Greetings! I am currently trying to understand how does the negative feedback in bjt current mirrors work. Here's what I think: The base voltage is always kept constant by the collector current of transistor 1. If the supply voltage increases, this will rise the voltage at the collector and this will increase the collector current, which will regulate the collector(or base) voltage back to what it was. Despite this, I realy have no idea

 

This link may be informative to some extent:
http://www.allaboutcircuits.com/tec...feedback-part-5-gain-margin-and-phase-margin/
There is a series of several articles ...

 

trying to understand how does the negative feedback in bjt current mirrors work.

depends on which negative feedback mechanism you are talking about. there are more than one.

 

re OP: take a look at the diagram on the page referenced above ... The bjt layout for a differential amplifier ...any help?

The grounded base, on the upper left transistor, is only for a demonstration of the case, with no feedback signal.

 

A two transistor current mirror has no negative feedback.
It works by having the base-emitter voltage of the driven transistor causing the output transistor to have the same current (for ideal matched transistor).
There is no negative feedback path.
And the output current in a real circuit can vary with collector voltage due to the non-ideal collector resistance being finite

A three-transistor mirror, such as a Wilson mirror(below), does have negative feedback to minimize the effect of the collector resistance on the output current.
If the Iout current increases due to the finite Rce of Q3, this current also goes through Q2, which increases the current through Q1 from the current-mirror effect between Q1 and Q2.
This negative feedback reduces the current available for Ib3 (since Iin is fixed), thus reducing Q3's current and minimizing the effect of Q3's Rce on Iout.

 

Oh. It's just that I watched a video on YouTube with two transistor current mirror and there the guy said that if a voltage spike appears at the base of the first transistor, this will increase the base current which will increase the collector current and since the collector is connected to the base, the voltage at the base will decrease, thus providing some negative feedback against voltage spikes. But if the supply voltage increases, I don't think there will be any negative feedback to save the base and collector currents to increase in magnitude.
Oh and Thanks for the answers everyone!

 

Oh. It's just that I watched a video on YouTube with two transistor current mirror and there the guy said that if a voltage spike appears at the base of the first transistor, ..................

Which one is the "first transistor"?

 

Which one is the "first transistor"?

The one at wich the base is connected to the collector. In your case that would be Q1

 

But if the supply voltage increases, I don't think there will be any negative feedback to save the base and collector currents to increase in magnitude.

yes. But that's a different question.

 

the guy said that if a voltage spike appears at the base of the first transistor, this will increase the base current which will increase the collector current and since the collector is connected to the base, the voltage at the base will decrease, thus providing some negative feedback against voltage spikes.

Since the base and collector of the "first" transistor are connected together it's not possible for a spike to appear at the base without it also appearing at the collector.
The two tied together look like a single diode, where any feedback is not possible.
What the guy is saying is nonsense.

 

I would surmise that the feedback mechanism he was talking about involves interaction with the current-setting component, which is usually a resistor connected between the positive supply (or other reference voltage) and the collector. If noise raises the base voltage, then this causes the transistor current to increase, but this causes the current in the resistor to increase which drops the collector/base current back down. A similar feedback reaction happens in reverse if noise causes the base voltage to drop.

 

I would surmise that the feedback mechanism he was talking about involves interaction with the current-setting component, which is usually a resistor connected between the positive supply (or other reference voltage) and the collector. If noise raises the base voltage, then this causes the transistor current to increase, but this causes the current in the resistor to increase which drops the collector/base current back down. A similar feedback reaction happens in reverse if noise causes the base voltage to drop.

The collector is connected to the base.
Functionally that acts as a diode in series with a resistor.
I fail to see how such a configuration can suppress any noise spikes.

 

I would surmise that the feedback mechanism he was talking about involves interaction with the current-setting component, which is usually a resistor connected between the positive supply (or other reference voltage) and the collector. If noise raises the base voltage, then this causes the transistor current to increase, but this causes the current in the resistor to increase which drops the collector/base current back down. A similar feedback reaction happens in reverse if noise causes the base voltage to drop.

Indeed

 

The collector is connected to the base.
Functionally that acts as a diode in series with a resistor.
I fail to see how such a configuration can suppress any noise spikes.

Well the just consider what happens with just a diode connected in series with a resistor if noise results in the voltage at the junction going up so as to increase the voltage across the diode. This tries to cause the current to go up while at the same time decreasing the voltage drop across the resistor which tries to cause the current to go down. The net effect is that the operating point is quickly reestablished.

 

Indeed

it is anything but.

the current mirror being discussed here has no supply voltage rejection. If you work out the math, you will find that the changes in current through the mirror is proportional to changes in the supply voltage -> ie, the supply voltage actually defines the current in the mirror.

Again, it goes back to the point I was making earlier: whether there is a negative feedback depends critically on the mechanism you want to talk about.

 

Well the just consider what happens with just a diode connected in series with a resistor if noise results in the voltage at the junction going up so as to increase the voltage across the diode. This tries to cause the current to go up while at the same time decreasing the voltage drop across the resistor which tries to cause the current to go down. The net effect is that the operating point is quickly reestablished.

I considered it and it does not compute.
Any noise will appear across the diode proportional to the voltage divider ratio between the resistance resistance and the diode dynamic resistance which will generate a corresponding noise current.
How is that reducing the noise??
How is that different from two resistor in series?

 

I considered it and it does not compute.
Any noise will appear across the diode proportional to the voltage divider ratio between the resistance resistance and the diode dynamic resistance which will generate a corresponding noise current.
How is that reducing the noise??
How is that different from two resistor in series?

I'm not saying it IS different that two resistors in series -- but what keeps the current in two resistors that are in series at the same value? Nor am I saying that it is some magical noise reduction mechanism. As already pointed out by others, the description of any negative feedback mechanism is within the context of what the cause and what the effect is. The TS has given no clue what that context is, so we are left having to guess what the context is in which the claim might make some sense.

 

but what keeps the current in two resistors that are in series at the same value?

Ohm's Law.
But I don't see that can be called negative feedback.

 

but what keeps the current in two resistors that are in series at the same value?

the two transistors are identical: if you look through the bases, the two paths to ground are identical so the base current for the two transistors must be the same -> current mirror.

 

the two transistors are identical: if you look through the bases, the two paths to ground are identical so the base current for the two transistors must be the same -> current mirror.

What does that have to do with the statement you quoted?