Electronic – How does this control loop work

boostcontrolpower electronics

The following description is taken from maximintegrated website (this link specifically), and after reading and trying to analyze the control technique it still does not make sense to me:

1 – In the figure it shows a boost converter, and it reads:

This duty cycle change then moves the output voltage to reduce the error signal to zero, thus completing the control loop.

From my point of view if that \$V_{ERROR}\$ is supposed to converge to zero for a steady state output voltage, the duty cycle should be kept steady at 0% since \$V_{RAMP}\$ will always be greater than \$V_{ERROR}\$ and the latter seemingly not subject to variations, which cannot be true since it meant the circuit would stop switching. A comparison I have used to understand it better is when I think of a PI controller, which brings the error to zero but has a memory in its output value, so it only responds to small variations once the steady state value is reached.

My question is: How is this control loop supposed to keep the output voltage steady regardless of for instance, load variations, if the error being reduced to zero would actually stop the switching?

By the way in the timing diagram they've inverted \$V_{SWITCH}\$ since it should be high when \$V_{ERROR} > V_{RAMP}\$, but it does not matter to the main question since it must've been a little mistake.

EDIT:

Following the main discussion, the figure below shows the circuit I have been inspired from to get this control loop done. It comes from this video in which the designer himself references maximintegrated's website.

I tried to replicate the idea using an op-amp, a behavioral voltage supply and a pulse source to generate the triangular wave in LTspice, trying to replicate the idea described in maxim's document and here are the results.

I have set a value of \$V = 5\,V\$ which is the reference for \$ 50\,V \$ in the output since \$V_{feed} = 0.1\,V_{out}\$.

Result: As you can see, my voltage level in the output is \$V_{out} \approx 25\,V\$, which is not what I have set the system for. Also, I tried changing values of the load and the output voltage also changes and that's not what I wanted either.

Finally, I'm probably mistaken in how this controller works and would really like to find out what is wrong.

Thank you in advance

Best Answer

How is this control loop supposed to keep the output voltage steady regardless of for instance, load variations, if the error being reduced to zero would actually stop the switching?

Basically, and in a nutshell, the error cannot reduce to zero. In a practical controller, the open loop gain may be numerically a thousand to 1 million so, if the output is (say) at 50 % of its range, then the error signal magnitude is a tiny fraction of 1 %. There has to be an error (no matter it being small), that drives the output to very, very close to the demanded level.

Exactly the same story for an op-amp; we say that both inputs are at the same voltage but that isn’t, strictly speaking, the precise truth.

And, in addition, if I address this point: -

the error being reduced to zero would actually stop the switching?

No it wouldn't because of the comparator and triangle wave shown below. They ensure that the system remains clocking out at the required duty cycle: -

Regards your simulation, now that I've had a good chance to look at it, you haven't got enough "gain" in the system to drive the error towards an acceptable value: -

Your error amplifier is only a gain of unity so, try adding a gain of ten stage between the op-amp output (U3) and the purple error box shown above. If things start to look unstable you might need a few pF across R11. If OK, proceed to increase the extra gain factor higher and higher. As you do so, the actual output will move ever closer to the target demand.

Related Solutions

Electronic – How to stabilize this control system

If you have huge transient error spikes, then you need a derivative part (i.e the D in PID) to handle these.

A simple PID system is not hard to write in code, it's the tuning that's the difficult part - you need to know roughly. The integral part will handle the DC offset, the proportional part the gain, and the derivative the reaction to change.

Here's some code I wrote for a temperature controlled etching tank I made with a PIC16F and a DS18B20 temp sensor ages ago (it used some nichrome wire driven via PWM wrapped round a couple of ceramic tiles for the heating element, so it had a large thermal capacity)
The results, once tuned, were an overshoot a of less than 2 degrees on power up, and a variation of less than a degree or so thereafter. It's very old code for a one off project thrown together quickly, so it can probably be improved a fair bit.
I found this document "PID without a PHD" quite helpful. Also, Matlab, Scilab or Octave can be used to simulate and tune PID systems.

/** Control variables **/

int DEADBAND = 0;   // Range where no change will be made

// PID gain values 
double Kp = 10;
double Ki = 0.02;
double Kd = 5;

// Accumulator variable for Integral calculation
double I_acc = 0;

// Returns out the proportional term
double get_prop(int error)
{
    double p;
    p = (Kp * error); 
    return p;
}

// Returns the integral term
double get_int(int error)
{
    double i;
    I_acc += (error);

    i = (double)(Ki * I_acc);
    if(i > 1000) i = 1000;
    if(i < 0) i = 0;
    return i;
}

// Returns the derivative term
double get_deriv(int error)
{
    double d;
    static int prev_error;
    d = (Kd * (error - prev_error));
    prev_error = error;
    return d;
}

double get_pid(int error)
{
    double result;
    double p, i, d;

    p = get_prop(error);
    i = get_int(error);
    d = get_deriv(error);
    result = p + i + d;
    return result;
}

Electronic – Open loop PWM using TL494

The TL494 can be used as a PWM generator. It's not clear what frequency you need, but the part is not fast. It might generate usable PWM up to about 30kHz. After that switch overlap can become a problem. But, following your questions:

It's probably not a good idea to drive the feedback pin directly with a pot wiper. Better would be to choose one of the error amps, make it unity gain, and then feed the pot wiper into the non-inverting input. That takes the loading off of the pot, and lets the amp do what it was meant to do.
The error amp you don't use needs to be biased on the input so that the output is low. Tie the inverting input to Vref through a 1KOhm and tie the non-inverting input to ground through a 1KOhm. DTC sets the dead time on the PWM waveform, with low being maximum PWM (minimal dead time), dead time increasing as DTC voltage rises. For maximal PWM, keep DTC below about 100mV.
Vref can be used to supply voltage, but has to be done carefully. Vref can't source more than 5 or 10 mA, so any more current than that and a buffer BJT (pnp) would be needed to supplement. Unless the extra load is very constant, more filtering would probably be needed. Vref is used inside the chip, so any noise added to it could mess up operation.
Normally to get a single output, one would connect E1 and E2 to ground and tie C1 and C2 together with a single pull-up resistor as a wire-or.

The TL494 is a bit unusual in that PWM duty cycle is inversely proportional to the feedback voltage, when feedback voltage is low PWM is maximal, as feedback voltage increases PWM duty cycle is reduced. If you haven't read "Designing Switching Voltage Regulators with the TL494" you should.

Best Answer

Related Solutions

Electronic – How to stabilize this control system

Electronic – Open loop PWM using TL494

Related Topic