# Input stage of a current mirror circuit

I'm trying to understand current mirror circuits and I've been reading this article. I'm unable to understand the input part of the circuit where the author says "Producing the opposite where the collector current controls the VBE is not possible in the conventional use of the device as a common emitter amplifier. The solution is to incorporate negative feedback." (refer to image). I do not understand how the feedback is actually 'controlling' base to emitter voltage (Vbe).

From what I understand, the Base-Emitter junction is just like a diode with a fixed ~0.7V across it. So, no matter what current you give to the base through the feedback connection, won't Vbe be equal to ~0.7 volts? How are we actually controlling the output(Vbe) here?

• Your assumption about Vbe is WRONG. Good enough for simple circuit analysis, but wrong. Read about the "Shockley rectifier equation" and come back to this. – Brian Drummond Jan 14 '17 at 11:01
• "From what I understand, the Base-Emitter junction is just like a diode with a fixed ~0.7V across it." It isn't fixed, the voltage varies (non linearly) with current and it also varies with temperature hence the need to match devices. – JIm Dearden Jan 14 '17 at 14:24
• Sumanth, do you still want to really understand how the current mirror input stage works? – Circuit fantasist Aug 5 '17 at 10:33
• @Circuit fantasist, I think jonk's answer has done an amazing job in clearing my doubts. Thank You! – Sumanth Aug 7 '17 at 3:35
• I mean a simple and clear explanation that shows the fundamental idea behind the current mirror... – Circuit fantasist Aug 8 '17 at 10:31

There are many different BJT models, with varying degrees of usefulness in varying circumstances. (See SIDEBAR at bottom.) I'm not going to delve into any of that as it's not necessary in this case. A nice simplification is quite sufficient for your use here.

Ignoring Nth order effects that aren't important here, a BJT's collector current is determined by its base-emitter voltage; here shown using the Shockley diode equation in a highly simplified (active mode) BJT model that ignores the Early Effect:

\begin{align*} I_\text{C}&=I_\text{SAT}\cdot\left(e^{\cfrac{V_\text{BE}}{V_T}}-1\right)\tag{Active Mode}\label{AM} \end{align*}

(In the above, $V_T=\frac{k \:T}{q}\approx 26\:\textrm{mV}$.)

The important thing to note here is that this equation expresses a relationship between $V_\text{BE}$ and $I_\text{C}$. A BJT is a voltage-controlled device, as a FET is. (The big difference is that in the BJT case you need to keep supplying additional current into the base in order to deal with recombination. See my explanation here, regarding a PNP, for some details.) This equation works both ways. If you can set $I_\text{C}$, then you also know $V_\text{BE}$ (see a derivation here):

\begin{align*} V_\text{BE}&\approx V_T\cdot \operatorname{ln}\left(\cfrac{I_\text{C}}{I_\text{SAT}}\right) \end{align*}

So it should be the case that $V_o$ is determined by $I_\text{C}\approx \frac{V-V_\text{BE}}{R_1}$. Plugging $I_\text{C}$ into the above equation can be solved by the LambertW function as:

$$V_\text{BE}=V-V_T\cdot\operatorname{LambertW}\left(\frac{R_1 I_\text{SAT}}{V_T}\cdot e^{\frac{V}{V_T}}\right)$$

(If you are interested in what the LambertW function is [how it is defined] and in seeing a fully worked example on how to apply it to solving problems like these, then see: Differential and Multistage Amplifiers(BJT).)

However, you could also just make a reasoned assumption about $V_\text{BE}$ to start, stick that in to get a refined value and then use that refined value one more time and you'd be close enough for all intents and purposes.

I still haven't mentioned the negative feedback. I didn't want to, at first, because I wanted you to instead focus on the fact that a collector current $I_\text{C}$ attempts to force a particular base-emitter voltage $V_\text{BE}$, just as a particular $V_\text{BE}$ would attempt to force a particular $I_\text{C}$.

In the current mirror, you drive a collector current into one BJT, which then forces a particular $V_\text{BE}$ for it, which is then passed along as a voltage to a second BJT as a $V_\text{BE}$ signal for it, which then attempts to cause a particular collector current for this second BJT's collector load.

So let's get down to the business of the negative feedback, now. Let's start with the schematic: simulate this circuit – Schematic created using CircuitLab

The equation is pretty simple:

\begin{align*} V_\text{BE}&= V-R_1\cdot I_{R_1}\\ &=V-R_1\cdot I_E\\\\ &= V-\frac{\beta+1}{\beta}\cdot R_1\cdot I_\text{C}\tag{Eq. 1}\label{eq1}\\\\ \textrm{or,}\\\\ I_\text{C}&=\frac{\beta}{\beta+1}\cdot\left(\frac{V}{R_1}-\frac{V_\text{BE}}{R_1}\right)\tag{Eq. 2}\label{eq2}\end{align*}

In equation $\ref{eq1}$ if $I_\text{C}$ increases, then clearly $V_\text{BE}$ decreases. But from the $\ref{AM}$ equation discussed earlier, a decreasing $V_\text{BE}$ must imply a lower $I_\text{C}$. The contrary relationship between these equations amounts to negative feedback. In equation $\ref{eq2}$, relating things the other way, you can see that if $V_\text{BE}$ increases then the collector current decreases as a result. But once again from the $\ref{AM}$ equation a lower collector current implies a smaller $V_\text{BE}$. So again, negative feedback.

In effect, $R_1$ (as arranged here) is providing negative feedback.

Now return back to the BJT current mirror case. If you ignore the fact that there are some errors caused by having to supply base currents, some errors caused by mismatched $\beta$ values in the two BJTs, some errors possibly caused by different temperatures in the two BJTs, some errors also caused by different saturation currents (and therefore different $V_\text{BE}$ given the same collector currents), and some further errors that may occur because of the Early Effect due to different $V_\text{CE}$ values between them...

If you can get past all that then sinking a current into one BJT leads to a necessary $V_\text{BE}$ for that BJT, which drives the other BJT's base-emitter voltage causing a corresponding collector current in it. The second BJT mirrors the collector current in the first BJT!

Similar arguments apply to the MOSFET case, too, though the equations and unaccounted errors and thermal behaviors are different.

SIDEBAR: These models include Ebers-Moll (three 1st-level DC models that are all equivalent and described in my answer elsewhere on EE.SE, plus follow-on modifications to Ebers-Moll in order to handle everything from lead resistance to base width modulation effects, to Gummel-Poon (provides a unified view of the Early and Late Effects and more, as well) and further modifications, then to VBIC, and still later ones which are used by FAB personnel for IC designs. All useful BJT models are based upon some kind of understanding of the the physics involved. But even that physics, itself, is yet another set of simplifications about a deeper reality. For example, even the most thorough attempts will incorporate an assumed gas cloud model of conduction band electrons, ideas of mean free path, and so on. The reality is vastly more complex than that. But once you move fully in that direction, you leave behind electronics entirely and have moved into profound physics.

• "If you can get past all that then sinking a current into one BJT leads to a necessary for that BJT, which drives the other BJT's base-emitter voltage causing a corresponding collector current in it. The second BJT mirrors the collector current in the first BJT!" - new way of understanding for me. Thanks – Umar Jan 15 '17 at 7:22
• @Umar Glad it helps. Sorry about writing so much. But it helps to have a fuller context even if the basic idea isn't hard, once acquired. – jonk Jan 15 '17 at 19:58
• I am following your answers and I see the same pattern. I find it useful as I am a slow learner. – Umar Jan 16 '17 at 3:24

To explain a circuit in such a way that the reader understands it, we need first to grasp the basic ideas on which it is built... "to see the forest for the trees". This will allow us not only to describe accurately HOW the circuit is made, but also WHY it is made exactly that way...

Following this approach, we can see that the current mirror performs two main functions - isolates the load from the source (current buffer) and inverts the current direction (current inverter). So the first powerful idea is to think of the current mirror as of an inverting current buffer (booster).

Let's compare the two kinds of buffers:

• The voltage buffer (e.g., an emitter follower) has extremely high input resistance (to not influence the imperfect input voltage source) and extremely low output resistance (to not be influenced by the imperfect voltage load).
• Conversely, the dual current buffer should have extremely low input resistance (to not influence the imperfect input current source) and extremely high output resistance (to be not influenced by the imperfect current load).

That is why, the next powerful idea is to implement the "inverting current buffer" by cascading two opposite (current-to-voltage and voltage-to-current) converters. They are deliberately made non-linear (logarithmic and antilogarithmic). As a result, the input voltage (across the circuit input) is compressed almost to zero while the output voltage (across the load) can be expanded up to the supply voltage.

So this is the next ingenious idea - to cascade a compressing (logarithmic) current-to-voltage converter with an expanding (antilogarithmic) voltage-to-current converter. Note that although the particular transfer characteristics are nonlinear, the whole transfer characteristic of the current mirror is linear... but only if the converters are identical. So let's see how we can implement them...

A diode can act as the input current-to-voltage converter and the transistor - as the output voltage-to-current converter. But these elements are not sufficiently identical. So, the next powerful idea is to implement both converters with the same element - direct and "reversed". We have two choices - to make a diode act as a transistor, or to make a transistor act as a diode. It turns out the latter is possible by means of the ubiquitous negative feedback. It can make the transistor adjusts its input base-emitter voltage so that to provide the desired collector current. Let's see how the negative feedback is implemented...

In an emitter follower, the negative feedback exists due to the internal connection between the input base-emitter and output collector-emitter junctions. So, if we change the emitter current as an input quantity, the transistor accordingly adjusts its base-emitter voltage as an output quantity. This technique is widely used to bias emitter-coupled circuits, differential amplifiers, op-amp stages...

But in the common-emitter stage we cannot simply pass the input current through the collector-emitter junction since the transistor cannot "sense" this "intervention" and will not react. It can do it if we connect its base to the collector. Thus the input base-emitter junction is connected in parallel to the output collector-emitter junction and this "circuit" behaves as a voltage stabilizer with negative feedback. When we change the collector current as an input quantity, this "system" will react to our intervention by adequate change of its input base-emitter voltage as an output quantity... as though the transistor is reversed.

So, the last insight about the current mirror is that its input part acts as a disturbed negative feedback system. The variation of the "input" collector current is the disturbance and the variation of the "output" base voltage is a reaction to this disturbance...