Let's start with the concept of a closed-loop feedback system. An amplifier is presented with signals from a summing junction. Output voltage is modified by a factor B, subtracted from the input voltage, and the result is the signal that the amplifer is given to amplify.

V_{2}= AV_{1}

V_{1}=Vin - BV_{2}

Substituting,

V_{2}= A(Vin - BV_{2})

Rearranging,

V_{2} (1+BA) = Avin

or

Op amp gains (A) are typically 50,000 to 100,000, at least at DC. Note, therefore, that the gain of this block is quite insensitive to A. For B = .01,

A=10,000 == 99.01 If A=100,000, =99.90 The term A (which could include other terms in addition to the gain of the amplifier as will be shown) changed by a factor of 10 but the overall gain changed by 0.9%

This is the funadamental relationship for feedback control systems, and it's very powerful. Note that in general, A and B are differential equations or LaPlace transforms that describe the behavior of these functions with frequency and time. Because the transfer function above is a ratio of polynomials, (A could be in series with some function G rather than just a gain block) and since the time and frequency nature of networks can similarly be described as ratios of polynomials in s (LaPlace transforms), many functions can be realized with this structure. Oscillators, filters, amplifiers, impedance changers, negative-impedance blocks comprise just a few. For now we'll confine ourselves to a simple DC case.

Because A is large and BA is therefore large compared to 1, the transfer function can often be simplified to

Note that if other functions are in series with A, their transfer functions could be lumped with A and would cancel out as they did above. This means that you can cancel or minimize the effects of functions you can't control simply by including them "inside the loop" - i.e., in series with A.

Now that we've dealt with the exact equations, we'll show some useful simpliying assumptions.

Op amps already have the summing junction built in; they have an inverting input and a non-inverting input . The amplifer's output is A times the __difference__ between the voltages on those two inputs. So a circuit that shows the situation shown above is :

B is the ratio of the output voltage to that fed back to the negative terminal of the summing junction, or

So

Note that if R1 is a short and R2 is missing (infinite resistance), then Vout = Vin

Generally, as a reasonably accurate approximation for most work, the input impedance of op amps is considered to be infinitely large compared to other circuit impedances. Therefore, they essentially draw no current at the input terminals. In addition, the gain is treated as infinite as an approximation, and the validity of that assumption was defended above.

Therefore, consider the following situation:

If the opamp draws no current at the inputs, then the voltage at the minus terminal must be zero. This is true because the voltage at the + terminal is zero - it's grounded. If the voltage at the minus terminal were anything other than zero, then the output voltage would be infinite.

A positive Vin causes the output to go negative until enough current is drawn thru Rf and Ri to bring the voltage at the minus input very close to zero. The input current from Vin is therefore Vin/Ri. But the opamp draws no current, so this current must be going thru Rf. Since the minus terminal is very near zero voltage, the output voltage must be . Thus, the block gain is the ratio of the two resistors, essentially independent of amplifier gain. And by the way, it's also independent of any output loading on the amplifier as long as the amplifier can supply enough current to satisfy the conditions. This illustrates the utility of the simplifying assumptions made.

Now let's extend the loop to contain more elements - like a DC motor and it's internal resistance. Let's say the transfer function of the tach is 1 volt per 1000 RPM. If we set Vset to 2 volts, then the amplifier output must become whatever it needs to be (if possible) to make the voltage at the minus terminal very close to 2 volts. If the motor is running too slow, the output voltage will be the maximum the amp can provide, accelerating the motor until the tachometer voltage reaches very nearly 2 volts.

At that point the system will settle out with the tach returning 2 volts, which means that the motor is running at 2000 RPM. (Note: see time response considerations later. It will settle out only if the amp is much quicker than the motor-which is very likely)

Ra is the armature resistance. Note that because it's now inside the feedback loop, it's nearly irrelevant. If the amplifier's output voltage needs to rise to offset drop in Ra, then it will do so as long as it's able to do so to satisfy the condition of very nearly zero difference in the voltage on its inputs.

In fact, opamps can't deliver enough current to drive a motor. An LM324 can deliver 5 to 10 mA. So we need current gain. A power transistor can do that. This is an emitter follower connection. The current delivered from the emitter (the terminal with the arrow on it) will be many times the current into the base, and the emitter voltage will follow the base voltage very closely except for a nearly constant drop, typically about 0.7 volts. For more gain, a "Darlington connection" can be used, which is just two tandem emitter followers.

In this case the voltage offset will be two base-emitter drops, or typically 1.4 volts or so. You can make your own darlington with a signal level transistor (like 2N2222A) followed by a power transistor (Like 2N3055) or you can use a part that has the darlington all in one package, like TIP126. Current gain of TIP126 is at least 1000; so if the load needs 5 amps to drive the tach to satisfy the opamp, the opamp need only deliver 5mA to the darlington to make it so - and that's within it's capability.

You do need enough voltage "overhead" from the supply to accommodate the drop in Ra, allow the darlington a volt or more of collector-emitter voltage to keep it happy, and allow a volt or so of headroom for the opamp. It's best to supply the opamp with regulated and filtered power, but it's only a few milliamps so that's easy to provide from the main supply. Just takes maybe a 100 uF (or larger) 50V capacitor and a 3-terminal regulator, like a 7824.

You also need to provide the darlington with a heatsink. If the motor is drawing 2 amps at 13 volts - because that's what it takes to drive the load and satisfy the opamp by driving the tach to commanded speed - and the supply voltage is 24 volts, then there's 24 - 13 or 11 volts from collector to emitter on the darlington and there's 2 amps flowing thru it. That power (V x I = 11 x 2 = 22 watts) is dissipated as heat in the transistor. If that heat isn't removed the transistor will overheat and fail. The transistor can run up to 100C with no problem. So the transistor must be mounted on a sheet of aluminum or other heatsink that can convey the dissipated power to the surrounding air without exceeding 100C. Extruded heatsinks with lots of fins can sometimes be found in the surplus stores, but just a sheet of aluminum works fine for home projects. A 6" square of 1/8" thick aluminum, vertically mounted, can dissipate about 35 watts at room temperature and stay below 100C without a fan. Adding a small fan makes a big difference.

The foregoing is not a power-efficient way to control a motor, but it works very well for small motors. This approach is usable at least to a couple of hundred watts, or 1/4 hp. (Might need more heatsink). Efficiencies approaching 90% (much less heat to worry about) are possible with switching controllers. - which are not much more complex in terms of electronics. But you need to understand this concept well before you can understand a switcher - it's still a feedback control system.

There are stability considerations that I've skipped here. Frequency response and time domain response of elements within a closed loop can cause instability such that the system goes completely bonkers. Seeing that happen in a system with 50 hp actuators can be exciting. However, the bandwidth of readily-available opamps is so much greater than the bandwidth of a motor - especially one with an inertial load - you're not likely to run into trouble. As long as one element in the loop is strongly dominant with regard to "slowness", things work out OK.

An intuitive illustration: you start the system, command 2000 RPM. The motor is at rest, tach voltage is zero. Op amp output hits the rail - all available voltage to the motor. Motor accelerates itself and its load as fast as it can. When it reaches speed, the loop equations are satisfied and the amp should back off the drive. But if the amp also has a lag in it, it keeps pouring the power out. Motor overspeeds. Amp finally responds, cuts power off completely. Motor slows, again runs thru the setpoint speed, but again the amp is a little slow to figure that out so speed drops below setpoint. Amp finally catches up, applies full power. And so on. You have an oscillator. Fortunately, opamps are much quicker than motors so probably not an issue here.

Email: Don S. Foreman

Rev: 04/05/98