Chuck D. Bones
Circuit Wizard
This is chapter where it all comes together: Feedback. There will be some math, it's unavoidable. I'll do my best to keep it simple and understandable.
The whole reason for creating opamps in the first place was to put negative feedback around them. With feedback, the gain, distortion and frequency response of an opamp circuit can be tailored and stabilized. Without negative feedback, an opamp is just a comparator. A comparator has a binary output: high or low. If the + input is greater than the - input, the output is high. If the + input is less than the - input, the output is low. Some of you may ask "what happens if the + and - inputs are equal?" That only happens if we have negative feedback to make it so. When we apply negative feedback to an opamp, we can make all sorts of wonderful circuits such as amplifiers, filters, limiters, oscillators, etc. Here is an ideal opamp with the simplest of feedback networks. By "ideal," I mean that there are no DC errors and for the moment, the gain and bandwidth are infinite. Assume that it's powered by ±9V.
The current into the - input is zero, which means that the currents in Ri and Rf are equal. We'll call that current "I". The voltage at the - input is zero because the negative feedback tries to maintain the voltage different between the + and - inputs at zero. Because of all that, Ohm's Law tells us that Vin / Ri = I and Vout /Rf = -I. In the second equation, the current is negative because positive current flows from left to right in both equations. If Vin is positive, then Vout must be negative for the - pin to be at zero volts. You can think of Ri & Rf as a teeter-totter with the fulcrum at the opamp's - pin. If one side goes up, the other side goes down. Still with me?
We can combine those two equations and say Vin / Ri = -Vout / Rf. The definition of voltage gain is H = Vout / Vin. We can rearrange the 3rd equation to be: H = Vout / Vin = -Rf / Ri. Because this is an inverting amplifier circuit, the gain will always be a negative number. In other words, the output will move in the opposite direction of the input. That's what "inverting" means. H has been traditionally used to represent gain mathematically.
This should look familiar because it is the simplest way to calculate the gain of an inverting amplifier. We can pick any resistor values we want and that's the gain we get, right? Well, it would be except for one little issue: real opamps do not have infinite gain or bandwidth. In the real world, we couldn't make Ri = 1Ω and Rf = 1MΩ and expect to get a gain of a million. To account for the opamp's limited gain, we'll introduce two new terms: feedback factor and open-loop gain.
Feedback factor is the transfer function (gain) of the feedback network. In electrical engineering, the feedback factor is symbolized by the Greek letter beta. Here, I'll use the letter "B." In this simple example, B = Ri / Rf. Note that in the vast majority of cases, Ri is less than Rf and therefore B is usually a number between 0 and 1.
The other factor, open-loop gain, is symbolized by the Greek letter alpha. Here', I'll use the letter "A." Open-loop gain is the actual gain of the opamp, from input to output, without any feedback. Now we can account for the fact that opamps have a finite open-loop gain. The correct way to express the closed-loop gain of an inverting opamp is this way:
H = -A / (1+A*B)
Remember, H is negative because this is an inverting amplifier. A*B is known at the loop-gain. It's the round-trip gain going thru the opamp, thru the feedback network, and back around to the opamp's - input. When A is very large (it can be over 100,000 in the opamps we use) and B is not too tiny, then the loop-gain (A*B) is also fairly large. In that case, we can drop the 1 in the denominator and approximate the last equation as:
H = -A / (A*B) which simplifies to H = -1 / B = -Rf / Ri.
And we're back to the simplified equation above.
Let's do a simple example. Ri = 10K and Rf = 1M. We'll use a TL072 opamp, which has a typical DC open-loop gain of 200,000. B = -Rf / Ri = -0.01. A = 200,000.
H = -200,000 / (1+(200,000 * 0.01)) = 200,000 / 2001 = -99.95. We expected -100, so there is a 0.05% error. I think we can live with that.
Maybe we want more gain for some heavy metal shredding. Let's make Rf = 10M. Now,
H = -200,000 / (1+(200,000 * 0.001)) = 200,000 / 201 = -995. The error is only 0.5%, still no biggie.
OK, that was the easy part. In the previous chapter, we touched on bandwidth. Opamps have limited bandwidth which means that as the operating frequency goes up, the open-loop gain goes down. The effect can be pretty dramatic. At 1KHz, the typical open-loop gain of a TL072 is 3,000. In the last example, we were calculating the gain at DC (zero Hz). Let's try it again at 1KHz.
H = -3,000 / (1+(3,000 * 0.001)) = 3,000 / 4 = -750. The gain is 25% low or about -2.5dB. Might be noticeable.
At 10KHz, the open-gain is around 300, so H = 300 / (1+(300*0.001)) = 231. Now the gain is less than 1/4 of what we wanted, a 12.7dB error. Note that the loop gain has fallen to 0.3. That's like almost no feedback at 10KHz. As a rough rule of thumb, if the loop-gain drops below 10, we're probably gonna start noticing the effects. Whether we care depends on the application. If we're building a phono preamp, it will matter big-time. If we're building a fuzz pedal, it might be ok.
A picture's worth 1,000 words, so here is a plot of the closed-loop gain of a TL072 with varying feedback factors. The feedback factor goes in 10X steps from 0.0000001 (one millionth) for the pink trace to 0.01 (one hundredth) for the green trace. Freq response is quite good for the bottom trace, but degrades as the loop-gain decreases. The loop-gain is so low for the pink trace that there's basically no feedback and we are essentially seeing the TL072's open-loop gain.
I'm going to pause here to let you all digest this. In part 6b, we'll consider capacitors in the feedback network. In that case, not only does the open-loop gain A vary with frequency, but the feedback factor B varies with frequency as well. That's where the magic happens.
The whole reason for creating opamps in the first place was to put negative feedback around them. With feedback, the gain, distortion and frequency response of an opamp circuit can be tailored and stabilized. Without negative feedback, an opamp is just a comparator. A comparator has a binary output: high or low. If the + input is greater than the - input, the output is high. If the + input is less than the - input, the output is low. Some of you may ask "what happens if the + and - inputs are equal?" That only happens if we have negative feedback to make it so. When we apply negative feedback to an opamp, we can make all sorts of wonderful circuits such as amplifiers, filters, limiters, oscillators, etc. Here is an ideal opamp with the simplest of feedback networks. By "ideal," I mean that there are no DC errors and for the moment, the gain and bandwidth are infinite. Assume that it's powered by ±9V.
The current into the - input is zero, which means that the currents in Ri and Rf are equal. We'll call that current "I". The voltage at the - input is zero because the negative feedback tries to maintain the voltage different between the + and - inputs at zero. Because of all that, Ohm's Law tells us that Vin / Ri = I and Vout /Rf = -I. In the second equation, the current is negative because positive current flows from left to right in both equations. If Vin is positive, then Vout must be negative for the - pin to be at zero volts. You can think of Ri & Rf as a teeter-totter with the fulcrum at the opamp's - pin. If one side goes up, the other side goes down. Still with me?
We can combine those two equations and say Vin / Ri = -Vout / Rf. The definition of voltage gain is H = Vout / Vin. We can rearrange the 3rd equation to be: H = Vout / Vin = -Rf / Ri. Because this is an inverting amplifier circuit, the gain will always be a negative number. In other words, the output will move in the opposite direction of the input. That's what "inverting" means. H has been traditionally used to represent gain mathematically.
This should look familiar because it is the simplest way to calculate the gain of an inverting amplifier. We can pick any resistor values we want and that's the gain we get, right? Well, it would be except for one little issue: real opamps do not have infinite gain or bandwidth. In the real world, we couldn't make Ri = 1Ω and Rf = 1MΩ and expect to get a gain of a million. To account for the opamp's limited gain, we'll introduce two new terms: feedback factor and open-loop gain.
Feedback factor is the transfer function (gain) of the feedback network. In electrical engineering, the feedback factor is symbolized by the Greek letter beta. Here, I'll use the letter "B." In this simple example, B = Ri / Rf. Note that in the vast majority of cases, Ri is less than Rf and therefore B is usually a number between 0 and 1.
The other factor, open-loop gain, is symbolized by the Greek letter alpha. Here', I'll use the letter "A." Open-loop gain is the actual gain of the opamp, from input to output, without any feedback. Now we can account for the fact that opamps have a finite open-loop gain. The correct way to express the closed-loop gain of an inverting opamp is this way:
H = -A / (1+A*B)
Remember, H is negative because this is an inverting amplifier. A*B is known at the loop-gain. It's the round-trip gain going thru the opamp, thru the feedback network, and back around to the opamp's - input. When A is very large (it can be over 100,000 in the opamps we use) and B is not too tiny, then the loop-gain (A*B) is also fairly large. In that case, we can drop the 1 in the denominator and approximate the last equation as:
H = -A / (A*B) which simplifies to H = -1 / B = -Rf / Ri.
And we're back to the simplified equation above.
Let's do a simple example. Ri = 10K and Rf = 1M. We'll use a TL072 opamp, which has a typical DC open-loop gain of 200,000. B = -Rf / Ri = -0.01. A = 200,000.
H = -200,000 / (1+(200,000 * 0.01)) = 200,000 / 2001 = -99.95. We expected -100, so there is a 0.05% error. I think we can live with that.
Maybe we want more gain for some heavy metal shredding. Let's make Rf = 10M. Now,
H = -200,000 / (1+(200,000 * 0.001)) = 200,000 / 201 = -995. The error is only 0.5%, still no biggie.
OK, that was the easy part. In the previous chapter, we touched on bandwidth. Opamps have limited bandwidth which means that as the operating frequency goes up, the open-loop gain goes down. The effect can be pretty dramatic. At 1KHz, the typical open-loop gain of a TL072 is 3,000. In the last example, we were calculating the gain at DC (zero Hz). Let's try it again at 1KHz.
H = -3,000 / (1+(3,000 * 0.001)) = 3,000 / 4 = -750. The gain is 25% low or about -2.5dB. Might be noticeable.
At 10KHz, the open-gain is around 300, so H = 300 / (1+(300*0.001)) = 231. Now the gain is less than 1/4 of what we wanted, a 12.7dB error. Note that the loop gain has fallen to 0.3. That's like almost no feedback at 10KHz. As a rough rule of thumb, if the loop-gain drops below 10, we're probably gonna start noticing the effects. Whether we care depends on the application. If we're building a phono preamp, it will matter big-time. If we're building a fuzz pedal, it might be ok.
A picture's worth 1,000 words, so here is a plot of the closed-loop gain of a TL072 with varying feedback factors. The feedback factor goes in 10X steps from 0.0000001 (one millionth) for the pink trace to 0.01 (one hundredth) for the green trace. Freq response is quite good for the bottom trace, but degrades as the loop-gain decreases. The loop-gain is so low for the pink trace that there's basically no feedback and we are essentially seeing the TL072's open-loop gain.
I'm going to pause here to let you all digest this. In part 6b, we'll consider capacitors in the feedback network. In that case, not only does the open-loop gain A vary with frequency, but the feedback factor B varies with frequency as well. That's where the magic happens.
