Electronic Circuits LaboratoryEE462GLab #4DC Power Supply Circuits Using DiodesInstrumentationThis lab requires the use of:¾ Various oscilloscope and function generator options, which you should be familiar with now from previous labs!¾ Don’t forget grounding issues! Both the function generators and oscilloscopes are earth grounded.Power SuppliesMost information processing circuits require DC power to power-up their embedded amplifiers and gates. Therefore, AC power from the local power company must be converted to DC for these circuits to function properly. There are 3 main operations to most power supplies:¾ Rectification (convert AC to pulsating DC)¾ Filtering (Smooth pulsating DC to reduce peaks and raise valleys)¾ Regulation (Control output voltage level to make it robust against input voltage fluctuations and load changes)RectificationA pulsating DC waveform can be created with the half-wave rectifier shown below: RL Vs Vout +-Time-10.0000.0+10.0000.0 +20.000m +40.000m +60.000m +80.000m +100.000mVsVoutRectificationA pulsating DC waveform can be created with the full-wave rectifier shown below.What potential problems exist in measuring output voltage?What are the advantages for full-wave over half wave rectification? RLVs Vout +- Time-10.0.0+10.0.0 +40.000m +80.000mVsVoutSmoothingA capacitor can be used to reduce the peaks and raise the valleys of the pulsating DC.0.03 0.04 0.05 0.06 0.07 0.080123456789SecondsVoltsT Vr Vs+Vout_SmoothingA performance measure of a DC power supply is the percent output ripple expressed as r in the equation below:peak-to-peak output voltagemean of the output voltageMultiply r by 100 for percent output ripple.Peak to peak ripple Average DCoVt popV−oVopopVVr−=What scope measurefunctions can be used to obtain an estimate of r?What would the average or mean of this signal be, if AC coupling was used?SmoothingA simple relationship between the capacitor value and Ripple voltage Vr can be derived with approximations.0.03 0.04 0.05 0.06 0.07 0.080123456789SecondsVoltsT Vr Vs+Vout_RLCFor diode-on state assume the time constant is much smaller than period T,so capacitor fully charges on pulse peaks (small cap-resistance product) and for diode-off state assume assume time constant is larger than period (large cap-resistance product) so capacitor discharges slowly.0.03 0.04 0.05 0.06 0.07 0.080123456789SecondsVoltsT Vr Smoothing Vs+Vout_RLCFor discharge cycle (diode off), the charge removed from the capacitor is:TIQL≈where T is the period of wave form and ILis load currentCharge is also related to the voltage change and capacitor value:CVQr≈From the above, the following relationships can be derived:rLVTIC ≈LrrmRVTVVC⎟⎠⎞⎜⎝⎛−≈2LmrCRTTVV+≈2VmRegulation:¾ A DC power supply provides constant DC voltage, independent of load and input voltage fluctuations.¾ A voltage regulator circuit is used to limit changes in output voltage.¾ A Zener diode can be used in a voltage regulator circuit by taking advantages of the Zener diode’s reverse breakdown characteristic. ¾ The regulator's resistor, Rreg, limits the current through the Zener to reduce dissipated power. Vin Vout+ - + - RregZener diodeForward BiasReverse BiasIdeal approximation:Short CircuitIdeal approximation:Open CircuitKneeVDIDVZRegulationAssume the Zener is operating in its breakdown region with a breakdown voltage of Vz. Show the power consumed by the Zener under a no-load condition is: Vin Vout+ - + - RregregzinzzRVVVP2−=Assuming an Ideal Zener diode (zero reverse bias current, and vertical breakdown region), what is the max power delivered to a load with output voltage Vz?Show that the load resistance at this point is (VzRreg)/(Vin-Vz).What happens when load resistance is less than this value?+Vz_Regulation¾ Performance of a regulator is typically measured in terms of regulation given by: Vin Vo + - + - RregoFLoFLoNLVVV −=Regulationis the average output no-load voltage (open circuit) is the average output full-load voltage (smallest rated resistive load) Percent regulation is obtained by multiplying Regulation by 100.What scope measurement functions would be helpful in assessing the regulation of the circuit? What coupling should be used in measuring the regulators output?oNLVoFLVPower Supply with All Stages Vin Vout+ - + - Rreg RL Vs + - RectifierFilter Regulator LoadAC source DC Power Supply Half-wave example:Critical SPICE Features for simulating these circuits in this lab include:¾The zener diode (if not in the main parts menu, use the browse for part option and type in zener as a key word and apply filter. Of the several choices, select the default or generic zener).¾Must set the breakdown voltage for the part through the edit simulation model¾For input, select a sinusoidal (do not check use AC) and set amplitude and frequency from the transient properties tab options.¾For simulation set up select Transient and Fourier Analysisoption, set start time, stop time and increment/step. These values must be selected to give you a clear plot in a reasonable amount of time. Do not enable Fourier.SPICEFinal Notes¾ Beware of grounding issues, especially on the full-wave rectifier!¾ Scope settings are important to describe in the procedures sections. These are not spelled out in the labs any more (so you are on your own). You should indicate probe placement, coupling, measurement functions, math settings, and any other setting critical for making the measurement.¾ Be sure to respond to all comments and questions in bold type inthe Discussion Section.¾ All measured values and requested values computed from the measured values should be in the Results Section. Describe how values were computed in the Results Section. Be explicit and brief …. use