PID Controller for Heating Resistive Element

8/12/2019 PID Controller for Heating Resistive Element

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BIOEN 337 Winter 2012

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Project4: PIDControllerforaResistiveHeatingElement

Objective: Develop a control system that uses feedback to maintain a target

temperature.

BackgroundJust as medicine involves both understanding a patients needs and acting on them,

a medical instrument is most useful when it can both read information and use thatinformation to establish some condition such as position, force, temperature, or

drug concentration.

Any piece of equipment that is intended to modify its environment ranging

from a hot plate to a computerized robot can be loosely divided into the part that

actually takes the action (the actuator) and the part that determines how hard the

actuator will work (the controller). The controller usually has two input signals, one

from a sensor that measures the condition of the system, and one from a user or

program that indicates the desired condition (the setpoint). Together, the

properties of the controller, the actuator, and the environment will determine how

the system will respond to a command from the user.

Control systems can be divided into two types.

Open loop systems supply an output in response to some user input but do notmonitor the result that the output is having on the system or environment. One

example is a burner on a typical stove. The user selects the power input to the

burner, but the actual temperature at the burner will depend on the burner

characteristics as well as the condition of the material that is being heated and

the temperature of the surrounding air. Effective openloop control requires

that the user understand how the input affects the system output.

Closed loop systems compare the user input to a value measured from thesystem to determine the output to send to the actuator. One simple example is

an oven with a thermostatic switch. The user needs only to set the desiredtemperature and the switch controls the average power input by turning the

heat on and off.

The use of feedback in a closedloop control system can accommodate a large

degree of ignorance of the system characteristics. However, in order to create a

control system that follows the desired output closely, we usually want to apply

some knowledge of how the system responds to actuation. In addition, high

performance control systems usually respond to the difference between the desired

and measured values (the error), not just to being over or under the setpoint.

TemperaturecontrolFor this weeks lab we will heat resistors by passing current through them. We

will assume that the resistors have a uniform temperature TR(t), so we can ignore

heat conduction and consider temperature only as a function of time. As you know

from BIOEN 325 and 327, the rate at which the temperature of an object rises is a

function of the heat capacity, mass, and applied power. Write the formula for rate of

change of temperature here, including units.

dT/dt = ?


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PrelabExercises

Write three formulas for the power P dissipated in a resistor as a function of

voltage v, current I, and resistance R.

P = ?

These equations show that we can control how fast a resistor heats by controlling

the applied power, and hence the applied voltage or current.

ProportionalcontrolFor the following discussion, let us define the error Eas the difference between

the desired and measured temperatures: E= TSET TMEAS. Each gain component is G

and the subscripts P, I and D, stand for proportional, integral and differential. The

gain can have whatever units you need to get the input and output to match up.

The simplest feedback control system is proportional, such that the controller

output equals GPE. For a thermal system we want the output to be power: the

farther the measured temperature is below the setpoint, the more power we want to

apply. On the other hand, if the measured temperature is above the setpoint, wewant to pull power out. This is very difficult (aka impossible) to do with a resistor.*

Therefore, we will control the heating and let the room air cool off the resistor

passively. Specifically, if the resistor heats the air so it rises and carries away the

heat, we are using the process of _________________ __________________.For this lab project you will set up an electrical circuit and write a control

program with LabView. Somehow the values that your control program produces

(numbers in a computer) need to be converted into actual power.

Lets start at the resistor and work backwards. Heating power is produced by

current passing through the resistor, so we are actually controlling the current. It

will be a lot more current than the computer can generate, and in addition, thecomputers DAQ module supplies a voltage, not a current. What we really need is a

hefty voltagecontrolled current source. For this we will use a transistor and one of

our big power supplies, such that the DAQ output is connected to the base of the

transistor (or the gate of a MOSFET).

PIDcontrolTo improve system performance, integral and differential terms can be added,

such that

This type of system is known as a proportionalintegraldifferential (PID)

controller. The flow chart for every PID controller is the same, and the gainsGP,I,Dare adjusted to produce the desired system behavior.

Note that the gains GP,I,Dare scalars, but the overall system gain VSENS/VCMDis

usually complex and is represented by a transfer function H(s), where s= + j.

*There are a few ways to remove heat at a controllable rate. One is to control the speed of a fan orcooling pump. Another is to use a Peltier cooler, which acts as a solidstate heat pump when avoltage is applied.

dt

dEGdtEGEGv

DIPout


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In today's lab exercise we will set up a heater circuit and start a process called

system identification, in which we model our system based on knowledge about its

physical construction, and on data that we have collected from it. Next week we will

use this model to design a controller and test it on our thermal system.

NomenclatureE Error: difference between measured and desired sensor valueGP Proportional gain

GI Integral gain

GD Differential gain

NTC Negative temperature coefficientVSENS Sensor voltage, hence the environmental value that the controller is trying to

control

VCMD Command voltage, the input from the user or program

TSET Temperature setpoint

TMEAS Measured or actual temperature

PID control theory assumes that we have a linear system. The implication here is

that if we double the value produced by our controller, we will double the power

into the resistor. However, this linearity assumption does not match our thermal

system in a few ways. First, the current through the transistor is not proportional

to the voltage at its base. Second, the power in the resistor is not proportional to the

current (or voltage). Of course, you could write a mathematical function in LabView

to adjust the DAQ output voltage, making the power proportional to the controller

output.

There is a more subtle problem, though. Unlike the resistor (for which V=IR), the

voltage across the transistor is almost independent of the current. In fact, the

voltage often goes downas the current goes up. Consider three cases in our heater

circuit.

1) The base voltage is low, the transistor current and resistor voltage are both

zero, and the voltage across the transistor is the supply voltage VCC. In this case, the

power dissipated in the transistor is P = IV = 0*Vpower = 0. Thats nice.

2) The base voltage is high, the transistor current and resistor voltage are both

high, and the voltage across the transistor is as low as it can get. In this case, the

power dissipated in the transistor is P = IV = high*low = not too bad.

3) The base voltage, transistor current, resistor voltage and transistor voltage are

all midrange. In this case, the power dissipated in the transistor is P = IV = mid

range*midrange >> 0. That means that a lot of the heat is dissipated in thetransistor, although we want it to heat up only the resistor. You can verify that

there is a maximum amount of power dissipated when VR= VCC/2.

We see that it is more efficient to operate with the transistor at zero or full

current. In order not to convert nice clean electricity into waste heat, we can use


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PulsewidthmodulationFor purposes of heating a resistor, it works just as well if we control the average

power rather than keeping the power exactly right all of the time. This power

averaging is the basis of a method of power control pulse width modulation (PWM).

PWM controllers provide an output that is either off or at full power, but with

timing that is determined by the error signal. Whereas an oven with a standardon/off thermostat would be at full power all of the time as the temperature rose

from Ttarget20 to Ttarget1, an oven with a proportional thermostat using PWM

might be at full output 80% of the time when the temperature deficit is 20, but only

5% of the time when the temperature deficit is 1.

The fluctuations are typically rapid, cycling several times per second. The

resistor has some heat capacity and heat resistance, so it behaves like a lowpass

filter that damps out the thermal oscillations. At locations away from the actual

point of heat application (e.g. at the thermistor), it is impossible to tell that the heat

has been applied in pulses. We calculate average power = max power * ontime

fraction, where ontime fraction is pulse width / pulse period.

When we use PWM, we use LabView to calculate a percent ontime, also calledthe duty cycle. Obviously the duty cycle must be between 0 and 100%. This

limitation does not seem so bad when we realize that the DAQ output is finite too,

ranging from 0 to 5 V. We use LabView to turn the the DAQ output on and off, where

off is usually 0 V and on is somewhere from 3 to 5 V, whatever is necessary to

permit maximum current through the transistor. Note that pulse width modulation

permits us to choose either the analog outputs or the digital outputs from our data

I/O device. The analog outputs can be varied in very fine increments over a wide

voltage range, but we dont need that here. The digital outputs are either 0 or 5 V, so

they are good for proportional control onlywith a PWM controller. Each digital

output is one bit of a binary number, so to use these outputs for thermal control we

simply turn the desired bit on or off at the appropriate rate.* In either case, thecurrent supplied by the data I/O device is small, so it is necessary to amplify the

current with a transistor before applying it to the heater.

*The analog outputs from an Arduino microprocessor board are actually digital, operated in pulsewidth mode.


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Lab Exercise

The goal of this lab project is to create a PID controller that regulates the

temperature of a resistorthermistor. This controller can then be the basis for the

thermal control system in an environmental control chamber. The major steps in

this project are:

Assemble a circuit that can be used to heat a resistor and measure itstemperature;

Acquire step response data and use it to create a Laplacedomain model of the

heater system;

Determine appropriate PID gains both analytically and by trial and error;

Create a feedback controller using Labview;

Test the behavior of the controlled system.

EquipmentBreadboard & connecting wire

TIP121 or TIP115 or 2N6292 or 2N6107 power transistor with heat sink

4.7, 5, or 10 resistor, rated for 1 W (2), called lowvalue resistor belowThermistor: 135202FAGJ01, 135103LAGJ01, or similar

1 kresistor (2)1 or 10 ktrim pot for Wheatstone bridge [we will actually just use a voltage divider set

up]

Power supply

USB6009 data I/O device

Computer with LabView

Circuitsetupa)Prepare the lowvalue resistor and thermistor by sliding wire insulation over

the leads; this provides electrical isolation.

b)Join the resistor and thermistor with heatshrink tubing. [this may be modified

epoxy instead]

c)Create a Wheatstone bridge for the thermistor, using the 1kresistors, the trim

pot, and the thermistor. The output from this bridge is the input to your data

I/O device. [we will actually just use a voltage divider]

d)Connect the lowvalue resistor(s) in series with the power transistor, i.e.

between the emitter and ground. The input to the transistor's base will be the

voltage from one of the analog outputs from Labview.

e)Test the operation of the circuit with a power supply and multimeter, or power

supply, function generator, and oscilloscope. Test the output voltage from the

bridge as you vary the voltage at the base of the transistor. Dont burn yourself!

Softwaresetupa)Plug the USB6009 data I/O device into an available USB port on your computer.

Complete the installation steps if necessary.

b)Create a LabView VI that contains one data acquisition channel and a pulse

width modulation output for one analog output channel. Create the front panel


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controls necessary to set the output duty cycle manually. This VI can be built

around the VI that you created for the colorimeter, but you are going to need a

lot of help with the PWM part. We will give it to you, and will provide an

example.

SystemidentificationOnce the circuit and software are complete, our first goal is to estimate an impulseresponse and a transfer function for the heater system.

a)Apply a step change to the duty cycle and measure both the input and output

voltages until an equilibrium voltage is reached. Estimate the time constant of

the transient response, e.g. by measuring the slope at t=0. This time constant

will probably be different for the cooling and heating cases; although you will

not control the cooling phase, it will be interesting to compare the two results.

Therefore, perform at least one experiment for each case. In each one, record

the command voltage VCMD(i.e. the voltage applied to the base of the transistor)

and the sensor voltage VSENS (i.e. the voltage from the bridge).

b)Use the input and output voltage signals with the VI that fits transfer functions

to response data. You will be shown how to do this in lab. Obtain three transfer

functions: 1) firstorder, 2) secondorder with no zeros, and 3) secondorder

with one zero. Choose one of the transfer functions with no zeros to proceed

with PID control development.

PIDcontrollerdevelopmenta)Plug your coefficients into the closedloop PID control simulation VI and

experiment with the PID gains. Your goal is to minimize the rise time and

settling time, and overshoot. The overshoot can be most important in a

biological control system, because excessive concentrations or temperatures

can kill.

b)Plug your coefficients into the general form for a transfer function in a PID

controlled system (assuming two poles and no zeros), and solve for PID

constants that produce a critically damped system. Test those constants in the

closedloop PID control simulation VI.

c)Add a PID control loop to your input/output VI. You may do these calculations

with mathematical icons, express (interactive) math functions, or by using a

MathScript node, which allows you to enter MATLAB script that operates on the

LabView variables. LabView also includes a variety of PID control functions thatdo not require you to do your own mathematical programming, but they do not

tend to work any better and you might not learn as much if you rely solely on

LabView functions. Your VI will need to include a conversion from the voltage

produced by your PID controller into output duty cycle.

d)Set the gains determined previously and test the response of the system to a

step input.


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Lab wri te-up:

1. See separate document for guidelines for this lab writeup.

InconclusionBy the end of the project, you should have learned...

In any system that needs some input just to stay at a constant value, a

proportional control system will produce a steadystate error.

The integral gain in a PID controller can eliminate the steadystate error.

The differential part of a PID controller can reduce overshoot, allowing you to

use higher proportional and integral gains.

A feedback controlled system with only one energy storage element (e.g., the

heat capacity of the resistor+thermistor) can act like a secondorder system.

Additional information can be found at

http://en.wikipedia.org/wiki/PID_controller and

http://en.wikipedia.org/wiki/Control_system


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NOT USED

[The original handout mentioned a linearity compensator and low pass filter. We

did not include these this year so you do not need to include them in the diagram.

The linearity compensator adjusts for the offset between the base and emitter

voltages at the transistor. A lowpass filter is needed if the data is noisy, because

voltage fluctuations cause rapid changes in the time derivative. Our signals looked

pretty smooth this year.]

PID Controller for Heating Resistive Element

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