# Tag Info

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Adapting Your Schematic Here's how I might adapt your schematic: I knew (because I looked) that the 2N2222 has $\beta\approx 200$. So I selected an appropriate base resistor, given the collector resistor I first selected. Other than that, there's not much to the adaptation except that I'm using two voltage sources instead of one so that I can sweep out ...

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You are correct, emitter current is base current + collector current. In fact, emitter current is I(emitter) = I(base) + I(collector) And you can calculate Beta from measured values... Beta = I(collector) / I(base) Then calculate exactly how much current is in each part of the transistor.

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The current in the emitter region is indeed larger than in the collector, because the emitter current has to be equal to the collector current plus the base current. That's just KCL. Just like $β$ is defined as $\frac{I_c}{I_b}$, you can also define $α=\frac{I_c}{I_e}$, which is always less than one--i.e., $I_c<I_e$. (α is also often seen defined ...

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Prelude In the section titled Question 1 in this answer of mine, I qualitatively discuss the rule of $\frac{\textbf{1}}{\textbf{10}}$ths. That answer was in response to a question on this very same page of the very same book (page 439 of "Practical Electronics for Inventors".) We recently engaged a short discussion (found in the comments at the ...

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It may be just slightly easier to see it drawn this way: simulate this circuit – Schematic created using CircuitLab (I've left things to use $V_{\!_\text{CC}}$ rather than $9\:\text{V}$ to keep it more abstract for now, despite the specific resistor and capacitor values still present. I've kept all your part name-designations, though.) Static ...

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To simulate BJT output curves like this: source I would use a schematic like: simulate this circuit – Schematic created using CircuitLab IB: constant current or stepped (parametric sweep). For example 1 uA or 1 uA to 10 uA in steps of 1 uA VCE: DC sweep 0 V to 12 V Note how I use the DC simulation (.DC) and not the transient (.tran) simulation! I don'...

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Capacitors aren't even connected to negative rail, how will they ever charge? They are. The negative terminal of each capacitor is connected to ground through the base-emitter junction of the transistors. The positive terminal is connected to +9 V via the LED and its current limiting resistor. Due to imbalances in the component tolerances and the ...

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This circuit is called an astable multivibrator. The very short description is, once the circuit gets going, each capacitor charges through the transistor collector and discharges through the opposite transistor base. The two transistors take turns doing this as one triggers the other into its opposite state. A more detailed breakdown of this circuit is here:...

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That circuit is a transistor astable multivibrator. Start with Q2 having just been switched on with its base at 0.7V and its collector at 0V. Q1’s base will be at -8.3V and will be slowly rising as C2 charges through R3. Simultaneously the collector of Q1 will be rising as C1 charges via R1. The collector of Q1 will reach 9V before its base reaches 0.7V ...

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You have a transistor with about 430uA flowing into the base. The transistor will act more-or-less like a constant current sink until the voltage across the transistor drops below 200mV or so. The current is hFE dependent, but somewhere around 100mA with a large possible variation. The resistor R1 fights that current sink, but only affects it by 5% or so ...

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