1
\$\begingroup\$

After I calculated that $$v_s=v_u\left(\frac{R_1}{R_1+R_2}\right)$$ I have to calculate the resistance seen by the voltage generator \$v_s\$.

My book, without any calculation, says it is: \$+\infty\$. Now, I am trying to figure out why.

I thought that the resistance seen by \$v_s\$ is $$\frac{v_s}{\frac{e_1-v_s}{R_s}}$$ (with the nodal analysis).

I tried to do algebraic calculations, but the results doesn't come out as \$+\infty\$. How should I calculate it?

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT: If ideal op. amplifiers have infinite input resistance, so should be in the following circuit:

schematic

simulate this circuit

My book says in this last circuit it's \$R_1\$ (without any calc. too). Why?

\$\endgroup\$
  • \$\begingroup\$ In the first circuit there is no current through Rs into the op-amp, hence input z is infinity. In the second circuit there is an input current, and that current flows through R1 and R2 to the op-amp output. Also, the -ve op-amp input is at virtual earth hence the magnitude of that current is Vs/R1, or in other words the input z is R1 \$\endgroup\$ – Chu May 11 '15 at 23:42
3
\$\begingroup\$

When an ideal op amp is connected with negative feedback, it obeys two rules:

  1. The voltages at the two input pins are equal.
  2. No current flows into either pin.

In your first circuit, \$V_S\$ is only connected to the non-inverting input. By rule #2, no current flows into that input. This lets us calculate the equivalent input resistance:

$$I_S = 0\ \mathrm A$$ $$R_{in} = \frac{V_S}{I_S} = \frac{V_S}{0\ \mathrm A} = \infty \ \Omega$$

Your second circuit is drawn incorrectly. You've connected R2 to one of the op amp's power pins, but it should be connected to the output. In this circuit, current can flow from \$V_S\$ to the output, although none flows into the inverting input. Here we need rule #1, which tells us that the voltage at the inverting input is equal to the voltage at the non-inverting input -- zero volts (ground). So the circuit acts like the right side of \$R_1\$ is grounded, which makes \$R_1\$ the input resistance. We can work out the math, too:

$$I_S = \frac{V_S - 0\ \mathrm V}{R_1} = \frac{V_S}{R_1}$$

$$R_{in} = \frac{V_S}{I_S} = \frac{V_S}{\frac{V_S}{R_1}} = R_1$$

\$\endgroup\$
  • \$\begingroup\$ The golden rules of Op Amps! Although, I believe a better way to say it is that the voltage of the two input pins will be equal. Your way is accurate, but might be misread at first. \$\endgroup\$ – Kurt E. Clothier May 12 '15 at 5:27
  • \$\begingroup\$ Yeah, that's a better way to phrase it. I edited the answer. \$\endgroup\$ – Adam Haun May 12 '15 at 13:43
2
\$\begingroup\$

Ideal op amps have infinite input resistance, and the voltage source is connected to only the input through Rs.

\$\endgroup\$
  • \$\begingroup\$ Could you please answer to my edit in the answer? \$\endgroup\$ – sl34x May 11 '15 at 17:04
1
\$\begingroup\$

At first, I assume that R2 (second picture) is connected to the output pin, OK?

1) First circuit (non-inverter): The input impedances of the opamp unit (without any external resistors) are very large (Mega-Ohm range) - and for most of the calculations they can be assumed to be infinite (∞). This large input resistance is even drastically enlarged due to the feedback effect (voltage feedback). For this reason, it is common practice to set, in this case, the input resistance for all calculations to an infinite value: Rin=Rs+∞=∞.

2.) The situation, however, is different for the second circuit (inverting amplifier). Now we have current feedback - and the input resistance (referred to the opamp input pin) is decreased due to the feedback effect (decreased by the large loop gain factor) and can be neglected if in series with Rs. Hence, the remaining input resistance (as seen by Vs) is only Rs.

Another explanation: For large values of the open-loop gain Ao (usually 1E5...1E6) the input differential voltage between both opamp inputs is in the microvolt range and can be neglected. Hence, we assume that the node voltage at the inv. input is at "virtual" ground - and the right side of R1 apprears to be grounded. Hence Rin=R1.

\$\endgroup\$
0
\$\begingroup\$

Rs is probably meant to be the source resistance. It is a property of (internal to) the voltage generator. Voltage sources are shown as Thevenin equivalent circuits consisting of an ideal voltage source in series with the source's internal resistance, as your diagram shows. The external load therefore is the op amp itself, not including any "Rs" external series resistor because it's not external. "The resistance seen by the voltage generator" would usually mean the resistance in the circuit external to the voltage generator.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.