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I have a simple question regarding the posted image.

Isn't the potential at point B = 0 or grounded?

In that case both the emitter and base will be effectively at ground and I don't see how the transistor would work then?

What am I missing?

enter image description here

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    \$\begingroup\$ Ground is where YOU define it to be. \$\endgroup\$
    – Rev
    Commented Aug 26, 2016 at 9:58

2 Answers 2

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There are many, many questions about ground on this website so I will be brief: Voltage is always measured between two points in a circuit. Ground, or 0, in a circuit like this is precisely where you want it to be - usually where it makes sense. Let's redraw your circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Does that make more sense? You can only pick one node in this circuit to be your ground, but you can pick any node. I have drawn it so that the emitter node would be the most reasonable place, because then the other nodes will be positive in reference to that point.

As you can now see, it is now quite clear that the base and the emitter is at different potentials.

A lot of confusion comes from a lousy schematic layout. Sometimes it is because the lecturer has limited experience actually designing real world circuits, but in this case I think it's deliberate - they want you to think about the current flow, and the relative voltages between every node. Mission accomplished. They got you confused, you started to think, and hopefully your first thought next time you find something like this is to redraw it so that it's sane: more positive nodes on top, inputs to the left, outputs to the right, current flows from top to bottom and left to right.

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  • \$\begingroup\$ Isn't your Avatar the very first WinUAE Icon? Good memories (no pun intended). \$\endgroup\$
    – Rev
    Commented Aug 26, 2016 at 9:59
  • \$\begingroup\$ @Rev1.0 Close enough I guess, it's the logo that's printed on the Amiga 1000 case. It's quite possible that it was used in WinUAE, but I mostly use the real deal. ;) \$\endgroup\$
    – pipe
    Commented Aug 26, 2016 at 10:02
  • \$\begingroup\$ Okay but isn't the base also in contact with $V_CC$ 's negative or ground plate? My thinking is that the minus means 0 potential whereever it is in the circuit. \$\endgroup\$
    – Weezy
    Commented Aug 26, 2016 at 10:57
  • \$\begingroup\$ @Weezy Nope. Minus just means the more negative part of the battery. Look into every device with AAA-batteries. They are connected in series to get a higher voltage. Obviously, the negative side of the "top" battery can then not be ground. \$\endgroup\$
    – pipe
    Commented Aug 26, 2016 at 11:00
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    \$\begingroup\$ @Weezy Exactly. It's important to remember that it's all relative. You only know that the base is at a potential that is Vee volts higher than the emitter, and that the load resistor is at a potential that is Vcc volts higher than the base. The voltage between the emitter and the top of the load is then Vee+Vcc volts. That's basically all that the circuit tells you. A battery or voltage source does not have an "intrinsic" ground. \$\endgroup\$
    – pipe
    Commented Aug 26, 2016 at 11:11
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I think you aren't very clear about how a BJT, or for that matter, an electric circuit functions. I suggest you to place a BJT device, instead of the symbol of BJT there, and then analyze the functioning.

To answer: The base is connected to the node B, and we can use it as reference; hence the base voltage is zero. You are right till there. But, see that the emitter and the node B are not directly connected; there's a voltage source between them. Hence, clearly, the emitter is at different voltage (negative, here) with respect to node B. The transistor would perfectly work. If you are wondering, then this is an npn-transistor, so try reading theory particularly of this device to perceive better.

I am still posting a picture for reference, but for an pnp-transistor to show that the circuit actually works. A similar nature is exhibited by npn-transistor. If you still face a confusion, let me know.

enter image description here

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  • \$\begingroup\$ I can't see any 'ground' in that circuit diagram, so I don't think "we can safely assume it [the base] is grounded"? \$\endgroup\$
    – Icy
    Commented Aug 26, 2016 at 10:35
  • \$\begingroup\$ @Icy Very true. In most of the books beginners start with, the base of the pnp or npn is generally grounded to ease the understanding of the reader. But, people tend to take the case for granted in every circuit, like the questioner here took. But the major blunder and confusion to be taken into cognizance here is the questioner's consideration of emitter voltage to be zero. So, I just tried to keep the focus there. But, yes, the node B can't be grounded, unless specified. \$\endgroup\$ Commented Aug 26, 2016 at 10:43
  • \$\begingroup\$ In many (probably most) NPN transistor circuits I see, it is the emitter not the base that is connected to the lowest potential - sometimes but not always, this is marked as 'ground' \$\endgroup\$
    – Icy
    Commented Aug 26, 2016 at 10:50
  • \$\begingroup\$ @Icy Yes, so you got it right? To beginners, they generally ground a node to clear the concept. Moreover, they start with grounding the base because common-base configuration is the easiest or most tangible to understand once you have learnt the basic internal functioning of a BJT. I've seen a lot of people stuck in this quagmire and I know where it generally comes from, hence I was trying to put it from that perspective. \$\endgroup\$ Commented Aug 26, 2016 at 10:57

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