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enter image description here

This may look notoriously trivial to ask. we can see the direction of electron flow as indicated in the picture. the down side of the arrow must be the positive side of the voltage drop in each resistor, so two positive voltages + v_be=0 it implies v_be is a negative (-0.7 approx.). but in every book it's written that v_be=0.7 volts, is not it incorrect? or they are just talking about the absolute value of v_be?

as indicated in answer section, it is just a matter of convention. now let us see a second picture : enter image description here a simple voltage divider JFET. Assuming these authors are following the same conventional current rules in all other circuits, here, the Gate-source voltage should be positive, as just like v_be, but this time it is negative. why?

Sayan
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    Your arrows correctly indicate the direction of the electrons flowing. However, current is defined as flowing in the opposite direction! So for circuits: only worry about current flow and forget about the electrons. Only when you're looking at how a diode or transistor works is it needed to look at electron flow. – Bimpelrekkie Jan 13 '21 at 15:55
  • @Bimpelrekkie And things like Lorentz's Law also use conventional current flow! Even if you have electrons flying through the air; No positive charges moving anywhere! I think...maybeI got it confused again. – DKNguyen Jan 13 '21 at 16:04
  • @DKNguyen You're right, I wrote "Only when you're looking at how a diode or transistor works" but that's not 100% correct, also for understanding how vacuum tubes etc. (klystrons, magnetrons) work the electron flow is important. So better would be: Only when looking at circuits (including electro magnets) consider current flow, for most other cases, use electron flow. – Bimpelrekkie Jan 13 '21 at 16:12
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    @Bimpelrekkie I used to follow the general rule that "EEs look at conventional current flow and physicists look at electron flow" but that lead me terrible astray when I started looking at the Lorentz Equation and it was a while before I realized what the problem was. Ugh. – DKNguyen Jan 13 '21 at 16:15
  • So went back to review...Lorentz's equation uses conventional current flow because q is negative for electron charge which flips the direction right back around again to electron flow on its own. So infuriating. – DKNguyen Jan 13 '21 at 16:42

1 Answers1

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Not notoriously trivial because the reason is notoriously silly.

Benjamin Franklin did not know what actually moved to carry electricity so had to guess a polarity for each charged component.

He guessed wrong but by the time we figured out it was the negative-assigned particle (electrons) that actually moved, it was too ingrained and difficult to change so most things continue to use so-called "conventional" current flow, not actual electron flow.

That is an N-channel JFET which looks like this: enter image description here

https://www.tutorialspoint.com/basic_electronics/basic_electronics_jfet.htm

Note the P and N-doped silicon. That forms a diode. If your gate voltage is more positive than the source terminal, that forward biases the diode and produces a short which blows the JFET.

So an N-channel JFET must have a gate voltage that is more negative than the source so that the PN-junction is reverse biased. In your circuit, the way the voltage dividers and resistors are set up, the gate voltage is lower than the source pin voltage, even though both voltages are still positive when measured relative to GND. Also, unlike the JFET circuit, the PN junction in the BJT is forward biased and heavily influences what voltage will appear between R1 and R2 since it effectively puts a diode and Re in parallel with R2.

It is not obvious in your diagram because it is using a single supply and dividers. It is much more obvious in a diagram like this:

http://www.learningaboutelectronics.com/Articles/P-channel-JFET

https://www.tutorialspoint.com/basic_electronics/basic_electronics_jfet.htm

DKNguyen
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  • please see the edited version. – Sayan Jan 13 '21 at 16:06
  • i think they are saying the terminal which is down that is as source. – Sayan Jan 13 '21 at 16:12
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    @Sayan Yes, I believe you are correct that the more negative terminal is usually considered the source terminal in spite of the JFET's symmetry. That's an N-channel JFET with the main body being a hunk of N-silicone and gate being a hunk of P-silicon which forms a diode. That means if your gate voltage is more positive than the source voltage you forward bias the resulting diode and cause a short which blows the JFET. – DKNguyen Jan 13 '21 at 16:17
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    That's why they say JFETs need a negative gate-source voltage: So that the PN junction remains reverse biased. It's not very obvious that is what is going on in the circuit since you have resistor dividers and a single supply source. If the circuit is functional, then the dividers must be producing a more negative/lower voltage at the gate than at the source. – DKNguyen Jan 13 '21 at 16:19
  • This is much more obvious: https://www.learningaboutelectronics.com/images/N-channel-JFET-setup.png – DKNguyen Jan 13 '21 at 16:20
  • "In your circuit, the way the voltage dividers and resistors are set up, the gate voltage is lower than the source pin voltage". But whatever value you select for the voltage divider circuit, voltage drop in r_2 and r_s would have same sign. but if i use conventional current, still how v_gs is negative? – Sayan Jan 13 '21 at 16:31
  • If Vg = 1V and Vs = 2V (both measured relative to GND), is Vgs negative or positive? What if Vg = 2V and Vs = 1V? – DKNguyen Jan 13 '21 at 16:32
  • "so had to guess a polarity for each charged particle." Actually, no. Franklin thought in terms of fluids - charged particles weren't yet a thing. But other than that, you're right. – WhatRoughBeast Jan 13 '21 at 16:33
  • @WhatRoughBeast Modified wording – DKNguyen Jan 13 '21 at 16:34
  • @DKNguyen obviously negative. but according to conventional current why this time taking it as negative? then why v_be is not negative? recall that time too it could be 2v AND 1 V respectievly – Sayan Jan 13 '21 at 16:36
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    @Sayan Perhaps what you are missing is how the forward biased diode in the BE junction affects the voltage at the divider on the base? It's not being controlled just by R1 and R2 like in the JFET circuit. – DKNguyen Jan 13 '21 at 16:38
  • @DKNguyen so the fact I understand is for a transistor I can select the voltage across R_2 according to my choice, but the voltage across R_E would be "automatically" adjusted by the circuit so that V_be=0.7 volts. On the other hand if it is a JFET, I can select the voltage R_2, thats not a problem, but my next task would be to set a voltage actross R_s SUCH that this latter voltage is more than the former. But how can i set a voltage of my choice across R_s ? – Sayan Jan 13 '21 at 17:56
  • @Sayan This is getting out of scope of the original question but you would have to find an appropriate balance between all the resistors in the circuit. – DKNguyen Jan 13 '21 at 20:21