I was checking out current mirrors which I understand enables almost the same amount of current flowing in one branch to flow through another branch. With reference to the image: http://sound.westhost.com/ab-f5-3.gif. When Q1 is in off state, the voltage at collector is 5v and this drives the base, turning on Q1. Thus, the collector voltage almost becomes oV which should turn the transistor off because of the drop in the voltage driving the base to 0V. Now doesn't this send the transistor into an oscillating state where it is turned on and off.....so will a continuous current flow through the branches????
No, that would only happen if the transistor has some kind of 'overshoot' with the gain. But the transistor doesn't have that, so it can't oscillate. The capacitance inside the transistor slows it also down. Q1 will find it's balanced level very smooth, and Q2 will follow.
Current mirror is an analog circuit, its not operating in either "on" or "off" mode, its in the (roughly) linear region of operation(*). For such circuits you apply the Ebers Moll equation for each device and solve simultaneous equations to get the response.
The possibility of oscillation is addressed by ac analysis and applying Nyquist's criterion - basically you have to do the maths to understand the operation properly, or just take it as given that its a current-mirror to a good approximation.
(*) The "on" state is saturation, where the collector voltage is well below the base voltage (typically 0.1V or so above the emitter). Here the left hand device has collector and base voltage the same, so definitely not saturated.
The way you normally think of a current mirror is that the left-hand resistor sets ("programs") the collector current, which determines the base voltage of both devices, which means they have identical collector currents.
In reality the right hand device collector current depends on the collector voltage somewhat.
Since the base and collector of Q1 are shorted, Q1 is essentially a diode. The base-emitter voltages of Q1 and Q2 are equal, this is what allows the configuration to act as a mirror. If Q2 is the same size as Q1 then whatever current is flowing in resistor Ri will flow in the load of Q2. This assumes that Q2 is not saturated and the gain Q2 is high enough that the base current of Q2 is negligible.
In short, the base-emitter voltages of Q1 and Q2 are equal therefor they carry the same current.
For such circuits you apply the Ebers Moll equation for each device and solve simultaneous equations to get the response.
So now you know!
The fact is, the negative feedback of the collector being connected to the base of Q1 results in a stable state, which is of course, exactly what negative feedback does unless there is a phase shift greater than 90° from resistors (or inductors) and capacitors, which there obviously is not in this case.
I must however query what you propose to use this for, as this is not the optimum circuit for a constant current LED driver because you would be wasting the “mirror” current. For constant current, a different circuit is used.
RImin and the “back” diode across the LED are not necessary in most cases.