May 2000 QST

By H. Ward Silver, N0AXTest Your Knowledge!No pain, no gain. Still more dreaded word problems…As Mark Twain once said, “Supposing is good, but finding outis better.” Let’s find out, shall we?1. An output driver package with eight 12-V outputs has the capabilityto handle a total of 150 mA. It’s connected to the followingloads:- a relay coil that consumes 600 mW- four LEDs that require 10 mA apiece- a display backlight that draws 200 mWCan you add an audio annunciator that needs 30 mA of drive withoutoverloading the chip?2. During emergency service a handheld rig could be used at a10% duty cycle of transmitting vs. total on-time. The rig draws 50mA while receiving and 1 A while transmitting. Each 8-V NiCdpack has a capacity of 2.5 A/h. How long will each pack last if 80%of its capacity can be used? How many fully-charged battery packsare needed for 24 hours of operation?3. A base-fed 160-meter inverted-L antenna has an impedance of 35+ j40 Ω. Find the value of a series capacitor attached to the feed pointthat will cancel the inductive component of the feed point impedanceat 1.830 MHz. Then, using an L-network with an inductor inseries with the feedpoint, transform the remaining 35 Ω resistiveimpedance to 50 Ω and calculate the values of L and C. (See theARRL Antenna Book for L-network design equations.)4. A transmitter requires 10.5 V or greater at its input terminals.At full load it draws 25 A. The power supply has an internalimpedance of 0.02 Ω and the cable and connectors have a totalresistance of 0.05 Ω. Will the rig have sufficient input voltagewhen the power supply output is set to 13.5 V? How much poweris dissipated by the cable and connectors? Can an additional splicingconnector with a resistance of 0.06 Ω be added to the cable andstill allow sufficient voltage at the rig under full load?5. Seeking to improve weak signal performance, you’re weighing thechoice of replacing your existing coax with hardline, or adding apreamp at the antenna. The coax has a total length of 275 feet and aloss of 3 dB per 100 feet. A preamp has a gain of 6 dB. Assuming thatthe hardline has a loss of 0.5 dB per 100 feet, which choice —hardlineor preamp—will result in better signal strength at the receiver input?Total Your Score!There are a total of nine possible answers. Give yourself onepoint for each correct answer.7-9 Hardly a tickle4-6 That didn’t hurt too much, did it?1-3 Ouch!of -40 + 23 = -17 Ω of reactance. This is equivalent to a single capacitorof 5120 pF. Thus, the matching network reduces to a 5120 pF capacitorin series with the antenna feedpoint and a 1133 pF capacitor connectedacross the feed line output.4. The voltage at the rig under full load is 13.5 V – 25 A × (0.02 + 0.05 Ω)= 11.8 V; enough for proper operation. The cables and connector dissipate25 2 × 0.05 = 31.3 W. Note that a loose connector can get mightyhot under loads like this! If an additional 0.06 Ω is added to the 0.07 Ωseries resistance, full load voltage at the rig drops to 13.5 – 25 (0.13) =10.3V—not enough! Note also that this single connector would dissipate37.5 W all by itself.5. The existing cable is causing 275 × 3/100 = 8.25 dB loss. Hardline of thesame length will result in 275 × 0.5/100 = 1.38 dB. Cable replacementresults in 8.25 – 1.38 = 6.88 dB of improvement, slightly better than thepreamp. Not considered in this problem, but still pertinent, is the factthat adding a preamp at the antenna would probably reduce the systemnoise figure and improvement performance more than a mere signalgain computation would indicate.Answers1. First, compute all current draws. The relay coil draws 600 mW / 12 V =50 mA. The LEDs consume a total of 40 mA. The backlight draws 200mW / 12 V = 16.7 mA. The total current already being handled is 50 +40 + 16.7 = 106.7 mA. An additional 30 mA brings the total to 136.7 mA,which is within the rated capacity of the driver.2. If 80% of a pack’s energy (expressed in amp-hours (Ah), which assumesconstant voltage) is available for use, that’s 2 Ah. The usageprofile given is an average current drain consisting of 10% at 1 A plus90% at 0.05 A = 145 mA. The total time this load can be supported is 2/ 0.145 = 13.8 hours. For 24 hours of operation you’ll need 2 packs.3. At 1.830 MHz, a capacitive reactance of 40 Ω requires 1/(6.28 ×1.830MHz × 40) = 2180 pF. This leaves an impedance of 35 Ω to be transformedto the desired impedance of 50 Ω. From the L-network designequations, Q is calculated to 0.66; XL = 0.66 × 35 = 23 Ω; X C = 50 / 0.66= 75 Ω; L = 23 / (1.83 MHz × 6.28) = 2 µH; C = 1 / (1.83 MHz × 6.28 ×75) = 1133 pF. Note that the feedpoint capacitor and matching networkinductor are in series, partially cancelling and leaving a net reactanceSTRAYSARRL FORUM ON-LINE◊ An on-line member-to-member ARRL e-mail discussion forum (notsponsored by the ARRL) has been established. You can elect to havethe daily messages summarized and sent to you in digest form once aday, or have all messages sent directly to your e-mail address as theyarrive. Please note that this is not a rant-and-rave anti-League forum.Uncivil discourse will be terminated and the offenders dropped. Constructivecomments and civil arguments are, of course, welcome. Youcan sign up for the forum on the Web at: http://www.onelist.com.—Jay Craswell, W0VNELOOKING FOR AN EXAM?◊ If you want to take an Amateur Radio exam, but don’t know whereto go, you can find out on the ARRLWeb at http://www.arrl.org/arrlvec/examsearch.phtml. At this handy site you can search forexam dates and times by state, ZIP code and even country (for USAmateur Radio exams offered overseas).Next StrayMay 2000 51