Transistor OFF but Voltage at emitter same as voltage at collector

[FreshmanNewbie]

This is simple simulation below of an NPN transistor :



In the above transistor, without giving any base voltage, whatever voltage I give at the collector of the transistor, I am getting at the emitter. The voltage drop across the resistor is ZERO.

From my understanding, the transistor is turned OFF when there is no base-emitter voltage. And from the above simulation, the base and emitter terminals of the transistor are floating. But still the voltage appears at the bottom end of the resistor. How is this possible? What is happening inside the N-P-N regions of the transistor for this behaviour to happen?

And If I connect the resistor to ground, there is a very small current of 10pA, flowing through the resistor? How is this possible and how can any current flow, when the transistor is supposed to be OFF (as there is no defined base-emitter voltage applied)

Can someone please explain?

 

[andre_luis]

A 'real world' transistor is not an ideal switch.
Most simulators attempt to model them as close as possible from the real device, which means even parasitics leakages are considered.
Since the open ended net has a load close to infinite, it is expected that any voltage from the collector side would be present at the emmiter side.

 

[d123]

Hi,

Not quite, IMO. The NPN is on as the base is floating, with no pull-down the NPN will conduct.

Reading the first chapters might answer your questions. I plan to re-read this to absorb all the info.

 

[betwixt]

From my understanding, the transistor is turned OFF when there is no base-emitter voltage. And from the above simulation, the base and emitter terminals of the transistor are floating. But still the voltage appears at the bottom end of the resistor. How is this possible? What is happening inside the N-P-N regions of the transistor for this behaviour to happen?

A transistor, when not biased into conduction still leaks a tiny current (Icbo in it's specification) therefore it still looks like a high value resistor to a DC circuit. Now when you look at your schematic and replace the transistor with a resistor (of any value) you will see with no current flowing there is no voltage drop.

Similarly, when you ground the emitter resistor, leakage current still flows. In Silicon transistors the leakage is normally quite small, normally uA or less, in Germanuium transistors it is usually much higher, several mA for big power devices. The leakage is also highly temperature dependent.

In physics terms, the leakage is mostly caused by impurity in the materials and by stray radiation displacing electrons.

Brian.

 

[KlausST]

Hi,

add a load.

without a load the simulation makes no sense. No current measurement and no voltage measurement makes sense.
With a path to GND (via load) you will see meaningful voltages and currents

When the base is not connected it is "floating" somehow. Avoid floating nodes. General rule for electronics design: don´t leave any (unused) input floating. This is true for analog inputs as well as digital inputs. Also for discretes (especially FETs) and ICs.
While for most bjt circuits a floating base is not critical (just cuses increased, b sill low emitter current) it may be critical for FETs and ICs.

The bjt datasheet shows a "cutoff current" where "V_BE = 0" is mentioned. Is is very low but may rise by decades when getting hot.
The "floating base cutoff current" may be much higher, because of (amplified) collector->base currents.

Klaus