# Understanding Summation of Currents in a Rail to Rail Differential Stage

In order to understand rail to rail OpAmps I consulted a (german) textbook. There, I found this principle schematics:

So far I understood, there are two complementary differential input stages (T1, T2 and T3, T4). When a differential signal is applied to the inputs, there is a current disbalance Iq, which is going to be amplified in the next stage.

This stage I do not understand.

It seems for me, that the top voltage source together with the resistors above T5, T6 constitute two current sources of 2I0.

The rest is not clear in detail: What is the idea behind, to put a current mirror T7, T8 at this place?

In particular, in a disbalanced situation with $$I_q>0$$, why are currents through T7 and T8 not the same: left we have $$I_0-I_q$$ , on the right $$I_0-3I_q$$ ? I thought the role of a current mirror is to keep both currents the same - here it isn't the case...

I just see, that finally I get an output current $$4I_q$$ to be available for the next stage. But the idea behind the summation circuitry in the middle is a miracle for me.

I appreciate any hints...Thanks!

• Why don't you find a document where this is clearer? Have you looked? Why does the circuit you embedded appear to have 5 power rails (+5, +, 0 volts, - and -5)? Feb 18 at 10:07
• This might be better: ti.com/lit/an/sloa039a/… Feb 18 at 10:12
• You are right: "Tietze Schenk" is an old classic book, but over over decades it got more and more confused. I couldn't find a good link. Maybe I study your link first... Feb 18 at 10:16
• However, the summation of the current signal after the input stage is not discussed in your link. I understand the differential stage, but not the postprocessing in the middle stage. Feb 18 at 10:20

$$\I_q\$$ appears to be a current imbalance from the first stage. Each branch follows kirchhoff's law, so $$\2Io-2Iq=Io+Iq+Io-3Iq\$$, the left branch follows similarly. If you take away the imbalance from the first stage, the currents in (the inner) output branches will be equal.

Think of the 1st stage output mirror, as like a mirror load in a one stage folded cascode (without rail to rail for conceptual simplicity). When current is imbalanced in any differential stage with a mirror load, the same current imbalance will happen at the output.

edit* I've attached a simulation right out of the text you mentioned (tietze-shenk). I do not have the book, but the authors share their simulations on their website.

If you look at the collector currents in $$\T_7\$$ and $$\T_8\$$, you notice they are practically identical. This is because the circuit is balanced at the operating point with the same currents dividing equally into all branches. And $$\I_o\$$ will just be 0 because they are balanced (not $$\4I_q\$$). If you do anything to disturb that point (i.e. differential input other than 0 or inject a difference current into input branches or sources/sinks - like the author shows), the branch currents will no longer be equal, but the node currents must still follow kirchoff's law at any time. It looks (from your diagram, again I don't have that text), that the author is using an explicit current imbalance at the input to show this.

edit** I looked at a preview of the book and there do appear to be typos/edit mistakes ($$\I_{T_3}\$$ should be $$\I_o+I_q\$$). This might possibly be a typo as well, without book in hand, I can't say for sure.

• But why are the currents in the marked red circles not equal? This is a current mirror and so I would expect, they are the same, Feb 18 at 20:27
• Study a simple current mirror load in any differential amplifier. Output node currents with a mirror will only be the same if the input currents are balanced (at the quiescent operating point). Any other deviation of current steering will result in a difference in currents at the branches. I don't have the book, but I think he showed it that way to emphasize how currents behave with imbalances from operating point.
– pat
Feb 18 at 20:53
• I think you didn't understand what I mean. The distribution of currents according to Kirchhoff law is clear. Also clear is, that in the balanced situation the current in T7 and T8 are the same. But according to the picture T7 and T8 do not carry the same current in the imbalanced case. Since T7 and T8 constitute a current mirror, I would expect them to be the same: it is the job of a current mirror to make both currents the same. Feb 19 at 8:34
• Please see the answer edit. Without book in hand, I agree it's odd. I would just chalk this up to a edit error or typo for now. The simulation they provide matches what we would expect. It does look like a good text, however. Thanks for sharing that. FWIW, Johan Huijsing texts cover rail to rail opamps in depth and without all the odd schematic notation.
– pat
Feb 20 at 22:50