& Y Connected resistors, Light emitting diode.


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1 & Y Connected resistors, Light emitting diode. Experiment # 02 Ojectives: To get some hndson experience with the physicl instruments. To investigte the equivlent resistors, nd Y connected resistors, nd KVL/ KCL lws. To lern out light emitting diodes (LED). To lern grphicl circuit nlysis. 1 Instruments nd Circuit Elements Used in This L 1.1 Resistors Resistors re devices tht not only conduct electricity ut lso dissipte electric energy s het. Therefore y dding resistnce, supply voltge my e reduced, or current e limited. There re two generl types of resistors: composition resistors nd wirewound resistors. You my get more informtion out vrious types of resistors in Appendix section of this l note. In order to find out the color code of the resistors you my consult with the Appendix. 1.2 Light Emitting Diode LED Chrge crrier recomintion occurs t pnjunction s electrons cross from the nside nd recomine with holes on the pside. When recomintion tkes plce, the chrge crriers give up energy in the form of het nd light. If the semiconductor mteril is trnslucent the light is emitted, nd the junction is light source, tht is, light emitting diode (LED). 1.3 Bredord ox A redord is used for mounting nd interconnecting components of circuit. Figure 1 shows the electricl connection scheme of the redord ox. The most importnt prts of redord ox re the superstrips, which re the white plstic strips with lots of holes for inserting ends of connecting wire. The nrrow superstrips, which hve two verticl sets of holes, re connected so tht ll holes lying on the sme verticl line re connected to ech other, i.e., horizontlly djcent holes re not connected to ech other. The two wide superstrips, which hve five horizontl sets of holes, re connected such tht ll holes lying on the sme horizontl line re connected to ech other, i.e., verticlly djcent holes re not connected to ech other. Some redords lso hve the following properties: A power switch, which is locted on the lefttop corner of the redord (see figure 1) Three knos for vrile 0 to ± 15 V nd fixed 5 Volt power supplies. The ground (GND) post on the top of redord is the ground of the power supplies nd lso internlly connected to the redord ox cse when the power is on, so tht the cse itself cn ct s ground. Directly elow ech power supply kno is hole tht wire ends cn e inserted into it nd ring the power supply to the terminl strips. It is recommended to wire end verticl strip to the GND post in order to provide sfe nd common ground for your circuit. At the ottom of the redord ox there re three coxil connectors. Coxil cles from signl genertors nd oscilloscopes re connected to these connectors. The coxil cle hs centrl inner signl wire nd n outer concentric shielding rided cylindricl wire. The outside shell of ech coxil connected is screwed to the cse of the redord ox nd therefore is t ground potentil.
2 ON OFF GND BNC terminls t the ottom of the redord Figure 1. Smple of Bredord ox. (The one we hve in the l is slightly different). 1.4 Digitl Multi meter (DM) Resistnce, voltge nd current mesurements cn e mde with this instrument. Connections to the instrument re mde through three leled inding posts. The instrument hs numer of uttons some of which re used to set the instrument function, i.e., to mke it red dc volts, nd others re for rnge control, i.e., to set it to red voltges tht under 2 volts. 1.5 Dul Power Supply (DPS) It is supply of vrile voltge tht hs two independent outputs. Both current nd voltge could e set. The two outputs could e connected either in series or in prllel. A fixed positive 5 volts is lso ville. The instrument hs n onoff switch, nd voltge control knos for ech power supply.
3 2 Bckground 2.1 ResistorsSets Equivlence Hving the vi reltionship of resistor s v(t) = i(t).r, we cn show tht the resistor equivlent to set of resistors connected in series, could e otined y simply summing the quntities of ech resistor in the set (Figure 2). For the set of resistors in prllel the replic of the equivlent resistor would e equl to sum of the repliction of ech resistor in the set (Figure 3). R 1 R 2 = R = R 1 R 2 R 3 R 3 Figure 2. The equivlent of the resistors connected in series. = R 1 R 2 R 3 1/R = (1/R 1 ) (1/R 2 ) (1/R 3 ) Figure 3. The equivlent of the resistors connected in prllel. A set of resistors could e replced y its equivlent without ffecting the KCL nd KVL equtions pplied t, nd etween the terminls where the replcement is done, respectively. For connected resistors, there is n equivlent Yconnected resistors set, y which the nlysis of the circuits contining connected resistors could e simplified, drmticlly. The resistors in the Yconnected set re relted to the comintion of resistors in the connected resistors set s follows: R R 2 R 1 c R 3 R R c c Figure 4. The nd Yconnected resistors sets. The ove two nd Yconnected resistors re equivlent when:
4 R1 R2 R2R3 R1R 3 R =, R =, Rc = = R1 R2 R3 2.2 LIGHT EMITTING DIODE CHARACTERISTICS (A Grphicl Anlysis Approch) In order to solve circuit in terms of its nodevoltges nd rnchcurrents, we hve to consider two kinds of equtions pplying to the circuit simultneously. One set of equtions is extrcted from the circuit topology, using KVL nd KCL lws. Theses equtions, therefore, reflects the specifictions of the circuit in terms of the wy in which the vrious components of the circuit re connected to ech other. The other group of equtions origintes from the intrinsic electricl ehvior of ech component, showing the physicl specifiction of the components, irrespective the wy tht they re connected to ech other. In the circuit nlysis procedure, there re mny cses in which solving the simultneous equtions is very difficult unless we use sophisticted computer progrms. There is still reltively simple grphicl pproch to solve the simultneous equtions tht works properly in mny cses. In Figure 5 the LED is connected in series with lod resistor R cross the power supply. Since the LED is nonliner device it is not esy to nlyticlly determine the voltges nd current in this single loop circuit. However, grphicl method using wht is clled lod line cn e pplied to give the LED operting voltge nd current (clled the operting point) for prticulr vlues of power supply voltge nd lod resistnce (Figure 6). 100 Ω V s Figure 5. A simple circuit with Light Emitting Diode. The lod line eqution is otined y pplying KVL to the single loop circuit (Figure 5), giving: vd=vs Rid This liner eqution cn e plotted on to the grph contining the LED vicurve. Note tht V s is the voltge xis intercept vlue nd the line hs the slope R, the intersection of the lod line with the vicurve yields solution for the LED operting point. Figure 6 illustrtes the method nd shows the LED operting voltge (v do ) nd operting current (i do ) for prticulr vlues of V s nd R. If either V s or R is chnged, the operting point will chnge too.
5 V s LED curve V d i d0 Slope R Lod line Figure 6. Lod line solution for the LED operting point. i d
6 PreL Hndout 1 Show tht the reltion of R = R 1 R 2 / (R1R 2 R 3 ) is vlid for the resistor network in Figure 4. 2 Otin the reltion tht is vlid etween R 1 nd the resistors of the Yconnected set. Hint: To otin the requested reltions, notice tht two resistor sets re equivlent if nd only if they show sme circuit ehvior, i.e., we will hve the sme result if we pply test voltge source on the sme terminl on resistor sets s shown elow. I T I T V T R 2 R 1 c V T R R 3 R R c c Figure 7. Circuit digrms used to clculte the resistors of ech set in terms of the resistors of the other set. Write the input resistnce seen y the test supply in ech circuit nd mke them equl to ech other, i.e. for the circuits shown in Figure 7 we hve: R R = R 2 (R 1 R 3 ) Apply the test supply to other terminls of oth circuits correspondingly nd otin the equtions similr to the ove one. Solving three simultneous equtions, otined in the wy explined, the resistors of ech set could e clculted in terms of the resistors of the other set. 3 ) Clculte the resistnce oserved in the terminl xx of the circuit shown in Figure 8 elow given tht R = 4.7 KΩ. Use the fct tht connection of three equl resistors R is equivlent to Yconnection of three equl resistors R Y when 3R Y =R. x c R/3 3R R Figure 8. Resistor network x R d R
7 R xx theoreticl = Ω ) Clculte the equivlent resistnce seen cross c terminls. R c = Ω The following section might e done using circuit nlysis or circuit simultion with workench. 4 Considering the circuit shown in Figure 8, ssume tht we pply dc power supply cross terminls xx hving 10 volts mgnitude. Clculte the voltges cross terminls. Hving known ll voltges of the circuit nodes, clculte the current pssing through ech resistor. V d (Theoreticl) = V d (Theoreticl) = I 3R (Theoreticl) = I R/3 (Theoreticl) = I RLeft (Theoreticl) = I RRight (Theoreticl) = I RBridge (Theoreticl) =
8 Experiment#2 Prt 1 1 Set up network shown in Figure 8 of prel. 2 Use the DM (set to the kω function nd 20kΩ rnge) to mesure the equivlent resistnce (R xx mesured ) t terminls xx of the ridgetype resistive network, shown in Figure 8. Record the mesured resistnce vlue nd report it in your l hndout. R xx mesured = Ω 3 Compre the mesured vlue of the resistor R xx with wht you clculted in the prel exercise. Justify the difference etween the two vlues, theoreticl nd mesured, if there exist more thn %5 error. Write the nswer in your l hndout. 4 Using the DM mesure the resistnce existing cross c terminls, record the mesured vlue elow nd report it in your l hndout. Why isn t it greter thn R/3? R mesured cross c terminls = Ω 5 Apply 10 volt power supply cross xx terminls of the circuit shown in Figure 8. Mesure the voltges cross the resistor Rridge (v ), with respect to the ground. Mesure the exct mount of the resistors shown in Figure 8. Clculte the current, flowing in ech rnch of the ridge circuit. Report the mesured vlues in your l hndout nd compre the results with wht you hve otined in the prt 4 of the prel hndout. Discuss ny difference in the otined results. V d (Mesured) = V d (Mesured) = I 3R (Mesured) = I R/3 (Mesured) = I RLeft (Mesured) = I RRightt (Mesured) = I RBridge (Mesured) =
9 Prt 2 1 Set up the circuit shown elow. The flt side of the light emitting diode (LED) identifies the node (A) led of the diode while the other led is the cthode (C). Before energizing the circuit hve the lortory instructor check your connection scheme. Note from the circuit digrm tht the DM (set to dc volts nd the 20V rnge) mesures the DPS voltge V s, one DT (rotry dil set to dc volts) mesures the LED voltge v d while the other DT (rotry dil set to dc 200mA) mesures the LED current i d. DPS DM V DT ma V s R 100ohms v d A C i d DT V 2 Energize the circuit y turning the DPS on with the power utton. Mesure the LED voltge s the diode current is set to ech of the vlues listed in Tle 1.1. The diode current is set y djusting the voltge nd current setting knos, (the control kno should initilly e set fully counterclockwise for zero volts), on the DPS until the desired current is otined. Enter the mesured voltges into Tle 1.1. TABLE 1.1 i d (ma) v d (V) 0.0 Before energizing the circuit let the TA check your connection scheme. 3 Plot the points in tle 1.1 onto grph to otin the LED voltgecurrent (vi) chrcteristic curve. Note the LED vicurve is highly nonliner for low voltge vlues. The diode chrcteristic α v d curve hs the exponentil form i = I e. Do fit to you dt to find oth α nd I. 0 d 0
10 4 Use the lod line method to predict the LED operting point vlues when V s =3.0V nd R=100Ω. Plot the lod line onto the grph contining the LED vicurve. Any two suitle vlues of i d cn e used in the lod line eqution, v d =V s Ri d, to clculte two v d vlues which when plotted yield the strightline lod line. (Note: In the eqution i d must e in the unit mperes nd not millimperes!). Report the results in your l hndout. 5 Now mesure the LED operting point for the conditions given in 4, i.e., in the experimentl setup djust the DPS so tht V s =3.0V with R=100Ω. Red the pproprite instruments to get v do nd i do. Report the otined vlues in your l report. 6 Insted of the DPS plug function genertor (V pp = 10 V, f = 1 khz), use the oscilloscope to oserve the voltge output cross the resistnce (CH1) nd cross the function genertor (CH2). Report you oservtions nd explin. (Do this experiment using the three different types of signls: sinusoidl, squre, nd swtooth) The lst prt of the experiment is optionl nd should only e completed if there is sufficient lortory time! 7 Repet prts 4 nd 5 for nother operting point where V s =2V nd R=150Ω. Plot the new lod line onto Figure 1.5. Mesure v do nd i do nd report them in your l report. Include the predicted vlues, s well.
11 L Hndout Nme: Section: Dte: Experiment#2 Prt 1 1 R xx mesured = Ω 2 Compre the mesured vlue of the resistnce R xx with wht you clculted in the Pre L exercise. Justify the difference etween two vlues, theoreticl nd mesured, if there exist more thn %5 error. 3 R mesured cross c terminls = Ω.Why isn t it greter thn R/3? 4 V d (Mesured) = V d (Mesured) = I 3R (Mesured) = I RLeft (Mesured) = I RBridge (Mesured) = I R/3 (Mesured) = I RRight (Mesured) = Compre the results with wht you hve otined in the prt 4 of the pre l section. Discuss ny difference in your results. Prt 2 1 For V s =3V nd R=100Ω: v do (predicted)= V i do (predicted)= ma
12 2 Is there minimum threshold for Vs such tht LED would not e lightening for Vs elow the threshold? How much is tht threshold in your circuit? 3 Wht is the effect of R=0 on the lod line slope? Is it physiclly possile to get lod line with tht slope? Why? 4 Wht hppens in terms of the lod line slope, if we mke the resistnce higher nd higher? Cn we reduce the effect of high resistnce y mking the Vs voltge higher? 5 For V s =3V nd R=100Ω: v do (predicted)= V v do (mesured)= V i do (predicted)= ma i do (mesured)= ma 6 Is it correct conclusion tht in order to hve desired working point for LED, we need to djust oth Vs nd R, within resonle limittions, simultneously? 7 For V s =2V nd R=150Ω: v do (predicted)= V v do (mesured)= V i do (predicted)= ma i do (mesured)= ma
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