For the love of circuits

Summing currents $$ I_{r1}=I_{r2}+I_{o} $$ $$ \frac{v_i-v_o}{R_1}=\frac{v_o}{R_2}+I_o $$ Rearrange for $$ v_o=\frac{R_2v_i-I_oR_1R_2}{R_1+R_2} $$ Assuming $ I_o$ is negligible $$ \frac{v_o}{v_i}=\frac{R_2}{R_1+R_2} $$ This derevation is common place and usually assumed in many calculations.

1.2 First-Order Standard Form

$ \tau=\frac{1}{\omega_0}$ and $ s=j\omega $
Low-pass $$ \frac{v_o}{v_i}=k\frac{1}{1+s\tau} = k\frac{1}{1+j\frac{\omega}{\omega_0}} $$ High-pass $$ \frac{v_o}{v_i}=k\frac{s\tau}{1+s\tau} = k\frac{j\frac{\omega}{\omega_0}}{1+j\frac{\omega}{\omega_0}} $$ Lead-Lag $$ \frac{v_o}{v_i}=k\frac{1+s\tau_1}{1+s\tau_2} = k\frac{1+j\frac{\omega}{\omega_1}}{1+j\frac{\omega}{\omega_2}} $$

1.3 Second-Order Standard Form

Low-pass $$ \frac{v_o}{v_i} = k\frac{1}{1+ \frac{s\tau}{Q}+s^2\tau^2 } = k\frac{1}{1+j\frac{\omega}{Q\omega_0}-\frac{\omega^2}{\omega_0^2} } $$ High-pass $$ \frac{v_o}{v_i} = k\frac{s^2\tau^2}{1+\frac{s\tau}{Q}+s^2\tau^2} = k\frac{-\frac{\omega^2}{\omega_0^2}}{1+j\frac{\omega}{Q\omega_0}-\frac{\omega^2}{\omega_0^2}} $$ Band-pass $$ \frac{v_o}{v_i} = k\frac{\frac{s\tau}{Q}}{1+\frac{s\tau}{Q}+s^2\tau^2 } = k\frac{j\frac{\omega}{Q\omega_0}}{1+ j\frac{\omega}{Q\omega_0}-\frac{\omega^2}{\omega_0^2} } $$ Band-stop $$ \frac{v_o}{v_i} = k\frac{1-s^2\tau^2}{1+ \frac{s\tau}{Q}+s^2\tau^2 } = k\frac{1+\frac{\omega}{\omega_0^2}}{1+ j\frac{\omega}{Q\omega_0}-\frac{\omega^2}{\omega_0^2} } $$

1.4 Low-pass C

Figure 2:First order low pass capacitor circuit

$$ I_{r1}=I_{c1}+I_{o} $$ $$ \frac{v_i-v_o}{R_1}=\frac{v_o}{\frac{1}{j\omega C_1}}+I_o $$ Rearrange for $$ v_o=\frac{v_i-I_oR_1}{1+j\omega C_1R_1} $$ Assuming $ I_o$ is negligible $$ \frac{v_o}{v_i}=\frac{1}{1+j\omega C_1R_1} $$ Convert into first-order Standard form $$ \frac{v_o}{v_i}=1\frac{1}{1+j\frac{\omega }{\frac{1}{C_1R_1}}} $$

1.5 Low-pass L

Figure 3:First order low pass inductor circuit

$$ I_{r1}=I_{l1}+I_{o} $$ $$ \frac{v_i-v_o}{R_1}=\frac{v_o}{j\omega L_1}+I_o $$ Rearrange for $$ v_o=\frac{v_ij\omega L_1-I_oR_1j\omega L_1}{R_1+j\omega L_1 } $$ Assuming $ I_o$ is negligible $$ \frac{v_o}{v_i}=\frac{j\omega L_1}{R_1+j\omega L_1 } $$ Convert into first-order Standard form $$ \frac{v_o}{v_i}=\frac{j\omega L_1}{R_1}\frac{1}{1+j\frac{\omega}{\frac{R_1}{L_1}} } $$

1.6 Low-pass LC

Figure 4:Second order low pass LC circuit

$$ I_{l1}=I_{c1}+I_{o} $$ $$ \frac{v_i-v_o}{j\omega L_1}=\frac{v_o}{\frac{1}{j\omega C_1}}+I_o $$ Rearrange for $$ v_o=\frac{v_i-I_oj\omega L_1}{1-\omega^2 C_1L_1} $$ Assuming $ I_o$ is negligible $$ \frac{v_o}{v_i}=\frac{1}{1-\omega^2 C_1L_1} $$ Convert into second-order Standard form $$ \frac{v_o}{v_i} = 1\frac{1}{1-\frac{\omega^2}{\frac{1}{C_1L_1}} } $$

1.7 High-pass C

Figure 5:First order high pass capacitor circuit

$$ I_{c1}=I_{r1}+I_{o} $$ $$ \frac{v_i-v_o}{\frac{1}{j\omega C_1}}=\frac{v_o}{R_1}+I_o $$ Rearrange for $$ v_o=\frac{v_ij\omega C_1R_1-I_oR_1}{1+j\omega C_1R_1} $$ Assuming $I_o$ is negligible $$ \frac{v_o}{v_i}=\frac{v_ij\omega C_1R_1}{1+j\omega C_1R_1} $$ Convert into first-order Standard form $$ \frac{v_o}{v_i}=1\frac{j\frac{\omega}{\frac{1}{C_1R_1}} }{1+j\frac{\omega}{\frac{1}{C_1R_1}}} $$

1.8 High-pass L

Figure 6:First order high pass inductor circuit

$$ I_{r1}=I_{l1}+I_{o} $$ $$ \frac{v_i-v_o}{R_1}=\frac{v_o}{j\omega L_1}+I_o $$ Rearrange for $$ v_o=\frac{v_ij\omega L_1-jI_oR_1\omega L_1}{R_1+j\omega L_1} $$ Assuming $I_o$ is negligible $$ \frac{v_o}{v_i}=\frac{j\omega L_1}{R_1+j\omega L_1} $$ Convert into first-order Standard form $$ \frac{v_o}{v_i}=1\frac{j\frac{\omega}{\frac{R_1}{L_1}} }{1+j\frac{\omega}{\frac{R_1}{L_1}}} $$

1.9 High-pass LC

Figure 7:Second order high pass LC circuit

$$ I_{c1}=I_{l1}+I_{o} $$ $$ \frac{v_i-v_o}{\frac{1}{j\omega C_1}}=\frac{v_o}{j\omega L_1}+I_o $$ Rearrange for $$ v_o=-\frac{v_i\omega^2 C_1 L_1+I_oj\omega L_1}{1-\omega^2 C_1 L_1} $$ Assuming $ I_o$ is negligible $$ \frac{v_o}{v_i}=-\frac{\omega^2 C_1 L_1}{1-\omega^2 C_1 L_1} $$ Convert into second-order Standard form $$ \frac{v_o}{v_i}=1\frac{-\frac{\omega^2}{\frac{1}{C_1 L_1}} }{1-\frac{\omega^2}{\frac{1}{C_1 L_1}}} $$

1.10 Wheatstone Bridge

Figure 8:Wheatstone bridge circuit

$$ v_+=v_i\frac{R_3}{R_3+R_1} $$ $$ v_-=v_i\frac{R_4}{R_4+R_2} $$ $$ v_o=v_+-v_-=v_i(\frac{R_3}{R_3+R_1}-\frac{R_4}{R_4+R_2} ) $$ $$ \frac{v_o}{v_i}= \frac{R_3}{R_3+R_1}-\frac{R_4}{R_4+R_2} $$ If $R_3=R_1$ and $R_4=R_2$ $$ \frac{v_o}{v_i}= \frac{R_1}{2R_1}-\frac{R_2}{2R_2}=0 $$ If $R_3=R_1+\Delta R$ and $R_4=R_2$ $$ \frac{v_o}{v_i}= \frac{R_1+\Delta R}{2R_1+\Delta R}-\frac{1}{2}= \frac{\Delta R}{4R_1+2\Delta R} $$

2 Active Circuits

A collection of key active circuits.

2.1 Simplified Hybrid-Pi

Figure 9:Simplified hybrid-pi equivalent NPN

The Hybrid-Pi circuit is the most common small signal transistor eqivalent circuit. The model assumes the transistor is correctly biased in the linear region and working with low frequency signals. When these assumptions are acceptable the hybrid-pi model is very accurate. For circuits where these assumptions cannot be made the full hybrid-pi model can be used.
Transistors are often described as transconductance amplifiers, that is because for a voltage on the base $ v_{be}$ a current is generated through the collector $ i_c $. This gain is therefore $ g_m=\frac{i_c}{v_{be}}$. As the input resistance of the transistor is purely resistive in the model the transconductance can be described as a current gain aka for an input $ i_b $ the collector current is $ i_c $ this current gain is given the value $ \beta=\frac{i_c}{i_b}$. These equations can be seen in the hybrid-pi model on the current source.

2.2 Emitter follower

Figure 10:Basic emitter follower circuit

Figure 11:Emitter follower equivaltent hybrid-pi circuit

Voltage over $ R_e $ $$ v_o=i_eR_e $$ $ i_c=\beta i_b$ $$ \therefore i_e=i_b+i_c=i_b(\beta+1) $$ $$ v_o=(\beta+1)i_bR_e $$ Summing $v_e$ and $v_{be}$ for $v_i$ $$ v_i=v_{be}+v_e=i_b R_\pi+(\beta+1)i_bR_e $$ $$ \frac{v_o}{v_i}=\frac{(\beta+1)R_e}{R_\pi+(\beta+1)R_e} $$

2.3 Common Emitter

Figure 12:Basic common emitter circuit

Figure 13:Common emitter equivalent hybrid-pi circuit

For small signal analysis sources are shorted $v_+=0$,looking at $v_c$ $$ v_o=-v_c=-i_cR_c $$ $$ i_c=g_mv_{be}=g_mv_i $$ $$ \therefore v_o=-g_mv_iR_c $$ Rearranged $$ \frac{v_o}{v_i}=-g_mR_c $$

2.4 Common Emitter with Degeneration

Figure 14:Basic common emitter circuit with degeneration

Figure 15:Common emitter with degeneration equivalent hybrid-pi circuit

For small signal analysis sources are shorted $v_+=0$,looking at $v_c$ $$ v_o=-v_c=-i_cR_c $$ $$ i_c=\beta i_b $$ $$ v_o=-\beta i_bR_c $$ Summing $v_e$ and $v_{be}$ for $v_i$ $$ v_i=v_e+v_{be}=i_eR_e+i_bR_\pi=i_b(\beta+1)R_e+i_bR_\pi=i_b((\beta+1)R_e+R_\pi) $$ Subsituting and re=arranging $$ \frac{v_o}{v_i}=- \frac{\beta R_c}{(\beta+1)R_e+R_\pi} $$

2.5 Common Base Common base circuits are generally used as a current buffer given the 1:1 input to output current.

Figure 16:Basic common base circuit

Figure 17:Common base equivalent hybrid-pi circuit

$$ v_{be}=-v_i $$ For small signal analysis sources are shorted $v_+=0$,looking at $v_c$ $$ v_o=-v_c=-i_cR_c $$ $$ i_c=g_m(-v_i) $$ Substituting $$ v_o=g_mv_iR_c $$ Rearranging $$ \frac{v_o}{v_i}=g_mR_c $$ To derive the current characteristics instead look at an input of $i_i$ $$ i_{be}+i_i+i_c=0 $$ $$ \therefore i_i=-i_b-i_c=0 $$ Where $\beta i_b=i_c$ $$ \therefore i_b=-\frac{i_i}{\beta+1} $$ $$ \therefore i_c=i_b\beta=-\frac{i_i\beta}{\beta+1}\approx -i_i $$ $$ \therefore \frac{i_c}{i_i}\approx-1 $$ Showing a current close to $i_b$ out from $i_c$

3 Opamp Circuits

A collection of key opamp circuits.

3.1 Opamp

Figure 18:Ideal opamp symbol.

For an input of $ v_+ $ and $ v_- $ the output $ v_o $ is the difference between the inputs multiplied by an open loop gain $ A_v $, usually in the range of hundreds of thousands. This simplification can be represented as the equation below. $$ v_o=A_v(v_+-v_-)$$ Opamp calculations commonly assume $ A_v=\infty $ which, when in negative feedback leads to the assumption $ v_-=v_+ $. It is also commonly assumed opamps take no current on the input, like all assumptions this is valid until this current draw effects the voltage at the input.

3.2 Voltage Follower

Figure 19:Opamp voltage follower

Given $$ v_o=A_v(v_+-v_-)$$ $$ v_- = v_o $$ $$ v_+ = v_i $$ Subsitute in $$ \therefore v_o=A_v(v_i-v_o)$$ Rearrange for $$ \frac{v_o}{v_i}=\frac{A_v}{A_v+1} $$ Assuming $ A_v >> 1 $ $$ \frac{v_o}{v_i}=1 $$

3.3 Inverting Amplifier

Figure 20:Opamp inverting amplifier

Given $ v_+=0 $ and $ v_o=A_v(v_+-v_-)$. $$ v_o=-A_vv_- $$ Assuming no current into the opamp and treating $ R_1 $ and $ R_2 $ as a potential divider. $$ (v_o-v_i)\frac{R_1}{R_1+R_2}+v_i=v_- $$ Subsituting for $$ (v_o-v_i)\frac{R_1}{R_1+R_2}+v_i=-\frac{v_o}{A_v} $$ $$ v_o(\frac{R_1}{R_1+R_2}+\frac{1}{A_v})=v_i(\frac{R_1}{R_1+R_2}-1) $$ Rearranging for $$ \frac{v_o}{v_i}=\frac{\frac{R_1}{R_1+R_2}-1}{\frac{R_1}{R_1+R_2}+\frac{1}{A_v}} = \frac{-R_2}{R_1+\frac{R_1+R_2}{A_v}} $$ Assuming assuming $ A_v = \infty $ and the opamp isn't saturated $$ \frac{v_o}{v_i}=-\frac{R_2}{R_1} $$

3.4 Schmitt Trigger

Figure 21:Opamp schmitt trigger

In a Schmitt Trigger the op-amp is setup in positive feedback meaning for an increase in voltage at $ v_+ $ will increase $ v_o $ and visa versa and therefore $ v_+ $ leading to a system always in saturation. Assuming $v_o$ is currently saturated to the positive rail $v_o=v_s$ and treating $ R_1 $ and $ R_2 $ as a potential divider. $$ v_+=(v_s-v_i)\frac{R_1}{R_1+R_2}+v_i $$ To stop the system being in positive saturation $ v_+<=0 $ Substituting in for $$ 0=(v_s-v_i)\frac{R_1}{R_1+R_2}+v_i $$ Rearranging for $$ v_i=-v_s\frac{R_1}{R_2}$$ This can be proven in the other direction similarly giving $$ v_i=v_s\frac{R_1}{R_2}$$ Giving a direction based resonse as seen below.

Figure 22:Schmitt Trigger response

3.5 Complex Inverting Amplifier

Figure 23:Opamp complex inverting amplifier

Similar to a typical inverting amplifier but with a complex impedances $ z_1 $ and $ z_2 $ instead of $ R_1 $ and $ R_2 $.
Given $ v_+=0 $ and $ v_o=A_v(v_+-v_-)$. $$ v_o=-A_vv_- $$ Assuming no current into the opamp and treating $ z_1 $ and $ z_2 $ as a complex potential divider. $$ (v_o-v_i)\frac{z_1}{z_1+z_2}+v_i=v_- $$ Subsituting for $$ (v_o-v_i)\frac{z_1}{z_1+z_2}+v_i=-\frac{v_o}{A_v} $$ $$ v_o(\frac{z_1}{z_1+z_2}+\frac{1}{A_v})=v_i(\frac{z_1}{z_1+z_2}-1) $$ Rearranging for $$ \frac{v_o}{v_i}=\frac{\frac{z_1}{z_1+z_2}-1}{\frac{z_1}{z_1+z_2}+\frac{1}{A_v}} = \frac{-z_2}{z_1+\frac{z_1+z_2}{A_v}} $$ Assuming assuming $ A_v = \infty $ and the opamp isn't saturated $$ \frac{v_o}{v_i}=-\frac{z_2}{z_1} $$

3.6 Noninverting Amplifier

Figure 24:Opamp non-inverting amplifier

Assuming no current into the opamp $$ i_1=i_2 $$ $$ \therefore v_-=v_o\frac{R_2}{R_1+R_2} $$ Given $$ v_o=A_v(v_+-v_-) $$ Substituing in for $$ v_o=A_v(v_i-v_o\frac{R_2}{R_1+R_2}) $$ Rearranging for $$ \frac{v_o}{v_i}=\frac{1}{\frac{1}{A_v}+\frac{R_2}{R_1+R_2}} $$ Assuming assuming $ A_v = \infty $ and the opamp isn't saturated $$ \frac{v_o}{v_i}=1+\frac{R_1}{R_2} $$

3.7 Complex Noninverting Amplifier

Figure 25:Opamp complex non-inverting amplifier

Similar to a typical noninverting amplifier but with a complex impedances $ z_1 $ and $ z_2 $ instead of $ R_1 $ and $ R_2 $.
Assuming no current into the opamp $$ i_1=i_2 $$ $$ \therefore v_-=v_o\frac{z_2}{z_1+z_2} $$ Given $$ v_o=A_v(v_+-v_-) $$ Substituing in for $$ v_o=A_v(v_i-v_o\frac{z_2}{z_1+z_2}) $$ Rearranging for $$ \frac{v_o}{v_i}=\frac{1}{\frac{1}{A_v}+\frac{z_2}{z_1+z_2}} $$ Assuming assuming $ A_v = \infty $ and the opamp isn't saturated $$ \frac{v_o}{v_i}=1+\frac{z_1}{z_2} $$

3.8 Integrator

Figure 26:Opamp integrator

It is possible to derive the integrating action using the equations already derived in the complex inverting amplifier but will be explained from scratch here. Assuming no current into the opamp $$ i_r+i_c=0 $$ Using $ I=C\frac{dv}{dt} $ $$ I_c=C_1\frac{dv_o-dv_-}{dt}=-\frac{v_i-v_-}{R_1} $$ Assuming $A_v = \infty $ then $v_- = v_+= 0 $ $$ C_1\frac{dv_o}{dt}=-\frac{v_i}{R_1} $$ Integrate for 0 to t where the opamp never saturates and $ R_1$ and $ C_1$ are constant $$ v_o=-\frac{1}{R_1C_1}\int_0^t v_i \,dt $$

3.9 Differentiator

Figure 27:Opamp differentiator

It is possible to derive the integrating action using the equations already derived in the complex inverting amplifier but will be explained from scratch here. Assuming no current into the opamp $$ i_r+i_c=0 $$ Using $ I=C\frac{dv}{dt} $ $$ I_c=C_1\frac{dv_i-dv_-}{dt}=-\frac{v_o-v_-}{R_1} $$ Assuming $A_v = \infty $ then $v_- = v_+= 0 $ $$ C_1\frac{dv_i}{dt}=-\frac{v_o}{R_1} $$ Rearrange for $$ v_o=-R_1C_1\frac{ dv_i}{dt} $$

3.10 Summing Inverting Amplifier

Figure 28:Opamp summing inverting amplifier

Assuming no current into the opamp $$ i_1+i_2+i_3=-i_4 $$ $$ \frac{v_{i1}-v_-}{R_1}+\frac{v_{i2}-v_-}{R_2}+\frac{v_{i3}-v_-}{R_3}=-\frac{v_o-v_-}{R_4} $$ $$ \frac{v_{i1}R_2R_3R_4+v_{i2}R_1R_3R_4+v_{i3}R_1R_2R_4+v_{o}R_1R_2R_3}{R_2R_3R_4+R_1R_3R_4+R_1R_2R_4+R_1R_2R_3}=v_- $$ $$ v_o=\frac{-A_v\frac{v_{i1}R_2R_3R_4+v_{i2}R_1R_3R_4+v_{i3}R_1R_2R_4}{R_2R_3R_4+R_1R_3R_4+R_1R_2R_4+R_1R_2R_3}}{1+A_v\frac{R_1R_2R_3}{R_2R_3R_4+R_1R_3R_4+R_1R_2R_4+R_1R_2R_3}} $$ Assuming $ A_v\frac{R_1R_2R_3}{R_2R_3R_4+R_1R_3R_4+R_1R_2R_4+R_1R_2R_3}>>1 $ $$ v_o=-\frac{v_{i1}R_2R_3R_4+v_{i2}R_1R_3R_4+v_{i3}R_1R_2R_4}{R_1R_2R_3} $$ if $ R_1=R_2=R_3=R_i$ $$ v_o=-\frac{R_4}{R_i}(v_{i1}+v_{i3}+v_{i3}) $$

3.11 Summing Non-Inverting Amplifier

Figure 29:Opamp summing non-inverting amplifier

Assuming no current into the opamp $$ i_1+i_2+i_3=0 $$ $$ \frac{v_{i1}-v_+}{R_1}+\frac{v_{i2}-v_+}{R_2}+\frac{v_{i3}-v_+}{R_3}=0 $$ if $ R_1=R_2=R_3=R_i$ $$ \frac{v_{i1}+v_{i2}+v_{i3}}{3}=v_{+} $$ looking at $ v_-$ and assuming no current into the opamp $$ v_-=v_o\frac{R_5}{R_4+R_5} $$ Subsitute into $ v_o=A_v(v_+-v_-)$ $$ v_o=A_v(\frac{v_{i1}+v_{i2}+v_{i3}}{3}-v_o\frac{R_5}{R_4+R_5}) $$ $$ v_o=\frac{A_v(\frac{v_{i1}+v_{i2}+v_{i3}}{3})}{(1+A_v\frac{R_5}{R_4+R_5})} $$ Assuming $ A_v\frac{R_5}{R_4+R_5}>>1 $ $$ v_o=\frac{R_4+R_5}{R_5}(\frac{v_{i1}+v_{i2}+v_{i3}}{3}) $$

3.12 Differential Amplifier

Figure 30:Opamp differential amplifier

Looking at $ v_- $ and assuming no current into the opamp. $$ i_1+i_2=0 $$ $$ \frac{v_{i1}-v_-}{R_1}+\frac{v_{o}-v_-}{R_2}=0 $$ $$ R_2v_{i1}+R_1v_{o}=v_-(R_1+R_2) $$ $$ \frac{R_2v_{i1}+R_1v_{o}}{R_1+R_2}=v_- $$ Looking at $ v_+ $ and assuming no current into the opamp. $$ v_+=v_{i2}\frac{R_4}{R_3+R_4} $$ Subsitute into $ v_o=A_v(v_+-v_-)$ $$ v_o=A_v(v_{i2}\frac{R_4}{R_3+R_4}-\frac{R_2v_{i1}+R_1v_{o}}{R_1+R_2}) $$ $$ v_o=\frac{A_v(v_{i2}\frac{R_4}{R_3+R_4}-\frac{R_2v_{i1}}{R_1+R_2})}{1+\frac{A_vR_1}{R_1+R_2}} $$ Assuming $ \frac{A_vR_1}{(R_1+R_2)}>>1 $ $$ v_o=\frac{v_{i2}\frac{R_4}{R_3+R_4}-\frac{R_2v_{i1}}{R_1+R_2}}{\frac{R_1}{R_1+R_2}} $$ $$ v_o=v_{i2}\frac{R_4(R_1+R_2)}{R_1(R_3+R_4)}-\frac{R_2v_{i1}}{R_1} $$ if $ R_1=R_3 $ and $ R_2=R_4 $ $$ v_o=\frac{R_2}{R_1}(v_{i2}-v_{i1}) $$