A few days after I read your article on wind-up error in my stack of delinquent reading (Electronic Design, Nov. 6, 2000, p. 146), I encountered an ad in a mechanical rag mentioning wind-up. My interest was sparked.
I haven't done a motion servo for at least 15 years. But I ran into a virtually identical problem with a 100-kW medical MRI gradient coil driver and a similar-size HV supply for a medical CT scanner a few years ago. Ultimately, output E/L limited the coil driver slew rate, and I/C limited the HV supply. A position or velocity servo is typically limited by force/mass or torque/moment of inertia. The medical coil driver and HV supply were absolutely unforgiving of discontinuities in the settling response or resultant overshoot.
Wind-up error results from accumulated charge in the error integrator when the output can't keep up with the demand. The ultimate solution was to keep the loop closed by rate-limiting the demand voltage to just below the natural slew limit of the system. The modified system slewed slower, but closed on equilibrium much faster—i.e., "slower is faster," or, "if it hurts to do that, don't do that."
Hello, Ron. I agree with your "slower is faster," or, "if it hurts to do that, don't do that." But for a truck with variable mass, one cannot foresee what may be the natural slew rate of the system. The acceleration rate on an upgrade will greatly differ from that on a downgrade, so you just about have to use anti-wind-up!—RAP
I found your "What's All This Tee Network Stuff, Anyhow?" column interesting (Electronic Design, June 4, p. 113). Like you, I learned the wye-delta, delta-wye transforms as an EE student over 30 years ago. My professor made an interesting observation, which he taught us. It has allowed me to remember how to do them over the years from memory. It is as follows: If you want to do the wye-delta transform, rewrite the impedances (resistances) as admittances (conductances), and the form of each transform expression will be the same as for the delta-wye direction. For instance, using the labels of your figure:
G1 = 1/R1, G2 = 1/R2, etc. Now:
GC = G1G2/(G1 + G2 + G3)
GA = G2G3/(G1 + G2 + G3) and
GB = G1G3/(G1 + G2 + G3). Then:
R1 = 1/G1, etc.
It's easy to remember to use admittances because Y is the symbol normally used for those quantities. The beautiful symmetry between series and parallel circuits makes this possible, no doubt. I teach this to my electronic instrumentation students through the example of calculating the transfer function or voltage-gain expression for an op-amp circuit with a Tee network in the feedback path, as in your example.
There are several ways to obtain the solution. But if you use the wye-delta transform to calculate RC, and recognize that the "−" op-amp input is a virtual ground, you can get the inverting voltage-gain expression as RC/RIN (not shown on your diagram). I have also used the expressions that you derived in a previous article to show the effect of finite open-loop voltage gain upon closed-loop voltage-gain accuracy.
Thanks for the comments, Elvin!—RAP
In the second paragraph of "What's All This Tee Network Stuff, Anyhow?" you stated something that has always been my belief, too: "You won't be able to tell them apart (except perhaps for their capacitive strays)." But as I read further, I realized that you have found a way! Imagine this. Build a test jig that consists of the circuit shown in the figure in a box with three terminals on the top of the box. Bring out the nodes where the R1, R2, R3 network connects to the circuit. Use a noisy op amp for convenience in measuring noise levels. Then, connect either the Tee network or the delta network. Using the values in your first example, the output noise will be 100 times greater (Nope. /rap) when the delta network is connected, which should be easily distinguishable.
Not at all. The noise gain and the noise will be exactly the same for both cases. Go ahead and try it!—RAP
I've made two AA-cell red LED "night-vision flashlights" over the past six years for sailing friends. I removed the glass incandescent element from the bulb base and mounted the LED and a 47-Ω series resistor in its place.
I just installed my homemade LED anchor light on my sailboat. It uses 12 white (5600 mcd) LEDs connected in four groups of three, which are in series, driven by the LM317L current source at 20 mA. The total consumption is 80 mA, for a consumption of 1 W at 12 V. The normal anchor-light lamp is 10 W. The four panels are mounted vertically in a square, and the three LEDs are angled 30° apart (/|\). This covers the full visible circle with the 12 LEDs. It may not be visible for the legal one-mile range, but it sure looks effective in the dark.
Thanks for the notes, Greg!—RAP
All for now. / Comments invited!
RAP / Robert A. Pease / Engineer
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