Minimum base current = 100 MA / 100 Minimum base current = 1 MA Minimum collector current = 100 mA / 100 Minimum base current = 1 mA In actual practice, it is best to calculate about 30% more current than we will need to guarantee our transistor switch is always saturated. In this case, we will use 1.3 mA. We must also select our supply voltage, so for this example we will use 12 volts. We can now calculate resistor R1 in the circuit as follows: Maximum Current Required = 100mA Supply Voltage = 12 Volts
R1 = Supply Voltage / ( Maximum Current Required / Minimum HFE * 1.3 ) R1 = 12 / (.1 / 100 * 1.3) R1 = 9230.7 or 10K for nearest standard value. R1 = 9230.7 or 10k for nearest standard value. Resistor R2 is not essential to this circuit but is generally used for stability and to insure that the transistor switch is completely turned off. This resistor insures that the base of the transistor does not go slightly negative which would cause a very small amount of collector current to flow. The value of this resistor is not critical but a value about 10 times R1 is normally chosen. For this circuit we will calculate R2 to be 10 times R1 as follows: R2 = 10 * 10000 R2 = 100K R2 = 100k To turn ON our transistor switch, all that is needed is to short resistor R1 to the negative ground. I don't like the word "short." It should be "take." I don't like "negative ground" Just use the word "ground." It is the negative of the battery, but if you use the word "negative" it will conjure up the possibility that the supply is both positive and negative, like a dual power supply: +12v and -12v. PNP buffers (BUFFER STAGE - INVERTER AMPLIFIER STAGE - "SWITCH" ) are very difficult to get working because the incoming signal has to rise to 12v (in this example) and if it just rises to 11.5v, the circuit will not turn OFF. Even a 555 will not work in this example because the output of a 555 on a 12v supply will only rise to 11v. The driver circuit MUST be on the same rail as the BUFFER because if they are different by as little as 0.5v, the BUFFER will not turn OFF.
While PNP transistors are normally used for a negative ground configuration, it is possible use a NPN What is a "negative ground configuration??" transistor if a positive ground configuration is desired as indicated in Figure 2. The calculation of resistor values is identical to the PNP version. However, in the NPN transistor, R1 must be shorted to the positive of the supply to turn the switch on. While our transistor switch can easily replace many mechanical relays, it does have a few drawbacks. The maximum design current must not be exceeded or the output voltage will be reduced. A short circuit of the output will overheat and destroy the transistor in many cases. Although the transistor is in saturation when turned on, about .3 volts is lost through the collector to the emitter of the transistor. We must also insure that the maximum power dissipation of the transistor is not exceeded. We can calculate the power dissipation by multiplying the current by .3 volts. In the case of 100 mA, the transistor must be able to withstand 30 milliwatts (.3 times .1).
Transistor switches are used for a wide variety of applications. Many amateurs will notice that the circuit in Figure 1 is used as the PTT in many transmitter circuits. Transistor switches are commonly used to turn on transmitter circuits, LED�s, cooling fans and even relays. However, when using a transistor to turn on a relay coil, it is very important to use a 1N4001 diode reversed biased in parallel with the relay coil as in Figure 3. This is to prevent the kickback voltage in the reverse polarity from destroying the transistor. This reverse voltage occurs momentarily when the normal current stops flowing through the coil. It is good practice to always use a diode when turning on any inductive load. Transistor switches are often used to take the low-level output from logic circuits to turn on or turn off a particular device.
The actual transistor used as a switch is not critical in these applications. Virtually any general purpose NPN or PNP transistor can be used as a switch. All that is needed is to know the minimum HFE and the power dissipation of the transistor. While most transistors in a TO-92 case will have HFE�s of at least 100, many power transistors in TO-220 cases often have an HFE no greater than 25. It is essential to know the HFE or Beta of a transistor, so that we can have a large enough base current to achieve saturation. If a power transistor is used to turn on a high current device, it may be necessary to use another lower current transistor switch to drive a transistor switch used in a high current application. This is especially important when using a low current logic output from a CMOS IC. One of the biggest traps in designing a BUFFER stage occurs when driving or illuminating a globe. A globe takes about 6 times more current when starting to illuminate because the cold filament has a very low resistance. The resistance increases enormously when it is fully glowing. This means the current required to get the globe to illuminate is 6 times the operating current. This even applies to a 12v 100mA globe and if you only consider 100mA when selecting a transistor and the base resistance, the globe will not turn ON. And you will wonder why. Now you know.
The author has gone to a lot of detail in how to work out the value of each resistor. He states it is best to assume a transistor has a gain of 50, and then states the circuit above has a very high gain. He does not understand the basis of the gain of a stage where 4 resistors are used to create the biasing. The final gain of the stage is simply the division of the load resistor by the resistance of the emitter resistor. For the first stage this is 1200/180 = 6 and for the second stage the gain is 560/100 = 5.6 This give the circuit a total gain of 33 and there will be lots of losses in the 3 coupling electrolytics. The 4u7 and 10u in the emitters will increase the gain enormously. The circuit has extremely low values of biasing components and with the low value of coupling components, it is doubtful if the output would give any results AT ALL. Let me give just one important way to look at how a circuit works. The 1k2 on the first stage "pulls" the electro UP and delivers a signal to the second stage. The transistor DOES NOT DO ANYTHING !!!! The input of the second stage is 1k (actually less). Replace the 4u7 with a link and you will see the voltage at the mid-point cannot be higher than half-rail voltage. A voltage divider is formed with the 1k2 and 1k. Thus the waveform can only extend from a low voltage (a few volts) to 6v. See how the low values of input resistance (impedance) can reduce the waveform considerably. SIGNIFICANT FIGURES Significant figures is a technical term for providing an answer that is more accurate than the figures given in the question. This applies very much to transistors. To turn ON a transistor may require a base current that is one-hundredth of the collector current. But if the resistance of the load is halved, it will need more than twice the base current. To make sure the transistor is fully turn ON, we often double the base current to make sure. This means talking about a base resistor with a value 9230.7 ohms is quite absurd: R1 = 9230.7 or 10K for nearest standard value. R1 = 9230.7 or 10k for nearest standard value. as the value will be in the range 10k or as low as 4k7 or even 2k2 and this is far from the "accurate value" of 9230.7 ohms !! In this example, the only significant figure is "9" and the other digits need to be "0's" as even the value of "9" will be changed to a value as low as "2." "9" is the only value of importance and the other figures are considered to provide a result that is of a higher quality than the original data. In other words, its accuracy cannot be justified and it is what we call: "going overboard." It is absurd, providing an accurate value when the final result will be completely different. This is just one thing to remember when providing "PHYSICAL RESULTS." It shows your understanding or lack of understanding of PHYSICS. The answer has to be in accord with the accuracy of the WORST value in the question (called the DATA).