# Issue 264: EQ Answers

The answers to the Circuit Cellar 264 (July) Engineering Quotient are now available. The problems and answers are listed below, along with a schematic.

Problem 1a—Is it possible to transmit on-off (DC) signals between two pieces of equipment in both directions simultaneously on the same wire, in much the same way that telephones do for audio?

Source: D. Tweed, CC264

Answer 1a—Why not? Hybrids work just as well at DC as they do for audio; you just need a receiver with balanced inputs, like an RS-422 buffer:

All resistors are the same value (e.g., 4,700 Ω) and the transmit driver needs to be a voltage source (low impedance).

If the transmitter switches between, say, 0 V and 5 V, the opposite receiver will see a voltage differential of 0 V and 2.5 V, respectively, while the local receiver will just see 0V.

For long lines, you’ll probably want to use lower resistances and you’ll want to limit the slew rate of the transmitter so that the receiver doesn’t produce glitches on the transitions of the local transmitter.

If the RS-422 receiver is replaced with an op-amp differential amplifier with a gain of 2, then any analog voltage transmitted by one end will be reproduced at the other end.

Problem 1b—But doesn’t a true hybrid use transformers, or at least some tricky transformer simulation with op amps to ensure the transmitted signal does not appear on the receive port?

Answer 1b—No. A hybrid is just a bridge circuit, with one arm of the bridge replaced by the line and the termination at the far end. The transmit signal is applied to two opposite corners of the bridge and the receive signal is taken from the other two corners.

In order to provide the Tx/Rx isolation, the bridge must be balanced, which in the example above, means that the lower resistor on each side must match the impedance of the line/far end combination. For DC and short lines, a simple resistor suffices. At audio frequencies and with the long unshielded twisted pairs used in telephony, a more complex matching impedance is required.

Transformers are used only because it’s the easiest way (and the only passive way) to get a balanced drive and/or receive signal — the transmit driver and receiver cannot share a ground. In order to mass produce phones that were dirt cheap, yet simple and reliable, the phone company figured out how to use a multi-winding transformer to provide the both the isolation and the balanced/unbalanced conversion in both directions, usually with a single resistor and capacitor to provide the line matching. As noted, modern electronic phones use active electronics to achieve the same things.

As always, the theory is simple, but the practical implementations can get complicated.

Problem 2a—The conventional way to calculate the magnitude (length) of a vector is to take the square root of the sum of the squares of its components. On small processors, this can be somewhat difficult (especially the square root operation), and various approximations are used instead.

One approximation that works surprisingly well for 2-D vectors and complex numbers is to take the absolute values of the two components, compare them, then add 1/3 of the smaller to the larger.

What is the maximum error using this method?

Answer 2a—If we restrict the discussion to unit vectors at various angles A, the x component is cos(A) and the y component is sin(A), and the correct magnitude is 1.

Furthermore, let’s concentrate on angles between 0 and 45° — then we know that both cos(A) and sin(A) are positive and that cos(A) > sin(A). (The absolute value and compare operations provide the symmetry that covers the rest of the unit circle.) The approximation then gives the result

Magnitude = cos(A) + sin(A)/3

Graphing this shows that this is most negative (0.943) at 45° and most positive (1.054) at approximately 18.4° (the actual angle is given by atan(1/3) —can you show why?). The peak error is therefore –5.7%, +5.4%.

Problem 2b—Is there a similar formula that gives even better results?

Answer 2b—Yes. One more multiplication operation gives a result that has less than 4% error:

Magnitude = 0.960433 × max(|x|, |y|) + 0.397826 × min(|x|, |y|)

This function is most negative at 0° and 45°, and most positive at 22.5°. The error is ± 3.96%. This form is well-suited to DSPs that have multiply-accumulate units. The two constants can be expressed as 62943/65536 and 26072/65536, respectively.

Contributor: David Tweed

# Issue 262: EQ Answers

Problem 1—The classic two-transistor astable multivibrator is shown below. Typically, R2 and R3 have at least 10 times the value of R1 and R4. This circuit oscillates, with Q1 and Q2 turning on alternately. From the point in time in a cycle where Q1 first switches on, describe what happens until Q2 switches on.

Source: D. Tweed, CC262

Answer 1—Right before the moment Q1 switches on, C1 is charged to VCC – VBE, with its left end positive, and the left end of C2 has just reached +VBE. The right end of C2 is being held at VCE(SAT) by Q2.

Source: D. Tweed, CC262

So, as Q1 begins to switch on, it pulls the left end of C1 low, and this also pulls the right end of C1 low, cutting off Q2. This in turn allows the right end of C2 to rise, emphasizing the turn-on of Q1 by increasing the voltage (and current) at the base of Q1.

Once Q1 is fully on, the right end of C1 is now at VCE(SAT)– (VCC – VBE) (a fairly substantial negative voltage), and C1 now begins to charge in the other direction, through R2. Once the right end of C1 reaches +VBE, Q2 begins to turn on, starting the second half of the cycle.

Problem 2—What determines the time of one half-cycle of the oscillation? Does this depend on VCC?

Answer 2—The time of the half-cycle described previously is the time that it takes the right end of C1 to charge from –(VCC – (VBE + VCE(SAT))) to +VBE.

Now, keep in mind that the capacitor is charging “toward” +VCC, but it gets halted by the B-E junction of Q2 at +VBE. This charging is occuring at a rate determined by the time constant C1 × R2, and we’re basically interested in the time that it takes to move halfway from its starting value to its final value. This works out to –ln(0.5), or 0.693 times the R-C time constant.

As long as VCC >> VBE, the time does not depend on VCC. That isn’t to say, however, that VCC can be arbitrarily large. If it exceeds the reverse-breakdown voltage of the transistors’ B-E junctions, current will flow and perturb the timing.

Problem 3—Recently, a different circuit appeared on the web, shown below. Again, R2 and R3 are significantly larger than R1 and R4. The initial reaction of one observer was that this circuit can’t work, because there’s no DC bias path for either transistor. Is this assessment correct?

Source: D. Tweed, CC262

Answer 3—No, it isn’t. This circuit can oscillate just fine. Again, look at how C1 charges and discharges.

Source: D. Tweed, CC262

If C1 starts out discharged, it will charge through R1 and the B-E junction of Q2. This current will turn on Q2, holding its collector at ground (really VCE(SAT)) and preventing Q1 from turning on.

However, as C1 reaches full charge, the current through it decays below a level that will keep Q2 turned on. When it starts to turn off, its collector voltage rises, which also forces current into Q1’s base through C2. As Q1 begins to turn on, it pulls its collector low, which also pulls the base of Q2 lower, emphasizing its turn-off. The circuit quickly “snaps” to the other state, with Q1 on and Q2 off. C1 is discharged through Q1 and D2 at the same time that C2 begins charging through R4 and Q1’s B-E junction.

Problem 4—What role do R2 and R3 play in this circuit?

Answer 4—R2 and R3 never have more than ±VBE across them; as a result, the current through them is negligible relative to the current through the capacitors. In other words, they’re superfluous.

Question 5—Does the timing of this circuit depend on VCC? If not, what does it depend on?

Answer 5—The time from when one of the transistors turns on to when it turns off is determined by the currents flowing into its base and collector. When the current into the base drops below the value needed to sustain the current into the collector, the transistor begins to turn off, and the circuit feedback then insures that this happens quickly.

Looking at Q2, and ignoring the transient associated with discharging C2 for now, the collector current is set by R4. The initial base current is set by R1, but this decays exponentially with a time constant of R1 × C1.

Therefore, the primary determinant of the half-cycle time period (in addition to the R-C time constant) is the current transfer ratio, or hFE of each transistor. When the base current drops to a value of 1/hFE of the collector current, the transistor begins to turn off.

Since both currents scale in the same way with VCC, it has no direct effect on the timing. There is only a secondary effect if the value of hFE changes with the value of the collector current.

Contributor:  David Tweed

# Issue 260: EQ Answers

These are the answers to the EQ questions that appeared in Circuit Cellar 260 (March 2012).

Problem 1—In an RS-232 interface, why is the idle or “mark” level a negative voltage?

Answer 1—RS-232 was developed in the days when people were connecting electromechanical teletypes to telephone lines with modems, and in fact, at that time the design of any equipment connected to a phone line was tightly controlled by the phone company.

The reason RS-232 uses a negative voltage for its idle state is the same reason the phone lines themselves use a negative voltage relative to ground for power—copper wires in long cables potentially exposed to moisture are significantly less likely to corrode if they have a negative DC bias on them.

Problem 2—Similarly, why does the “mark” level correspond to a logical “high” level on the TTL side of the interface?

Answer 2—Again, back in the days when RS-232 was developed, the primary logic families in use were DTL and TTL. Both of these technologies draw significantly less power when a signal is in the high state than in the low state, so the high state is preferred for the inactive state of any signal.

Problem 3—What does the following C function compute? You may assume the input argument is a positive integer.

Answer 3—Remember the algorithm that computes integer square roots by subtracting successive odd numbers from the input value? This function extends that concept to computing integer cube roots.

The reason this works is that taking the differences between successive values is the discrete equivalent of taking a derivative in the continuous world. The derivative of a cubic curve is a quadratic, and the derivative of a quadratic is a straight line. To generate a “straight line” in the discrete world, you just add a constant to a variable.

When you look at successive squares—0, 1, 4, 9, 16, 25, etc.—the differences are 1, 3, 5, 7, 9, etc. This is why the algorithm that subtracts odd numbers works for computing square roots.

When you look at the successive cubes—0, 1, 8, 27, 64, 125, etc.—the first set of differences is the sequence 1, 7, 19, 37, 61, etc. This doesn’t look very useful until you take the differences between those numbers, which are: 6, 12, 18, 24, etc., which is obviously another straight line.

Problem 4—Suppose you are given some calibration constants for a sensor in the form of four-digit hexadecimal (16-bit) integers, and you are told that the format of these numbers is “7 integer bits and 9 fractional bits.” How would you go about converting these constants to floating-point so that you could, for example, work with them in a spreadsheet?

Answer 4—The direct way to convert numbers in an arbitrary fixed-point representation to the equivalent floating-point value is to figure out what the representation for “1.000” would be in the fixed-point notation and then divide the given numbers by that constant.

In this case, with 7 integer bits and 9 fraction bits, “1.000” would be represented as binary 0000001.000000000, or 0x0200. So, if you are given a constant of, say, 0x5453, just divide it by 0x0200 to find out that it represents the value “42.162.”

Contributed by David Tweed (eq at circuitcellar.com)

# Issue 258: EQ Answers

These are the answers to the EQ questions that appeared in Circuit Cellar 258(January 2012).

Answer 1—Q1 and Q2 are wired as a differential amplifier, but in this digital application, it’s probably better to think of them as a current switch. Whenever VIN < VTH, the current through R1 is shunted to ground, and whenever VIN > VTH, the current passes instead through Q2 to the base of Q3, turning it on. As long as Q1 and Q2 are reasonably well matched, VTH is the input voltage at which the switchover occurs, and this is relatively stable with respect to temperature. For example, if VIN is being driven by standard TTL logic, you might set VTH to 1.5 V.

Answer 2—R1, combined with VTH, sets the amount of current flowing through Q2 to drive the base of Q3. This current should be sufficient to drive Q3 into saturation, given the expected load on it (R3 plus the external circuit). R2 serves to make sure that Q3 isn’t turned on by any leakage current through Q2 when it’s supposed to be off. For example, a value of 100 kΩ would bypass currents of up to 6 µA or so.

Answer 3—When VIN is high, there is virtually no current flowing through the input terminal—just the leakage current through Q1’s base-collector junction. When VIN is low, the driving circuit must sink the current set by R1 divided by the beta (current gain) of Q1. For example, if R1 is 4,300 Ω, giving a current through Q1 of about 1 mA, and the beta of Q1 is 50, then the driving circuit must sink about 20 µA.

Answer 4—Better: The component count is reduced by one transistor. Better: VTH is stabilized by the forward drop of a diode, making it less dependent on the exact value of the positive supply. Worse: The switching voltage is now determined by the combination of D1 and the base-emitter drop of Q2, both of which vary with temperature. Worse: Now the driving circuit must supply all of the base current for Q3 when it is in the high state.

Contributed by David Tweed (eq at circuitcellar.com)