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So I am little weak when it comes to circuits and i am confused in transistors. There are two diagrams for an npn transistor and both seem different.. enter image description here

enter image description here

Now in the first picture the base is connected to positive terminal through Vee and negative terminal through Vcc. How is this possible? In the second diagram however the base is only positive.

1) If the first one is correct. Then how does the base know to choose the + and become forward bias for base-emitter and when to choose negative to become reverse biased for base-collector?

2) If the second one is correct then how is the base-collector reverse bias as they are both connected to +ve terminal. The depletion region may still widen because the electrons in collector are attracted by the positive terminal (Vcc) and the positive ions left behind will repel the positive holes from the base. Also the +ve terminal (Vee) is also repelling the holes so how does this work out?

3) If both are correct then this means both diagrams are equivalent? How is this possible? Do note i am a computer science student so try to give the answer in simple words.

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  • \$\begingroup\$ The problem with the first diagram is that the emitter is at the right side and the collector at the left side. No problem - everything correct, but it is a rather uncommon arrangement. \$\endgroup\$
    – LvW
    Apr 6, 2016 at 11:46
  • \$\begingroup\$ Both are correct. The first diagram illustrates "common base" or "grounded base" configuration, the second "common emitter". (which you will see more often) \$\endgroup\$
    – user_1818839
    Apr 6, 2016 at 12:18

2 Answers 2

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Both circuits are same. I mean the whole effect produced will be same

Firstly, it is meaningless to talk about just voltage. It is a relative quantity.

Let me tell in some other context. The statement "I am moving at 10Km/hr w.r.t to the cycle behind me" has more and clear information than saying "I am moving at 10Km/hr". Of course, in the later statement, the speaker is moving at 10Km/hr w.r.t the road. The phrase "w.r.t the road" is hidden here and is understood without explaining explicitly.

If you have two points say A and B on a circuit(any two points), it is meaningless to talk about potential of point A. Instead it must be "potential of point A w.r.t point x is .......". Remember potential difference matters and not just potential. You use two terminals of a voltmeter to find voltage across a device. Not just one terminal.

Note: In all the cases where one says just "Potential of point A is ......", he/she means potential of that point A w.r.t another point in the circuit which as fixed voltage zero(GND). Just like how "w.r.t the road" is hidden in the previous case.

So if you take a battery of 1V, it means that, the negative terminal is at potential 'x' Volts(some random value - we haven't defined our reference)and the positive terminal is at 'x+1" Volts. So on the whole, the battery is {(x+1)-x}=1V.

One necessary point for a BJT to conduct is that potential difference across Base-Emitter (Vb-Ve >0.7 V for Silicon) must be more than 0.7V.

1) As long as VEE>0.7V in first circuit and VB>0.7V in second circuit, the BE junction will be forward biased.

2) Refer 1)

3) Yeah both are same

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  • \$\begingroup\$ Ok it was somewhat helpful for clearing my voltage concept but there is still something unanswered. Check my comment on Jlm's answer. Thanks \$\endgroup\$
    – gallickgunner
    Apr 6, 2016 at 13:57
  • \$\begingroup\$ Hm, yeah both are connected to positive terminals. So what? As long as the collector terminal is more positive than base terminal, base-collector junction will be reverse biased in case of npn transistor. I told you, remember, the difference of potential matters. If Vc-Vb > 0, then base collector is reverse biased. Fir eg, Vcc=10 V and VB=1v would do that. \$\endgroup\$
    – user3219492
    Apr 6, 2016 at 14:06
  • \$\begingroup\$ Ok that's what i was expecting. But when i think by visualizing i am still confused. IF both are connected to positive terminals. Then collector electrons get attracted by the positive terminal and base holes are repelled by the +ve terminal while in a proper reverse bias the holes are attracted to widen the depletion region \$\endgroup\$
    – gallickgunner
    Apr 6, 2016 at 14:12
  • \$\begingroup\$ You are doing the same mistake I guess. Assume two people are pulling you in either side. One with 1 N and the other with 19 Newtons of force. Both are pulling you, but the resultant is 9 Newton in the direction of 10 Newton force. Similarly, holes will move towards the battery connected to the base inspite of whether it is positive terminal or not. Remember that the positive terminal if the battery VB is negative w.r.t positive terminal of VCc \$\endgroup\$
    – user3219492
    Apr 6, 2016 at 14:35
  • \$\begingroup\$ Wait... wouldn't the resultant force be 18N in the direction of 19N of force? That aside your last line is what i don't get. "The +ve terminal of Vb is -ve w.r.t +ve terminal of Vcc" How? I know that Vcc is of higher potential and Vb is of low but being low potential doesn't make it negative ( or does it?), it's still connected to +ve terminal. I know that electrons tend to move from low potential to high potential which means holes must move from high to low.. so thinking this way it makes sense that the holes will be attracted by +ve terminal of Vb. \$\endgroup\$
    – gallickgunner
    Apr 6, 2016 at 15:57
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Both are correct.

In diag 1 the batteries are not 'real' they simply represent the polarity of the voltages present at the base with respect to the collector and emitter.

In other words the emitter is more negative than the base and the collector is more positive.

In diag 2 the batteries are real with the resistors limiting the current into the base and through the transistor.

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  • \$\begingroup\$ Ok, if the diag 2 has real batteries then how does reverse biasing in Base-Collector works. Both are connected to the positive terminal. \$\endgroup\$
    – gallickgunner
    Apr 6, 2016 at 13:56
  • \$\begingroup\$ @wandering-warrior, look closely. The second diagram shows multiple cells for Vcc. \$\endgroup\$
    – The Photon
    Apr 6, 2016 at 14:50

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