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I am an engineering student and this question is one of my assignments this week and I need a more professional opinion.

I don't see how the transistor is even operational with the base connected to ground. With no voltage applied to the base to overcome barrier potential both collector-base and emitter-base diodes will remain reverse biased. Also collector current (\$ I_{C} \$) is labeled on the emitter side and emitter current (\$ I_{E} \$) is labeled on the collector side.

I'm not sure if this is a trick question or if I am over looking something. Am I wrong about my assessment of the question?

  1. For the circuit shown in Figure 3 below, draw the DC load line and locate its quiescent or DC working point. Show all work. week 4 assignment diagram C

Picture of the transistor circuit

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  • \$\begingroup\$ Is Vee +30 V or -30 V? \$\endgroup\$ – The Photon Feb 25 '17 at 18:29
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    \$\begingroup\$ Yes, you missed something. Vee is -30, which is below ground. Therefore the base collector (BC) junction is forward biased. The base emitter junction is reverse biased. Since this is a school problem, I will just ask you, what is the mode of operation of a bipolar transistor when the BC junction is forward biased, and the BE junction is reverse biased? \$\endgroup\$ – mkeith Feb 25 '17 at 18:33
  • \$\begingroup\$ After looking closely it is -30V \$\endgroup\$ – 1fastk Feb 25 '17 at 18:33
  • \$\begingroup\$ With the BC forward biased and the BE reversed bias a large amount of current wont be allowed to flow from the collector to the emitter which would cancel the amplification of the transistor. Effectively making the circuit useless really. \$\endgroup\$ – 1fastk Feb 25 '17 at 18:49
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    \$\begingroup\$ @1fastk Look up 'reverse active' mode. \$\endgroup\$ – jonk Feb 25 '17 at 19:43
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It is possible that the circuit is just drawn wrong. Is a mistake. But more likely, this is an example of a transistor being operated in the reverse active mode. In reverse active mode, the collector and emitter are basically swapped. Compared to forward active mode, reverse active mode has much lower beta and usually is considered to be not very useful.

Legends circulate that in the old days, BJT's might be used in reverse mode as switches. The saturation voltage (Vec) could be as low as a few mV, which could be useful for voltage sensing or other analog applications.

See also: BJT in Reverse Active Mode of Operation

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2N1711 is an NPN transistor. As you have stated, the currents are marked at wrong places. Along with that, the voltage source names must also be interchanged. I mean, replace VEE with VCC and vice-versa.

Coming to the point:

I don't see how the transistor is even operational with the base connected to ground. With no voltage applied to the base to overcome barrier potential both collector-base and emitter-base diodes will remain reverse biased.

No. Not necessary. To make an NPN transistor work, what you need is a positive voltage difference across Base-Emitter junction which is actually present (but after proper labeling). If VEE and VCC are interchanged (also Ie and Ic), the potential difference across Base-emitter junction = Voltage at Base - Voltage at Emitter = 0-(-30) = 30V (not exactly as I omitted the voltage across 50KOhm resistor). In is not the voltage that matters, it is actually the voltage difference(potential difference)

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    \$\begingroup\$ Since it is an NPN transistor, shouldn't the collector (N-type) and emitter (also N-type) be interchangeable? So does the OP really need to reverse the transistor? I am asking questions to avoid giving away too much information, since this is a school problem. \$\endgroup\$ – mkeith Feb 25 '17 at 18:42
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This is an NPN bi-polar transistor: The circuit shown has the Emitter MORE POSITIVE than the Base. THEREFORE THE TRANSISTOR IS OFF. PERIOD.

The Base-Collector junction is forward biased. This is something that happens when using this type of transistor in a switching mode and the Base is driven very hard: saturation. This is generally not a good idea because it takes a while to get the transistor out of saturation. So the load line that may be requested here is a diode (CB junction) with RB as the resistance and VEE as the bias. I will take a wild guess that there is a "polarity" mistake/typo with VEE or VCC. I have been designing analog circuits for 47 years and have also taught at the university level. I would call this a very silly circuit/problem. No wonder is is confusing to you.

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    \$\begingroup\$ Not off. It is in reverse active mode. All three terminals of the transistor will have non-zero current. Not sure how you would call that "off." During the 47 years of teaching at university level, did you ever crack open a textbook that covered "reverse active mode?" \$\endgroup\$ – mkeith Feb 28 '17 at 8:00
  • \$\begingroup\$ But it is very possible that the professor meant to draw a PNP. Problem would make more sense that way. \$\endgroup\$ – mkeith Feb 28 '17 at 8:03
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Is this NPN with base grounded and collector at negative voltage functional?

depends on your definition of "functional". it is not going to be an amplifier. But possibly an oscillator.

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