![]() Which also gives you some flexibility about handling C_miller_effect. If you must have a high Rin, then consider a Darlington. Pick a very large Ce, to ensure Zc << 2.6 ohms.įor high bandwidth, you might worry about C_Miller_Effect.Īt high frequencies, that charge consumption will suck up lots of input energy. With 130 ohms as Rcollector, we will mostly ignore the Rload value in computing the Voltage Gain. For a gain of 50, set Rc = 50 * 2.6 ohms, or 130 ohms. You have no required Vout_peakPeak, despite that Rload.īias the transistor to 10mA (why not? power is free). Since 12/0.026 is about 480, then yes, we can move ahead.īias the transistor emitter to VDD/3. The maximum gain a single bipolar can provide, with resistive load, is Hang in there.įor a gain of 50X, using a single transistor, with resistive load, first let us be sure we can achieve that 50X. ![]() This resistance is known as the Emitter Resistance, R E. Your instructor is letting you immerse, and learn quickly. The amplifiers bias voltage can be stabilised by placing a single resistor in the transistors emitter circuit as shown. If you have NO emitter bypassing, then Rin is beta * (1/gm Re). The transistor Rin, assuming excellent bypassing of Re, will be Beta/gm where gm is the transistor transconductance. Input Resistance will be parallel combination of R1, R2, and the transistor Rin. These 2 Rs allow stabile biasing and stable Gain.Īnd thanks to LvW for error detection and system thinking. ![]() [For best design, add a 2nd Re in series with first R, with no bypassing. In practice I suspect that designers tend to take more of a heuristic approach to circuit design rather than solve equations like this. I find that my courses at university are very theoretical and so I tend to over complicate the designs. I'd also be interested to know how more experienced designers would approach this simple design problem. So biasing the BJT will set the emitter-base junction in a forward-biased state. In this case, the process enables a transistor to amplify an AC input signal in a transistor circuit. Could anyone explain what is happening here? Generally speaking, transistor biasing involves applying a specific amount of voltage to a BJT’s base and emitter terminals, improving its efficiency and performance. I would expect these values to be the same as what I worked them out to be previously from the voltage gain equation ( \$92 \Omega \$)? In simulation this method seems to work however I can't understand how the input and thevenin resistances have seemingly changed value. I calculate \$ \beta = \frac = 5357 \Omega\$ I want to keep that low so I choose \$2mA\$ and plot the IC vs VCE curves for this specific transistor in a spice simulator:Ĭhoosing a point half way on the load line for symmetrical swing, I obtain: I began my design by picking a maximum current. I've drawn the small-signal model as follows, assuming that in the midband coupling capacitors are treated as shorts and bypass and load capacitors are treated as open circuits: Using a 2N2222 BJT transistor and the following CE configuration: Load \$5k\Omega\$ with a coupling capacitor (not shown below).I am trying to design a Common-Emitter amplifier to the following specifications:
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