Dear Genius Ladies and Gentlemen,
Suppose two circuits of bjt connections (e.g. pnp, see attached figure below):
a) open base
The base is floating (no external leakage current).
The supply current is Iceo.
The base voltage Vb is settled to one diode below VDD,
let say Vb~2.5V at VDD=3V.
b) open emitter
Now, the emitter is floating (no external leakage current).
The supply current is Icbo:
Icbo=Iceo/(beta+1)
The question is:
To which voltage Ve is settled in the circuit b)?
Is it equal to VDD, to GND, or to what, and why?
I'm not sure that available models represent well such region of operation.
If the Ve is higher than Vb, the emitter behaves like a current source generating current Iceo. Iceo is >> Ibco, so Ve will drop towards GND.
Who will stop it at Vb?
collector injects electrons, thus Icbo is formed
some electrons rich the emitter and make its potential more negative
if Ve is < Vb, the emitter is inverse biased and a small reverse current will start to flow in its area, the emitter area is
Dear Dobromir,
let me ask a couple of questions:
1. It is well known that in a pn junction at termal equilibrium, the n region is at a higher potential than the p region. Moreover the space charge region is positive in the n-region and negative in p region, thus the built-in potential.
If you connect two probes and a voltmeter between the n-region and p-region, what potential difference you will measure?
2. If the emitter is floating and no current can flow trough it, the emitter-base junction is in termal equilibrium, isn't it?
3. Why do electrons flow towards emitter from base if the built-in potential between base and emitter it is built-in order to prevent it?
Hi Simone,
1. 0V for the external world (I=0, V=0, P=0)
2. emitter is floating, but it is not alone. it is a part from the transistor, and is in relations with another junction CB.
3. the CB is reverse biased and "consume" current (energy) from the supply. The collector is the most negative potential.
let talk for holes.
the holes from emitter area "jump to collector", making emitter more negative.
In practice the voltage will depend on how you measure it. Whether you measure from emitter to ground, or from emitter to Vdd (consider GND to be -3V, Vdd to be GND - perfectly legitimate!). In either case you are adding a load resistance and getting some current flow through the emitter. To be purist you would have to measure the potential that needed to be applied to the emitter in order to have a zero current flow through the emitter - so what happens if the transistor is n illuminated OCP71?
Justin, what is "OCP71"?
Ve direct measurement is impossible, because the currents are in fA-nA range.
So, introducing 10Mohm voltmeter resistance will fix the Ve.
Firstly Ve must be buffered (with mos transistor gate without ESD protection) and after that measured.
aaa OCP71 is a photo transistor...
phototransistor simply is a BJT transistor wherein CB junction is a photodiode.
the photocurrent is like Icbo, so when CB junction is illuminated Icbo will be increased hundred or thousand times, and the floating emitter will be stick to the collector.
Hi Dobromir,
2. But while emitter current = 0 and No other source of energy (illumination, etc.) feed the junction, the diode is in thermal equilibrium => 1. => same voltage of the base.
3. yes CB is reversed biased. But, why would the holes jump FROM emitter, if the built-in potentiial is high enough to make this net flow = 0 ? Holes could jump TO the emitter, with the effect of lowering the built-in potential with the result of causing an equal and opposite flow of holes that will null it.
I think that untill the emitter is not connected to something it is in termal equilibrium with the base, so Fermi level is constant. => 1.
Hi Simone,
In all cases, the experiments exclude any doubts in the theory of the knowledge!
The general purpose PNP BC560C was measured in the circuit below.
The emitter drops from 3V to 0.7V! What is the explanation?
In the simulation ...
Given J is open,
then what is I gs
then what is Power gs ?
If it is nano nano Watts,
then the Vbe may rise to some calculated number.
Energy is not destroyed in the "real world".
If "leakage is possible to the base"
then how is it "not to gnd" ?
In summary:
Are you assuming that "calculated Simulations"
are the same as "real world circuits" ?
Sounds like you are asking about two issues:
1) the simulation calculated response
2) the real world expectations for comparison.
This is a simulation, right ?
What is the Simulation programmed to do ?
Is is an infinite impedance (calculated response) meter ?
That would not be a real world result.
Are you assuming that "calculated Simulations"
are the same as "real world circuits" ?
This is the bane of my projects,
starting from calculations and Simulations,
and then bread-boarding the circuit
to bring in real world signals where Vnoise = Vsignal.
Then the real story begins.
Glen
Glen,
I'm sorry but I can not understand what you are wanting to say me.
The calculations are from real measurement and not from simulation!
The emitter drops towards GND and the question is why?
Dob,
"The calculations are from real measurement and not from simulation!"
To me, they appeared to be from a Simulator.
Thank you for clarifying. My post is not helpful.
Glen
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