Can someone tell me why my 2n3904 transistor sim not working as expected
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Gary Gao 4 months ago
Hi, I am trying to use EasyEDA and created a test project using 2n3904 transistor. But the sim results is way off my expectation. Here is the URL to the project: [https://easyeda\.com/editor\#id=\|453b6f20177b458490e819c012281c6c\|a73489478bf84f94955a83eed5dd728f](https://easyeda.com/editor#id=%7C453b6f20177b458490e819c012281c6c%7Ca73489478bf84f94955a83eed5dd728f) . According to the 2n3904 datasheet, it should have BetaDC at around 100. However, no matter how I tune the resistors, I always got the following result from the simulator. I particularly don't understand how the Vbe(<span class="colour" style="color: rgb(85, 85, 85);">-8.11e+00) and </span>Ic(-4.16e-04) could be negative as showing below. Can someone shed a light on this problem? Any advice would be appreciated! Direct Newton iteration for .op point succeeded. Semiconductor Device Operating Points:                         --- Bipolar Transistors --- Name:       q1 Model:    2n3904 Ib:       8.37e-05 Ic:      -4.16e-04 Vbe:     -8.11e+00 Vbc:      6.29e-01 Vce:     -8.73e+00 BetaDC:  -4.97e+00 Gm:      -1.29e-02 Rpi:      1.00e+12 Rx:       2.00e+01 Ro:       7.77e+01 Cbe:      3.54e-12 Cbc:      3.24e-09 Cjs:      0.00e+00 BetaAC:  -1.29e+10 Cbx:      0.00e+00 Ft:       6.31e+05
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andyfierman 4 months ago
Sorry about the incovenience but the collector and emitter spice pins had been reversed in the symbol. It is correct now and your simulation runs OK.
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Gary Gao 4 months ago
@andyfierman Thank you for the help. The transistor works as expected now. I do have another question though. It seems that the resistors has direction. I figured a component rotates anticlockwise when pressing space. So to get a right order of the pin for a resistor when rotating from horizontal to vertical, the space needs to be pressed 3 times. I am wondering if there is an easier way to tell the direction of the resistor other than looking at the sim value?  Also, can you explain why the Ie(G1) is still negative?
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Gary Gao 4 months ago
Sorry, should be Ie(Q1), i.e. the emitter current of the transistor.
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andyfierman 4 months ago
I don't know of a way to identify the current direction associated with resistors. For the US symbol it is possible if you memorise the asymmetry of the symbol but not if you're using the EU symbols. It annoys the heck out of me but all spice simulators do it. As for " the Ie(Q1) is still negative?". That's because the current flows out of the emitter. Ic(Q1) is positive because it flows into the collector.
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