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I like Serena
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Femme_physics said:Oh yea, you'll get 2 flowers
It looks very well presented. Good work. https://www.physicsforums.com/images/icons/icon14.gifFemme_physics said:I think I got it... full solution here:
Thank you, but my classmate disagrees with me. He says that we don't know that "I" that goes down, which I marked as "Iamp", because it splits to the op-amp and to the ground. Do you see what I mean?It looks very well presented. Good work.
Is the determination of Ra, Rb and Rx the full requirement for this exercise?
Is this particular brand/number of OP-AMP one that you can tell us anything about, i.e., have you studied its data sheet?
P.S. To your classmates keeping watch on this thread...what will the teacher think when you all submit identical homework
I like Serena said:
M Quack said:Now that you've worked out all the currents and voltages, do you have any clue what the circuit does? And what would you need to change to adapt it to a different output value?
How do you know it is not an inverter? Or how could you tell whether it is? It has either no inputs, or it has two, depending on how you want to view it. If you say it has two, then one input is non-inverting, the other is inverting.Femme_physics said:I still don't understand a few things. In the original drawing, we see a current that goes INTO the op-amp, but it is not an inverter, so what gives?
No. But I conjecture that you are seeing the labelling arrows denoting Vout on the op-amp output and mistakenly concluding it represents a short circuit to ground https://www.physicsforums.com/images/icons/icon4.gif I ascribed to it a current of zero since we are told nothing about a load there. Consider it an open circuit; if it isn't, we would have been told.Thank you, but my classmate disagrees with me. He says that we don't know that "I" that goes down, which I marked as "Iamp", because it splits to the op-amp and to the ground. Do you see what I mean?
No. But I conjecture that you are seeing the labelling arrows denoting Vout on the op-amp output and mistakenly concluding it represents a short circuit to ground I ascribed to it a current of zero since we are told nothing about a load there. Consider it an open circuit; if it isn't, we would have been told.
What is the number written on the op-amp in your first schematic?
How do you know it is not an inverter? Or how could you tell whether it is? It has either no inputs, or it has two, depending on how you want to view it. If you say it has two, then one input is non-inverting, the other is inverting.
In amplifiers, the term "inverting" has two meanings. One is that if the input is +x volts, the output will be -x volts. That behaviour isn't possible here, for the simple reason that the OP-AMP output cannot go negative because the OP-AMP isn't powered by a negative supply; it is powered by only a single positive supply, that's the 25v.
The other meaning of "inverting" is that if the input level should rise slightly it will cause the output to fall. So if the input rose, say, from 3.4
It definitely could not be a short circuit to ground; this would contradict the specification that Vout = 12V.Femme_physics said:You mean that Vout = 12V?
And well, yes, it could be a short circuit to the ground,
It's obviously the circuit's output, but (in the absence of information to the contrary) just regard it as a node where we can measure a voltage.it could be anything..-- we did make the assumption so figure it's illegal to touch it.
Unless we can identify something to call an "input" we can't compare input vs. output to see whether there is a inversion or not! (Suppose you have an arrangement where the input current is negative so flows "out" of the input, then a non-inverting amplifier would have output current flowing "in" as here.)BUT according to the fact the current goes into the op-amp, that's our key in telling it's an inverter, right?
We needs to know!NascentOxygen said:What is the number written on the op-amp in your first schematic?
It's not clear to whom you direct your question, M Quack. Are you offering to guide Femme_physics deeper into the workings of this intriguing circuit? Or are you the first of her classmates to discover you can tap into Femme_physics online homework solutions, and you're angling for bonus marks?? http://img593.imageshack.us/img593/2293/starwarssmiley010.gifM Quack said:Why is Ra connected to the output of the op-amp, and not to the +25V supply? (obviously one would have to change the value of Ra to keep the current through the Zener at 2mA)
NascentOxygen said:It's not clear to whom you direct your question, M Quack. Are you offering to guide Femme_physics deeper into the workings of this intriguing circuit? Or are you the first of her classmates to discover you can tap into Femme_physics online homework solutions, and you're angling for bonus marks?? http://img593.imageshack.us/img593/2293/starwarssmiley010.gif
Ah, that old qurious disease. Is that what they call Q Fever?M Quack said:But I have become curious (professional disease among physicists).
You think that if the output of the op-amp were to be momentarily shorted to ground, it will stay there? Maybe you are right, or maybe not. How would you go about trying to show that 0v is a stable point, without actually constructing the circuit?would there not be a second operating point at V+=V-=Vout=0V, not currents flowing except into the op-amp?
A technique sometimes used in power supplies is to feed the zener from the regulated line so that the current into the zener is exceptionally stable. I don't know whether that thinking was behind the reason for doing it here.M Quack said:my point about feeding Ra from +25V. You have less control over the current through the Zener, though if there are variations on the +25V line.
Femme_physics said:You mean that Vout = 12V?
True, we can't really tell. Can we? It's rather confusing. BUT according to the fact the current goes into the op-amp, that's our key in telling it's an inverter, right?
Actually, your formula is perfectly correct! I was relying on memory and in my mind had Rb and R1 swapped around. :(NascentOxygen said:Femme_physics, I notice in your calculation of V(-) you used an erroneous formula for the resistive divider involving Rb:
True, but that's only half the story...rude man said:The circuit is definitely a non-inverting amplifier, gain = +2.4V/V. If you were to change the input from +5V to say + 5.5V you would get 2.4*5.5 = +2.64V output.
I guess that's right. Op-amps usually are powered by double supplies, so your Vs/2 term is unusual, but I can't see it could be anything else.M Quack said:The op-amp as a huge open-loop gain G such that Vout=G(V+ - V-) + Vs/2.
That's right.looking at the characterstic curve of the Zener diode: Around 0V, the Zener diode behaves like a resistor with a fairly large impedance Rz. I guess Rz>RI/4 (you will see later why).
So there is no stable state with Vout=0. In practice, you usually won't get the op-amp delivering an output equal to either supply rail, in any case, so you wouldn't be able to obtain Vout=0.situation is unstable and Vout will drift up until the Zener no longer behaves as a
resistor.