# Practical Electronics/Operational amplifiers

## Intro

Op Amp is a short hand term for Operational Amplifier. An operational amplifier is a circuit component that amplifies the difference of two input voltages:

Vo = A (V2 - V1)

Op Amps are usually packaged as an 8-pin integrated circuit.

Pin Usage
1 Offset Null
2 Inverted Input
3 Non-Inverted Input
4 -V Supply
5 No use
6 Output
7 +V Supply
8 No use

Op Amp symbol

• V+: non-inverting input
• V: inverting input
• Vout: output
• VS+: positive power supply
• VS−: negative power supply

Op amps amplify AC signal or AC Voltage better than a simple bipolar junction transistor.

## Op Amp Functions

From above

V0 = A (V2 - V1)

### Voltage Comparator

V2 > V1 , V0 = +Vss
V2 < V1 , V0 = -Vss
V2 = V1 , V0 = 0

### Inverting Amplifier

With one voltage is grounded

If V2 = 0 , V0 = -A V1 . Inverting Amplifier

### Non-Inverting Amplifier

With one voltage is grounded

If V1 = 0 , V0 = A V2 . Non-Inverting Amplifier

## Linear Configurations

### Differential amplifier

$V_{\mathrm {out} }=V_{2}\left({\left(R_{\mathrm {f} }+R_{1}\right)R_{\mathrm {g} } \over \left(R_{\mathrm {g} }+R_{2}\right)R_{1}}\right)-V_{1}\left({R_{\mathrm {f} } \over R_{1}}\right)$ • Differential $Z_{\mathrm {in} }$ (between the two input pins) = $R_{1}+R_{2}$ #### Voltage Difference Amplifier

Whenever $R_{1}=R_{2}$ and $R_{\mathrm {f} }=R_{\mathrm {g} }$ ,

$V_{\mathrm {out} }={R_{\mathrm {f} } \over R_{1}}\left(V_{2}-V_{1}\right)$ #### Voltage Difference

When $R_{1}=R_{\mathrm {f} }$ and $R_{2}=R_{\mathrm {g} }$ (including previous conditions, so that $R_{1}=R_{2}=R_{\mathrm {f} }=R_{\mathrm {g} }$ ):

$V_{\mathrm {out} }=V_{2}-V_{1}\,\!$ ### Inverting Amplifier

$V_{\mathrm {out} }=-V_{\mathrm {in} }\left({R_{f} \over R_{1}}\right)$ Inverting Amplification is dictated by the ratio of the two resistors

### Non-Inverting Amplifier

$V_{\mathrm {out} }=V_{\mathrm {in} }\left(1+{R_{2} \over R_{1}}\right)$ Non-Inverting Amplification is dictated by the ratio of the two resistors plus one

### Voltage Follower

From Non-Inverting Amplifier's formula. If the resistors has the same value of resistance then output voltage is exactly equal to the input voltage

$V_{\mathrm {out} }=V_{\mathrm {in} }\!\$ From Inverting Amplifier's formula. If the resistors has the same value of resistance then output voltage is exactly equal to the input voltage and inverted

$V_{\mathrm {out} }=-V_{\mathrm {in} }\!\$ ### Summing amplifier

$V_{\mathrm {out} }=-R_{\mathrm {f} }\left({V_{1} \over R_{1}}+{V_{2} \over R_{2}}+\cdots +{V_{n} \over R_{n}}\right)$ When $R_{1}=R_{2}=\cdots =R_{n}$ , and $R_{\mathrm {f} }$ independent

$V_{\mathrm {out} }=-\left({R_{\mathrm {f} } \over R_{1}}\right)(V_{1}+V_{2}+\cdots +V_{n})\!\$ When $R_{1}=R_{2}=\cdots =R_{n}=R_{\mathrm {f} }$ $V_{\mathrm {out} }=-(V_{1}+V_{2}+\cdots +V_{n})\!\$ ### Integrator

Integrates the (inverted) signal over time

$V_{\mathrm {out} }=\int _{0}^{t}-{V_{\mathrm {in} } \over RC}\,dt+V_{\mathrm {initial} }$ (where $V_{\mathrm {in} }$ and $V_{\mathrm {out} }$ are functions of time, $V_{\mathrm {initial} }$ is the output voltage of the integrator at time t = 0.)

### Differentiator

Differentiates the (inverted) signal over time.

The name "differentiator" should not be confused with the "differential amplifier", also shown on this page.

$V_{\mathrm {out} }=-RC\left({dV_{\mathrm {in} } \over dt}\right)$ (where $V_{\mathrm {in} }$ and $V_{\mathrm {out} }$ are functions of time)

### Comparator

• $V_{\mathrm {out} }=\left\{{\begin{matrix}V_{\mathrm {S+} }&V_{1}>V_{2}\\V_{\mathrm {S-} }&V_{1} Từ V0 = A (V2 - V1)

• Vo = 0 khi V2 = V1
• Vo > 0 khi V2 > V1
Vo = Vss
• Vo < 0 khi V2 < V1
Vo = V-ss

When two input voltages equal. The output voltage is zero . When the two input voltages different and if one is greater than or less than the other

1. Vo = Vss khi V2 > V1
2. Vo = V-ss khi V2 < V1

### Instrumentation amplifier

Combines very high input impedance, high common-mode rejection, low DC offset, and other properties used in making very accurate, low-noise measurements

### Schmitt trigger

A comparator with hysteresis

Hysteresis from ${\frac {-R_{1}}{R_{2}}}V_{sat}$ to ${\frac {R_{1}}{R_{2}}}V_{sat}$ .

### Gyrator

A gyrator can transform impedances. Here a capacitor is changed into an inductor.

$L=R_{\mathrm {L} }RC$ ### Zero level detector

Voltage divider reference

• Zener sets reference voltage

### Negative impedance converter (NIC)

Creates a resistor having a negative value for any signal generator

• In this case, the ratio between the input voltage and the input current (thus the input resistance) is given by:
$R_{\mathrm {in} }=-R_{3}{\frac {R_{1}}{R_{2}}}$ ## Non-linear configurations

### Rectifier

Behaves like an ideal diode for the load, which is here represented by a generic resistor $R_{\mathrm {L} }$ .

• This basic configuration has some limitations. For more information and to know the configuration that is actually used, see the main article.

### Peak detector

When the switch is closed, the output goes to zero volts. When the switch is opened for a certain time interval, the capacitor will charge to the maximum input voltage attained during that time interval.

The charging time of the capacitor must be much shorter than the period of the highest appreciable frequency component of the input voltage.

### Logarithmic output

• The relationship between the input voltage $v_{\mathrm {in} }$ and the output voltage $v_{\mathrm {out} }$ is given by:
$v_{\mathrm {out} }=-V_{\gamma }\ln \left({\frac {v_{\mathrm {in} }}{I_{\mathrm {S} }\cdot R}}\right)$ where $I_{\mathrm {S} }$ is the saturation current.

• If the operational amplifier is considered ideal, the negative pin is virtually grounded, so the current flowing into the resistor from the source (and thus through the diode to the output, since the op-amp inputs draw no current) is:
${\frac {v_{\mathrm {in} }}{R}}=I_{\mathrm {R} }=I_{\mathrm {D} }$ where $I_{\mathrm {D} }$ is the current through the diode. As known, the relationship between the current and the voltage for a diode is:

$I_{\mathrm {D} }=I_{\mathrm {S} }\left(e^{\frac {V_{\mathrm {D} }}{V_{\gamma }}}-1\right)$ This, when the voltage is greater than zero, can be approximated by:

$I_{\mathrm {D} }\simeq I_{\mathrm {S} }e^{V_{\mathrm {D} } \over V_{\gamma }}$ Putting these two formulae together and considering that the output voltage $V_{\mathrm {out} }$ is the inverse of the voltage across the diode $V_{\mathrm {D} }$ , the relationship is proven.

Note that this implementation does not consider temperature stability and other non-ideal effects.

### Exponential output

• The relationship between the input voltage $v_{\mathrm {in} }$ and the output voltage $v_{\mathrm {out} }$ is given by:
$v_{\mathrm {out} }=-RI_{\mathrm {S} }e^{v_{\mathrm {in} } \over V_{\gamma }}$ where $I_{\mathrm {S} }$ is the saturation current.

• Considering the operational amplifier ideal, then the negative pin is virtually grounded, so the current through the diode is given by:
$I_{\mathrm {D} }=I_{\mathrm {S} }\left(e^{\frac {V_{\mathrm {D} }}{V_{\gamma }}}-1\right)$ when the voltage is greater than zero, it can be approximated by:

$I_{\mathrm {D} }\simeq I_{\mathrm {S} }e^{V_{\mathrm {D} } \over V_{\gamma }}$ The output voltage is given by:

$v_{\mathrm {out} }=-RI_{\mathrm {D} }\,$ 