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Prelab 8

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In lab this week we will be building the final parts of the Doppler ultrasound system, namely:

  • A boost converter, to take the 5 V USB supply and turn it into 30 V, so that we don't need multiple power supplies anymore.

  • A comparator to turn the signal at the output of the S-K filter into a square wave, so that we can measure it's frequency using the Teensy.

1) Boost converter

In Lecture 20 we discussed Boost converters and their operation. A basic "theoretical" Boost Converter will take in a low voltage v_{IN}, and produce a higher output voltage v_{OUT}:

This phenomenon is based on the circuit going through a three-stage cyclic operation depicted below which occurs because of the state of the diode and FET. Both of these devices act as switches depending on some external conditions. Specifically:

  • The diode will be closed (conduct) when v_D and i_D are positive and will be open (not conduct) when v_D and i_D are negative
  • The FET will be closed (conduct) when v_{control} is high (a sufficiently high non-zero voltage) and open (not conduct) when v_{control} is low.

A PWM signal applied to v_{control} is what pushes the system through its three states, which is based on a period T and a duty cycle D (which varies from 0.0 to 1.0).

In brief summary (see lecture notes for more details!) the system operation is as follows:

  1. In Stage 1, the FET is ON, connecting the inductor L_1 to ground. As a result, the inductor begins to build up a current through it based on the integral of the voltage across it. The longer the system is in stage 1, the more current builds up! While the inductor builds up current (energy), the capacitor(s) on the output provide energy to any load (external devices) attached to the system. This stage will last for DT seconds, where D is the duty cycle and T is the period.
  2. In Stage 2, the FET turns off. The fact that a positive current has built up through the inductor (and that inductors require continuous current through them) means that current will be pushed through the diode and into the RC circuit. This current is flows directly into the load or it goes into the capacitor, charging it up in the process. In the process of charging it up, the voltage across the inductor will decrease (become smaller in size if positive or more negative if negative), and in the process resulting in a drop in inductor current. This stage will last for D\left(1-T\right) seconds provided the current through the inductor stays positive. Otherwise the system transfers to Stage 3.
  3. In Stage 3 (which the system may or may not get to depending on how far current through the inductor decays), the FET remains off, and the diode opens up, preventing any back-flow of energy. This stage will last for however long is neccessary until the cycle repeats (back to Stage 1). If the Boost is operated in what is called "Continuous Current Mode", the circuit will not normally end up in Stage 3.

1.1) Boost Controller Chip

While a Boost Converter is theoretically pretty easy to conceptualize, there's a lot of details that go into its design, and deploying one is far from trivial. For example, how do you pick your duty cycle? How do you protect against dangerous edge cases in operation like the buildup of too much current?. For this reason and others, a control chip is usually used to manage the v_{control} signal appropriately. For our ultrasonic board we use the MAX770 Boost Chip! We deploy it in a circuit below, and you should see many of the parts present match those found in our "theoretical boost" converter are present in this actual boost converter.

  • Inductor is L1
  • FET is Q1
  • Diode is D1
  • Output capacitors are C11 and C12

A few other parts are present that help our Boost converter operate closer to an ideal situation. Specifically:

  • Capacitors C9, C7, and C8 are used to "smooth out" the input voltage to the entire system (making v_{IN} appear more like a constant, which we assume!)
  • Resistor R14 is used to provide current-feedback to limit the current built-up in the coil
  • Resistors R15 and R16 are used to provide feedback about the output voltage to automatically adjust the duty cycle!

1.2) Deciding the Duty Cycle

As discussed in lecture, the output voltage from a boost converter will be based off of the Duty Cycle D in the following manner (where D varies from 0.0 to 1.0):

v_{OUT} = \frac{v_{IN}}{1-D}

This "solution" assumes a lot of ideal conditions and various other things (the capacitor is large enough to assume no ripple on v_{OUT}, the inductor current has both increasing and decreasing phases, etc. But if you deviate from these assumptions, things go wonky. For example, if the output current of the inductor doesn't stay positive for the entire duration of Stage 2, the expression for v_{OUT} is closer to:

v_{OUT} = v_{IN}\cdot \frac{1+\sqrt{1+\frac{4D^2RT}{2L}}}{2}

If the current from v_{IN} is limited, this equation can change even more...and the list goes on.

Therefore, while it is useful to get a rough idea of what a Boost converter will provide, the reality is that in real-life the Duty Cycle will need to vary depending on the system's load. As a result it is very hard to approach the Duty-Cycle design in a set-it-and-forget-it approach. Instead the MAX770 utilizes information about the output voltage fed back to it in order to decide how to adjust the duty cycle.

The MAX770 senses the output voltage with the FB pin (for "feedback"). This FB pin connects to an internal comparator that is roughly working as so.

The signal generated by this comparator v_{CON} is used to "teach" the boost converter the correct duty cycle D to apply to the FET. If v_{CON} is high, some other internal circuitry gradually increases the duty cycle. If v_{CON} is low, it does the opposite, decreasing the duty cycle. In this way, the system exhibits negative feedback and that allows it to lock onto a target voltage at the output. What is that target voltage?

If v_{OUT} is attached directly to the v_{FB}, at what voltage will the system tend to target its v_{OUT} to approximately (in Volts)?

v_{OUT} =

What if this target from above isn't what we want? We want 30V for our circuit, you know. How do we get higher voltages using this feedback mechanism? In other words how could we make this chip drive the output voltage higher? The answer is to "lie" to it. We build a resistor network that takes the output and doesn't tell the chip the whole truth so that it thinks it is further off than it needs to be. Let's call this voltage v_{LIE}. If we treat the output voltage v_{OUT} as a voltage signal below, what will v_{LIE} be in terms of v_{OUT}, R_{15} and R_{16}? (Use vOUT,R15,and R16) in your answer?

v_{LIE} =

Now let's put everything together and lie to the MAX770. Let's have v_{FB} be based on v_{LIE}. What will v_{OUT} end up being approximately in terms of R15, R16 and other known constants? (answer in terms of volts).

v_{OUT} =

Given a value for R16, generate an expression for R15 as a function of R16 so that v_{OUT} is set to 30V (which is what we need.)

R_{15} =

Verify that the actual values of R15 and R16 in the schematic match up with this.

1.3) Protecting Over-Current

Another interesting phenomenon that can happen in a Boost converter is if the load on the output is sufficiently low in resistance such that v_{OUT} can never build up to a large voltage, the potential for current-run-away exists. This is a phenomenon, where v_{OUT} never gets high enough to set up things so the inductor current decreases during stage 2. As a result, cycle-to-cycle the inductor can build up more and more of a current, potentially damaging the circuit!

To protect against this, the MAX770 can also monitor the current through the inductor during Stage 1 with an in-series resistor R14. One side of this resistor is connected to ground and the other is the bottom side of the FET. The voltage at the bottom side of the FET is called v_{CS}. Generate an expression for the voltage at the current sense pin (CS) as a function of R14 and inductor current iL.

v_{CS} =

The voltage present on the CS pin is fed into another internal comparator in the MAX770 that has an approximate circuit below:

This circuit works in the following way:

  • If v_{CL} is low, the circuit works as usual during Stage 1, building up energy in the inductor.
  • If v_{CL} is high, the circuit prematurely ends Stage 1 to protect too much current from building up.

Using the circuit above, determine the current limit i_{LIM} targeted with our the value of R14 in our schematic. Provide your answer in mA:

i_{LIM} =

2) Comparator

The output of our Sallen-Key Filter from Lab 10 is a low-frequency signal based on the doppler frequency. Our Teensy microcontroller is going to measure this frequency using a digital pin, but as a result we need to digitize this low-frequency signal. In order to do that we'll run our low frequency signal through a comparator. This is going to be a similar design problem/need as our first comparator from exercise 06 and Lab 7. Make sure to review those two assignments since the procedure is nearly identical!