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I'm having a lot of trouble with more complex transistor DC circuits (can't wait for the AC, heh). Basically, what I want to ask is does KVL holds "around" a transistor as well?

For example, if I have a p-type MOSFET, and I successfully calculate Vsd and Vdg, can I find Vsg like Vsg=Vsd+Vdg? I think I remember seeing that this is true only in saturation mode, but I can't seem to find it again anywhere for the life of me. Similarly, can I go "through" a wire when doing BJT circuits? What I mean is, if I know that BJT is in saturation (and Vbe=0.7, Vce=0.2), is it allowed to have both of them in the equation while doing KVL (effectively going through the emitter wire).

TL;DR: Is it valid to write Ib*Rb+Vbe-Vce=0 regardless of the mode of operation. Same for the MOSFETs?

Example circuit

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  • \$\begingroup\$ KVL and KCL pretty much "just work." (KVL sometimes won't apply in the case of changing magnetic fields where Faraday's law may be required.) In the saturated BJT case, the collector "looks like" a voltage source. But in active mode, the collector "looks like" a current source. So your analysis choices will vary, depending. \$\endgroup\$ – jonk May 20 '19 at 19:56
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I won't say a blanket "yes," since I'm not totally sure what you are asking. Looking at your examples, though:

  1. Yes, if you know \$V_{sd}\$ and \$V_{dg}\$, you can easily calculate \$V_{sg}\$ as you describe. To check the signs, expand each of the terms \$V_{xy}\$ into \$(V_x - V_y)\$. This gives: $$V_{sg} = V_s - V_g = (V_s - V_d) + (V_d - V_g) = V_{sd} + V_{dg}$$. I wouldn't call it KVL though (I don't really ever call anything KVL though, so maybe it actually is).

  2. I'm not quite sure what you mean by going "through" the BJT emitter. You can form different voltage loops through the BJT depending on which path you take, and if your problem tells you what the voltage drop is between those two terminals, then you should be able to use those values. The more complicated case is when you aren't given specific voltages and need to use various transistor equations to solve for the voltages in the circuit.

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  • \$\begingroup\$ That very much is a "blanket yes" but with even more explanation, so thank you very much! Just to clarify a bit further (because this is insanely important to me at the moment) what I meant in the second question: If we look at this circuit for example (image.slidesharecdn.com/transistorbjtdcandac-170518092345/95/…) is Ib*Rb+Vbe-Vce=0 valid? On this picture they did "left" ground to -Vee, and "top" ground to -Vee but not "left" ground to "top" ground, which is what I'm asking. \$\endgroup\$ – slamdunker May 20 '19 at 19:59
  • \$\begingroup\$ In hte circuit in your link, the two ground symbols are connected, and considered "Zero Volts". Neither is connected to -Vee. \$\endgroup\$ – Peter Bennett May 20 '19 at 20:09
  • \$\begingroup\$ @slamdunker Yes, that is valid. You can make KVL loops through ground symbols, and through the BJT if you know the voltage drops. I would tend to try to solve for node voltages relative to ground myself, but that's a personal preference. E.g., Vb = -Ib*Rb from Ohm's law, Vc = 0, and Ve = Ie*Re-Vee again from Ohm's law, then subtract the different values if Vce, Vbe are needed. \$\endgroup\$ – Justin May 20 '19 at 20:10
  • \$\begingroup\$ That pretty much answers my questions completely. Thank you very much! \$\endgroup\$ – slamdunker May 20 '19 at 20:14
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KVL is invariant in the absence of changing magnetic fields (see @Jonk's comment). KCL is just invariant (as long as you account for charge injection phenomenon like corona discharge).

So, yes, you can use KVL around a transistor. If it helps, just relabel the terminals as "point A", "point B", etc. If you know the voltage from point A to point B, and you know the voltage from point B to point C, then you know the voltage from point A to point C (or visa-versa, if you don't scramble your brains on the sign changes).

What you cannot do (and I don't think you're confused on this point, but I'm gonna say it anyway) is assume that the transistor has an I-V characteristic that in any way shape or form resembles a resistor or collection of resistors.

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  • \$\begingroup\$ And that's regardless of the mode of operation or the type of transistor? BJT, MOSFET, saturation, active regime..? Cool. Thank you very much! \$\endgroup\$ – slamdunker May 20 '19 at 20:10
  • \$\begingroup\$ KVL actually works around any closed path (in the absence of those pesky varying magnetic fields). Transistors, vacuum tubes, Klingons with disruptor beams -- it doesn't matter. \$\endgroup\$ – TimWescott May 21 '19 at 0:55

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