Calculating the relationship between Vout and Vin in a non-inverting op-amp

Question

Circuit:

I have to determine the relationship between V_out and V_in. When analyzing the circuit I realized it was quite similar to a non-inverting op-amp circuit. I know from my past homework that I can calculate Vout in the following way: $$V_\text{out} = (1 + R_2/R_1)V_\text{in}$$ This ends up giving me: $$V_\text{out} = (1 + 9\text{ k}/1\text{ k})V_\text{in} = 10V_\text{in}$$

However, to use this formula I had made the assumption in the past that \$V_{+} = V_\text{in}\$. But now there is a resistor in front of the positive entrance, so I do not think this still holds true.

However, I am not sure what value I'd give it now. I think I have to use the 5 V power source to decide this, but I also get the feeling it will have an effect on the negative side, but I do not know what sort. Any tips or hints to get me to the right direction?

Answer 1

The voltage, \$V_-=V_+=V_\text{in}\$ is produced by the voltage divider between \$V_\text{out}\$ and the \$ 5\text{ V}\$ source.

Thus, \$V_\text{in}=5+0.1(V_\text{out}-5) \$

Answer 2

As a practical matter, you know that an opamp's output voltage doesn't rise above its DC supply voltage (+5V in this case). And since it's negative DC supply is ground, output voltage cannot go below 0V. In-between these extremes, it operates linearly.

So it may be helpful to examine these two cases. What input voltages on +in would yield an output of 0V?...an output of +5V? As a start, assume the opamp has infinite gain: this means that for linear operation, +in voltage = -in voltage.

Here's the case where Vout rests at 0V...solve for V+in and for V-in:

^{simulate this circuit – Schematic created using CircuitLab}

• V-in is simply a resistor voltage divider 9k, 1k...so V-in=+4.5V
• V+in must also be +4.5V, if gain is infinite.

You can go through the same type calculation, setting Vout=+5V and finding where V-in and V+in rests. These are very quick, easy calculations that give you the extreme limits of linear operation for a rail-to-rail opamp. Any input voltage on +in pin outside this range saturates the output at 0V or at +5V. So it should be clear that any Vin outside the range from +4.5V up to +5.0 would result in an output outside the linear range of 0V up to +5V. When you come up with an equation describing Vout as a function of Vin, check that a Vin of +4.5V results in Vout of 0V.

Be aware that not every opamp has a linear range rail-to-rail. Output voltage might only approach its DC supply voltage to within a volt (or two) before its linear range is exceeded.
And linear input voltage range for -in and +in pins for non-rail-to-rail opamps are very often similarly limited.

So regard these limit calculations as extreme limits - real world may be more limited.

Answer 3

Anyone has any tips or hints to get me to the right direction?

Now that you understand that you can largely ignore the 1 kΩ resistor in series with the non-inverting input, you can begin. So, whatever voltage is the input (non-inverting input pin), the op-amp tries to force onto the inverting input pin via the 9 kΩ feedback resistor.

So, if the input is (say) 3 volts, 3 volts is the voltage at the inverting input. Then you solve this: -

^{simulate this circuit – Schematic created using CircuitLab}

There are 2 volts across R1 hence, the current is 2mA. And, 2 mA also flows towards Vout hence, Vout is 18 volts lower than 3 volts = -15 volts. Of course this breaks the rule that says the op-amp output has to be within the power rails but, nevertheless, that is how you solve it.

Without an in-depth knowledge of the op-amp it's just an exercise in numbers with little practical use unless the op-amp is a rail-to-rail type or capable of working with inputs at the positive rail (some op-amps will be OK but, only a few percent).

Answer 4

Welcome Sydon: An operational amplifier characteristics can only be considered ideal if all of its inputs and outputs are properly connected. Ideally we call the negative input a virtual ground or a virtual input depending on the configuration.

Vin is not properly connected unless it is connected to a voltage source of some specified value. If open does this mean that the negative input is a virtual open? I ask this so viewers can carefully speculate on the concept.

If the question is to calculate the voltages around the circuit with Vin open then the answer is, "It can't be done." If you build this circuit or simulate with different real amplifiers, that resulting values will range from almost 0 to almost 5V.

I speculate that there is an assumption to be made based on the indication that the noninverting input is connected to a voltage source. So the circuit should be redrawn as:

^{simulate this circuit – Schematic created using CircuitLab}

We can now proceed to analyze this with superposition. We solve for each input separately with the other set to zero then add the results together.

So we short the voltage source we don't want to use. But it is the supply voltage as well. So disconnect NODE1 and ground it. Then\$ V_{OUT} = 10V_{IN}\$.

Next, set Vin to 0V and reconnect NODE1.In this case, \$V_{OUT}=-9(5)=-45V\$. Since this is nonsensical, the output will likely be zero.

So we can ask, "What values of \$V_{IN}\$ will produce an output between 0 and 5V?"

I leave that for you to complete.

All op-amps have a different input structure. If one input is left open then i/2 of the input amplifier will not function.

Cheers

Answer 5

Hint: the resistance of the resistor on the non-inverting input is negligible compared to the resistance of the actual input.

Darn, I meant this as a comment. So, anyway, now it's here, that means you can basically ignore the resistor's value. The full value of Vin can be used in your calculation. But note that your inverting input is referenced to V+.