Issue 284: EQ Answers

PROBLEM 1
Can you name all of the signals in the original 25-pin RS-232 connector?

ANSWER 1
Pins 9, 10, 11, 18, and 25 are unassigned/reserved. The rest are:

Pin Abbreviation Source Description
1 PG - Protective ground
2 TD DTE Transmitted data
3 RD DCE Received data
4 RTS DTE Request to send
5 CTS DCE Clear to send
6 DSR DCE Data Set Ready
7 SG - Signal ground
8 CD DCE Carrier detect
12 SCD DCE Secondary carrier detect
13 SCTS DCE Secondary clear to send
14 STD DTE Secondary transmitted data
15 TC DCE Transmitter clock
16 SRD DCE Secondary received data
17 RC DCE Receiver clock
19 SRTS DTE Secondary request to send
20 DTR DTE Data terminal ready
21 SQ DCE Signal quality
22 RI DCE Ring indicator
23 - DTE Data rate selector
24 ETC DTE External transmitter clock

 

PROBLEM 2
What is the key difference between a Moore state machine and a Mealy state machine?

ANSWER 2
The key difference between Moore and Mealy is that in a Moore state machine, the outputs depend only on the current state, while in a Mealy state machine, the outputs can also be affected directly by the inputs.

 

PROBLEM 3
What are some practical reasons you might choose one state machine over the other?

ANSWER 3
In practice, the difference between Moore and Mealy in most situations is not very important. However, when you’re trying to optimize the design in certain ways, it sometimes is.

Generally speaking, a Mealy machine can have fewer state variables than the corresponding Moore machine, which will save physical resources on a chip. This can be important in low-power designs.

On the other hand, a Moore machine will typically have shorter logic paths between flip-flops (total combinatorial gate delays), which will enable it to run at a higher clock speed than the corresponding Mealy machine.

 

PROBLEM 4
What is the key feature that distinguishes a DSP from any other general-purpose CPU?

ANSWER 4
Usually, the key distinguishing feature of a DSP when compared with a general-purpose CPU is that the DSP can execute certain signal-processing operations with few, if any, CPU cycles wasted on instructions that do not compute results.

One of the most basic operations in many key DSP algorithms is the MAC (multiply-accumulate) operation, which is the fundamental step used in matrix dot and cross products, FIR and IIR filters, and fast Fourier transforms (FFTs). A DSP will typically have a register and/or memory organization and a data path that enables it to do at least 64 MAC operations (and often many more) on unique data pairs in a row without any clocks wasted on loop overhead or data movement. General-purpose CPUs do not generally have enough registers to accomplish this without using additional instructions to move data between registers and memory.

Experimenting with Dielectric Absorption

Dielectric absorption occurs when a capacitor that has been charged for a long time briefly retains a small amount of voltage after a discharge.

“The capacitor will have this small amount of voltage even if an attempt was made to fully discharge it,” according to the website wiseGEEK. “This effect usually lasts a few seconds to a few minutes.”

While it’s certainly best for capacitors to have zero voltage after discharge, they often retain a small amount through dielectric absorption—a phenomenon caused by polarization of the capacitor’s insulating material, according to the website. This voltage (also called soakage) is totally independent of capacity.

At the very least, soakage can impair the function of a circuit. In large capacitor systems, it can be a serious safety hazard.

But soakage has been around a long time, at least since the invention of the first simple capacitor, the Leyden jar, in 1775. So columnist Robert Lacoste decided to have some “fun” with it in Circuit Cellar’s February issue, where he writes about several of his experiments in detecting and measuring dielectric absorption.

Curious? Then consider following his instructions for a basic experiment:

Go down to your cellar, or your electronic playing area, and find the following: one large electrolytic capacitor (e.g., 2,200 µF or anything close, the less expensive the better), one low-value discharge resistor (100 Ω or so), one DC power supply (around 10 V, but this is not critical), one basic oscilloscope, two switches, and a couple of wires. If you don’t have an oscilloscope on hand, don’t panic, you could also use a hand-held digital multimeter with a pencil and paper, since the phenomenon I am showing is quite slow. The only requirement is that your multimeter must have a high-input impedance (1 MΩ would be minimum, 10 MΩ is better).

Figure 1: The setup for experimenting with dielectric absorption doesn’t require more than a capacitor, a resistor, some wires and switches, and a voltage measuring instrument.

Figure 1: The setup for experimenting with dielectric absorption doesn’t require more than a capacitor, a resistor, some wires and switches, and a voltage measuring instrument.

Figure 1 shows the setup. Connect the oscilloscope (or multimeter) to the capacitor. Connect the power supply to the capacitor through the first switch (S1) and then connect the discharge resistor to the capacitor through the second switch (S2). Both switches should be initially open. Photo 1 shows you my simple test configuration.

Now turn on S1. The voltage across the capacitor quickly reaches the power supply voltage. There is nothing fancy here. Start the oscilloscope’s voltage recording using a slow time base of 10 s or so. If you are using a multimeter, use a pen and paper to note the measured voltage. Then, after 10 s, disconnect the power supply by opening S1. The voltage across the capacitor should stay roughly constant as the capacitor is loaded and the losses are reasonably low.

Photo 1: My test bench includes an Agilent Technologies DSO-X-3024A oscilloscope, which is oversized for such an experiment.

Photo 1: My test bench includes an Agilent Technologies DSO-X-3024A oscilloscope, which is oversized for such an experiment.

Now switch on S2 long enough to fully discharge the capacitor through the 100-Ω resistor. As a result of the discharge, the voltage across the capacitor’s terminals will quickly become very low. The required duration for a full discharge is a function of the capacitor and resistor values, but with the proposed values of 2,200 µF and 100 Ω, the calculation shows that it will be lower than 1 mV after 2 s. If you leave S2 closed for 10 s, you will ensure the capacitor is fully discharged, right?

Now the fun part. After those 10 s, switch off S2, open your eyes, and wait. The capacitor is now open circuited, at least if the voltmeter or oscilloscope input current can be neglected, so the capacitor voltage should stay close to zero. But you will soon discover that this voltage slowly increases over time with an exponential shape.

Photo 2 shows the plot I got using my Agilent Technologies DSO-X 3024A digital oscilloscope. With the capacitor I used, the voltage went up to about 120 mV in 2 min, as if the capacitor was reloaded through another voltage source. What is going on here? There aren’t any aliens involved. You have just discovered a phenomenon called dielectric absorption!

Photo 2: I used a 2,200-µF capacitor, a 100-Ω discharge resistor, and a 10-s discharge duration to obtain this oscilloscope plot. After 2 min the voltage reached 119 mV due to the dielectric absorption effect.

Photo 2: I used a 2,200-µF capacitor, a 100-Ω discharge resistor, and a 10-s discharge duration to obtain this oscilloscope plot. After 2 min the voltage reached 119 mV due to the dielectric absorption effect.

Nothing in Lacoste’s column about experimenting with dielectric absorption is shocking (and that’s a good thing when you’re dealing with “hidden” voltage). But the column is certainly informative.

To learn more about dielectric absorption, what causes it, how to detect it, and its potential effects on electrical systems, check out Lacoste’s column in the February issue. The issue is now available for download by members or single-issue purchase.

Lacoste highly recommends another resource for readers interested in the topic.

“Bob Pease’s Electronic Design article ‘What’s All This Soakage Stuff Anyhow?’ provides a complete analysis of this phenomenon,” Lacoste says. “In particular, Pease reminds us that the model for a capacitor with dielectric absorption effect is a big capacitor in parallel with several small capacitors in series with various large resistors.”

Build an Inexpensive Wireless Water Alarm

The best DIY electrical engineering projects are effective, simple, and inexpensive. Devlin Gualtieri’s design of a wireless water alarm, which he describes in Circuit Cellar’s February issue, meets all those requirements.

Like most homeowners, Gualtieri has discovered water leaks in his northern New Jersey home after the damage has already started.

“In all cases, an early warning about water on the floor would have prevented a lot of the resulting damage,” he says.

You can certainly buy water alarm systems that will alert you to everything from a leak in a well-water storage tank to moisture from a cracked boiler. But they typically work with proprietary and expensive home-alarm systems that also charge a monthly “monitoring” fee.

“As an advocate of free and open-source software, it’s not surprising that I object to such schemes,” Gualtieri says.

In February’s Circuit Cellar magazine, now available for membership download or single-issue purchase, Gualtieri describes his battery-operated water alarm. The system, which includes a number of wireless units that signal a single receiver, includes a wireless receiver, audible alarm, and battery monitor to indicate low power.

Photo 1: An interdigital water detection sensor is shown. Alternate rows are lengths of AWG 22 copper wire, which is either bare or has its insulation removed. The sensor is shown mounted to the bottom of the box containing the water alarm circuitry. I attached it with double-stick foam tape, but silicone adhesive should also work.

Photo 1: An interdigital water detection sensor is shown. Alternate rows are lengths of AWG 22 copper wire, which is either bare or has its insulation removed. The sensor is shown mounted to the bottom of the box containing the water alarm circuitry. I attached it with double-stick foam tape, but silicone adhesive should also work.

Because water conducts electricity, Gualtieri sensors are DIY interdigital electrodes that can lie flat on a surface to detect the first presence of water. And their design couldn’t be easier.

“You can simply wind two parallel coils of 22 AWG wire on a perforated board about 2″ by 4″, he says. (See Photo 1.)

He also shares a number of design “tricks,” including one he used to make his low-battery alert work:

“A battery monitor is an important feature of any battery-powered alarm circuit. The Microchip Technology PIC12F675 microcontroller I used in my alarm circuit has 10-bit ADCs that can be optionally assigned to the I/O pins. However, the problem is that the reference voltage for this conversion comes from the battery itself. As the battery drains from 100% downward, so does the voltage reference, so no voltage change would be registered.

Figure 1: This is the portion of the water alarm circuit used for the battery monitor. The series diodes offer a 1.33-V total  drop, which offers a reference voltage so the ADC can see changes in the battery voltage.

Figure 1: This is the portion of the water alarm circuit used for the battery monitor. The series diodes offer a 1.33-V total drop, which offers a reference voltage so the ADC can see changes in the battery voltage.

“I used a simple mathematical trick to enable battery monitoring. Figure 1 shows a portion of the schematic diagram. As you can see, the analog input pin connects to an output pin, which is at the battery voltage when it’s high through a series connection of four small signal diodes (1N4148). The 1-MΩ resistor in series with the diodes limits their current to a few microamps when the output pin is energized. At such low current, the voltage drop across each diode is about 0.35 V. An actual measurement showed the total voltage drop across the four diodes to be 1.33 V.

“This voltage actually presents a new reference value for my analog conversion. The analog conversion now provides the following digital values:

EQ1Table 1 shows the digital values as a function of battery voltage. The nominal voltage of three alkaline cells is 4.75 V. The nominal voltage of three lithium cells is 5.4 V. The PIC12F675 functions from approximately 2 to 6.5 V, but the wireless transmitter needs as much voltage as possible to generate a reliable signal. I arbitrarily coded the battery alarm at 685, or a little above 4 V. That way, there’s still enough power to energize the wireless transmitter at a useful power level.”

Table 1
Battery Voltage ADC Value
5 751
4.75 737
4.5 721
4.24 704
4 683
3.75 661

 

Gaultieri’s wireless transmitter, utilizing lower-frequency bands, is also straightforward.

Photo 2 shows one of the transmitter modules I used in my system,” he says. “The round device is a surface acoustic wave (SAW) resonator. It just takes a few components to transform this into a low-power transmitter operable over a wide supply voltage range, up to 12 V. The companion receiver module is also shown. My alarm has a 916.5-MHz operating frequency, but 433 MHz is a more popular alarm frequency with many similar modules.”

These transmitter and receiver modules are used in the water alarm. The modules operate at 916.5 MHz, but 433 MHz is a more common alarm frequency with similar modules. The scale is inches.

Photo 2: These transmitter and receiver modules are used in the water alarm. The modules operate at 916.5 MHz, but 433 MHz is a more common alarm frequency with similar modules. The scale is inches.

Gualtieri goes on to describe the alarm circuitry (see Photo 3) and receiver circuit (see Photo 4.)

For more details on this easy and affordable early-warning water alarm, check out the February issue.

Photo 3: This is the water alarm’s interior. The transmitter module with its antenna can be seen in the upper right. The battery holder was harvested from a $1 LED flashlight. The box is 2.25“ × 3.5“, excluding the tabs.

Photo 3: This is the water alarm’s interior. The transmitter module with its antenna can be seen in the upper right. The battery holder was harvested from a $1 LED flashlight. The box is 2.25“ × 3.5“, excluding the tabs.

Photo 4: Here is my receiver circuit. One connector was used to monitor the signal strength voltage during development. The other connector feeds an input on a home alarm system. The short antenna reveals its 916.5-MHz operating frequency. Modules with a 433-MHz frequency will have a longer antenna.

Photo 4: Here is my receiver circuit. One connector was used to monitor the signal strength voltage during development. The other connector feeds an input on a home alarm system. The short antenna reveals its 916.5-MHz operating frequency. Modules with a 433-MHz frequency will have a longer antenna.

 

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)

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