notes/Resources/electricity/active-components/op-amp-circuits.md

Non-Inverting Amplifier

$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
a 192 240 304 240 9 15 -15 1000000 4.9999000019999595 5 100000
r 192 320 192 400 0 1000
r 304 320 192 320 0 1000
w 192 320 192 256 0
w 304 240 304 320 0
O 304 240 368 240 1 0
g 192 400 192 432 0 0
v 96 352 96 224 0 0 40 5 0 0 0.5
w 96 224 192 224 2
g 96 352 96 432 0 0
b 144 288 289 401 0
x 264 386 278 389 4 24 β
x 240 345 252 348 4 12 Rf
x 160 363 176 366 4 12 Rg

What is the closed loop gain of this circuit?

\begin{flalign}
&V_- = V_+ = V_s&\\\
&V_- \text{is the output of a voltage divider}\\
\end{flalign}

Because V_- is equal to V_+ and

V_- = V_s = V_o (\frac{R_g}{R_G+R_F})

If we solve that equation for \frac{V_O}{V_s} we get the following formula:

Formula

\displaystyle Gain =\frac{V_o}{V_s} = 1+\frac{R_F}{R_G}

Buffer (Voltage-Follow)

This circuit is useful, because the output always replicates the voltage at the input. For example if you connect the output of a voltage divider you can drive a load wth V_o and the impedance in the Load will not change the V_o

$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
a 192 240 304 240 9 15 -15 1000000 4.999950000499995 5 100000
w 192 320 192 256 0
w 304 240 304 320 0
O 304 240 368 240 1 0
v 96 304 96 224 0 0 40 5 0 0 0.5
w 96 224 192 224 2
g 96 304 96 352 0 0
w 192 320 304 320 0

Inverting Amplifier

When the output of the op-amp is connected to its own inverting input, and the non-inverting input is connected to ground then it is in a closed loop inverting configuration.

Because the op-amp tries make sure that the voltage of both its inputs pins are the same, it will try to create a voltage which cancels out the voltage on the inverting input. This means if the input is 5V then the output voltage is -5V.

Another way to think about this is, that we know that the op-amp inputs do not consume current. So all the current goes through Ri and Rf that means that they both have the same exact current flowing through them. Through ohms law we can then figure out the voltage drops across them.

$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
v 64 304 64 192 0 0 40 5 0 0 0.5
r 176 192 112 192 0 1000
w 176 224 176 272 0
g 176 272 176 320 0 0
g 64 304 64 320 0 0
r 176 128 288 128 0 1000
w 288 128 288 208 0
w 176 192 176 128 0
g 288 272 288 320 0 0
p 288 208 288 272 1 0 0
x 230 94 242 97 4 12 Rf
x 129 166 141 169 4 12 Ri
a 176 208 288 208 8 15 -15 1000000 0.00004999900001999959 0 100000
w 112 192 64 192 2

\begin{flalign}
&A_V = -\frac{R_F}{R_I}&\\\
\end{flalign}

Single Ended Inverting Amplifier

When we connect V_- of the op-amp to ground, the output signal can't go below 0V this means that if the input signal goes below 0V it is cut of to fix this problem, we need to raise the signal by half of the V_+ voltage so that it does not get cut off.

y=\sin(x*5)|5>x>0|y>0
y = \sin(x*5)|x<0
y=\sin(x*5)+1|5<x

To do that we create a voltage divider which takes half of the V_+ voltage and routes it to the non-inverting input of the op-amp.

$ 1 0.000005 6.450009306485578 50 5 50 5e-11
O 352 224 432 224 0 0
w 352 160 352 224 0
r 224 160 352 160 0 3000
w 224 208 224 160 0
r 160 160 224 160 0 1000
g 224 320 224 352 0 0
r 224 240 224 320 0 1000
r 80 240 224 240 0 1000
R 64 160 16 160 0 1 40 2 0 0 0.5
R 80 240 16 240 0 0 40 15 0 0 0.5
a 224 224 352 224 8 15 0 1000000 7.499985233224565 7.5 100000
c 112 160 160 160 0 0.00001 -6.787036831414538 0.001
s 64 160 112 160 0 0 false
o 0 16 0 12294 13.0746715037385 0.0001 0 1 output
o 8 16 0 12294 2 0.0001 0 2 8 3 source

Difference Amplifier

$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
R 144 192 96 192 0 0 40 4 0 0 0.5
R 144 224 96 224 0 0 40 5 0 0 0.5
r 144 192 224 192 0 1000
r 144 224 224 224 0 1000
a 224 208 352 208 8 15 -15 1000000 2.499990000199996 2.5 100000
r 224 128 352 128 0 1000
w 352 128 352 208 0
w 224 192 224 128 0
w 352 208 400 208 0
p 400 208 400 320 1 0 0
g 400 320 400 352 0 0
r 224 224 224 320 0 1000
g 224 320 224 352 0 0
x 176 171 192 174 4 12 R1
x 176 244 192 247 4 12 R1
x 199 279 215 282 4 12 R2
x 279 148 295 151 4 12 R2

\displaystyle V_O = \frac{R2}{R1}(V_2-V_1)

Calculate Non-Inverting Amplifier Bandwidth

Let's calculate the bandwidth for the following non-inverting op-amp with a GBP of 1Mhz.

! The formula for calculating the gain of a non-inverting op-amp is: !#Non-Inverting Amplifier#Formula

Now with the numbers in the graph, that:

A_V = 1+\frac{99*10^3}{10^3} = 100

So now we now the gain of our circuit, but we did not check if it is even in the bandwidth, so lets to that now:

!formulas#Gain Bandwidth Product

\displaystyle f_c = \frac{1*10^6}{100} = 10*10^3 = 10kHz

That means that over 10kHz the gain of our op-amp is not at 100 anymore.

We can then lower the gain of a single op-amp to increase its bandwidth like so: !op-amp-bandwidth-example-2.jpg Where our new gain is now:

A_V = 1+\frac{9*10^3}{10^3} = 10

So our new bandwidth is:

f_c = \frac{GBP}{A_V} = \frac{10^6}{10} = 100kHz

That means the gain of 10 will persist even for our signal of 50kHz.

To achieve our old gain of 100 we can connect multiple of those amplifiers in series.

To calculate the total bandwidth of all the op-amps we can use a formula which is very similar to the one of multiple low-pass filters.

!formulas#Bandwidth of Multiple OpAmps

Lets calculate the gain for two of our op-amp configurations in series:

A_V = 100kHz \sqrt{2^\frac{1}{2} - 1} \approx 64.35kHz

For three op-amps the formula is (dont know where the 215.44$kHz$ come from...)

A_V = 215.44kHz\sqrt{2^\frac{1}{3}-1} \approx 109.84kHz