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I was wondering what happens when VCE voltage becomes negative? suppose I am forwarding both junctions (no resistors). I know that saturation happens when VCE is between 0.1V- 0.3V but what happens below that range (V1 > V2) as in the following scheme?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Your question cannot be answered unless you tell us what you do with the base terminal. In an NPN Common Emitter configuration taking down the Collector will mean the Base-Collector junction will go into forward mode. Also, how is making Vce negative useful in any way? \$\endgroup\$ – Bimpelrekkie Jan 12 '18 at 14:51
  • \$\begingroup\$ @Bimpelrekkie Sorry, edited, (base is grounded). I don't know whether it's useful actually, I just don't know what would happen then. \$\endgroup\$ – muhzi Jan 12 '18 at 14:55
  • \$\begingroup\$ Draw a circuit of the actual situation the NPN is in using ground and voltage sources. A grounded base is not a CE circuit so still unclear what you're after. You could also just picture the NPN as two diodes in anti-series, then if one PN is shorted (for example Base-Emitter) then Collector-Emitter will simply behave as a silicon diode. \$\endgroup\$ – Bimpelrekkie Jan 12 '18 at 15:02
  • \$\begingroup\$ @Bimpelrekkie I think that's what I mean (Edited). Anyway, I'm a total noob at solving the circuit. \$\endgroup\$ – muhzi Jan 12 '18 at 15:29
  • \$\begingroup\$ Not exactly what you asked, but you may want to read about the "reverse active" mode of transistor operation. \$\endgroup\$ – mkeith Jan 12 '18 at 15:48
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The definition of saturation is that both junctions be in forward bias. So, with appropriate voltages for V1 and V2, your transistor would be operating in the saturation region. It would be important for the voltages to be not too large, otherwise very large currents would flow and destroy the device.

The circuit is difficult and confusing because of the way it is drawn. For example, you chose to label the common positive node of the two supplies as GND. It may have been better to leave GND out of it.

Depending on the specific voltages of V1 and V2, net current flow might be from emitter to collector or collector to emitter. Because of the differences in the two junctions, detailed predictions about current and voltage are somewhat difficult.

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  • \$\begingroup\$ I think I understand ur point. But, would adding a resistor in the collector branch help in the case of V1 and V2 being too large? like making it in proper saturation mode? \$\endgroup\$ – muhzi Jan 12 '18 at 16:30
  • \$\begingroup\$ Yes, that should help. \$\endgroup\$ – mkeith Jan 12 '18 at 23:53
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Assuming that V1, V2 > 0:
Both internal diodes (from B to E and from B to C) are forward biased and will act just like discrete diodes. If the voltages are high enough (and that doesn't have to be much above 0.6V) you will just destroy the transistor.

BTW: the title of the question "Negative voltage for Vce - BJT NPN" doesn't fit to the schematic: Vce would only be negative if V1 > V2.

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  • \$\begingroup\$ Thanks for your answer. yeah, I wrote that (V1 > V2) in the question to clarify that. I'd like to ask though, by putting a resistor in the collector branch. now does the transistor adjust itself to get to the value of Vcesat and rest of voltage goes on that resistor? \$\endgroup\$ – muhzi Jan 12 '18 at 15:48
  • \$\begingroup\$ V1 > V2 and a resistor in (any) of the branches doesn't alter the fact that both diodes are forward biased. So basically the same answer... the transistor will be broken even if only the BE diode is destroyed. \$\endgroup\$ – Curd Jan 12 '18 at 15:52
  • \$\begingroup\$ Yes I know, I'm trying to understand whether putting a resistor in such situation would be a sort of protection for the transistor not to go below the saturation voltage. \$\endgroup\$ – muhzi Jan 12 '18 at 15:56
  • \$\begingroup\$ How can it be a protection if the transistor will still be broken? \$\endgroup\$ – Curd Jan 12 '18 at 16:00
  • \$\begingroup\$ why would it be broken in that case if I divide the voltage so that not all of it drops on the junctions? making it in proper saturation? \$\endgroup\$ – muhzi Jan 12 '18 at 16:26
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schematic

simulate this circuit – Schematic created using CircuitLab

Redrawn with proper conventions for orientation in a logic diagram.

In your question, Both BE and BC are forward conducting diodes, not the same, with different Vr breakdown characteristics when reverse biased due to doping differences.

...meaning yours is not logical.

below correct application as a switch shows both junctions with forward voltage but Vbc has a forward drop is due to transistor function with collector current.

We consider both junctions saturated and thus Vce=Vce(sat) saturated.

schematic

simulate this circuit

A transistor does not behave like 2 diodes yet the forward voltage differences indicate Vce{sat} at rated current and often rated at Ic/Ib=10.

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  • \$\begingroup\$ you mean, of no use in practical purposes, right? \$\endgroup\$ – muhzi Jan 12 '18 at 15:54
  • \$\begingroup\$ Correct. No practical use. and violates spec. \$\endgroup\$ – Sunnyskyguy EE75 Jan 12 '18 at 15:55
  • \$\begingroup\$ @Tony Stewart. EE since '75: airomyst said he want's to add a resistor in the collector path \$\endgroup\$ – Curd Jan 12 '18 at 15:57
  • \$\begingroup\$ Yet when using correct supply polarity BC junction is almost same forward voltage but current flows from Collector to Emitter for this NPN. \$\endgroup\$ – Sunnyskyguy EE75 Jan 12 '18 at 16:05
  • \$\begingroup\$ Vce = V2-V1. Vbe = V2. So if V2 is 0.55V, and V1 is 0.35V, then Vce = 0.2V, and the device may very well be in saturation, with a net current flowing from collector to emitter. As V1 is increased from 0.35, at some point, the net current flow will stop and reverse and be from emitter to collector. \$\endgroup\$ – mkeith Jan 12 '18 at 20:02
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When we apply a positive voltage between collector and emitter a tiny depletion region forms between the np junction of collector and base and the pn junction between base and emittor is biased directly and electrons are driven from emitter to the collector. I think that in case of applying the reverse voltage the depletion region between collector and base disappears allowing the electrons of n side of collector to move down toward emittor, on the other hand the electrons of the heavily dopped n , in the emittor, is flowing up side to the collector! Therefore two currents want to flow inversely. I think the sum would be a really small current which is not considerable. So, I think no current will flow!

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