Operational Amplifier

The circuit schematic of the typical 741 op-amp is shown below:

A component-level diagram of the common 741 op-amp. Dotted lines outline:

• Current mirrors/current source (red)
• Differential amplifier (blue)
• Class A gain stage (magenta)
• Voltage level shifter (green);
• output stage (cyan)

Like all op-amps, the circuit basically consists of three stages:

• Differential amplifier with high input impedance that generates a voltage signal, the amplified voltage difference $v^+-v^-$.
• Voltage amplifier (class A amplification) with a high voltage gain to further amplify the voltage.
• Output amplifier (class AB push-pull emitter follower) with low output impedance and high current driving capability.

The op-amp requires two voltage supplies $\pm V_{cc}$ of both polarities (typically $V_{CC}=15$ V).

Although the op-amp circuit may look complicated, the analysis of its operation and behaviors can be simplfied based on the following assumptions:

• The huge input resistance $r_{in}$ can be treated as infinity $r_{in}\rightarrow \infty$.
• The input current drawn by an op-amp is small ( $10^{-9}\sim10^{-12}\;A$), and could be approximated to be zero $I^+=I^-=0$.
• The small output impedance $r_{out}$ can be treated as zero $r_{out}\approx 0$, i.e., the output $V_{out}$ is not affected by the load $R_L$ (so long as it is much greater than $r_{out}$).
• The bandwidth is large ( $1 \sim 20MHz$), i.e., the property of the op-amp remain unchanged for all frequencies of interest.

Based on these approximations, an op-amp can be modeled in terms of the following three parameters:

• Input impedance $r_{out}$: very large, typically a few mega-Ohms or higher ( $10^6\sim 10^{12}\,\Omega$, e.g., 741 $6\;M\Omega$), depending on the frequency and specific components used (e.g., BJT or FET).

• Output resistance $r_{out}$:, very small, typically a few tens of ohms, e.g., 75 $\Omega$.

• Open-circuit gain $A$:, based on both the inverting input $v^-$ and the non-inverting input $v^+$:

  $$v_{out}=A_d (v^+ - v^-)+A_c \frac{1}{2}(v^+ + v^-)\approx A_d (v^+ - v^-)$$ (1)

  where $A_d$ is the differential-mode gain and $A_c$ is the common-mode gain. It is desired that $A_d\rightarrow \infty$ and $A_c\rightarrow 0$, i.e., the output is only proportional to the difference $v^+-v^-$ between the two inputs. The common-mode rejection ratio (CMRR) is defined as the ratio between differential-mode gain and common-mode gain:

  $$CMRR=20\;\log_{10} \left(\frac{A_d}{A_c}\right)>100\;dB, \;\;\;\;\;\;\;\;(A_d>10^5 A_c)$$ (2)

Also, as the output $v_{out}=A(v^+-v^-)$ is in the range between $-V_{CC}$ and $V_{CC}$ and $A>10^5$ is large, $v^+-v^-=v_{out}/A$ is small (in the micro-volt range), i.e., $v^-\approx v^+$. If, as in some op-amp circuits, $v^+=0$ is grounded, then $v^-\approx v^+=0$ is very close to zero, i.e., it is almost the same as ground, or virtual ground. The analysis of various op-amp circuits can be much simplified by this virtual ground assumption.

As $A$ is large, $V_{out}=A(v^+-v^-)$ is usually saturated, equal to either $V_{CC}$ or $-V_{CC}$ (called the “rails”), depending on whether or not $v^+$ is greater than $v^-$. For $v_{out}$ to be meaningful, some kind of negative feedback is needed. In the following, we consider some typical op-amp circuits to show how to analyze an Op-amp circuit to find its input resistance $R_{in}$, output resistance $R_{out}$, and open-circuit voltage gain $G_{oc}$.

• Voltage follower: The input is connected to the positive input while the output is directly connected to the negative input (100% negative feedback), as shown in (A) in the figure below. The op-amp can be modeled by its input impedance $r_{in}$, output impedance $r_{out}$ and voltage gain $A$, as shown in (B). Then the voltage follower can be modeled by its input impedance $R_{in}$, output impedance $R_{out}$, and voltage gain $G_{oc}$, as shown in (C).

  

  Specifically, $R_{in}$, $R_{out}$ and $G_{oc}$ can be found below. Here the voltage source in the op-amp is $A(v^+-v^-)=A \;i_{in} r_{in}$.

  • Input impedance $R_{in}$: Applying KVL to the loop we get

    $$v_{in}=i_{in}(r_{in}+r_{out})+A\;(v^+-v^-) =i_{in}(r_{in}+r_{out})+A\;i_{in}r_{in} =i_{in}((A+1)r_{in}+r_{out})$$ (3)

    Dividing both sides by $i_{in}$ we get the input impedance:

    $$R_{in}=\frac{v_{in}}{i_{in}}=(A+1)r_{in}+r_{out} \approx Ar_{in}$$ (4)

  • Open-circuit voltage gain $G_{oc}$: The open-circuit output voltage is

    $$v_{out}=v_{oc}=A(v^+-v^-)+i_{in}r_{out}=Ai_{in}r_{in}+i_{in}r_{out} =i_{in}(Ar_{in}+r_{out})$$ (5)

    The open-circuit gain is

    $$G=\frac{v_{out}}{v_{in}}=\frac{Ar_{in}+r_{out}}{(A+1)r_{in}+r_{out}}\approx 1$$ (6)

    The approximation is due to the fact that $(A+1)r_{in}\gg r_{out}$ and $A\gg 1$. We therefore have

    $$v_{out}=v_{oc}\approx v_{in}$$ (7)

  • Output impedance $R_{out}$: With a short-circuit load, we have $v^+-v^-=v_{in}$, and the short-circuit current can be found by superposition:

    $$i_{sc}=\frac{v_{in}}{r_{in}}+\frac{A(v^+-v^-)}{r_{out}} =\frac{v_... ...ft(\frac{r_{out}+Ar_{in}}{r_{in}r_{out}}\right) \approx v_{in}\frac{A}{r_{out}}$$ (8)

    As we also know $v_{oc}=v_{out}\approx v_{in}$, we get the output impedance (Thevenin's model):

    $$R_{out}=\frac{v_{oc}}{i_{sc}}\approx \frac{v_{in}}{i_{sc}} =\left(\frac{r_{in}r_{out}}{r_{out}+Ar_{in}}\right) \approx \frac{r_{out}}{A}$$ (9)

    The approximation is due to the fact that $Ar_{in}\gg r_{out}$

  In summary, we see that the voltage follower has a unit voltage gain, but much increased input resistance $R_{in}\approx A r_{in}$ (e.g., $10^{10}\Omega$) and much reduced output resistance $R_{out}\approx r_{out}/A$ (e.g., $10^{-3} \Omega$). In practice we could simply assume $R_{in}=\infty$ and $R_{out}=0$.

  Example:

  

  The figure on the left shows a circuit represented by an ideal voltage source $V_s$ in series with an internal resistance $R_s$ (Thevenin's theorem), with a load $R_L$. The voltage delivered to the load by this non-ideal source is

  $$v_{out}=v_s \;\frac{R_L}{R_L+R_s} =v_s - v_s\left(\frac{R_s}{R_L+R_s} \right)<v_s$$ (10)

  We see that the output voltage across the load $R_L$ is only a fraction of the voltage due to the voltage drop $v_s R_s/(R_L+R_s)$ across the internal resistance $R_s$. If it is desired for the output voltage to be as close to the source as possible, the internal resistance $R_s$ needs to be small compared to the load resistance $R_L$.

  Next consider inserting a voltage follower (buffer), in between the source and the load, as shown in the middle figure. The follower is modeled by its input and output resistances $R_{in}$ and $R_{out}$, as well as its voltage gain $G_{oc}$, as shown in the right figure. The output voltage can be obtained after two levels of voltage dividers:

  $$v_{out}=G_{oc} v_{in} \left(\frac{R_L}{R_{out}+R_L}\right)= G_{oc... ...frac{R_{in}}{R_s+R_{in}}\right)\left(\frac{R_L}{R_{out}+R_L}\right) \approx v_s$$ (11)

  We see that the output voltage across $R_L$ can be very close to source voltage, i.e., $v_{out}\approx v_s$, due to the nature of the op-amp:
  • $G_{oc} \approx 1$
  • $R_{in}\gg R_s \;\;\Longrightarrow \;\;R_{in}/(R_s+R_{in})\approx 1$
  • $R_{out}\ll R_L\;\;\Longrightarrow \;\;R_L/(R_{out}+R_L) \approx 1$

• Inverting Amplifier

  

  As the analysis of the circuit using full model of the op-amp is very involved, certain approximation is made to simplify the analysis.

  • Open-circuit voltage gain $(R_s=0,\;R_L=\infty)$:

    As $r_{in}\gg R_1,\,R_f$ and $r_{out}\ll R_1,\,R_f$, we approximate $r_{in}\approx \infty$ and $r_{out}\approx 0$. Also, we have $v^+=0$, $v^-=v_s-i_{in} R_1$ and $v^+-v^-=-v^-=i_{in} R_1 -v_s$. Now we have

    $$i_{in}=\frac{v_s-A(v_+-v_-)}{R_1+R_f}=\frac{v_s-A(i_{in} R_1-v_s)}{R_1+R_f}$$ (12)

    Solving for $i_{in}$ we get

    $$i_{in}=\frac{(A+1)v_s}{(A+1)R_1+R_f}$$ (13)

    The output voltage is:

    $$v_{out}=v_s-(R_1+R_f) i_{in} =v_s\left[1-\frac{(A+1)(R_1+R_f)}{(A+1)R_1+R_f}\right] =v_s\frac{-AR_f}{(A+1)R_1+R_f}$$ (14)

    Now we have the open-circuit voltage gain:

    $$G_{oc}=\frac{v_{out}}{v_s}=\frac{-AR_f}{(A+1)R_1+R_f} \approx - \frac{R_f}{R_1}$$ (15)

    The approximation is due to the fact that $A\gg 1$.

  • Input resistance:

    We assume $R_s=0$, and find the input resistance $R_{in}$ as the ratio of $v_s$ and the input current $i_{in}$. By KCL applied to the node of $v^-$:

    $$\frac{v^--v_s}{R_1}+\frac{v^--(-Av^-)}{R_f}=0$$ (16)

    Solving for $v^-$:

    $$v^-=v_s\frac{R_f}{R_f+(A+1)R_1}$$ (17)

    The input current is
    

    $$i_{in} = \frac{v_s-v^-}{R_1} =\frac{v_s}{R_1}\left[1-\frac{R_f}{R_f+(A+1)R_1}\right]$$

    $$= v_s\frac{1}{R_1}\frac{(A+1)R_1}{R_f+(A+1) R_1} =v_s\;\frac{A+1}{R_f+(A+1) R_1}$$ (18)

    
    Dividing $v_s$ by $i_{in}$ we get

    $$R_{in}=\frac{R_f+(A+1) R_1}{A+1}\approx R_1$$ (19)

    Note that this input resistance is significantly smaller than that of the voltage follower with $R_{in}\approx A r_{in}$!

  • Output resistance: Here we assume $r_{in}\rightarrow \infty$ to simplify the analysis.
    • Find short-circuit output current: Applying KCL to the output node we get

      $$i_{sc}=\frac{-A v^-}{r_{out}}+\frac{v^-}{R_f} =v^-\;\left(\frac{r_{out}-AR_f}{r_{out}R_f} \right)$$ (20)

      but as $v^-=v_s \,R_f/(R_s+R_1+R_f)$, the above becomes

      $$i_{sc}=v_s\left(\frac{R_f}{R_s+R_1+R_f}\right) \left(\frac{r_{out}-AR_f}{r_{out}R_f} \right) =v_s\,\frac{r_{out}-AR_f}{(R_s+R_1+R_f)r_{out}}$$ (21)

    • Find open-circuit output voltage: Applying KCL to the node of $v^-$ we get

      $$\frac{v_s-v^-}{R_s+R_1}+\frac{(-Av^-)-v^-}{R_f+r_{out}}=0$$ (22)

      Solving for $v^-$ we get

      $$v^-=v_s \left(\frac{R_f+r_{out}}{(A+1)(R_s+R_1)+R_f+r_{out}} \right)$$ (23)

      The open-circuit output voltage can be found to be (voltage divider)
      

      $$v_{oc} = [v^--(-Av^-)]\frac{r_{out}}{R_f+r_{out}}-Av^- =v^- \left(\frac{r_{out}-AR_f}{R_f+r_{out}}\right)$$

      $$= v_s \;\frac{r_{out}-AR_f}{(A+1)(R_s+R_1)+R_f+r_{out}}$$ (24)

      
      
    • Find output resistance:
      

      $$R_{out} = \frac{v_{oc}}{i_{sc}} =\frac{(R_s+R_1+R_f)r_{out}}{(A+1)(R_s+R_1)+R_f+r_{out}}$$

      $$\approx \frac{(R_s+R_1+R_f)r_{out}}{A(R_s+R_1)} \approx \left(\frac{R_1+R_f}{R_1}\right) \frac{r_{out}}{A}$$ (25)

      
      The approximation is based on $A\gg 1$ and $R_s\ll R_1,\,R_f$.

  In summary,

  • open-circuit voltage gain:

    $$G_{oc}\approx -\frac{R_f}{R_1}$$ (26)

  • input resistance:

    $$R_{in}\approx R_1$$ (27)

  • output resistance:

    $$R_{out}\approx \left(\frac{R_1+R_f}{R_1}\right)\;\frac{r_{out}}{A}$$ (28)

• Non-Inverting Amplifier (Homework)

  

  The three parameters of this non-inverting amplifier can be found to be (see here):

  • open-circuit voltage gain:

    $$G_{oc} \approx \frac{R_1+R_f}{R_1}>1$$ (29)

  • input resistance:

    $$R_{in} \approx \left(\frac{R_1}{R_1+R_f}\right)A r_{in} \approx\frac{A}{G_{oc}}r_{in}<A r_{in}$$ (30)

  • output resistance:

    $$R_{out}\approx \left(\frac{R_1+R_f}{R_1} \right)\,\frac{r_{out}}{A} \approx\frac{G_{oc}}{A}r_{out}>r_{out}/A$$ (31)

  Comparing these results with those of the voltage follower, we see that $G_{oc}>1$ is a little better, but both $R_{in}$ and $R_{out}$ are a little worse. In particular if $R_f=0$, this non-inverting amplifier becomes a voltage follower with $G_{oc} \approx 1$, $R_{in}\approx A r_{in}$, and $R_{out}\approx r_{out}/A$.