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I am attaching the following image of a flip-flop circuit.

enter image description here

I find myself constantly trying to visualise what is occurring at the exact moment the circuit is closed, and the current starts (electron flow: negative to positive) I really don't want a super complicated explanation at this stage of my learning, I'm just trying to understand how the current is moving. I visualise current flow as like a 'conveyor belt', in the sense that electrons at every point in the circuit start moving at the same time.

My question is, what occurs at the BJT's? Where do the electrons go first?

From emitter to base like here? ->

enter image description here

Or does it somehow move from emitter to collector ->

enter image description here

Although I don't really see how this circuit gets started, since the NPN transistors would prohibit current from flowing from emitter to collector (with seemingly no voltage applied to the base)

I just keep getting stuck on how electrons start to move through circuits when first 'fired up' (closed).

Any help for this noob would be appreciated.

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This circuit can be understood without worrying about what happens first and what happens second. You can imagine that the base-emitter current and the collector-emitter current change simultaneously. – The Photon May 11 '14 at 21:23

3 Answers 3

up vote 2 down vote accepted

Just because watching electron flow is generally more difficult doesn't mean you can't do it that way. I started that way, but switched over after I found out it was accepted practice and it did not cause us to make errors.

Since you specifically want see it this way, let's take a look.

Not only does an excess of electrons flow from battery - into the emitters, but another excess of electrons at the collectors flow out through the resistors and LED's into the electron-deficient + end of the battery. The result is a 9V potential across the transistors. Electrons also flow out of the bases through their resistors toward the battery. Now there is a potential across the Emitter Base junctions, as well. You can probably figure out what happens next.

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Many thanks - so it is possible although more difficult to perceive it from an electron flow perspective, although easier to use conventional flow and as you said, doesn't cause errors. Thank you. – Maikl May 12 '14 at 6:38
Hello gbarry, so when there is a potential created across the emitter base junctions .... actually I cannot figure out what happens next, I'm embarressed to say. Could you shed light? Many thanks. – Maikl May 12 '14 at 10:58
Well, once that exceeds .6 volts, the electrons flow from emitter to base, and so on through the base resistors as before. Now that there's current flow through the E-B junction, the transistors start to turn on; thus, more electrons flow from E to C and out through their series components. Of course, with current through the resistors, there is now a voltage across them as well. We're well past the "immediate start" of this, but the other answers should help explain what comes next if you want to go through the whole cycle. – gbarry May 12 '14 at 16:39
Thanks gbarry! I've been reading this stack post with interest too ->… – Maikl May 13 '14 at 8:32

Forget about electrons, in this case it is much easier to think about electricity (flowing from + to -).

Also forget about what 'happens in exactly which order'. It does not matter.

At start, assume neither transistor is conducting, hence the collectors will be at ~ 7V. This will cause base current to flow, which in turn causes collector current to flow, which in turn cause the collector voltage to drop. This happens simultaneously for both transistors. They both 'try' to cut of the other one's base current. Due to asymmetries in the circuit (the B of the two transistors will not be the same) one will win, which results in the stable situation where one transistor conducts and the other does not. (until the user preses the button that reverses the situation.)

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Hello thanks for responding. I appreciate it is much easier to understand how this works from a conventional current perspective, I could see that, in fact I often find it easier to view things from that perspective, however I just wanted to understand it from the electron flow perspective, and I can't seem to understand it from that perspective. – Maikl May 11 '14 at 17:34
You are using the wrong mental model. A semiconductor is not a metal, and thinking about electron flow is not all that useful. In addition to electrons, there are holes, which also move. And the current gain of a BJT is due in part to the statistics of the recombination of holes and electrons. – WhatRoughBeast May 11 '14 at 18:19

First of all, keep in mind that physical circuits have stray and intrinsic capacitance and inductance thus, voltages and currents do not change instantly though it is often the case that they effectively change instantly.

Initially, both transistors are off since the respective base voltages are initially zero.

Since both bases are connected through resistors and diodes to the positive terminal of the battery, both base voltages begin to rise as the stray and intrinsic capacitances charge.

However, it is almost impossible that the voltages will rise at the same rate. Thus, one of the base voltages will increase faster and will 'turn on' earlier. By "turn on", I mean that, due to the base voltage, there will be significant electron flow into the emitter of the transistor.

Most of the electrons will flow out of the collector with a small proportion flowing out of the base.

This current out of the collector will cause a voltage drop across the collector resistor which will act to reduce the voltage applied to the base of the other transistor.

Thus, once one transistor 'gets ahead', it acts to turn the other transistor off.

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Thank you very much, greatly appreciate your answer there. – Maikl May 12 '14 at 6:36

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