Arduino-Based DIY Voltage Booster (EE Tip #117)

If your project needs a higher voltage rail than is already available in the circuit, you can use an off-the-shelf step-up device. But when you want a variable output voltage, it’s less easy to find a ready-made IC. However, it’s not complicated to build such a circuit yourself, especially if you have a microcontroller board that’s as easy to program as an Arduino. And this also lets you experiment with the circuit so you can get a better understanding of how it works.

Source: Elektor, April 2010

Source: Elektor, April 2010

No surprises in the circuit—a largely conventional boost converter. The MOSFET is driven by a pulse width modulated (PWM) signal from the microcontroller, and the output voltage is measured by one of the microcontroller’s analog inputs. The driver adjusts the PWM signal according to the difference between the output voltage measured and the voltage wanted.

We don’t have enough space here to go into details about how this circuit works, but it’s worth mentioning a few points of special interest.

The small capacitor across the diode improves the efficiency of the circuit. The load is represented by R3. The components used make it possible to supply over 1 A (current limited by the MSS1260T 683MLB inductor from Coilcraft), but maximum efficiency (89%) is at around 95 mA (at an output voltage of 10 V). To avoid damaging the controller’s analog input (≤5 V), the output voltage may not exceed 24 V. For higher voltages, the values of resistors R1 and R2 would need to be changed.

The MOSFET is driven by the microcontroller, which is nothing but a little Arduino board. The Arduino’s default PWM signal frequency is around 500 Hz—too low for this application, which needs a frequency at least 100 times higher. So we can’t use the PWM functions offered by Arduino. But that’s no problem, as the Arduino can also be programmed in assembler, allowing a maximum frequency of 62.5 kHz (the microcontroller runs at 16 MHz). To sample the output voltage, a frequency of 100 Hz is acceptable, which means we can use Arduino’s standard timers and analog functions. The Arduino serial port is very handy: we can use it for sending the output voltage set point (5–24 V) and for collecting certain information about the operation. Thanks to the Arduino environment, it only took about half an hour to program. Software is available. — Clemens Valens (Elektor, April 2010)

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Issue 282: EQ Answers

PROBLEM 1
Construct an electrical circuit to find the values of Xa, Xb, and Xc in this system of equations:

21Xa – 10Xb – 10Xc = 1
–10Xa + 22Xb – 10Xc = –2
–10Xa – 10Xb + 20Xc = 10

Your circuit should include only the following elements:

one 1-Ω resistor
one 2-Ω resistor
three 10-Ω resistors
three ideal constant voltage sources
three ideal ammeters

The circuit should be designed so that each ammeter displays one of the values Xa, Xb, or Xc. Given that the Xa, Xb, and Xc values represent currents, what kind of circuit analysis yields equations in this form?

ANSWER 1
You get equations in this form when you do mesh analysis of a circuit. Each equation represents the sum of the voltages around one loop in the mesh.

PROBLEM 2
What do the coefficients on the left side of the equations represent? What about the constants on the right side?

ANSWER 2
The coefficients on the left side of each equation represent resistances. Resistance multiplied by current (the unknown Xa, Xb, and Xc values) yields voltage.
The “bare” numbers on the right side of each equation represent voltages directly (i.e., independent voltage sources).

PROBLEM 3
What is the numerical solution for the equations?

ANSWER 3
To solve the equations directly, start by solving the third equation for Xc and substituting it into the other two equations:

Xc = 1/2 Xa + 1/2 Xb + 1/2

21Xa – 10Xb – 5Xa – 5Xb – 5 = 1
–10Xa + 22Xb – 5Xa – 5Xb – 5 = –2

16Xa – 15Xb = 6
–15Xa + 17Xb = 3

Solve for Xa by multiplying the first equation by 17 and the second equation by 15 and then adding them:

272Xa – 255Xb = 102
–225Xa + 255Xb = 45

47Xa = 147 → Xa = 147/47

Solve for Xb by multiplying the first equation by 15 and the second equation by 16 and then adding them:

240Xa – 225Xb = 90
–240Xa + 272Xb = 48

47Xb = 138 → Xb = 138/47

Finally, substitute those two results into the equation for Xc:

Xc = 147/94 + 138/94 + 47/94 = 332/94 = 166/47

PROBLEM 4
Finally, what is the actual circuit? Draw a diagram of the circuit and indicate the required value of each voltage source.

ANSWER 4
The circuit is a mesh comprising three loops, each with a voltage source. The common elements of the three loops are the three 10-Ω resistors, connected in a Y configuration (see the figure below).

cc281_eq_fig1The values of the voltage sources in each loop are given directly by the equations, as shown. To verify the numeric solution calculated previously, you can calculate all of the node voltages around the outer loop, plus the voltage at the center of the Y, and ensure they’re self-consistent.

We’ll start by naming Va as ground, or 0 V:

Vb = Va + 2 V = 2 V

Vc = Vb + 2 Ω × Xb = 2V + 2 Ω × 138/47 A = 370/47 V = 7.87234 V

Vd = Vc + 1 Ω × Xa = 370/47 V + 1 Ω × 147/47A = 517/47 V = 11.000 V

Ve = Vd – 1 V = 11.000 V – 1.000 V = 10.000 V

Va = Ve – 10 V = 0 V

which is where we started.

The center node, Vf, should be at the average of the three voltages Va, Vc, and Ve:

0 V + 370/47 V + 10 V/3 = 840/141 V = 5.95745 V

We should also be able to get this value by calculating the voltage drops across each of the three 10-Ω resistors:

Va + (Xc – Xb) × 10 Ω = 0 V + (166 – 138)/47A × 10 Ω = 280/47 V = 5.95745 V

Vc + (Xb – Xa) × 10 Ω = 370/47V + (138-147)/47A × 10 Ω = 280/47 V = 5.95745 V

Ve + (Xa – Xc) × 10 Ω = 10 V + (147-166)/47 A × 10 Ω = 280/47 V = 5.95745 V

Electrical Engineering Crossword (Issue 283)

The answers to Circuit Cellar’s February electronics engineering crossword puzzle are now available.

283-crossword-key

Across

2. LITZWIRE—Separately insulated strands woven together [two words]
4. LINKFIELD—First in a message buffer’s line [two words]
6. PETAFLOPS—Measures a processor’s floating point unit performance
8. ANION—negatively charged atom
9. LISP—Used to manipulate mathematical logic
11. STATCOULOMB—i.e., franklin (Fr)
12. AMBISONICS—Typically requires a soundfield microphone
15. BROUTER—This device can send data between networks and it can forward data to individual systems in a network
16. TRINITRON—This CRT technology was originally introduced the 1960s
17. OXIDE—The “O” in CMOS
18. DETENT—Used to prevent or stop something from spinning

Down

1. ELECTRICSUSCEPTIBILITY—XE [two words]
3. RADECHON—A barrier-grid storage tube
5. COLPITTS—This oscillator uses two-terminal electrical components to create a specific oscillation frequency
7. BEAGLEBOARD—TI’s open-source SBC
10. PICONET—A network that is created using a wireless Bluetooth connection
13. BETATRON—Designed to accelerate electrons
14. TEBIBYTE—More than 1,000,000,000,000 bytes

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