5 ms·

Linear Circuit Design Handbook (2008)

101 points by mindcrime 5y ago

JKCalhoun 5y ago
No one will ever be able to explain op-amps to me satisfactorily. I am too dense. But I am endlessly fascinated by them. Analog circuits have (re)kindled a love for electronics in me.
- jareklupinski 5y ago
  The Art of Electronics introduced me to Transistor Man, who helpfully explained some of the more opaque topics with "there's a demon doin it" until the theory made more sense after learning more details https://electronics.stackexchange.com/a/476734 https://electronics.stackexchange.com/a/476734
- GeorgeTirebiter 5y ago
  This is one of the best explanations of op amps: https://www.ti.com/lit/an/sboa092b/sboa092b.pdf?ts=1626461495778 https://www.ti.com/lit/an/sboa092b/sboa092b.pdf?ts=162646149...
- ur-whale 5y ago
  Strange, op-amps (if you don't dive into pushing them to their limits, and if you steer clear of trying to understand how they are implemented) are actually a lot simpler to grok than - say - an NPN transistor.
  - GeorgeTirebiter 5y ago
    Well.... yes, and. If you think of an NPN transistor as a current-operated switch (e.g. to drive an LED or relay), then you can have a pretty simple model of said NPN transistor. If you want to make a low-noise PIN diode photodetector amplifier with an op amp, you want to understand noise per root Hz, input offset voltages, variations with temperature, gain, frequency and phase response, and so much more. So, as is often the case in Engineering: It Depends...
  - analog31 5y ago
    Bipolar (NPN and PNP) circuits can be analyzed by applying the following rules: 1. If current flows through the base-emitter junction, then there is a voltage of roughly 0.6 V across that junction. 2. The collector current is roughly proportional to the base current, by the beta constant, which is about 10 for big transistors and 100 for little ones. There are some rare textbook cases such as logarithmic amplifiers that just have to be looked up, following the Ebers-Moll equation. As for FETs... once you get the type correct (P or N type, enhancement or depletion mode, which I still have to look up or find an example circuit), gate-source voltage is constant, gate current is zero, and the drain current equals the source current. As for vacuum triodes... they behave much like N type JFETs. You can check your analysis of any of these circuits using the wonderful LTSpice.
    - exmadscientist 5y ago
      > As for FETs... once you get the type correct (P or N type, enhancement or depletion mode, which I still have to look up or find an example circuit) I used to have this problem too. Turns out it's easier to work with than you think: the symbol actually describes things pretty well. (This ought to be illustrated, but unfortunately I haven't got the time for that.) JFETs are pretty easy. They're always depletion devices, so their channel always starts conducting: it's solid on the symbol. The gate is directly connected to the channel, so it touches, it's symmetrical, so we draw the gate in the center (though I will draw it near one terminal if I am trying to communicate "hey reader, this one is the source, look at it first"), and there's a "diode arrow" that has either N-type material at the channel end (N-channel part!) or P-type material at the channel end (P-channel part!). MOSFETs are more confusing, but it still isn't bad. They're actually four-terminal devices; we'll add the body in a moment. But their defining feature is that insulated gate, so we can show that by pulling the gate away from the channel, drawing it as a solid line running parallel to the channel. The gate comes off it next to the source, to reinforce the idea that V_GS is the controlling voltage here. The body (also called bulk or substrate, though be careful with that last one as it is not actually the substrate in all processes) is what the channel gets formed out of, so let's add the body terminal as coming out of the middle of the channel. The body diode is now formed between the channel and the body proper. We draw it relative to the channel: for N-channel parts, draw the "diode arrow" with the N-type material at the channel and the P-type material at the body terminal, or vice versa for a P-channel part. The body gets tied off to the source in the vast, vast majority of MOSFETs (and indeed this sometimes is part of what singles out one terminal as the source), so we can do that right now by drawing a line between them. We've now made everything we need to draw the symbol of or to explain a depletion MOSFET that works rather like the JFETs previously discussed. But most MOSFETs aren't depletion mode, they're enhancement mode. And so we indicate that by drawing the channel as a dashed line, rather than solid. Since there are three terminals, attached to the channel, D, B, and S, we draw the channel as three dashes, one for each terminal. Now we're done. Walking through that exercise really helped me understand just what these critters are inside and why we draw them the way we do. I hope it helps someone else, too. And I especially hope it might inspire any analog designer who doesn't draw things correctly, perhaps preferring the simplified symbols created by the simplified disciplines (cough digital cough) who can ignore that such things as "linear mode" operation even exist, to draw things using the correct symbols, which are complicated because they tell a rather complicated device's story. (Bonus factoid: ever wonder why P-channel depletion MOSFETs don't exist? As far as I've been able to put together, they do exist. It's just that when you try to manufacture them in the usual processes, the process inevitably causes you to end up with charge trapped under the gate... which turns the poor things off for you. So they're back to being enhancement mode! This effect doesn't bother N-channel parts, since it happens to try to turn them on. I believe this effect can be overcome with careful process control, but no one thinks it would be profitable to bring discrete devices into mass manufacturing. A few integrated parts with P-channel depletion FETs inside have shipped in full volume for decades.)
- analog31 5y ago
  My college didn't have an engineering department, but we took electronics as physics majors, using The Art of Electronics. The rules that we learned in class work for a large number of op amp circuits: 1. The inputs draw no current 2. If there is negative feedback present, the output will do whatever is necessary to render the two inputs equal So you can write the equations for the voltages at the two inputs, include an equation that sets them equal, and solve for the output. To handle circuits like filters, you have to solve using complex numbers. This will be good enough to derive the design formulas in the handbooks. These days, a third good rule is: 3. Buy an op amp that works as well as needed to obey rules 1 and 2 in your application. Op amps have made huge strides in terms of functionality and ease of design since I was in college 4 decades ago.
- ms-fellag 5y ago
  If watching videos is compatible with your learning style then i highly recommend this whole series on Youtube about Op-Amps, the channel is called RSD-Academy. This dude explained-it so well that i was astonished as to why i didn't 'get' before. [the first video you can skip to 1:45] <https://www.youtube.com/watch?v=7OuUiwI8fDk https://www.youtube.com/watch?v=7OuUiwI8fDk> [the whole series] <https://www.youtube.com/playlist?list=PLL_nf1OmixTTd7rEoqoM6aJ5ZdI-V47vs https://www.youtube.com/playlist?list=PLL_nf1OmixTTd7rEoqoM6...> There is also Dave Jones' op-amp tutorial at EEVBlog (if you like Aussie accents and a bit more detailed/technical explanation yet still targeted at beginners) <https://www.youtube.com/watch?v=7FYHt5XviKc https://www.youtube.com/watch?v=7FYHt5XviKc>
  - margor 5y ago
    Thank you for posting the RSDAcademy channel, you just earned him a new paid subscriber :) His content is a true gem and has finally helped me understand basic concepts like in no other place.
- sasaf5 5y ago
  The way I keep it in my pocket is: The op-amp subtracts the voltage of both inputs, multiplies the result by infinite, and presents it as the output voltage (limited by the power supply). When you calculate this with the feedbacks, it makes sense.
  - dragontamer 5y ago
    I never liked this explanation alone. The full explanation is that the + and - terminals are a virtual short. Negative feedback will try to set the + and - terminals to the same voltage but no current flows between the two locations. The amount of current (and the limitations of voltages) are set by the quality of the opamp. Better opamps have closer voltages at faster speeds at better currents. Cheaper opamps will have a bit of error and diverge from the model. The delay can be largely modeled by capacitance over the + and - terminals (among other models of opamp errors) But even a cheap opamp will keep the + and - sides within .05 Volts or so under typical circumstances.
    - sasaf5 5y ago
      All true, but I never liked this explanation. How do you use it to analyze a given circuit? With the infinite gain concept you can use a limit to calculate the behavior of any circuit.
      - dragontamer 5y ago
        With virtual short method you set + and - terminals to the same voltage but 0 current between them. This is only true for negative feedback cases but that's the most common case. It is easier in most cases than infinite gain model IMO. Once you have voltage and current of that location figured out, everything else in the circuit is solvable by freshman level analysis.
    - exmadscientist 5y ago
      One small trap: current never (in normal cases; yes, I know about the antiparallel diodes and how to deal with/avoid them...) actually flows between the input terminals, it flows to the rails. So you will not, except by sheer luck, get 10nA into the (+) terminal and 10nA out of the (-) terminal. You will get 10nA into the (+) terminal and 4nA into the (-) terminal, both returned to VEE.
- halayli 5y ago
  If you want to understand the fundamental principles of how op-amps and electric circuits work, I recommend learning the basic principles of elementary particles, electric charge, electric fields, and electromagnetic force, without diving into the mathematics behind them. It's easier to understand analog circuits without knowing the above because you are dealing with concrete components that abstract a lot of details away from you. What drove me down this path is when one day I asked myself what is voltage? Why do they use the water pressure example to explain it? We are familiar with V = IR but that doesn't explain what voltage is just like an apple falling on the ground doesn't explain gravitational force. Don't be surprised if this journey takes you down to quantum mechanics.
  - elevation 5y ago
    > learning the basic principles of elementary particles, electric charge, electric fields, and electromagnetic force, without diving into the mathematics behind them I cannot emphasize this enough for those who have a "water in a pipe" mental model of DC electric circuits but struggle to intuitively grasp AC and RF concepts like filters, fields, transformers, antennas, or even buck/boost converters. Once you're able to mentally reduce some of these circuits to an understanding of the forces upon a single electron at any point in the circuit at any point in time, you being to be able to create your own designs rather than relying on the examples of others.
    - halayli 5y ago
      exactly. this field of education needs a big revamp.
  - bewbaloo 5y ago
    can you recommend any books on this subject?
- xyzzy21 5y ago
  The key thing: the standard working model for circuit design with Op Amps (see below) are approximations that become good approximations because of their high gain and the typical use of negative feedback when using them. These rules are non-intuitive if you are thinking raw circuits but are very easy to use if you hold your nose and just take the model on faith (because the model pretty much works all the time). Only if you design the innards of an Op Amp does the model become less shiny and reality creeps in. So when you design with an op amp you use certain "simplifying magical assumptions: 1. There is zero input leakage so you can KCL both input and feedback current pretending that the op amp isn't affecting either of these. That's pretty much a valid assumption until you design ultra low current circuits with an op amp. But then you'll likely want to switch to an "instrumentation op amp" of a specific design for such applications. That's <<1% of anyone. 2. The input voltages (+ and -) "magically" can be assumed to always be equal without any external current or voltage seeming to cause it to happen. This is due to the innards of the op amp design plus feedback. These two assumptions mean that you can apply simply circuit analysis and you quickly attain accurate predictions of how the circuit will work (gain, bandwidth, etc.). Even most EEs work with this. Only analog IC designers go beyond this.
ur-whale 5y ago
One thing I've never managed to wrap my head around is why many analog designers insist on calling op-amp based analog circuit design "linear design". If there's one thing these circuits are absolutely not it's being linear. Every interesting circuit using op-amps basically is interesting because of the various non-linear behaviors you have to deal with when you optimize the circuit.
- dreamcompiler 5y ago
  Op-amps are (theoretically) just linear differential amplifiers with extremely high gain (like 10,000x or greater). They appear nonlinear in open-loop configuration because their gain is so high that any realistic input signal immediately pegs the output against the rails. But with negative feedback they start to appear linear again because you can set the gain to something more reasonable. They're marvelous devices.
- tuatoru 5y ago
  This is a somewhat contrarian statement--I'd like some explanation and examples of what you mean. I find a x10 amplifier with high Zin, low Zout, low noise, low error, and constant delay to be interesting enough. Every circuit is non-linear when used outside its design limits, but this hardly needs asserting. And of course real devices are not perfect. Resistors have inductance and capacitance and a thermal coefficient, capacitors likewise have flaws, wires have resistance and self-inductance, and circuit layout gives you stray capacitance, crosstalk, and instability. Op amps themselves produce noise, have input currents and offset voltages and limited slew rate and bandwidth. And so on, and on. But again, none of that needs asserting. While you certainly can design lot of deliberately non-linear circuits with op amps and non-linear devices in their feedback paths, it's difficult to design linear circuits without them, or instead with circuits, the core of which end up looking like simplified op-amps. So, what did you have in mind when you said "the various non-linear behaviours"?
- mhh__ 5y ago
  The one thing I wouldn't expect from electrical engineers is a focus on nomenclature - I do love electronics but I am perpetually annoyed by similar niggles as you are (it seems) + a tendency to teach the same equation 3 times in 3 different regimes even though I have spent years training myself to be able to identify these approximations myself.

101 points by mindcrime 5y ago

No one will ever be able to explain op-amps to me satisfactorily. I am too dense. But I am endlessly fascinated by them. Analog circuits have (re)kindled a love for electronics in me.

jareklupinski 5y ago
The Art of Electronics introduced me to Transistor Man, who helpfully explained some of the more opaque topics with "there's a demon doin it" until the theory made more sense after learning more details https://electronics.stackexchange.com/a/476734 https://electronics.stackexchange.com/a/476734
GeorgeTirebiter 5y ago
This is one of the best explanations of op amps: https://www.ti.com/lit/an/sboa092b/sboa092b.pdf?ts=1626461495778 https://www.ti.com/lit/an/sboa092b/sboa092b.pdf?ts=162646149...
ur-whale 5y ago
Strange, op-amps (if you don't dive into pushing them to their limits, and if you steer clear of trying to understand how they are implemented) are actually a lot simpler to grok than - say - an NPN transistor.
- GeorgeTirebiter 5y ago
  Well.... yes, and. If you think of an NPN transistor as a current-operated switch (e.g. to drive an LED or relay), then you can have a pretty simple model of said NPN transistor. If you want to make a low-noise PIN diode photodetector amplifier with an op amp, you want to understand noise per root Hz, input offset voltages, variations with temperature, gain, frequency and phase response, and so much more. So, as is often the case in Engineering: It Depends...
- analog31 5y ago
  Bipolar (NPN and PNP) circuits can be analyzed by applying the following rules: 1. If current flows through the base-emitter junction, then there is a voltage of roughly 0.6 V across that junction. 2. The collector current is roughly proportional to the base current, by the beta constant, which is about 10 for big transistors and 100 for little ones. There are some rare textbook cases such as logarithmic amplifiers that just have to be looked up, following the Ebers-Moll equation. As for FETs... once you get the type correct (P or N type, enhancement or depletion mode, which I still have to look up or find an example circuit), gate-source voltage is constant, gate current is zero, and the drain current equals the source current. As for vacuum triodes... they behave much like N type JFETs. You can check your analysis of any of these circuits using the wonderful LTSpice.
  - exmadscientist 5y ago
    > As for FETs... once you get the type correct (P or N type, enhancement or depletion mode, which I still have to look up or find an example circuit) I used to have this problem too. Turns out it's easier to work with than you think: the symbol actually describes things pretty well. (This ought to be illustrated, but unfortunately I haven't got the time for that.) JFETs are pretty easy. They're always depletion devices, so their channel always starts conducting: it's solid on the symbol. The gate is directly connected to the channel, so it touches, it's symmetrical, so we draw the gate in the center (though I will draw it near one terminal if I am trying to communicate "hey reader, this one is the source, look at it first"), and there's a "diode arrow" that has either N-type material at the channel end (N-channel part!) or P-type material at the channel end (P-channel part!). MOSFETs are more confusing, but it still isn't bad. They're actually four-terminal devices; we'll add the body in a moment. But their defining feature is that insulated gate, so we can show that by pulling the gate away from the channel, drawing it as a solid line running parallel to the channel. The gate comes off it next to the source, to reinforce the idea that V_GS is the controlling voltage here. The body (also called bulk or substrate, though be careful with that last one as it is not actually the substrate in all processes) is what the channel gets formed out of, so let's add the body terminal as coming out of the middle of the channel. The body diode is now formed between the channel and the body proper. We draw it relative to the channel: for N-channel parts, draw the "diode arrow" with the N-type material at the channel and the P-type material at the body terminal, or vice versa for a P-channel part. The body gets tied off to the source in the vast, vast majority of MOSFETs (and indeed this sometimes is part of what singles out one terminal as the source), so we can do that right now by drawing a line between them. We've now made everything we need to draw the symbol of or to explain a depletion MOSFET that works rather like the JFETs previously discussed. But most MOSFETs aren't depletion mode, they're enhancement mode. And so we indicate that by drawing the channel as a dashed line, rather than solid. Since there are three terminals, attached to the channel, D, B, and S, we draw the channel as three dashes, one for each terminal. Now we're done. Walking through that exercise really helped me understand just what these critters are inside and why we draw them the way we do. I hope it helps someone else, too. And I especially hope it might inspire any analog designer who doesn't draw things correctly, perhaps preferring the simplified symbols created by the simplified disciplines (cough digital cough) who can ignore that such things as "linear mode" operation even exist, to draw things using the correct symbols, which are complicated because they tell a rather complicated device's story. (Bonus factoid: ever wonder why P-channel depletion MOSFETs don't exist? As far as I've been able to put together, they do exist. It's just that when you try to manufacture them in the usual processes, the process inevitably causes you to end up with charge trapped under the gate... which turns the poor things off for you. So they're back to being enhancement mode! This effect doesn't bother N-channel parts, since it happens to try to turn them on. I believe this effect can be overcome with careful process control, but no one thinks it would be profitable to bring discrete devices into mass manufacturing. A few integrated parts with P-channel depletion FETs inside have shipped in full volume for decades.)
analog31 5y ago
My college didn't have an engineering department, but we took electronics as physics majors, using The Art of Electronics. The rules that we learned in class work for a large number of op amp circuits: 1. The inputs draw no current 2. If there is negative feedback present, the output will do whatever is necessary to render the two inputs equal So you can write the equations for the voltages at the two inputs, include an equation that sets them equal, and solve for the output. To handle circuits like filters, you have to solve using complex numbers. This will be good enough to derive the design formulas in the handbooks. These days, a third good rule is: 3. Buy an op amp that works as well as needed to obey rules 1 and 2 in your application. Op amps have made huge strides in terms of functionality and ease of design since I was in college 4 decades ago.
ms-fellag 5y ago
If watching videos is compatible with your learning style then i highly recommend this whole series on Youtube about Op-Amps, the channel is called RSD-Academy. This dude explained-it so well that i was astonished as to why i didn't 'get' before. [the first video you can skip to 1:45] <https://www.youtube.com/watch?v=7OuUiwI8fDk https://www.youtube.com/watch?v=7OuUiwI8fDk> [the whole series] <https://www.youtube.com/playlist?list=PLL_nf1OmixTTd7rEoqoM6aJ5ZdI-V47vs https://www.youtube.com/playlist?list=PLL_nf1OmixTTd7rEoqoM6...> There is also Dave Jones' op-amp tutorial at EEVBlog (if you like Aussie accents and a bit more detailed/technical explanation yet still targeted at beginners) <https://www.youtube.com/watch?v=7FYHt5XviKc https://www.youtube.com/watch?v=7FYHt5XviKc>
- margor 5y ago
  Thank you for posting the RSDAcademy channel, you just earned him a new paid subscriber :) His content is a true gem and has finally helped me understand basic concepts like in no other place.
sasaf5 5y ago
The way I keep it in my pocket is: The op-amp subtracts the voltage of both inputs, multiplies the result by infinite, and presents it as the output voltage (limited by the power supply). When you calculate this with the feedbacks, it makes sense.
- dragontamer 5y ago
  I never liked this explanation alone. The full explanation is that the + and - terminals are a virtual short. Negative feedback will try to set the + and - terminals to the same voltage but no current flows between the two locations. The amount of current (and the limitations of voltages) are set by the quality of the opamp. Better opamps have closer voltages at faster speeds at better currents. Cheaper opamps will have a bit of error and diverge from the model. The delay can be largely modeled by capacitance over the + and - terminals (among other models of opamp errors) But even a cheap opamp will keep the + and - sides within .05 Volts or so under typical circumstances.
  - sasaf5 5y ago
    All true, but I never liked this explanation. How do you use it to analyze a given circuit? With the infinite gain concept you can use a limit to calculate the behavior of any circuit.
    - dragontamer 5y ago
      With virtual short method you set + and - terminals to the same voltage but 0 current between them. This is only true for negative feedback cases but that's the most common case. It is easier in most cases than infinite gain model IMO. Once you have voltage and current of that location figured out, everything else in the circuit is solvable by freshman level analysis.
  - exmadscientist 5y ago
    One small trap: current never (in normal cases; yes, I know about the antiparallel diodes and how to deal with/avoid them...) actually flows between the input terminals, it flows to the rails. So you will not, except by sheer luck, get 10nA into the (+) terminal and 10nA out of the (-) terminal. You will get 10nA into the (+) terminal and 4nA into the (-) terminal, both returned to VEE.
halayli 5y ago
If you want to understand the fundamental principles of how op-amps and electric circuits work, I recommend learning the basic principles of elementary particles, electric charge, electric fields, and electromagnetic force, without diving into the mathematics behind them. It's easier to understand analog circuits without knowing the above because you are dealing with concrete components that abstract a lot of details away from you. What drove me down this path is when one day I asked myself what is voltage? Why do they use the water pressure example to explain it? We are familiar with V = IR but that doesn't explain what voltage is just like an apple falling on the ground doesn't explain gravitational force. Don't be surprised if this journey takes you down to quantum mechanics.
- elevation 5y ago
  > learning the basic principles of elementary particles, electric charge, electric fields, and electromagnetic force, without diving into the mathematics behind them I cannot emphasize this enough for those who have a "water in a pipe" mental model of DC electric circuits but struggle to intuitively grasp AC and RF concepts like filters, fields, transformers, antennas, or even buck/boost converters. Once you're able to mentally reduce some of these circuits to an understanding of the forces upon a single electron at any point in the circuit at any point in time, you being to be able to create your own designs rather than relying on the examples of others.
  - halayli 5y ago
    exactly. this field of education needs a big revamp.
- bewbaloo 5y ago
  can you recommend any books on this subject?
xyzzy21 5y ago
The key thing: the standard working model for circuit design with Op Amps (see below) are approximations that become good approximations because of their high gain and the typical use of negative feedback when using them. These rules are non-intuitive if you are thinking raw circuits but are very easy to use if you hold your nose and just take the model on faith (because the model pretty much works all the time). Only if you design the innards of an Op Amp does the model become less shiny and reality creeps in. So when you design with an op amp you use certain "simplifying magical assumptions: 1. There is zero input leakage so you can KCL both input and feedback current pretending that the op amp isn't affecting either of these. That's pretty much a valid assumption until you design ultra low current circuits with an op amp. But then you'll likely want to switch to an "instrumentation op amp" of a specific design for such applications. That's <<1% of anyone. 2. The input voltages (+ and -) "magically" can be assumed to always be equal without any external current or voltage seeming to cause it to happen. This is due to the innards of the op amp design plus feedback. These two assumptions mean that you can apply simply circuit analysis and you quickly attain accurate predictions of how the circuit will work (gain, bandwidth, etc.). Even most EEs work with this. Only analog IC designers go beyond this.

ur-whale 5y ago

One thing I've never managed to wrap my head around is why many analog designers insist on calling op-amp based analog circuit design "linear design". If there's one thing these circuits are absolutely not it's being linear. Every interesting circuit using op-amps basically is interesting because of the various non-linear behaviors you have to deal with when you optimize the circuit.

dreamcompiler 5y ago
Op-amps are (theoretically) just linear differential amplifiers with extremely high gain (like 10,000x or greater). They appear nonlinear in open-loop configuration because their gain is so high that any realistic input signal immediately pegs the output against the rails. But with negative feedback they start to appear linear again because you can set the gain to something more reasonable. They're marvelous devices.
tuatoru 5y ago
This is a somewhat contrarian statement--I'd like some explanation and examples of what you mean. I find a x10 amplifier with high Zin, low Zout, low noise, low error, and constant delay to be interesting enough. Every circuit is non-linear when used outside its design limits, but this hardly needs asserting. And of course real devices are not perfect. Resistors have inductance and capacitance and a thermal coefficient, capacitors likewise have flaws, wires have resistance and self-inductance, and circuit layout gives you stray capacitance, crosstalk, and instability. Op amps themselves produce noise, have input currents and offset voltages and limited slew rate and bandwidth. And so on, and on. But again, none of that needs asserting. While you certainly can design lot of deliberately non-linear circuits with op amps and non-linear devices in their feedback paths, it's difficult to design linear circuits without them, or instead with circuits, the core of which end up looking like simplified op-amps. So, what did you have in mind when you said "the various non-linear behaviours"?
mhh__ 5y ago
The one thing I wouldn't expect from electrical engineers is a focus on nomenclature - I do love electronics but I am perpetually annoyed by similar niggles as you are (it seems) + a tendency to teach the same equation 3 times in 3 different regimes even though I have spent years training myself to be able to identify these approximations myself.