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Suppose when the collector is connected we get \$I_b=10 mA\$, and \$I_c=1000mA\$, so naturally we have \$I_E=1010mA\$.

What will happen when the collector is disconnected? Will the base current change from 10 to 1010mA? In other words, can we say the emitter current is solely determined by the base configuration? In this case it is V_BB-0.7/1K (assuming the transistor is not operating in the breakdown region).

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    \$\begingroup\$ In this case, what you have is IB = IE and an "ordinary" diode voltage drop between base and emitter. \$\endgroup\$ – G36 Dec 20 '20 at 17:56
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    \$\begingroup\$ So when I connect the collector just this current base current would reduce , and the collector would draw the excess current right? \$\endgroup\$ – Sayan Dec 20 '20 at 17:59
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    \$\begingroup\$ That's 1010V across the emitter resistor! So the base is still supplied from a HV supply; the emitter voltage can't drop much; neither can the current, so unless you also reduce the 1.01kV base bias, it must all come from Ib. But this is not a practical circuit. \$\endgroup\$ – user_1818839 Dec 20 '20 at 18:01
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    \$\begingroup\$ Yes, when you connect a collector then the BJT can "amplifier" the current, thus the collector would draw the excess current. But do not be confused and assumed that the base current is being somehow magically magnified to form the collector current. This is not the case (not true). What is happening is that the base current is controlling the amount of current that collector-emitter drawn from a supply source. electronics.stackexchange.com/questions/355899/… \$\endgroup\$ – G36 Dec 20 '20 at 18:05
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    \$\begingroup\$ upload.wikimedia.org/wikipedia/commons/7/73/… \$\endgroup\$ – G36 Dec 20 '20 at 18:09
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The emitter current is (Vbase - 0.7)/1k with or without the collector connected to a positive supply voltage (assuming the battery has zero internal resistance).

With the collector connected, this emitter current will be split in the ratio of hFE (beta) between the base and collector.

Now disconnect the collector supply and the emitter current doesn't change (assuming an ideal battery which has zero output resistance) but now all the emitter current is pulled from the battery through the base. There is no collector current and the emitter current has remained unchanged throughout.

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    \$\begingroup\$ There will be some difference in the voltage drop across the base-emitter junction because the current through the junction is quite different in the two cases. So there will be some difference in the emitter current. \$\endgroup\$ – Circuit fantasist Dec 20 '20 at 23:24
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    \$\begingroup\$ But usually we assume the base-emitter junction voltage 0.7 volts if the BE junction is working correctly. \$\endgroup\$ – Sayan Dec 21 '20 at 6:39
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    \$\begingroup\$ @Sayan, That would be (absolutely) right but if the current through the junction was (absolutely) constant. This is not the case here - once it is 10 mA and other times 1010 mA. So there is some although small voltage difference. Look at the input transistor IV curve what is nothing else than an IV curve of a diode. Do you have an interest in a more general and in-depth explanation of the circuit? \$\endgroup\$ – Circuit fantasist Dec 21 '20 at 9:19
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    \$\begingroup\$ @Circuitfantasist I did forsee that there was a possibility of someone picking me up on the point about Vbe changing with Ib but took the risk not to mention it in order to keep the answer simple and clear rather than introducing the added complexities of Ebers-Mol, Early effect etc. Obviously, when providing an answer, the responder must decide upon the complexity of the information to be given. I hope in this case I didn't make things too simple in order to communicate the main concepts of the situation in assuming that Vbe is equal to 0.7V. It is a good approximation in many situations. \$\endgroup\$ – James Dec 21 '20 at 10:24
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    \$\begingroup\$ @ Circuit fantasist so you want to say there should be slight increment tn the emmiter current after adding the collector circuit? my point was whether connecting the collector would affect the existing emitter current which was before when the collector was not connected \$\endgroup\$ – Sayan Dec 21 '20 at 10:28
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To complement James' answer, it might be helpful to approach the open collector condition by gradually increasing a collector resistance connected to a positive power supply.

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Here I have used a reasonably small Vb and an initial value for Rc that won't cause the transistor to fall into saturation. With these values, we set the emitter current via Re to the (approximate but reasonable accurate) value of (5-0.7)/1k = 4.3 mA, which will give a nearly equal collector current through Rc. Voltage drop across Rc=8 ohms will be a handful of mV and Vce will be nearly 7.7V. The transistor is in active zone and hFE is the full value 200, so Ib needs to be only 20 uA. Basically, all the emitter current is due to the collector current.

When we increase Rc, the transistor will do its job to keep the current in Rc almost constant as long as it stays out of saturation. But we cannot force 4.3 mA forever into increasing values of Rc. At one point the voltage drop caused by Rc*4.3mA will exceed the supply voltage minus the voltage drop across Re. The transistor cannot drop Vce under zero volts - actually it will reach 0.1V or so, so the maximum value of Rc for which we can pull the current 'programmed' via Ie is about (12V-4.3V)/4.3mA = 1.8 kohm (rough estimate).
After that, we will assist to the demise of the transistor action, and the reinstatement of Ohm's law in all its glory.
Increasing Rc will decrease Ic as in (Vcc-fixed value)/Rc, and we can see this trend in the following graph

enter image description here

The decrease in the ratio Ic/Ib is reflected by the forced hFE that characterize saturation.
Since Ie is roughly set by the BE mesh, the decrease in Ic will be associated with a corresponding increase in Ib. In the limit for Rc->infinity, all the emitter current will have to be supplied by the base.

We can try to visualize how this plays out on the output V-I characteristic of the transistor if we acknowledge the fact that being Ie constant, the term Re*Ie is also constant and we can subtract it from the expression of the load line. As Rc varies we will see this set of curves

VI chars and quasi-load line

where the operating point, intersection of the output characteristic with the increasingly tilted load line, is seen moving from the active region into to the saturation region, down to the point where Ic = 0 and all Ie is contributed by Ib. Notice that we are crossing characteristics here: we do not stay on the same Vce-Ic curve for a fixed Ib as would happen with a common emitter configuration with base limiting resistor, but as Ib increases, we move to different characteristic curves.

A comparison with the CE configuration will be given in your other question: BJT current distribution.

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    \$\begingroup\$ I have suggested a small edit; I hope this does not offend you. BTW I suppose that it would be interesting for OP to see how "the transistor will do its job to keep the current in Rc almost constant when we increase Rc"... \$\endgroup\$ – Circuit fantasist Dec 21 '20 at 9:40
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    \$\begingroup\$ I know that the transistor would try to push a constant current I_c until R_c*I_c is exceeding the collector supply voltage, isnt it? \$\endgroup\$ – Sayan Dec 21 '20 at 9:45
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    \$\begingroup\$ @Circuitfantasist no problem, but all I see is the correction of a letter. Maybe later I will add a picture showing how the operating point slides down the knee into saturation as Rc increases. \$\endgroup\$ – Sredni Vashtar Dec 21 '20 at 9:45
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    \$\begingroup\$ @ Sredni Vashtar, Thanks... this is an example of a goodwill; we need it here. I have added also "minus the voltage drop across Re" to "the voltage drop caused by Rc*4.3mA will exceed the supply voltage". Your circuit is very suitable for explaining the mechanism for maintaining a constant current by the transistor through the concept of "dynamic resistance". Judging by the OP comments, I think this will be useful ... \$\endgroup\$ – Circuit fantasist Dec 21 '20 at 9:58
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    \$\begingroup\$ @Circuitfantasist oops, you are right, I had forgotten about that. I corrected the estimate for maximum Rc before reaching saturation. (I will add the rest after a few hours sleep). \$\endgroup\$ – Sredni Vashtar Dec 21 '20 at 10:26
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You're correct; in this case the base current would change. With the collector source added, assuming a collector resistor is also added, this would become what's known as a common-base amplifier, unusual at low frequencies but a common sight in RF electronics, at least those circuits that still use BJTs.

This answer originally said the exact opposite, because I am apparently too tired and misread the circuit; I thought there was a base resistor.

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    \$\begingroup\$ How? If you apply the KVL at the base side when collector is connected. You would get I_E=1010, after disconnection of collector, you basically do not changing the base circuit, so after applying kvl, you would again get I_E=1010, even the collector is not connected. Am i missing anything? \$\endgroup\$ – Sayan Dec 20 '20 at 18:10
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    \$\begingroup\$ If base and collector are connected to supply voltage and the current gain is \$\beta =100\$ and the collector current is 1000mA. Thus, the base current is 10mA and the emitter is 1010mA. But if you disconnect the collector the base current will increases its value to 1010mA thus to have Ie = Ib \$\endgroup\$ – G36 Dec 20 '20 at 18:16
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    \$\begingroup\$ @Sayan Ah, you're actually correct. I missed that there was no base resistor, an unusual configuration for a BJT. \$\endgroup\$ – Hearth Dec 20 '20 at 18:19
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    \$\begingroup\$ @Sayan The emitter current will be constant as long as you don't do anything to change the base-emitter voltage, yes. Obviously, changing Vb or the resistor will change the current. \$\endgroup\$ – Hearth Dec 20 '20 at 18:28
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    \$\begingroup\$ I am trying to see a "common-base amplifier" here... but I can not... \$\endgroup\$ – Circuit fantasist Dec 20 '20 at 23:31
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Тhe answers so far are perfect but specific; they explain how the circuit is made. For the true understanding, however, this is not enough. It is also necessary to explain why (for what purpose) this was done in this way and to consider its applications in practice. I will try to satisfy this need with my answer below.

The idea

I was pleasantly surprised to see that someone, willingly or unwillingly, came up with the same idea as me many years ago when I used it to make a battery backup power supply. It is based on an inherent BJT property sometimes causing strange problems - when the base-emitter junction is fully on, it actually connects, like a forward biased diode, the base to the emitter... and the input voltage source is directly connected to the emitter. This situation can happen in an emitter follower (such as here), if the emitter ceases, for some reason, to follow the base. Usually, this is an undesired situation but there are also its useful applications like in the link above.

So, the OP's question is about an emitter follower considered in two situations - without supply voltage ("damaged") and with supply voltage ("restored"). Considering them one by one, we can explain the meaning of this circuit when used as a buffer.

I will use two figures from the link above to illustrate my explanations. There the invisible electrical quantities are visualized as follows. Voltages are represented by vertical segments (voltage bars) with proportional height in red. They are summed (subtracted) geometrically, according to KVL (this clearly shows the relationship between voltages). Thus the set of voltage bars on the circuit diagram can be considered as a snapshot of the voltage relief. Current paths are shown by closed lines (current loops) starting from the positive terminal of the power supply and ending at its negative terminal. Current magnitudes are hinted by the line thickness.

"Damaged" emitter follower

If there is no supply voltage (the collector is unconnected), there is no emitter follower since the transistor is not a transistor but rather a diode - Fig. 1. Only its base-emitter junction is used as such a "diode"... and the input voltage VB is transferred directly through it to the load in the emitter (the 1 k resistor in the OP's picture).

Battery supply

Fig. 1. An emitter follower without supply voltage

Since the battery supplies the load through the base-emitter junction the whole load current IL flows through the junction. As though the input voltage source is directly connected to the load (only a small voltage drop VBE is subtracted from the input voltage). Usually, this is not a desired situation since the load can be too heavy for the input voltage source and can overload it.

"Restored" emitter follower

So let's restore the supply voltage (connect the collector) - Fig. 2.

Battery backup supply

Fig. 2. An emitter follower with supply voltage

The transistor "becomes a transistor" and begins acting as a device with negative feedback... or as a "being" implementing some goal. "Seeing" the significant base-emitter voltage, it begins increasing its collector current (IL) so the voltage drop across the load (VL = IL.RL) increases. Since the base-emitter voltage is a difference between the constant input voltage VB and the increasing output voltage VL, it begins decreasing (VBE = VB - VL)... and finally the equilibrium is reached by the negative feedback. VL will be (slightly) higher than in the case of the "damaged" follower; so the load current (IL = VL/RL) will be slightly higher as well.

Now let's look at the situation from another perspective. The transistor passes its collector current through the load in parallel to the base current thus "helping" it. The input battery provides only the small base current needed to set the beta times higher collector current and, accordingly, the load current needed. As though, the battery tries to supply the load by injecting the small base current through the base-emitter junction to the load... and the power supply, by the help of the transistor, "helps" it by adding beta times bigger collector current. Figuratively speaking, you can think of the collector-emitter part as a "big brother" that helps the "little brother" (base-emitter junction:) This metaphor can help to intuitively understand the main idea behind the emitter follower used as a buffer.

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    \$\begingroup\$ @Sayan, I am sorry but I do not have inspiration at the moment... I had conceived this story as an exciting "tale" about the emitter follower but something does not work out. Maybe you will help me with some interesting questions to regain my inspiration? \$\endgroup\$ – Circuit fantasist Dec 21 '20 at 19:29
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    \$\begingroup\$ @Sayan, I got inspired and wrote what I wanted to tell. Now it's your turn... \$\endgroup\$ – Circuit fantasist Dec 22 '20 at 4:41
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    \$\begingroup\$ your circuit is too innovative for me to follow. Perhaps I should try again when you have an updated version of the V-bar, I-loop diagram. Cheers. \$\endgroup\$ – tlfong01 Dec 22 '20 at 14:53
  • \$\begingroup\$ @tlfong01, Thanks for the response. I think this is the most common intuitive idea of ​​the voltage as a height and the current as a loop. It has the advantages of the geometrical way of representation. Here is a more sophisticated picture about ECL gate at zero input voltage. I created it, ten years ago, for the Wikipedia page about ECL. \$\endgroup\$ – Circuit fantasist Dec 22 '20 at 16:02

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