The supply divider resistors feeding the base of the transistor are of equal value, so that the transistor base is fed with half the supply voltage. You can use this formula to re-calculate the component values for other supply voltages.įormula for calculating emitter resistor: Enter the battery supply voltage (Vbatt) and your chosen transistor standing emitter current (Ie). If you want an on-line formula then here it is. I select the next lowest common value of 330 Ohms because the battery will have a tendency to fall, not rise. Re is therefore equal to E / I = 3.8 / 0.01 = 380 Ohms. The emitter voltage is equal to the base voltage (Vbatt = 9v / 2 = 4.5v), minus 0.7v = 3.8v. The emitter resistor Re is therefore selected to give an emitter current of 10mA (0.01A). The transistor will not get hot, but there is enough current to allow it to work, but not enough to flatten a battery in a couple of hours. The BC547 is rated at 300mA, so 3% of this is a nice general figure. I will also run the transistor at about 10mA (0.01A). Small voltage changes will therefore NOT affect the transistor capacitance changes by much. In the interest of stability I will run the transistor base at 50% of the supply voltage. This gives us a current amplification, but no voltage amplification. The basic circuit is an emitter follower transistor amplifier. So I will use a 9v battery for this project (calculate a little lower than 9v). Another consideration is the box - many commercially available hobby boxes (enclosures) have an integral 9v PP3 type battery compartment. Small voltage across a transistor cause the internal CB capacitance to rise, in effect it is a small varicap diode. Nearly all my circuits are drawn with a positive and negative supply rail connected to a supply, of some description. Some of the formulas given are just rough, in the interest of simplicity. So I shall start with a simple one-transistor device and build up the circuit. Of course, this page is not for the experienced constructor, but I am please that so many people seem to take an interest in my homepages and the techniques I use. Ok, this I can do, and I shall try to give clickable calculations whenever possible. This is the smallest possible value of S and leads to the maximum possible thermal stability.This has become a sort of regular periodic question - " Harry can you please explain how the FM Wireless microphone functions?", ". As R B can be found directly, this method is called as fixed bias method. We know that V CC is a fixed known quantity and I B is chosen at some suitable value. Since V BE is generally quite small as compared to V CC, the former can be neglected with little error. Therefore,Ĭonsidering the closed circuit from V CC, base, emitter and ground, while applying the Kirchhoff’s voltage law, we get, Let I C be the required zero signal collector current. The figure below shows how a base resistor method of biasing circuit looks like. The required value of zero signal base current and hence the collector current (as I C = βI B) can be made to flow by selecting the proper value of base resistor RB. The base emitter junction is forward biased, as base is positive with respect to emitter. The required zero signal base current is provided by V CC which flows through R B. In this method, a resistor R B of high resistance is connected in base, as the name implies. Biasing with Collector feedback resistorĪll of these methods have the same basic principle of obtaining the required value of I B and I C from V CC in the zero signal conditions.The commonly used methods of transistor biasing are It is economical to minimize the DC source to one supply instead of two which also makes the circuit simple. The biasing in transistor circuits is done by using two DC sources V BB and V CC. Emitter Follower & Darlington Amplifier.Transformer Coupled Class A Power Amplifier.
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