feat: add some more op-amp stuff
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Areas/electricity/active-components/op-amp-circuits.md
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Areas/electricity/active-components/op-amp-circuits.md
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# Non-Inverting Amplifier
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```circuitjs
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$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
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a 192 240 304 240 9 15 -15 1000000 4.9999000019999595 5 100000
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r 192 320 192 400 0 1000
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r 304 320 192 320 0 1000
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w 192 320 192 256 0
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w 304 240 304 320 0
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O 304 240 368 240 1 0
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g 192 400 192 432 0 0
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v 96 352 96 224 0 0 40 5 0 0 0.5
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w 96 224 192 224 2
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g 96 352 96 432 0 0
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b 144 288 289 401 0
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x 264 386 278 389 4 24 β
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x 240 345 252 348 4 12 Rf
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x 160 363 176 366 4 12 Rg
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```
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What is the closed loop gain of this circuit?
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$$
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\begin{flalign}
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&V_- = V_+ = V_s&\\\
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&V_- \text{is the output of a voltage divider}\\
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\end{flalign}
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$$
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Because $V_-$ is equal to $V_+$ and
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$V_- = V_s = V_o (\frac{R_g}{R_G+R_F})$
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If we solve that equation for $\frac{V_O}{V_s}$ we get the following formula:
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## Formula
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$\displaystyle Gain =\frac{V_o}{V_s} = 1+\frac{R_F}{R_G}$
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# Buffer (Voltage-Follow)
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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 $V_o$
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```circuitjs
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$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
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a 192 240 304 240 9 15 -15 1000000 4.999950000499995 5 100000
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w 192 320 192 256 0
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w 304 240 304 320 0
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O 304 240 368 240 1 0
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v 96 304 96 224 0 0 40 5 0 0 0.5
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w 96 224 192 224 2
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g 96 304 96 352 0 0
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w 192 320 304 320 0
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```
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# Inverting Amplifier
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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.
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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 $-5V$.
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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.
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```circuitjs
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$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
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v 64 304 64 192 0 0 40 5 0 0 0.5
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r 176 192 112 192 0 1000
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w 176 224 176 272 0
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g 176 272 176 320 0 0
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g 64 304 64 320 0 0
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r 176 128 288 128 0 1000
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w 288 128 288 208 0
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w 176 192 176 128 0
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g 288 272 288 320 0 0
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p 288 208 288 272 1 0 0
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x 230 94 242 97 4 12 Rf
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x 129 166 141 169 4 12 Ri
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a 176 208 288 208 8 15 -15 1000000 0.00004999900001999959 0 100000
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w 112 192 64 192 2
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```
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$$
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\begin{flalign}
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&A_V = -\frac{R_F}{R_I}&\\\
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\end{flalign}
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$$
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# Single Ended Inverting Amplifier
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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.
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```desmos-graph
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y=\sin(x*5)|5>x>0|y>0
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y = \sin(x*5)|x<0
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y=\sin(x*5)+1|5<x
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```
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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.
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```circuitjs
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$ 1 0.000005 6.450009306485578 50 5 50 5e-11
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O 352 224 432 224 0 0
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w 352 160 352 224 0
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r 224 160 352 160 0 3000
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w 224 208 224 160 0
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r 160 160 224 160 0 1000
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g 224 320 224 352 0 0
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r 224 240 224 320 0 1000
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r 80 240 224 240 0 1000
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R 64 160 16 160 0 1 40 2 0 0 0.5
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R 80 240 16 240 0 0 40 15 0 0 0.5
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a 224 224 352 224 8 15 0 1000000 7.499985233224565 7.5 100000
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c 112 160 160 160 0 0.00001 -6.787036831414538 0.001
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s 64 160 112 160 0 0 false
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o 0 16 0 12294 13.0746715037385 0.0001 0 1 output
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o 8 16 0 12294 2 0.0001 0 2 8 3 source
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```
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# Difference Amplifier
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```circuitjs
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$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
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R 144 192 96 192 0 0 40 5 0 0 0.5
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R 144 224 96 224 0 0 40 4 0 0 0.5
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r 144 192 224 192 0 1000
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r 144 224 224 224 0 1000
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a 224 208 352 208 8 15 -15 1000000 2.000009999800004 2 100000
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r 224 128 352 128 0 1000
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w 352 128 352 208 0
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w 224 192 224 128 0
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w 352 208 400 208 0
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p 400 208 400 320 1 0 0
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g 400 320 400 352 0 0
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r 224 224 224 320 0 1000
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g 224 320 224 352 0 0
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x 176 171 192 174 4 12 R1
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x 176 244 192 247 4 12 R1
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x 199 279 215 282 4 12 R2
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x 279 148 295 151 4 12 R2
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```
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$\displaystyle V_O = \frac{R2}{R1}(V_2-V_1)$
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# Calculate Non-Inverting Amplifier Bandwidth
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Let's calculate the bandwidth for the following non-inverting op-amp with a $GBP$ of $1Mhz$.
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![[op-amb-bandwidth-example.jpg]]
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The formula for calculating the gain of a non-inverting op-amp is:
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![[#Non-Inverting Amplifier#Formula]]
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Now with the numbers in the graph, that:
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$$
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A_V = 1+\frac{99*10^3}{10^3} = 100
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$$
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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:
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![[formulas#Gain Bandwidth Product]]
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$\displaystyle f_c = \frac{1*10^6}{100} = 10*10^3 = 10kHz$
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That means that over $10kHz$ the gain of our op-amp is not at 100 anymore.
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We can then lower the gain of a single op-amp to increase its bandwidth like so:
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![[op-amp-bandwidth-example-2.jpg|400]]
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Where our new gain is now:
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$A_V = 1+\frac{9*10^3}{10^3} = 10$
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So our new bandwidth is:
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$f_c = \frac{GBP}{A_V} = \frac{10^6}{10} = 100kHz$
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That means the gain of 10 will persist even for our signal of $50kHz$.
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To achieve our old gain of 100 we can connect multiple of those amplifiers in series.
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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.
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![[formulas#Bandwidth of Multiple OpAmps]]
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Lets calculate the gain for two of our op-amp configurations in series:
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$A_V = 100kHz \sqrt{2^\frac{1}{2} - 1} \approx 64.35kHz$
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For three op-amps the formula is (dont know where the 215.44$kHz$ come from...)
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$A_V = 215.44kHz\sqrt{2^\frac{1}{3}-1} \approx 109.84kHz$
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@ -16,12 +16,19 @@ $$
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&X_2 = \text{Input V2 (Inverting Input)}\\
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\end{flalign}
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$$
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# Rules
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**1. No Current flows in or out of the outputs**
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**2. The op-amp tries to keep the input voltages the same**
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The second rule only applies when the op-amp is in closed loop configuration
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# Regions
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Op Amps functions in different regions, just like diodes, and transistors.
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![[op-amp-regions.png|400]]
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# Regions
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**Linear Region**
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This is how the Op-Amp normally functions.
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@ -30,7 +37,7 @@ When the output of the op-amp would be higher than $+V_{CC}$ or lower than $-V_{
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In real life OpAmps have $A_V$ values as high as $10^8$ or $10^9$ due to this even very small input voltages would quickly leave the linear region. That is why we need
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**Negative Feedback**
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# Negative Feedback
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To use negative feedback we connect the output of the OpAmp to one of its inputs. This connection is modified by a *feedback factor* ($\beta$) which can be in the range $0 \le \beta \le 1$.
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Due to this feedback the new formula for the output $V_O$ is now:
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@ -47,107 +54,40 @@ $$
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$$
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# Configurations
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# Non-Inverting Amplifier
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**Open Loop**
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When the output of the Op Amp is not connected to any of its inputs, it is in the so called "open loop configurations"
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```circuitjs
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$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
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a 192 240 304 240 9 15 -15 1000000 4.9999000019999595 5 100000
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r 192 320 192 400 0 1000
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r 304 320 192 320 0 1000
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w 192 320 192 256 0
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w 304 240 304 320 0
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O 304 240 368 240 1 0
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g 192 400 192 432 0 0
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v 96 352 96 224 0 0 40 5 0 0 0.5
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w 96 224 192 224 2
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g 96 352 96 432 0 0
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b 144 288 289 401 0
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x 264 386 278 389 4 24 β
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x 240 345 252 348 4 12 Rf
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x 160 363 176 366 4 12 Rg
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```
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**Closed Loop**
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When we connect the output of the OpAmp to either $V_+$ or $V_-$ the OpAmp is in the "closed loop configuration".
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What is the closed loop gain of this circuit?
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# Bandwidth Limitations
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$$
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\begin{flalign}
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&V_- = V_+ = V_s&\\\
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&V_- \text{is the output of a voltage divider}\\
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\end{flalign}
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$$
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Real op-amps behave differently depending on the input signals frequency. Usually the internal open-loop gain gets lower as the input frequency gets higher like this.
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Because $V_-$ is equal to $V_+$ and
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The op-amps bandwidth is the frequency range in which the voltage gain is above 70.7% ($3dB$) of its maximum output. The point at which it is below that gain, is called the **breakpoint**.
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$V_- = V_s = V_o (\frac{R_g}{R_G+R_F})$
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t
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If we solve that equation for $\frac{V_O}{V_s}$ we get the following formula:
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![[op-amp-bandwidth.png|400]]
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$\displaystyle Gain =\frac{V_o}{V_s} = 1+\frac{R_F}{R_G}$
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# Buffer (Voltage-Follow)
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This is also one of the reason we use op-amps in closed loop configuration. Because it allows is to trade maximum gain for a larger bandwidth.
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This circuit is usefull, 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 $V_o$
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![[op-amp-bandwidth-closed-loop.png|400]]
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```circuitjs
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$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
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a 192 240 304 240 9 15 -15 1000000 4.999950000499995 5 100000
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w 192 320 192 256 0
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w 304 240 304 320 0
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O 304 240 368 240 1 0
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v 96 304 96 224 0 0 40 5 0 0 0.5
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w 96 224 192 224 2
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g 96 304 96 352 0 0
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w 192 320 304 320 0
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```
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If we want to find out the bandwidth of an op-amp, we can check the datasheet. The *LM741 OpAmp* for example:
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# Inverting Amplifier
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![[lm741-datasheet-bandwidth.png]]
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```circuitjs
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$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
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v 48 304 48 192 0 0 40 5 0 0 0.5
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r 176 192 96 192 0 1000
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w 176 224 176 304 0
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g 176 304 176 352 0 0
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g 48 304 48 352 0 0
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r 176 112 336 112 0 1000
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w 336 112 336 208 0
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w 176 192 176 112 0
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w 336 208 416 208 0
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g 416 320 416 352 0 0
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p 416 208 416 320 1 0 0
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x 249 86 261 89 4 12 Rf
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x 129 166 141 169 4 12 Ri
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a 176 208 336 208 8 15 -15 1000000 0.00004999900001999959 0 100000
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w 96 192 48 192 2
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```
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The thing is that this frequency only applies when the op-amp has a gain of 1, this frequency point is also called **unity gain**. It is called the **Gain Bandwidth Product**, which is calculated as follows:
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$$
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\begin{flalign}
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&\frac{V_o}{V_S} = -\frac{R_F}{R_I}&\\\
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\end{flalign}
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$$
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# Difference Amplifier
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$GBP = A_V * f_c$
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```circuitjs
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$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
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R 144 192 96 192 0 0 40 5 0 0 0.5
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R 144 224 96 224 0 0 40 4 0 0 0.5
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r 144 192 224 192 0 1000
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r 144 224 224 224 0 1000
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a 224 208 352 208 8 15 -15 1000000 2.000009999800004 2 100000
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r 224 128 352 128 0 1000
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w 352 128 352 208 0
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w 224 192 224 128 0
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w 352 208 400 208 0
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p 400 208 400 320 1 0 0
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g 400 320 400 352 0 0
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r 224 224 224 320 0 1000
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g 224 320 224 352 0 0
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x 176 171 192 174 4 12 R1
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x 176 244 192 247 4 12 R1
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x 199 279 215 282 4 12 R2
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x 279 148 295 151 4 12 R2
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```
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Where $A_V$ is the voltage gain, and $f_c$ is the cutoff frequency.
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$\displaystyle V_O = \frac{R2}{R1}(V_2-V_1)$
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With this equation we can also solve for $f_c$ like so:
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$\displaystyle f_c = \frac{GBP}{A_V}$
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![[lm741.pdf]]
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Areas/electricity/assets/yjs-architecture.excalidraw.md
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---
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excalidraw-plugin: parsed
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tags: [excalidraw]
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---
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==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠==
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# Text Elements
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Server ^WkbqBiLT
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Client ^ZHPBLTHz
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adapter ^onSLaWJU
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adapter ^4E2JLjbz
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signal-server ^H1H62P9W
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WebRTC/WebSocket Provider Singletons ^EmE2WToq
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WebRTC/WebSocket Provider Singletons ^WzmDhRB1
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%%
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# Drawing
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%%
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BIN
Areas/electricity/datasheets/lm741.pdf
Normal file
BIN
Areas/electricity/datasheets/lm741.pdf
Normal file
Binary file not shown.
@ -144,4 +144,16 @@ $Z = \sqrt{R^2 + (X_L - X_C)^2}$
|
||||
|
||||
# Current through a transistor
|
||||
|
||||
$\displaystyle I_{EQ} = \frac{V_{BB}-{V_{BE}}}{\frac{R_B}{(\beta+1)}+R_E}$
|
||||
$\displaystyle I_{EQ} = \frac{V_{BB}-{V_{BE}}}{\frac{R_B}{(\beta+1)}+R_E}$
|
||||
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||||
# Gain Bandwidth Product
|
||||
$GBP = A_V * f_c$
|
||||
$\displaystyle f_c = \frac{GBP}{A_V}$
|
||||
|
||||
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||||
# Bandwidth of Multiple OpAmps
|
||||
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||||
Where $n$ = number of stages
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||||
and $BW$ = Bandwidth of single op-amp
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||||
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||||
$BW_E = BW\sqrt{2^\frac{1}{n}-1}$
|
||||
Loading…
Reference in New Issue
Block a user