Hobby servos, such as the one pictured at right, are wonderfully useful little devices. You’ll find them moving control surfaces on model planes, in steering linkages on RC cars, and even in the feeding mechanism of an automatic ping-pong ball launcher (one of my simpler college design projects).
Anytime you need something to rotate to a specific position, think of the hobby servo. They’re fairly low cost, and come in a variety of torque sizes, from tens to hundreds of ounce-inches.
So let’s say you’ve bought yourself a servo from Tower Hobbies (or wherever). How are you going to control it? Well, you could purchase a radio and receiver, but if you’re not planning on building your servo into a vehicle of some sort, that’s really overkill (and expensive). You could program a microcontroller to generate the control signals, but that could get complicated if you’ve never worked with MCUs before. Instead, what I’d like to discuss today is a purely analog circuit for PWM servo control.
First, let me give you a little background information. Hobby servos are typically connected by three wires: power (red), ground (black), and signal (yellow/white). The power and ground lines are typically hooked directly to your battery or power supply. The signal line, however, is used to command the servo to move to a specific angular position. This signaling is normally accomplished via pulse-width modulation (PWM). That is, a digital pulse is sent to the servo on a routine basis (e.g. at 100Hz, or 100 times per second). The width or duration of this pulse determines the position of the servo’s horn. For instance, a pulse width of 1ms commands a fully clockwise rotation, a width of 2ms commands a fully counter-clockwise rotation, and a width of 1.5ms will center the horn.
Now the question is, how do we generate such a signal? Why, we simply use the following pulse-width modulator circuit (adapted from Maxim Application Note 3201):
But then again, those three op-amps could actually all be housed inside a single 14-pin DIP/SOIC package. And beyond that, all you need are eight resistors, one capacitor, and one potentiometer (a variable resistor). So while this may be physically more complex than just plopping down a microcontroller, there’s no software required.
So just how does this circuit create our PWM signal? Well let’s start with the “Integrator” section. This group of components (R1, C1, and U1) mathematically integrate or sum the voltage wired into the left terminal of R1 (line label #5). Put simply, the capacitor C1 is summing up this input voltage over time. To see how this happens, let’s start by analyzing the node between R1 and C1 (label #2). Now assuming all of our op-amps are ideal (a fair assumption in most cases), no current will enter or leave their inverting (-) and non-inverting (+) terminals. Since the current flowing through a series connection of electrical components (R1 and C1) must be equal, we can write the following:
The first half of this formula may look familiar; it’s ohm’s law (V/R = I). However, we’ve defined the voltage across R1 as (V5 – 2.5). Why? This is because the voltage at the input terminals (+ and -) of our ideal op-amp must be equal since we have a negative feedback path (a connection from output to inverting (-) terminal) through the capacitor. And we know that the voltage at the non-inverting (+) terminal of the op-amp must be 2.5V because of the voltage divider at node #1. Thus, since C1 is providing a feedback path for the op-amp, we can safely assume that the inverting terminal is also at 2.5V. The second half of this equation comes from the I-V relationship for capacitors, I = C*dv/dt.
So if we solve this equation for V3, with an initial capacitor voltage of zero, we get:
If we keep the voltage V5 constant, we’re left with just an equation for a straight line. Basically, the output V3 starts at 2.5V, then ramps linearly up/down, depending on V5, as time (t) goes on. If left unchecked, the output of U1 would eventually hit a supply limit (either 0V or 5V). However, the second half of the above circuit, labeled “Oscillator Comparator” ensures that this does not happen by switching V5 between 0V and 5V.
Let’s take a look at U2, the second op-amp pictured above. You’ll notice there’s no feedback path between its output and its inverting (-) terminal. So what we’ve got is a comparator. That is, if the voltage on its non-inverting (+) terminal is greater than that on its inverting (-) terminal, the output of U2 will be roughly 5V (our positive supply voltage). Otherwise, the output will be roughly 0V. I say “roughly” because this op-amp (TL072) can’t operate “rail-to-rail”, which means its output can’t quite reach our supply voltages.
In order to understand this comparator a little better, let’s take a look at the point at which it switches between its high (5V) and low (0V) output. Since the inverting terminal of U2 is fixed at 2.5V by the voltage divider at node #1, this switching must take place when node #4 passes 2.5V. Let’s determine the voltage V3 necessary for this to occur. To begin, I’ll equate the currents through R2 and R3 (since again, no current flows into the + terminal):
Now don’t be confused about where that 2.5V is coming from. This is the switching voltage for U2. We’re not saying that U2’s non-inverting (+) terminal is fixed at 2.5V. It’s not, because we don’t have negative feedback. This voltage will vary based on V3 and V5. Anyway, solving for the switching-point voltage V3 we obtain the following:
So we’re going to have two switching points, based on the two possible values of V5. When V5 is 5V, V3 will be decreasing linearly, and a switch will occur at V3 = 1.325V. However, when V5 is 0V, V3 will be increasing linearly, and switching will occur at V3 = 3.675V. So this is how the oscillation happens: V3 ramps linearly in one direction until it reaches a switching threshold, at which point the integration reverses and V3 ramps backwards. So what does this give you? A triangle wave, as seen in green in this PSpice simulation:
Of course, we can’t just use a triangle wave to signal our servo. What we need now is a third comparator to generate a PWM signal using this triangle wave and a variable threshold voltage (the red line pictured above). This is where the components around U5 come into play. Again, since U5 has no negative feedback path, it operates as a comparator. Thus, its output can only be 5V or 0V (roughly). So if we feed our triangle wave into its non-inverting (+) input, and a DC threshold voltage into its inverting (-) input, what we get at the output is a square wave (the blue line) whose pulse width is inversely proportional to our threshold (i.e. a higher threshold yields a shorter pulse).
The last trick here is that we can’t just hook up a potentiometer (pot) between power and ground. That would give us a threshold voltage variable from 0V to 5V. What we actually need is a threshold voltage that varies from about 3.2V to 3.5V, for a pulse width ranging from 1-2ms (based on the form of triangle wave shown above). Well in order to accomplish this, I’ve placed two additional resistors (Rx and Ry) in series with the potentiometer (Rpot). In order to determine appropriate values for these resistors, I’ll start with two voltage divider formulae which are based on the two limits of the potentiometer:
So when the pot’s screw is turned fully clockwise, the pot’s entire 10kΩ resistance will be placed between Rx and the inverting (-) input of U5. This will produce our maximum threshold voltage, VH. However, when the screw is turned fully counter-clockwise, the pot will act as a short between Rx and U5, yielding our lowest threshold voltage, VL. If we now combine and solve these two formulae, we can determine values for Rx and Ry:
Note: To determine resistances for different potentiometer values, you’d just need to replace the 10k in the first set of equations with your updated value and re-solve.
Now of course, all we really need here is a means of controlling the threshold voltage at U5’s inverting (-) terminal. Back when I was working on my automatic ping-pong ball launcher, I wanted to use my laptop and a DAQ card to control my servo. The DAQ card I had available at the time didn’t allow me to generate precisely-timed digital signals. However, it did provide several analog outputs, which I could have connected directly to U5 in order to control this circuit’s pulse width. But I didn’t know about this circuit back then, so I actually just used a microcontroller programmed to generate the appropriate signals based on an ADC input connected to my DAQ hardware.
Finally, you may also be wondering, how can I calculate the frequency of this PWM signal? (Or maybe you’re getting sick of all these equations?) Well, given the above formulae, it’s actually quite simple to calculate. We just need to set the integration formula equal to the switching voltage formula, like so:
We now solve for the time t, then multiply by four (since this equation gives you the time required by one quarter of a full cycle), and invert to find the frequency:
Alright, enough of these crazy formulae. Pictures of the final circuit? Yes, please!
You’ll notice that Rx is actually a series combination of three resistors, while Ry is a series combination of two resistors. This is because I didn’t have suitable values for Rx and Ry just lying around. Oh well, it just makes things a little messier! Here’s the full setup:
Finally, here’s a screenshot of the IOBoard oscilloscope VI I used to test out my PWM circuitry. You’ll notice that, as I mentioned earlier, the comparator’s output doesn’t quite reach 0V and 5V because these op-amps (TL072) do not have rail-to-rail outputs:
One final note on the schematic above. The resistance R10 should have been unnecessary. I initially included it because PSpice wouldn’t run my simulation with U5’s output floating. However, after constructing this circuit I found it necessary for reliably servo operation. I’m not entirely sure why this was the case; perhaps the voltage levels without R10 were slightly outside of the servo’s acceptable range? The signal on the screen certainly didn’t appear much different with or without it. Perhaps if I had a higher frequency/resolution scope I’d see something more telling… Oh well, it may not be an issue with your servo.
Anyway, if you have any questions on this circuit or would like to make suggestions, feel free to leave a comment. I’d love to hear about your experience.
Also, please help yourself to my PSpice files (from the Orcad 16.0 student demo). These were used to create the schematic shown here as well to perform simulations.
Here’s my final bill of materials (BOM):
- 2x TL072 Operational Amplifier
- 1x Hitec HS-81 Servo
- 1x 4.7uF Capacitor (Can be electrolytic, despite the slight negative voltage)
- 1x 10kΩ Potentiometer
- 2x 20kΩ Resistor
- 2x 1kΩ Resistor
- 1x 470Ω
- 1x 119kΩ Resistor
- 1x 166kΩ Resistor
Update (11/2/2010): It was pointed out that the TL072 may require a minimum supply voltage of 7V, so 5V could be cutting it a little close here. Now I’ve looked through the datasheet and don’t see a specific limit mentioned, but most of the graphs do only go down to ±3.5V (which I suppose you could interpret as 7V). Regardless, the circuit works fine with a single 5V supply, although the outputs don’t go rail-to-rail (which is normal operation), as I mentioned earlier. The real concern with most op-amps is their upper supply limit, as you don’t want to fry anything. Also, you might be more interested in the TL074 for this project, which contains four op-amps in one package. I didn’t happen to have a quad op-amp lying around when I built this circuit, hence the two duals.