Nonlinear Limitations of a Real Op Amp

[dudforreal]

Homework Statement
A certain op amp has a maximum output voltage range of ±9 V. The maximum output current magnitude is 20 mA. The slew-rate limit is SR = 300 kV/s. The op amp is used in the amplifier circuit, shown in the diagram below. For a frequency of 10 kHz and RL = 2.2 kΩ, what peak input voltage is possible without distortion?

The attempt at a solution

I'm not sure how to do this question and don't know how to relate the frequency and tried to multiply the current with the resistor RL but this didn't get the correct answer.

 

[rude man]

You've got three worries:
1. max output current spec
2. max output voltage spec
3. max slew rate spec

Your job is to find which of these three specs takes effect first (by 'first' I mean you slowly increase the input voltage & record the output voltage and current peaks until the first of the three limitations takes effect).

 

[dudforreal]

You've got three worries:
1. max output current spec
2. max output voltage spec
3. max slew rate spec

Your job is to find which of these three specs takes effect first (by 'first' I mean you slowly increase the input voltage & record the output voltage and current peaks until the first of the three limitations takes effect).

What is spec?
Are talking about placing an input test voltage and then record the output voltage and current peaks? And which formulaes do I use and how do I know which limitation is in effect?

 

[rude man]

'Spec' is short for 'specification'. Every op amp has specifications delineating the limitations of that particular amplifier.

Yes.

As I said, you have to consider each effect separately, then pick the one that limits the amplifier with the lowest input voltage.

For I(out) and E(out), just use Kirchhoff's laws.

For slew rate - what is the max slew rate for a voltage V0*sin(wt)? (Slew rate = dV/dt).

 

[dudforreal]

I just can't get the Kirchoff's Laws for the circuit.

 

[rude man]

You only have one unknown node: V(out). That's because V(in) = V1 = V2 of the op amp.

So now write the sum of currents = 0 at V(out) with an input voltage Vi = V0sin(wt).

 

[dudforreal]

where is V₁ and V₂? and how are they equal?

 

[rude man]

V1 is the - input to the op amp. V2 is the + input.

Since an ideal op amp has infinite gain, the two have to be equal or the output is infinite, right? That's called "saturation". A saturated op amp is totally useless.


(That statement is incorrect if we are including finite op amp gain, and offset voltage & input currents, but here we are not.)

 

[dudforreal]

(V_o - V_i)/R_f + V_o/R_L = 0?

 

[rude man]

How about the op amp's output current? Where's it going?

 

[dudforreal]

Isn't that it?

 

[rude man]

No, those are the currents flowing OUT OF the node, into RL and Rf. Where's the current flowing INTO the node?

 

[dudforreal]

Current flowing into the node equals zero.

 

[rude man]

Not hardly.
What about the op amp output pin?

 

[dudforreal]

Where is the output pin? Isn't that equal to zero as I said?

 

[rude man]

The output pin is Vo. There is current flowing out of that pin, or into it.

If an op amp couldn't produce output current then it couldn't produce any output voltage either (with your RL connected as shown, for example). What made you think that the op amp output voltage was identically zero?

 

[dudforreal]

no I was referring to the op amp's output current from the other side which is zero.

 

[rude man]

"The other side" is known as inputs, not output. The input currents of an ideal op amp are always zero. The output current is whatever it takes to satisfy the output voltage requirement, assuming the output is not saturated.