![]() ![]() This means one of the transistor terminals must be common to both the input and output circuits. Representing the basic amplifier as a two port network as in figure 9.1, there would need to be two input and two output terminals for a total of four. Apply algebraic substitution to this formula so that alpha is defined as a function of beta: f () f ( ). The transistor, as we have seen in the previous chapter, is a three-terminal device. It is defined as the ratio between collector current and emitter current: I C I E I C I E. I really cannot figure out what I'm doing wrong here. A bipolar junction transistor parameter similar to is alpha, symbolized by the Greek letter. I'm really confused.įor those old designs, the ones I initially thought should work, I duplicated the same circuit with my old 2N3904s, and the values were right on the money. ![]() What gives guys? I thought maybe the spec for the transistor drifts a lot when it's outside of the measured conditions (5V 2mA Ic), but I used a 5V supply as well and the results are the same. This time, the values agree! The emitter voltage is about 940 mV, and the base voltage is about 1.54 V. I worked out values of 18k for R1 and 2.8k for R2 (bias resistance of about 2.42k). So, with that, the expected resistance looking into the base is about 20k. Eventually, I tried to work out the beta value backwards, and it looks like the transistor has a real beta value of about 43-44, not even close to the claimed 420-800. In other words the current flowing int he base circuit controls. Alpha is the relationship of collector current (output current) to. Bipolar transistors are what are termed current controlled devices. The current gain in the common-base circuit is calculated in a method similar to that of the common emitter except that the input current is I E not I B and the term Alpha () is used in place of beta for gain. I tried to reduce the total base resistance to reduce the loading effect, but that made it worse. To calculate beta, use the following formula : I BI C. I've built a few more circuits with different values. for a realistic case of large but finite beta, it can be anything. Needless to say, I'm definitely not getting 2mA from this source. What happens when bjt beta infinity Is it in saturation or cut off or active mode if it is indeed infinity, it can be either in saturation (for any Ib>0 -> Ic infinity) or undefined (for any Ib0 -> Ic 0infinity undefined). I expected to be at least somewhat close to a volt, but this is less than half. The emitter voltage worked out to be 447 mV. The divider above should work out to a Thevenin Equivalent of about 28.7k, pretty dang close.Īfter the quick sketch, I built the thing and to my surprise my design did not work out at all. Michael Shur, in Encyclopedia of Physical Science and Technology (Third Edition), 2003. This means I'll have to aim for a base resistance of about 28k and a base voltage of about 1.5-1.6 To bias the transistor, I multiplied the emitter resistance by beta (assumed 600) and divided by 10 as a rule of thumb. ![]() I want to put the emitter at about 1V, so at 1V / 2 mA gives 500 Ohms, I went with 470 Ohms as the closest value I have. The actual range is actually like 420 - 800 or something.Īnyway, I was trying to build a current source for a differential pair and wanted to pump out about 2mA of current through the transistor, so I set it up like this for a quick test: ![]() Reading through the datasheet, (and on digikey's page), I am led to believe that the DC gain for this transistor is at minimum 420 2mA, 5V]. If you do, the BJT will likely operate in "soft" saturation-i.e., in the transition region between the BJT's forward-active mode and its "hard" saturation (fully ON) mode.I bought these transistors on digikey: BC549CTA Transcribed image text: Consider the circuit shown in Figure 1, where the BJT transistor has a beta 100, VA infinity, The MOSFET parameters are mun Cox (W/L) 0.5 mA/V2, Vt 1V, lambda 0V-1 Using DC analysis, calculate the following DC voltages: V1, V2, and V3. \$\beta\$, \$\beta_\$ to perform saturation calculations. And for what it's worth there are multiple \$\beta\$ factors to consider when designing BJT circuits: ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |