# Fundamental Physics/Electricity/Electric circuits/Electric Circuits

## Electric Circuits

### Resistor Circuit

 Entities Symbol Mathematical Formula Unit Voltage ${\displaystyle V}$ ${\displaystyle IR}$ Volt . v Current ${\displaystyle I}$ ${\displaystyle {\frac {V}{R}}}$ Ampere . a Resistance ${\displaystyle R}$ ${\displaystyle {\frac {V}{I}}}$ Ohom . Ω Conductance ${\displaystyle G}$ ${\displaystyle {\frac {I}{V}}}$ 1/Ω Power provided ${\displaystyle P_{V}}$ ${\displaystyle IV}$ Watt . W Power loss ${\displaystyle P_{R}}$ ${\displaystyle I^{2}R(T)}$ W Power transmitted ${\displaystyle P}$ ${\displaystyle P_{V}-P_{R}}$ W

### RL Circuit

RL Circuit refers to a circuit having combination of resistance(s) and inductor(s). They are commonly used in chokes of luminescent tubes. In an A.C. circuit, inductors helps in reducing voltage, without the loss of energy. Due to the inductive reactance, the higher the AC frequency, the greater the impeadence of the inductor. Under DC conditions, an inductor acts as a static resistance.

Like RC circuit , with one resistor and one coil can be connected to form a low pass filter or a high pass filter. A high-pass filter allows frequencies above the cut-off frequency to pass, while a low-pass filter allows frequencies beneath the cut-off frequency to pass. The arrangement of the resistor and the capacitor is what determines their behaviour.

Note that at a particular frequencly, called the cut-off frequency, the Inductive Reactance is equal to the Resitance value. (There is also an associated phase shift of 45 degrees.)

${\displaystyle R=X_{L}}$

Substituting ${\displaystyle X_{L}={2\pi fL}}$ we then have:

${\displaystyle R={2\pi fL}}$

The cut-off frequency, defined as the frequency at which the signal power is attenuated by 50% (or 3.01 dB), is a function of the resistive and capacitive values. We can rearrange the above formula to solve for ${\displaystyle f}$ as follows:

${\displaystyle f_{cut-off}={\frac {R}{2\pi L}}}$

#### RL Series

A circuit of 2 component a resistor and an inductor connected in series

${\displaystyle V_{L}+V_{R}=0}$
${\displaystyle L{\frac {d}{dt}}i(t)+Ri(t)=0}$
${\displaystyle {\frac {d}{dt}}i(t)=-{\frac {R}{L}}i(t)}$
${\displaystyle \int {\frac {di(t)}{i(t)}}=-\int {\frac {R}{L}}dt}$
${\displaystyle Lni(t)=-{\frac {R}{L}}t+c}$
${\displaystyle i(t)=e^{-{\frac {R}{L}}t+c}}$
${\displaystyle i(t)=Ae^{-{\frac {R}{L}}t}}$

#### High pass filter

When the inductor is in parallel with the load while the resistor is in series with the inductor and load, this creates a high pass filter.

High pass filter has a transfer function

${\displaystyle H(j\omega )={\frac {v_{o}}{v_{i}}}={\frac {j\omega T}{1+j\omega T}}}$
${\displaystyle T={\frac {L}{R}}}$

Frequency response of High pass filter

${\displaystyle \omega =0.v_{o}=0}$
${\displaystyle \omega =\omega _{o}.v_{o}={\frac {v_{i}}{2}}}$
${\displaystyle \omega =00.v_{o}=v_{i}}$

Cut off frequency, ${\displaystyle \omega _{o}}$ , frequency at which ${\displaystyle v_{o}={\frac {1}{2}}v_{i}}$

${\displaystyle \omega _{o}={\frac {1}{T}}}$

#### Low pass filter

When the resistor is in parallel with the load while the inductor is in series with the resistor and load, a low pass filter is created.

Low pass filter has a transfer function

${\displaystyle H(j\omega )={\frac {v_{o}}{v_{i}}}={\frac {1}{1+j\omega T}}}$
${\displaystyle T={\frac {L}{R}}=RC}$

Frequency response of Low pass filter

${\displaystyle \omega =0.v_{o}=0}$
${\displaystyle \omega =\omega _{o}.v_{o}={\frac {v_{i}}{2}}}$
${\displaystyle \omega =00.v_{o}=v_{i}}$

Cut off frequency, ${\displaystyle \omega _{o}}$ , frequency at which ${\displaystyle v_{o}={\frac {1}{2}}v_{i}}$

${\displaystyle \omega _{o}={\frac {1}{T}}}$

A single RL circuit creates a filter with a 20.0 dB/decade, or 6.02 dB/octave, slope.

### RC Circuits

RC circuits are circuits that contain a resistor and a capacitor. These circuits are primarily used as frequency filters. There are two basic arrangements: high-pass and low-pass. A high-pass filter allows frequencies above the cut-off frequency to pass, while a low-pass filter allows frequencies beneath the cut-off frequency to pass. The arrangement of the resistor and the capacitor is what determines their behaviour.

Note that at a particular frequencly, called the cut-off frequency, the Capactive Reactance is equal to the Resistance value. (There is also an associated phase shift of 45 degrees.)

${\displaystyle R=X_{c}}$

Substituting

${\displaystyle X_{C}={\frac {1}{2\pi fC}}}$ we then have:
${\displaystyle R={\frac {1}{2\pi fC}}}$

The cut-off frequency, defined as the frequency at which the signal power is attenuated by 50% (or 3.01 dB), is a function of the resistive and capacitive values. We can rearrange the above formula to solve for ${\displaystyle f}$ as follows:

${\displaystyle f_{cut-off}={\frac {1}{2\pi RC}}}$

#### RC series

A circuit of 2 component a resistor and a capacitor connected in series

${\displaystyle V_{C}+V_{R}=0}$
${\displaystyle C{\frac {d}{dt}}v(t)+{\frac {v}{R}}v(t)=0}$
${\displaystyle {\frac {d}{dt}}v(t)=-{\frac {1}{RC}}v(t)}$
${\displaystyle \int {\frac {dv(t)}{v(t)}}=-\int {\frac {1}{RC}}dt}$
${\displaystyle Lnv(t)=-{\frac {1}{RC}}t+c}$
${\displaystyle v(t)=e^{(-{\frac {1}{RC}}t+c)}}$
${\displaystyle v(t)=Ae^{-{\frac {1}{RC}}t}}$

#### Low pass filter

When the capacitor is in parallel with the load while the resistor is in series with the capacitor and load, this creates a low pass filter.

Low pass filter has a transfer function

${\displaystyle H(j\omega )={\frac {v_{o}}{v_{i}}}={\frac {1}{1+j\omega T}}}$
${\displaystyle T={\frac {L}{R}}=RC}$

Frequency response of Low pass filter

${\displaystyle \omega =0.v_{o}=v_{i}}$
${\displaystyle \omega =\omega _{o}.v_{o}={\frac {v_{i}}{2}}}$
${\displaystyle \omega =00.v_{o}=0}$

Cut off frequency, ${\displaystyle \omega _{o}}$ , frequency at which ${\displaystyle v_{o}={\frac {1}{2}}v_{i}}$

${\displaystyle \omega _{o}={\frac {1}{T}}}$

#### High pass filter

When the resistor is in parallel with the load and the capacitor is in series with the resistor, a high pass filter is created.

High pass filter has a transfer function

${\displaystyle H(j\omega )={\frac {v_{o}}{v_{i}}}={\frac {j\omega T}{1+j\omega T}}}$
${\displaystyle T={\frac {L}{R}}=RC}$

Frequency response of High pass filter

${\displaystyle \omega =0.v_{o}=0}$
${\displaystyle \omega =\omega _{o}.v_{o}={\frac {v_{i}}{2}}}$
${\displaystyle \omega =00.v_{o}=v_{i}}$

Cut off frequency, ${\displaystyle \omega _{o}}$ , frequency at which ${\displaystyle v_{o}={\frac {1}{2}}v_{i}}$

${\displaystyle \omega _{o}={\frac {1}{T}}}$

A single RC circuit creates a filter with a 20.0 dB/decade, or 6.02 dB/octave, slope.

### RLC Circuit

A circuit of 3 component a resistor and a capacitor and an inductor connected in series

#### Circuit at equilibrium

${\displaystyle V_{C}+V_{L}+V_{R}=0}$
${\displaystyle L{\frac {d}{dt}}i(t)+{\frac {1}{C}}\int i(t)dt+Ri(t)=0}$
${\displaystyle {\frac {d^{2}}{dt^{2}}}i(t)+{\frac {R}{L}}{\frac {d}{dt}}i(t)+{\frac {1}{LC}}i(t)=0}$
${\displaystyle {\frac {d^{2}}{dt^{2}}}i(t)+2\alpha {\frac {d}{dt}}i(t)+\beta i(t)=0}$
${\displaystyle {\frac {d^{2}}{dt^{2}}}i(t)=-2\alpha {\frac {d}{dt}}i(t)-\beta i(t)}$

Lapalce transform yield

${\displaystyle s^{2}i(t)=-2\alpha si(t)-\beta i(t)}$

Solution to the above equation

• 1 real root
${\displaystyle s=-\alpha }$
${\displaystyle i(t)=Ae^{-\alpha t}}$
• 2 real roots
${\displaystyle s=-\alpha \pm {\sqrt {\alpha -\beta }}}$
${\displaystyle i(t)=Ae^{(-\alpha \pm {\sqrt {\alpha -\beta }})t}}$
• 2 complex roots
${\displaystyle s=-\alpha \pm j{\sqrt {\beta -\alpha }}}$
${\displaystyle i(t)=Ae^{(-\alpha \pm j{\sqrt {\beta -\alpha }})t}=A(\alpha )Sin\omega t}$
${\displaystyle A(\alpha )=Ae^{\alpha t}}$
${\displaystyle \omega ={\sqrt {\beta -\alpha }}}$
${\displaystyle \beta ={\frac {1}{T}}={\frac {1}{LC}}}$
${\displaystyle \alpha =\beta \gamma ={\frac {R}{2L}}}$
${\displaystyle T=LC}$
${\displaystyle \gamma =RC}$

#### Circuit at resonant

${\displaystyle V_{L}+V_{C}+V_{R}=R}$
${\displaystyle V_{L}+V_{C}=0}$
${\displaystyle i(\omega =0)=0}$
${\displaystyle i(\omega =\omega _{o})={\frac {v}{R}}}$
${\displaystyle i(\omega =0)=0}$

### LC Circuit

A circuit of 2 component a capacitor and an inductor connected in series

#### Circuit at equilibrium

${\displaystyle V_{C}+V_{L}=0}$
${\displaystyle L{\frac {d}{dt}}i(t)+{\frac {1}{C}}\int i(t)dt=0}$
${\displaystyle {\frac {d^{2}}{dt^{2}}}i(t)+{\frac {1}{LC}}i(t)=0}$
${\displaystyle {\frac {d^{2}}{dt^{2}}}i(t)=-{\frac {1}{T}}i(t)}$
${\displaystyle s^{2}i(t)=-{\frac {1}{T}}i(t)}$
${\displaystyle i(t)=ASin\omega t}$
${\displaystyle \omega ={\sqrt {\frac {1}{T}}}}$
${\displaystyle T=LC}$

#### Circuit at Resonant

${\displaystyle Z_{L}=-Z_{C}}$
${\displaystyle V_{L}=-V_{C}}$
${\displaystyle V(\theta )=ASin(\omega _{o}t+2\pi )-ASin(\omega _{o}t-2\pi )}$

### Diode's circuits

#### An electronics switch

Diode is switch On when ${\displaystyle V_{i}>V_{d}}$
Diode is switch Off when ${\displaystyle V_{i}

### Transistor's circuits

With the right connection, by connecting transistor with resistor(s) Transistor can acts as

#### An electronics amplifier

##### Current amplifier
${\displaystyle i_{b}>0}$
${\displaystyle i_{e}>\alpha i_{b}}$
${\displaystyle i_{c}>\beta i_{b}}$

Provided that, transistor should be turned ON with biased voltage at the base must be greater than diode's break over voltage

${\displaystyle v_{b}>v_{be}}$ . ${\displaystyle v_{be}=v_{d}}$
##### Non inverting voltage amplifier
${\displaystyle {\frac {v_{o}}{v_{i}}}={\frac {R_{2}}{R_{2}+R_{1}}}{\frac {R_{4}}{R_{3}}}}$

With

${\displaystyle R_{4}=R_{3}}$ and ${\displaystyle R_{2}=0}$ . ${\displaystyle v_{o}=0}$ . Transistor acts as Switch OFF (open circuit)
${\displaystyle R_{4}=R_{3}}$ and ${\displaystyle R_{1}=0}$ . ${\displaystyle v_{o}=1}$ . Transistor acts as Switch ON (close circuit)

With

${\displaystyle R_{1}=0}$ and ${\displaystyle R_{4}=nR_{3}}$ . ${\displaystyle v_{o}=nv_{i}}$ . Transistor acts as Non inverting amplifier
##### Inverting voltage amplifier
${\displaystyle {\frac {v_{o}}{v_{i}}}=1-{{\frac {R_{2}}{R_{2}+R_{1}}}{\frac {R_{4}}{R_{3}}}}}$

With

${\displaystyle R_{4}=R_{3}}$ and ${\displaystyle R_{2}=0}$ . ${\displaystyle v_{o}=1}$ . Transistor acts as Switch ON (close circuit)
${\displaystyle R_{4}=R_{3}}$ and ${\displaystyle R_{1}=0}$ . ${\displaystyle v_{o}=0}$ . Transistor acts as Switch OFF (open circuit)

With

${\displaystyle R_{1}=0}$ and ${\displaystyle R_{4}=(n+1)R_{3}}$ . ${\displaystyle v_{o}=-nv_{i}}$ . Transistor acts as Inverting amplifier

#### An electronics switch

Transistor is switch On when ${\displaystyle V_{b}>V_{be}}$
Transistor is switch Off when ${\displaystyle V_{b}

#### A buffer

${\displaystyle v_{c}=v_{b}}$

### Operational amplifier's circuits

#### Amplifiers

We begin these examples with that of the differential amplifier, from which many of the other applications can be derived, including the inverting, non-inverting, and summing amplifier, the voltage follower, integrator, differentiator, and gyrator.

##### Differential amplifier (difference amplifier)

Amplifies the difference in voltage between its inputs.

The name "differential amplifier" must not be confused with the "differentiator," which is also shown on this page.
The "instrumentation amplifier," which is also shown on this page, is a modification of the differential amplifier that also provides high input impedance.

The circuit shown computes the difference of two voltages, multiplied by some gain factor. The output voltage:

${\displaystyle V_{\text{out}}={\frac {\left(R_{\text{f}}+R_{1}\right)R_{\text{g}}}{\left(R_{\text{g}}+R_{2}\right)R_{1}}}V_{2}-{\frac {R_{\text{f}}}{R_{1}}}V_{1}=\left({\frac {R_{1}+R_{\text{f}}}{R_{1}}}\right)\cdot \left({\frac {R_{\text{g}}}{R_{\text{g}}+R_{2}}}\right)V_{2}-{\frac {R_{\text{f}}}{R_{1}}}V_{1}.}$

Or, expressed as a function of the common mode input Vcom and difference input Vdif

${\displaystyle V_{\text{com}}=(V_{1}+V_{2})/2;V_{\text{dif}}=V_{2}-V_{1}\,,}$

the output voltage is

${\displaystyle V_{\text{out}}{\frac {R_{1}}{R_{\text{f}}}}=V_{\text{com}}{\frac {R_{1}/R_{\text{f}}-R_{2}/R_{\text{g}}}{1+R_{2}/R_{\text{g}}}}+V_{\text{dif}}{\frac {1+(R_{2}/R_{\text{g}}+R_{1}/R_{\text{f}})/2}{1+R_{2}/R_{\text{g}}}}.}$

In order for this circuit to produce a signal proportional to the voltage difference of the input terminals, the coefficient of the Vcom term (the common-mode gain) must be zero, or

${\displaystyle R_{1}/R_{\text{f}}=R_{2}/R_{\text{g}}}$

With this constraint If you think of the left-hand side of the relation as the closed-loop gain of the inverting input, and the right-hand side as the gain of the non-inverting input, then matching these two quantities provides an output insensitive to the common-mode voltage of ${\displaystyle V_{1}}$ and ${\displaystyle V_{2}}$.</ref> in place, the common-mode rejection ratio of this circuit is infinitely large, and the output

${\displaystyle V_{\text{out}}={\frac {R_{\text{f}}}{R_{1}}}V_{\text{dif}}={\frac {R_{\text{f}}}{R_{1}}}\left(V_{2}-V_{1}\right).}$

where the simple expression Rf / R1 represents the closed-loop gain of the differential amplifier.

The special case when the closed-loop gain is unity is a differential follower, with:

${\displaystyle V_{\text{out}}=V_{2}-V_{1}.\,}$
##### Inverting amplifier

An inverting amplifier is a special case of the differential amplifier in which that circuit's non-inverting input V2 is grounded, and inverting input V1 is identified with Vin above. The closed-loop gain is Rf / Rin, hence

${\displaystyle V_{\text{out}}=-{\frac {R_{\text{f}}}{R_{\text{in}}}}V_{\text{in}}\!\,}$.

The simplified circuit above is like the differential amplifier in the limit of R2 and Rg very small. In this case, though, the circuit will be susceptible to input bias current drift because of the mismatch between Rf and Rin.

To intuitively see the gain equation above, calculate the current in Rin:

${\displaystyle i_{\text{in}}={\frac {V_{\text{in}}}{R_{\text{in}}}}}$

then recall that this same current must be passing through Rf, therefore (because V = V+ = 0):

${\displaystyle V_{\text{out}}=-i_{\text{in}}R_{\text{f}}=-V_{\text{in}}{\frac {R_{\text{f}}}{R_{\text{in}}}}}$

A mechanical analogy is a seesaw, with the V node (between Rin and Rf) as the fulcrum, at ground potential. Vin is at a length Rin from the fulcrum; Vout is at a length Rf. When Vin descends "below ground", the output Vout rises proportionately to balance the seesaw, and vice versa.

##### Non-inverting amplifier

A non-inverting amplifier is a special case of the differential amplifier in which that circuit's inverting input V1 is grounded, and non-inverting input V2 is identified with Vin above, with R1R2. Referring to the circuit immediately above,

${\displaystyle V_{\text{out}}=\left(1+{\frac {R_{\text{2}}}{R_{\text{1}}}}\right)V_{\text{in}}\!\,}$.

To intuitively see this gain equation, use the virtual ground technique to calculate the current in resistor R1:

${\displaystyle i_{1}={\frac {V_{\text{in}}}{R_{1}}}\,,}$

then recall that this same current must be passing through R2, therefore:

${\displaystyle V_{\text{out}}=V_{\text{in}}+i_{1}R_{2}=V_{\text{in}}\left(1+{\frac {R_{2}}{R_{1}}}\right)}$

Unlike the inverting amplifier, a non-inverting amplifier cannot have a gain of less than 1.

A mechanical analogy is a class-2 lever, with one terminal of R1 as the fulcrum, at ground potential. Vin is at a length R1 from the fulcrum; Vout is at a length R2 further along. When Vin ascends "above ground", the output Vout rises proportionately with the lever.

The input impedance of the simplified non-inverting amplifier is high, of order Rdif × AOL times the closed-loop gain, where Rdif is the op amp's input impedance to differential signals, and AOL is the open-loop voltage gain of the op amp; in the case of the ideal op amp, with AOL infinite and Rdif infinite, the input impedance is infinite. In this case, though, the circuit will be susceptible to input bias current drift because of the mismatch between the impedances driving the V+ and V op amp inputs.

##### Voltage follower (unity buffer amplifier)

Used as a buffer amplifier to eliminate loading effects (e.g., connecting a device with a high source impedance to a device with a low input impedance).

${\displaystyle V_{\text{out}}=V_{\text{in}}\!\ }$
${\displaystyle Z_{\text{in}}=\infty }$ (realistically, the differential input impedance of the op-amp itself, 1 MΩ to 1 TΩ)

Due to the strong (i.e., unity gain) feedback and certain non-ideal characteristics of real operational amplifiers, this feedback system is prone to have poor stability margins. Consequently, the system may be unstable when connected to sufficiently capacitive loads. In these cases, a lag compensation network (e.g., connecting the load to the voltage follower through a resistor) can be used to restore stability. The manufacturer data sheet for the operational amplifier may provide guidance for the selection of components in external compensation networks. Alternatively, another operational amplifier can be chosen that has more appropriate internal compensation.

##### Summing amplifier

A summing amplifier sums several (weighted) voltages:

${\displaystyle V_{\text{out}}=-R_{\text{f}}\left({\frac {V_{1}}{R_{1}}}+{\frac {V_{2}}{R_{2}}}+\cdots +{\frac {V_{n}}{R_{n}}}\right)}$
• When ${\displaystyle R_{1}=R_{2}=\cdots =R_{n}}$, and ${\displaystyle R_{\text{f}}}$ independent
${\displaystyle V_{\text{out}}=-{\frac {R_{\text{f}}}{R_{1}}}(V_{1}+V_{2}+\cdots +V_{n})\!\ }$
• When ${\displaystyle R_{1}=R_{2}=\cdots =R_{n}=R_{\text{f}}}$
${\displaystyle V_{\text{out}}=-(V_{1}+V_{2}+\cdots +V_{n})\!\ }$
• Output is inverted
• Input impedance of the nth input is ${\displaystyle Z_{n}=R_{n}}$ (${\displaystyle V_{-}}$ is a virtual ground)
##### 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

#### Oscillators

##### Wien bridge oscillator

Produces a very low distortion sine wave. Uses negative temperature compensation in the form of a light bulb or diode.

#### Comparator

An operational amplifier can, if necessary, be forced to act as a comparator. The smallest difference between the input voltages will be amplified enormously, causing the output to swing to nearly the supply voltage. However, it is usually better to use a dedicated comparator for this purpose, as its output has a higher slew rate and can reach either power supply rail. Some op-amps have clamping diodes on the input that prevent use as a comparator.

#### Integration and differentiation

##### Inverting integrator

The integrator is mostly used in analog computers, analog-to-digital converters and wave-shaping circuits.

Integrates (and inverts) the input signal Vin(t) over a time interval t, t0 < t < t1, yielding an output voltage at time t = t1 of

${\displaystyle V_{\text{out}}(t_{1})=V_{\text{out}}(t_{0})-{\frac {1}{RC}}\int _{t_{0}}^{t_{1}}V_{\text{in}}(t)\,\operatorname {d} t}$

where Vout(t0) represents the output voltage of the circuit at time t = t0. This is the same as saying that the output voltage changes over time t0 < t < t1 by an amount proportional to the time integral of the input voltage:

${\displaystyle -{\frac {1}{RC}}\int _{t_{0}}^{t_{1}}V_{\text{in}}(t)\,\operatorname {d} t}$

This circuit can be viewed as a low-pass electronic filter, one with a single pole at DC (i.e., where ${\displaystyle \omega =0}$) and with gain.

A slightly more complex circuit can ameliorate the second two problems, and in some cases, the first as well.

Here, the feedback resistor Rf provides a discharge path for capacitor Cf, while the series resistor at the non-inverting input Rn, when of the correct value, alleviates input bias current and common-mode problems. That value is the parallel resistance of Ri and Rf, or using the shorthand notation ||:

${\displaystyle R_{\text{n}}={\frac {1}{{\frac {1}{R_{\text{i}}}}+{\frac {1}{R_{\text{f}}}}}}=R_{\text{i}}||R_{\text{f}}}$

The relationship between input signal and output signal is now:

${\displaystyle V_{\text{out}}(t_{1})=V_{\text{out}}(t_{0})-{\frac {1}{R_{i}C_{f}}}\int _{t_{0}}^{t_{1}}V_{\text{in}}(t)\,\operatorname {d} t}$

##### Inverting differentiator

Differentiates the (inverted) signal over time.

${\displaystyle V_{\text{out}}=-RC\,{\frac {\operatorname {d} V_{\text{in}}}{\operatorname {d} t}}\,\qquad {\text{where }}V_{\text{in}}{\text{ and }}V_{\text{out}}{\text{ are functions of time.}}}$
• This can also be viewed as a high-pass electronic filter. It is a filter with a single zero at DC (i.e., where angular frequency ${\displaystyle \omega =0}$ radians) and gain. The high-pass characteristics of a differentiating amplifier (i.e., the low-frequency zero) can lead to stability challenges when the circuit is used in an analog servo loop (e.g., in a PID controller with a significant derivative gain). In particular, as a root locus analysis would show, increasing feedback gain will drive a closed-loop pole toward marginal stability at the DC zero introduced by the differentiator.

#### Synthetic elements

##### Inductance gyrator

Simulates an inductor (i.e., provides inductance without the use of a possibly costly inductor). The circuit exploits the fact that the current flowing through a capacitor behaves through time as the voltage across an inductor. The capacitor used in this circuit is smaller than the inductor it simulates and its capacitance is less subject to changes in value due to environmental changes.

This circuit is unsuitable for applications relying on the back EMF property of an inductor as this will be limited in a gyrator circuit to the voltage supplies of the op-amp.

##### 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:

${\displaystyle R_{\text{in}}=-R_{3}{\frac {R_{1}}{R_{2}}}}$

In general, the components ${\displaystyle R_{1}}$, ${\displaystyle R_{2}}$, and ${\displaystyle R_{3}}$ need not be resistors; they can be any component that can be described with an impedance.

#### Non-linear

##### Precision rectifier

The voltage drop VF across the forward biased diode in the circuit of a passive rectifier is undesired. In this active version, the problem is solved by connecting the diode in the negative feedback loop. The op-amp compares the output voltage across the load with the input voltage and increases its own output voltage with the value of VF. As a result, the voltage drop VF is compensated and the circuit behaves very nearly as an ideal (super) diode with VF = 0 V.

The circuit has speed limitations at high frequency because of the slow negative feedback and due to the low slew rate of many non-ideal op-amps.

##### Logarithmic output

• The relationship between the input voltage Vin and the output voltage Vout is given by:
${\displaystyle V_{\text{out}}=-V_{\text{T}}\ln \left({\frac {V_{\text{in}}}{I_{\text{S}}\,R}}\right)}$
where IS is the saturation current and VT is the thermal voltage.
• If the operational amplifier is considered ideal, the inverting input 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:
${\displaystyle {\frac {V_{\text{in}}}{R}}=I_{\text{R}}=I_{\text{D}}}$
where ID is the current through the diode. As known, the relationship between the current and the voltage for a diode is:
${\displaystyle I_{\text{D}}=I_{\text{S}}\left(e^{\frac {V_{\text{D}}}{V_{\text{T}}}}-1\right).}$
This, when the voltage is greater than zero, can be approximated by:
${\displaystyle I_{\text{D}}\simeq I_{\text{S}}e^{\frac {V_{\text{D}}}{V_{\text{T}}}}.}$
Putting these two formulae together and considering that the output voltage is the negative of the voltage across the diode (Vout = −VD), the relationship is proven.

This implementation does not consider temperature stability and other non-ideal effects.

##### Exponential output

• The relationship between the input voltage ${\displaystyle V_{\text{in}}}$ and the output voltage ${\displaystyle V_{\text{out}}}$ is given by:
${\displaystyle V_{\text{out}}=-RI_{\text{S}}e^{\frac {V{\text{in}}}{V_{\text{T}}}}}$

where ${\displaystyle I_{\text{S}}}$ is the saturation current and ${\displaystyle V_{\text{T}}}$ is the thermal voltage.

• Considering the operational amplifier ideal, then the negative pin is virtually grounded, so the current through the diode is given by:
${\displaystyle I_{\text{D}}=I_{\text{S}}\left(e^{\frac {V_{\text{D}}}{V_{\text{T}}}}-1\right)}$

when the voltage is greater than zero, it can be approximated by:

${\displaystyle I_{\text{D}}\simeq I_{\text{S}}e^{\frac {V_{\text{D}}}{V_{\text{T}}}}.}$

The output voltage is given by:

${\displaystyle V_{\text{out}}=-RI_{\text{D}}.\,}$