years ago |
| Comment | Post #279010 |
You made me begin thinking about the "3-input diff amp"... It looks to me as a sort of a "summing-subtracting circuit". R3-R1 network is a passive voltage summer with weighted inputs. So Vp and Vref (0 V here since R1 is grounded) are summed according to the superposition principle - Vp.R1/(R1 + R3) ... (more) |
— | about 4 years ago |
| Comment | Post #279005 |
Indeed, I know there are applications where negative impedance converters (NIC) are used as amplifiers but I am not sure if this is the case here... Also, if the current is DC and it enters R6 from the left (as shown), the A1A output voltage should be negative. But the circuit is single-supplied only... (more) |
— | about 4 years ago |
| Comment | Post #279005 |
@2kind, It is a pleasure for me to think about such problems. Just to clarify that I have no special practical experience in this problem, I am just guided by a common sense. I think there is no need of TVS between the op-amp inputs since they are connected in series (through the ground) and this ne... (more) |
— | about 4 years ago |
| Edit | Post #279005 |
Post edited: Minor edit |
— | about 4 years ago |
| Edit | Post #279005 |
Post edited: Another problem |
— | about 4 years ago |
| Comment | Post #279005 |
OK, there was some problem with edit... I have deleted the duplicate. (more) |
— | about 4 years ago |
| Edit | Post #279005 | Initial revision | — | about 4 years ago |
| Answer | — |
A: ESD Protection - Differential Amplifier Although the voltage drop across the current-sensing resistor R6 is floating and it can "move" between the 28 V rails, the A1A op-amp input voltages can (have to) vary within the 5 V power supply. So, you can protect the op-amp (and the whole circuit) against any overvoltages by connecting a bidir... (more) |
— | about 4 years ago |
| Comment | Post #278933 |
@tlfong01, Exactly! (1) is true... It is an unregulated power supply consisting of (a) + (b) + (c) cascaded... I frequently use the battery symbol to represent a voltage source... And (2) is completely true. I will write you a message later in the chat... (more) |
— | about 4 years ago |
| Comment | Post #278933 |
@tlfong01, Thanks for the attention to my humble "invention". "Bin is replaced by Bb" is said in the sense that, when the mains supply fails, already Bb but not Bin supplies the load. Bin stays at its place but it does not produce power (that is why it is drawn in light gray). (more) |
— | about 4 years ago |
| Comment | Post #278933 |
@tlfong01, BIN actually represents a diode rectifier (although it can be battery as well). The diodes inside this rectifier would be backward biased by the backup battery voltage Vb applied through the resistor R. Really, R will waste power... but what do we do? We can only decrease the difference b... (more) |
— | about 4 years ago |
| Comment | Post #278933 |
@tlfong01, Thanks for the response. It is strange why others keep silence... Browse the web for diode rectifiers (e.g., https://en.wikipedia.org/wiki/Diode_bridge) and you will see that, in this case, the diodes inside the rectifier are backward biased... and they stop the return current. (more) |
— | about 4 years ago |
| Edit | Post #278933 |
Post edited: Minor edit |
— | about 4 years ago |
| Edit | Post #278933 |
Post edited: Minor edit |
— | about 4 years ago |
| Edit | Post #278933 |
Post edited: Minor edit |
— | about 4 years ago |
| Edit | Post #278933 |
Post edited: Added link |
— | about 4 years ago |
| Edit | Post #278933 |
Post edited: Added link |
— | about 4 years ago |
| Edit | Post #278933 |
Post edited: Refining |
— | about 4 years ago |
| Comment | Post #278836 |
Just to clarify ... In inverting circuits, e.g. op-amp inverting amplifier with resistors R1, R2, the op-amp output acts as a negative resistor with "resistance" -R2. It neutralizes the positive resistance R2 and the result is virtual zero resistance (R2 - R2 = 0) aka "virtual ground". (more) |
— | about 4 years ago |