Ideal Op Amp HW: KCL, V/R, V+ = V-

[CoolDude420]

Homework Statement

Homework Equations

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KCL
I = V/R
V+ = V- (ideal op amp)

The Attempt at a Solution

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I really don't think my voltages for ia and ib are correct. I am just not sure how to use the V- = 4V in those some how?

 

[Charles Link]

The $V_-$ terminal is equal to $V_+=+4V$. The three current sources feed into (or draw current from) the feedback resistor and feed through the feedback resistor. No current goes across the infinite input impedance. The output voltage is determined by the voltage drop through the feedback resistor. Any necessary additional output current out of the op-amp will be supplied through the load resistor (or extra current from the feedback resistor will go into the output pin) to make the voltage drop across the load resistor equal the voltage drop from the output pin of the op-amp to ground. (The output pin of the op-amp has its voltage determined by the $V_-$ virtual ground and the feedback loop, and can supply extra current to the load resistor freely as needed, with basically an infinite supply.) $\\$ Note: Your equation should read $i_d=i_a+i_b+i_c$. ($i_b$ and $i_c$ are both negative but the equation still reads how I just gave it.)

 

[CoolDude420]

The $V_-$ terminal is equal to $V_+=+4V$. The three current sources feed into (or draw current from) the feedback resistor and feed through the feedback resistor. No current goes across the infinite input impedance. The output voltage is determined by the voltage drop through the feedback resistor. Any necessary additional output current out of the op-amp will be supplied through the load resistor (or extra current from the feedback resistor will go into the output pin) to make the voltage drop across the load resistor equal the voltage drop from the output pin of the op-amp to ground. (The output pin of the op-amp has its voltage determined by the $V_-$ virtual ground and the feedback loop, and can supply extra current to the load resistor freely as needed, with basically an infinite supply.) $\\$ Note: Your equation should read $i_d=i_a+i_b+i_c$. ($i_b$ and $i_c$ are both negative but the equation still reads how I just gave it.)

Hm. I understand the KCL equation you gave. At the moment my issue is relating to filling in that KCL equation with values.

id = ia +ib + ic

Ia = 16/16000? Is that correct? My question is do I have to take the V- into account when calculating the currents for Ia and Ib. I know I have to use it for Ic.
I keep having trouble with voltages in op-amps related circuits. I've tried going over them and reviewing KCL and the basics but there's always a different sort of "trick" in each question with voltages.

 

[Charles Link]

Meanwhile to compute $i_c$, $i_c=(-5-4)V/(12 \, k \Omega)=-.75 mA$. The others should be easy to compute in a similar manner. And yes, you need to use the $V_-=4V$ in those as well. Note: $i_d$ should be fairly easy to compute. Once you have that, you can determine $V_o$.

 

[CoolDude420]

Meanwhile to compute $i_c$, $i_c=(-5-4)V/(12 \, k \Omega)=-.75 mA$. The others should be easy to compute in a similar manner. And yes, you need to use the $V_-=4V$ in those as well.

Is this correct?

One more question. In calculating ia and ib, my intuition tells me that it should be (16V - Voltage at A)/16000 and (2V- Voltage at A)/8000. However we are using -4V instead. Is the entire vertical line which contains A a single node? If so shouldn't the node voltage at A be 4V as well?

This is what my intuition tells me: V = eb - ea for the voltage across 16k resistor

 

[Charles Link]

You could say $i_d=-.25 mA$, but yes your -1/4000 is correct. What do you get for $V_o$? (That should be easy to get.) Your -6V voltage drop is correct. Suggest you write out the algebraic result to solve for $V_o$. If you consider $V_o$ as generating a positive +.25mA current back through the feedback resistor, it might be easier to compute.

 

[CoolDude420]

You could say $i_d=-.25 mA$, but yes your -1/4000 is correct. What do you get for $V_o$? (That should be easy to get.)

I think you apply current division to get V0:

= current through 12k resistor.
That turns out to be -1/6000.

V = IR
V = (-1/6000) x (12000)
V = -2V

which.. isn't the right answer. The answer given to us is 10V

 

[Charles Link]

The full algebra is $4-V_o=-6$. Thereby $V_o=10 \, V$. Alternatively, you could write $V_o-(.25mA)(24k \Omega)=4V$

 

[CoolDude420]

The full algebra is $4-V_o=-6$. Thereby $V_o=10 \, V$.

Where are you getting the 4 from?

 

[Charles Link]

$4V=V_-$ which is also point "A". Your diagram is probably confusing you slightly, but these 3 currents(considering them as positive) feed into $V_-$ And they can't go inside the op-amp because the input impedance is infinite. The current has nowhere else to go but through the feedback resistor.

 

[Charles Link]

@CoolDude420 Please read the edited version of post #10 above.

 

[CoolDude420]

$4V=V_-$ which is also point "A".

Ah so the 4V at point A plus the 6V drop across the 24k resistor gives 10V node voltage at the point to the right of i0. And that 10V is between The 13k resistor and ground so 10v across that resistor.

Perfect. I got the right answer for part b too. Awesome. thank you so much!