How to measure collector current in BJT? - Electrical Engineering Stack Exchange most recent 30 from electronics.stackexchange.com 2019-09-15T16:29:43Z https://electronics.stackexchange.com/feeds/question/403102 https://creativecommons.org/licenses/by-sa/4.0/rdf https://electronics.stackexchange.com/q/403102 1 How to measure collector current in BJT? Bee https://electronics.stackexchange.com/users/181200 2018-10-25T05:06:30Z 2018-10-25T11:12:10Z <p>In the following circuit:</p> <p><a href="https://i.stack.imgur.com/lACtx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lACtx.png" alt="enter image description here"></a></p> <p>I know the expression for the collector current is <span class="math-container">$$I_C =\frac{V_{CC}-V_C}{R_L}$$</span> Say I have a potentiometer for <span class="math-container">\$R_L\$</span> and I decide to set it to 0 Ω. What would <span class="math-container">\$I_C\$</span> be in this case?</p> https://electronics.stackexchange.com/questions/403102/-/403107#403107 3 Answer by Rohat Kılıç for How to measure collector current in BJT? Rohat Kılıç https://electronics.stackexchange.com/users/103420 2018-10-25T05:40:37Z 2018-10-25T11:12:10Z <p>Let me show the voltages and collector current (I<sub>C</sub>).</p> <p><img src="https://i.stack.imgur.com/3rPSb.png" alt="schematic"></p> <p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2f3rPSb.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p> <p>As you can see, <span class="math-container">\$V_B = V_{BE} + V_E = V_{BE} + I_E \cdot R_E\$</span>. </p> <p>Since <span class="math-container">\$I_C \approx I_E\$</span>, the equality above turns into <span class="math-container">\$V_B \approx V_{BE} + I_C \cdot R_E\$</span>. Finally, <span class="math-container">$$I_C = \frac{V_B - V_{BE}}{R_E}$$</span> <em>(I will not show what VB is since it can be found with simply voltage divider rule.)</em></p> <p>See? Collector current is independent from collector load!</p> <p>NOTE: Placing a resistor to emitter creates a constant current source. It also guarantees the thermal stability.</p> https://electronics.stackexchange.com/questions/403102/-/403108#403108 -1 Answer by analogsystemsrf for How to measure collector current in BJT? analogsystemsrf https://electronics.stackexchange.com/users/132064 2018-10-25T05:44:11Z 2018-10-25T05:44:11Z <p>Lets assume Beta = infinity. And assume the Ic is near 1mA, so the Vbe is approximately 0.6 volts (0.5 volts near 10uA, 0.4 volts near 100 nanoAmps).</p> <p>Ic = (Vbase - 0.6) / Re</p> <p>Where Vbase = VDD * the voltage divider ratio</p> <p>thus Vbase = VDD * R2 / (R1 + R2)</p> <p>and now, finally, </p> <p>Ic = { [VCC * R2/(R1 + R2)] - 0.6} / RE</p> <p>Again, assuming beta = infinity, Vbe = 0.6, and Vearly = infinity</p> <p>With all 3 resistors being 1KOhm, and VCC = 3 volts, we have</p> <p>Ratio = 0.5, </p> <p>Vbase = 3*0.5 = 1.5 volts, </p> <p>Vemitter = Vbase - 0.6 = 1.5 - 0.6 = 0.9 volts</p> <p>Ic = 0.9 / 1K = 0.9 milliAmps</p>