My Troubleshooting Textbook by Max Robinson Chapter 1 Introduction to Troubleshooting. Chapter 2 Test Equipment. 2.1 The Volt-Ohm-Milliammeter (VOM). 2.2 The Electronic Voltmeter. 2.3 The Digital Multimeter (DMM). 2.4 Choosing The Correct Test Meter. 2.5 Analog Versus Digital Meters. 2.6 The Oscilloscope. 2.7 The Signal Tracer. 2.8 Miscellaneous Test Equipment. 2.9 Instruction Manuals. Chapter 3 Failure Modes. 3.1 Generalized Failure Modes. 3.2 Electrolytic Capacitors. Chapter 4 Troubleshooting Techniques. 4.1 Check The Obvious First. 4.2 Do Not Make Modifications. 4.3 The Power Supply Section. 4.4 Half-splitting. 4.5 Signal Tracing. 4.6 Signal Injection. 4.7 Disturbance Testing. 4.8 Static Testing. 4.9 Shotgunning. Chapter 5 Faults in Power Supplies. 5.1 Rectifier-Filter Circuits. 5.2 Analog Voltage and current Regulator Circuits. 5.3 Switching Mode Power Supplies. Chapter 6 Faults in Transistor Circuits. 6.1 Common Emitter Amplifier. 6.2 The Emitter-follower's Fatal Flaw. 6.3 AC Coupled Amplifiers. 6.4 DC Coupled Amplifiers. 6.5 Radio Frequency Amplifiers. 6.6 Switching Circuits. Chapter 7 Transistorized Consumer Equipment. 7.1 Audio Amplifiers. 7.2 Radios and tuners. 7.3 Things you should leave alone. Chapter 8 Faults in Vacuum Tube Circuits. 8.1 Audio Amplifiers. 8.2 Radio Receivers. Chapter 9 Antique Equipment. 9.1 Before Turning on the Power. 9.2 Pre 1930 Radios. 9.3 Pre World War Two Radios. 9.4 The All American Five. 9.5 Three Way Portable Radios. 9.6 Phonographs and Record Changers. 9.7 Consoles and High Fi Components. 9.8 Wire and Tape Recorders. 9.9 Why TV Sets Are Not covered. Chapter 10 Things That Have Never Worked. 10.1 Power Supplies. 10.2 Audio Amplifiers. 10.3 Radio Receivers. 10.4 Simple Test Equipment. hapter 0 Electrical Fundamentals. 0.1 DC Circuits. 0.2 AC Circuits. 0.3 Power Supplies. 0.4 Bipolar Transistor Fundamentals. 0.5 Bipolar Transistor Circuits. 0.6 Operational Amplifiers. 0.7 Field Effect Transistors. 0.8 FET Circuits. 0.9 Vacuum Tube Fundamentals. 0.10 Vacuum Tube Circuits. Chapter 0 Review of Fundamentals. In order to understand the material in this book, the reader must understand certain fundamentals of electricity and electronics. The teaching of these fundamentals is far beyond the scope of this book. This chapter is merely a review. If the reader is not familiar with any of what is presented in this chapter, it is strongly urged that he or she supplement his or her knowledge by studying Electronics for Physicists. Back to Fun with Transistors. Back to Fun with Tubes." Back to Table of Contents. Back to top. 0.1 DC CIRCUITS. Kirchhoff's voltage law states that the algebraic sum of all voltage drops around a closed loop is equal to zero. Kirchhoff's current law states that the algebraic sum of all currents flowing into a node is equal to zero. The consequences of these laws for series circuits are a) the sum of all voltage rises is equal to the sum of all voltage drops, and b) the current anywhere in a series circuit is equal to the current anywhere else in the same circuit. The consequences of Kirchhoff's laws for parallel circuits are a) the sum of all upward flowing currents is equal to the sum of all downward flowing currents, and b) the voltage across any element in the circuit is equal to the voltage across any other element in the same circuit. Ohm's law states the relationship between voltage, current and resistance in an electric circuit. V = I x R (Eq. 0.1) I = V/R (Eq. 0.2) R = V/I (Eq. 0.3) where V is voltage in volts, I is current in amperes and R is resistance in ohms. To properly apply Ohm's law it is necessary to use the resistance of a particular resistor, the current through that same resistor and the voltage across that same resistor. When resistors are connected in series the total resistance is RT = R1 + R2 + R3 + ... + Rn for any number of resistors in series. When resistors are connected in parallel the equivalent resistance is 1/Req = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn for any number of resistors in parallel. Figure 0.1, Voltage Divider Circuit. For a verbal description click here. Figure 0.1 is of a voltage divider. RA and RB may be the equivalent of other resistors in series and/or parallel. The output voltage Vo is given by Vo = E RB/(RA + RB) (Eq. 0.4) Thevenin's theorem states that a network with output terminals and is of any level of complexity, using resistors, can be reduced to one ideal voltage source in series with a single resistance. The ideal (Thevenin) voltage source is equal to the no- load output voltage of the original network. The Thevenin resistance is equal to the resistance looking in at the terminals of the original network with all voltage sources replaced by shorts and all current sources replaced by opens. Example 0.1. Determine the Thevenin equivalent circuit of the network of figure 0.2. Figure 0.2, Circuit for Example 0.1. For a verbal description click here. Solution: With no load connected to the output, there will be no voltage drop across R3. Therefore, the voltage across R2 will be the same as the output voltage. R1 and R2 make up a voltage divider. By equation 0.4 we have Vo = 48 v x 1 k ohm / (3 k ohm + 1 k ohm ) = 12 volts. Figure 0.3, Network of Figure 0.2 with the voltage source replaced by a short. For a verbal description click here. If we replace the battery in figure 0.2 with a short and redraw the circuit we have the circuit of figure 0.3. The resistance looking in at the terminals is R1 in parallel with R2 and this combination in series with R3. R1 in parallel with R2 is R12 = 1/( 1/(1 k ohm ) + 1/(3 k ohm )) = 750 ohm. Now adding R3 in series we have 750 ohm + 500 ohm = 1250 ohm. The Thevenin equivalent circuit is shown in figure 0.4. Figure 0.4, Thevenin's Equivalent Circuit of Figure 0.2. For a verbal description click here. There are many other methods of solving electric networks. A few of these are Norton's theorem, superposition theorem, loop equations, node equations and Y-delta transformation. If you do not understand these techniques, it will be necessary to take an introductory level course in electrical engineering. RC Time Constant. If a capacitor C is being charged or discharged through a resistor R the time constant is given by T = R x C (Eq. 0.5) where T is time in seconds, R is resistance in ohms and C is capacitance in farads. T is the time required for a charging capacitor to charge up to 63.2 percent of its final voltage or for a discharging capacitor to discharge to 36.8 percent of its starting voltage. Theoretically a capacitor will never get fully charged or discharged. As a matter of practicality a capacitor is considered to be fully charged or discharged after 5 x T seconds have elapsed. Back to Fun with Transistors. Back to Fun with Tubes." Back to Table of Contents. Back to top. 0.2 AC CIRCUITS. All of the laws and theorems which apply to DC also apply to AC. In theorems stated in words replace the word "resistance" by the word "impedance". In equations, replace the symbol R by the symbol Z where Z = R + jX or Z = Z /_ Theta A capacitor is an open circuit for DC but has reactance for AC. The reactance of a capacitor is XC = 1 / (2 x Pi x f x C) (Eq. 0.6) where XC is the reactance of the capacitor in ohms, f is the frequency in hertz and C is the capacitance in farads. An inductor has a very low resistance for DC (ideally zero) but has reactance for AC. The reactance is given by XL = 2 x Pi x f x L (Eq. 0.7) where XL is the reactance of the inductor in ohms, f is the frequency in hertz and L is the inductance in henrys. Back to Fun with Transistors. Back to Fun with Tubes." Back to Table of Contents. Back to top. 0.3 POWER SUPPLIES. Power supplies serve the purpose of changing the 120 volt AC line voltage to one or more useful DC voltage(s). Figure 0.5, Full-Wave Bridge Rectifier Power Supply. For a verbal description click here. The circuit of a full-wave bridge rectifier is shown in figure 0.5. The no-load output voltage of this circuit is given by EM = EAC x Square Root (2) (Eq. 0.8) where EM is the no-load DC output voltage of the circuit and EAC is the RMS AC voltage across the entire transformer secondary. Figure 0.6, Full-Wave Center-Tapped Rectifier Power Supply. For a verbal description click here. The circuit of a full-wave center-tapped rectifier is shown in figure 0.6. The no-load voltage of this circuit is given by EM = EAC x Square Root (2)/2 (Eq. 0.9) where EM is the no-load DC output voltage of the circuit and EAC is the RMS AC voltage across the entire transformer secondary. The ripple output from a full-wave rectifier power Supply operating on 60 Hz is given by VR = EM / (200 x RL x C (Eq 0.10) where VR is the peak to peak ripple voltage of the power supply, EM is the no-load DC output voltage as given by equation 0.8 or 0.9, RL is the equivalent load resistance on the power supply in ohms and C is the capacitance of the filter capacitor in farads.