2N3904 Beginner Question

[Newbie]

Hello All

I'm learning about transistors and have a simple question to ask. I
have set up a 2N3904 as below

--- 5V
|
V LED
---
|
\
/ 220 ohm
\
|
| /
220 |
5V -/\/---|
|\
V
|
| Gnd

The circuit works fine and the LED lights up. The question I have is
when I take the 3904 out, reverse it, and put it back into the circuit
so that the emitter is where the collector was and vice versa the
circuit still works (the base does not change). The LED lights up,
grounding the base shuts it off as you would expect. My grasp of the
theory is not too good yet but why should it do this? I thought maybe
it had something to do with the Emitter base breakdown voltage (listed
at 6v) but I dropped the supply voltage to 3 volts and the effect
persisted. Replacing the 220 ohm base resistor with a 47K resistor
makes the circuit work in the first configuration but not in the
second.

Where is the current going? And how is it getting throught the
transistor? Is this expected behaviour if a transistor is reversed? I
would appreciate any light that anyone might shed on this.

Many Thanks
Dale

 

[John Larkin]

On Wed, 4 Feb 2004 21:31:22 +0000 (UTC), Newbie
<NoSpam@DontEmailHere.com> wrote:

Hello All

I'm learning about transistors and have a simple question to ask. I
have set up a 2N3904 as below

--- 5V
|
V LED
---
|
\
/ 220 ohm
\
|
| /
220 |
5V -/\/---|
|\
V
|
| Gnd

The circuit works fine and the LED lights up. The question I have is
when I take the 3904 out, reverse it, and put it back into the circuit
so that the emitter is where the collector was and vice versa the
circuit still works (the base does not change). The LED lights up,
grounding the base shuts it off as you would expect. My grasp of the
theory is not too good yet but why should it do this? I thought maybe
it had something to do with the Emitter base breakdown voltage (listed
at 6v) but I dropped the supply voltage to 3 volts and the effect
persisted. Replacing the 220 ohm base resistor with a 47K resistor
makes the circuit work in the first configuration but not in the
second.

Where is the current going? And how is it getting throught the
transistor? Is this expected behaviour if a transistor is reversed? I
would appreciate any light that anyone might shed on this.

Many Thanks
Dale

If you swap the emitter and collector of an NPN transistor, it's still
an NPN structure. Some very old germanium transistors were truly
symmetric... emitter and collector were interchangable.

A modern silicon transistor is not symmetric, but it still has
"reverse beta", finite current gain if e-c are swapped. This beta is
very low, on the order of 1, as opposed to normal beta (base-collector
current gain) that is more like 100.

John

 

[John Popelish]

Newbie wrote:

Hello All

I'm learning about transistors and have a simple question to ask. I
have set up a 2N3904 as below

--- 5V
|
V LED
---
|
\
/ 220 ohm
\
|
| /
220 |
5V -/\/---|
|\
V
|
| Gnd

The circuit works fine and the LED lights up. The question I have is
when I take the 3904 out, reverse it, and put it back into the circuit
so that the emitter is where the collector was and vice versa the
circuit still works (the base does not change). The LED lights up,
grounding the base shuts it off as you would expect. My grasp of the
theory is not too good yet but why should it do this?

The emitter and collector are both n type semiconductor and they lay
on opposite sides of the P type base region. Either can act as
emitter or collector. They have been doped differently (emitter
highly doped, collector lightly doped, to optimize the current gain
and breakdown voltage in one of these orientations. But it will work
with poorer specs in the other orientation.

I thought maybe
it had something to do with the Emitter base breakdown voltage (listed
at 6v) but I dropped the supply voltage to 3 volts and the effect
persisted.
That rating implies that if your supply was more than 6 volts, the

emitter acting as collector would not be able to turn off and hold
back the full supply voltage. The base to collector junction has
something like 40 volts of reverse breakdown capability.

Replacing the 220 ohm base resistor with a 47K resistor
makes the circuit work in the first configuration but not in the
second.

You just proved that reverse beta (current gain) is not as great as
forward beta. The high dopant concentration in the emitter is one of
the big factors that make the forward beta high. When you use the
lightly doped collector as emitter, most of the base current is holes
from the base crossing to the emitter (the functional emitter), rather
than electrons that would have been injected into the thin base layer
by a highly doped emitter. It is these electrons dumped into the P
type base layer that wonder into the collector region and get swept
out by the collector positive voltage that gives the transistor
current gain.

Where is the current going? And how is it getting throught the
transistor? Is this expected behaviour if a transistor is reversed? I
would appreciate any light that anyone might shed on this.

I hope I have gotten you started.

Many Thanks
Dale

--
John Popelish

 

[Newbie]

Thank you for your kind and well thought out replys to my question.
You have cleared up the issue considerably. I will go back to the
theory and see if I can clarify in my own mind exactly what the
electrons and holes are doing and where they are going.

Looking at the diagrams your explanations now make a good deal of
sense. One wonders why the textbooks make no mention of this. I guess
a reversal of the supply voltage through the transistor just doesn't
happen in "real life".

Many thanks
Dale

On Wed, 4 Feb 2004 21:31:22 +0000 (UTC), Newbie
<NoSpam@DontEmailHere.com> wrote:

Hello All

I'm learning about transistors and have a simple question to ask. I
have set up a 2N3904 as below

--- 5V
|
V LED
---
|
\
/ 220 ohm
\
|
| /
220 |
5V -/\/---|
|\
V
|
| Gnd

The circuit works fine and the LED lights up. The question I have is
when I take the 3904 out, reverse it, and put it back into the circuit
so that the emitter is where the collector was and vice versa the
circuit still works (the base does not change). The LED lights up,
grounding the base shuts it off as you would expect. My grasp of the
theory is not too good yet but why should it do this? I thought maybe
it had something to do with the Emitter base breakdown voltage (listed
at 6v) but I dropped the supply voltage to 3 volts and the effect
persisted. Replacing the 220 ohm base resistor with a 47K resistor
makes the circuit work in the first configuration but not in the
second.

Where is the current going? And how is it getting throught the
transistor? Is this expected behaviour if a transistor is reversed? I
would appreciate any light that anyone might shed on this.

Many Thanks
Dale