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In this differential amplifier:

enter image description here

The PNP transistor Q6 acts as an active load resistor and as a current mirror. Since current source Q6 has nearly infinite impedance, we will be able to have very large voltage gain Rc/r'e in which Q6 is the Rc.

However, intuitively speaking, I don't get how this current source will generate large gain.

For example, if v1(noninverting input) is at its negative peak, current passing through diode Q5 will decrease and this decrease in current will be mirrored to Q6. At the same time, Q2 collector current is increasing. How will that generate a negative peak AC voltage at Q2?

I mean, the Q6 has only a voltage drop of Vce at that point, and I believe that won't be a sufficient drop to decrease the Vout to its negative peak. Just how does this current source act as a replacement to a normal resistor RC?

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    \$\begingroup\$ However, intuitively speaking, I don't get how this current source will generate large gain. You need to draw the small signal equivalent circuit of this circuit and then you will (should) see that the signal current coming out of Q2 goes into the high output impedance of Q6 and that will result in a high gain. Using the current mirror Q5 + Q6 like this is called an active load. If you search for that term I'm sure you'll find helpful information. \$\endgroup\$ – Bimpelrekkie Jun 25 at 10:36
  • \$\begingroup\$ If I'm basing on the fact that current source has infinite impedance, I understood it no problem. It is only when I tried to simulate it(meaning "intuitively speaking" ) on my mind just like my example in the post that I don't get it. I want to understand it without relying on the mathematical formula of Rc/r'e=infinite(since impedance is large) \$\endgroup\$ – Iwatani Naofumi Jun 25 at 10:41
  • \$\begingroup\$ "If I'm basing on the fact that current source has infinite impedance". Can you explain what you imagine when saying "infinite impedance"? Because, if you can't imagine it, you can't understand through it what the so-called "dynamic load" is... \$\endgroup\$ – Circuit fantasist Jun 25 at 13:10
  • \$\begingroup\$ Notice that the AC resistance seen from Q2 collector is Q6 output resistance (r_o due to Early effect) thus the gain is r_o/re. and because r_o is big the gain can also be big. electronics.stackexchange.com/questions/333502/… \$\endgroup\$ – G36 Jun 25 at 13:13
  • \$\begingroup\$ @G36,sorry I don't have much knowledge of R_o that is computed through early effect since the book said Rout is so big that it does not change much the voltage gain, so I will try to read more of that later. But are you saying that this Rout is the one that act as the real resistor and act as a replacement to Rc? \$\endgroup\$ – Iwatani Naofumi Jun 25 at 13:50
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The PNP transistor Q6 acts as an active load resistor and as a current mirror. Since current source Q6 has nearly infinite impedance, we will be able to have very large voltage gain Rc/r'e in which Q6 is the Rc.

However, intuitively speaking, I don't get how this current source will generate large gain.

For example, if v1(noninverting input) is at its negative peak, current passing through diode Q5 will decrease and this decrease in current will be mirrored to Q6. At the same time, Q2 collector current is increasing. How will that generate a negative peak AC voltage at Q2?

As the amplifier output is very high impedance, it's better to think of it as a current output amplifier, with the output voltage determined by the load it's driving. In normal use, the load will maintain the output voltage such that neither Q2 nor Q6 is saturated.

As Q6 current decreases and Q2 current increases, both of those effects mean that the output source current decreases, to the point of becoming an increasing sink current.

I mean, the Q6 has only a voltage drop of Vce at that point,

Q6 voltage drop is not Vce, the voltage is substantially determined by the load.

and I believe that won't be a sufficient drop to decrease the Vout to its negative peak. Just how does this current source act as a replacement to a normal resistor RC?

With a resistor in circuit to source the current, Q2 is working into the parallel impedance of the load and RC. Using Q6 as a current mirror, not only is the current gain doubled, but also the high output impedance of Q6 means that they are working into the load resistance only.

The voltage gain of the amplifier is given by its current gain times the load impedance.

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  • \$\begingroup\$ I cannot agree with "As Q6 current decreases and Q2 current increases..." since the current through two elements in series (here, the collector-emitter parts of Q6 and Q2) is the same. This fact was established in the 19th century... \$\endgroup\$ – Circuit fantasist Jun 25 at 13:30
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    \$\begingroup\$ @Circuitfantasist You're forgetting there's a third connection to the Vout node, which is to the load. If you read my answer, you'll see that the load plays a vital part in the analysis of this amplifier, as it establishes the output voltage, is a linear component of the voltage gain, and provides the path for whatever current imbalance there is between Q2 and Q6. Love your answer by the way, though I wonder if you're a little over the top? \$\endgroup\$ – Neil_UK Jun 25 at 14:47
  • \$\begingroup\$ Neil_UK, Thanks! Only I wonder, if we rely on the external load to explain the circuit operation, how can we do it in the case of an open circuit? In my opinion, such a circuit of two elements in series (voltage divider), which "produces" voltage, is designed to work without load (open circuit). Otherwise, it becomes a "current divider". Similarly, the output of a current divider should be short-circuited; otherwise, it becomes a "voltage divider"... \$\endgroup\$ – Circuit fantasist Jun 25 at 16:19
  • \$\begingroup\$ @Circuitfantasist In the limit of operation with no load resistor, the effective load is provided by the Early-voltage effect output impedance of Q2 and Q6, and the voltage gain becomes very high. Only a very small input voltage is required to saturate the output in either output direction. Don't forget that it stops working as a linear amplifier when it saturates, there's little need to analyse a not-working amplifier, except to define when it is, and isn't, working. Often, the output load will be capacitive, and we see an input voltage dependent slew rate on the output. \$\endgroup\$ – Neil_UK Jun 25 at 16:42
  • \$\begingroup\$ @ Neil_UK, There is no problem with the very high gain... and this structure is designed exactly with the purpose to have an enormously high gain... since it always work with negative feedback... or, if there is no feedback, as a comparator. I have nothing against the load; I only would like to say that this stage will work also without a load. A similar example is an op-amp inverting amplifier with R1 = R2; so VOUT = -VIN. There, VIN and VOUT vary in opposite directions and the common current stays unchanged (here, R1 and R2 vary in opposite directions and the common current stays unchanged. \$\endgroup\$ – Circuit fantasist Jun 25 at 17:10
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As I see you r confused about linearity and saturation. To make it easier assume we have a Rload at Vout. The difference between Q6 and Q2 current would follow into the Rload. So the voltage gain of the circuits depends on Rload, because Vout = Rload * I_diff(between Q2 & Q6) . Now increase the Rload to infinity, what would happen? Yes, the voltage gain increases to infinity. But that doesn't mean you will have an infinite voltage at Vout. It operates under the limitations, the upper bound of Vout is when Q6 is saturated and the bottom bound of Vout is when Q2 is saturated. As long as the Vout is somewhere between these two bound, your voltage-gain is infinite. Even a very little difference between Q2 and Q6 current would result in one of them be in saturation mode.

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"If I'm basing on the fact that current source has infinite impedance, I understood it no problem."

"Infinite impedance" is abstract and formal concept; so it is not suitable for intuitive understanding.

The idea behind this circuit solution is brilliant and it has been conceived in a beautiful mind. What the OP needs is not some formal explanation but the pure idea shown in a simple, clear and intuitive way. This is what I will try to do.

"Static" voltage divider as amplifier. Generally speaking, transistor amplifier stages are voltage dividers… but "static" voltage dividers. The one of the resistors (usually, the grounded R2) is replaced by a variable "resistor" (transistor) and the other (R1) is constant. The variable "resistor" is controlled by the small input voltage… so the output voltage changes between supply rails.

"Dynamic" voltage divider as amplifier. We can really understand the idea of "dynamic load" (in the version, implemented in this circuit) by the concept of "dynamic voltage divider". It means to replace the ordinary "static" (ohmic, constant, steady…) resistors R1 and R2 of the ordinary voltage divider with "dynamic resistors" (Q6 and Q2 in the OP's circuit diagram) and vary them in opposite directions. The result is that the output voltage vigorously changes.

Electronic implementation. To implement this idea, we have to move Q1 above Q2 and join their collectors). But since it is impossible (Q1 is needed where it is), we clone (mirror) it by the p-n-p Q6. Thus V1 makes Q6 (through Q5) change its "dynamic resistance" in one direction while V2 makes Q2 (directly) change its "dynamic resistance" in the opposite direction… and VOUT vigorously changes as we will see in the pictures below. I have used them in other discussions to explain, in such an intuitive manner, the exotic current-feedback amplifier (CFA) - Fig. 1.

Current feedback amplifier

Fig. 1. A dynamic load stage in the output of a current feedback amplifier (Wikipedia)

Potentiometer analogy. The ordinary 19's century potentiometer is the simplest (not exact) example of this arrangement. An interesting phenomenon in its operation is that when we move the wiper, the one partial resistance increases but the other decreases so their sum stays constant (see the graphical representation in Fig. 2). So, the current through them does not change... only the output voltage changes (not vigorously, since the resistances are "static").

Potentiometer analogy

Fig. 2. "Static-potentiometer analogy" of the dynamic load (graphical representation)

Dynamic potentiometer. The operation of the CFA output stage is presented graphically in Fig. 3 by two oppositely moving intersecting lines - the IV output curves of the transistors Q4 and Q6. Their intersection (operating) point moves along a horizontal line in a perpendicular direction.

CFA as a "dynamic potentiometer

Fig. 3. "Dynamic-potentiometer analogy" of the dynamic load (graphical representation)

We can intuitively understand and explain this phenomenon if we think in terms of static (instant, chordal) collector-emitter resistances instead of currents flowing through them. This means to think of the two collector-emitter junctions (CE4 and CE6) as of two partial resistances (RCE4 and RCE6) of the potentiometer above.

When the input base-emitter voltages (VBE4 and VBE6) change differentially - e.g., the magnitude of VBE4 increases while of VBE6 decreases, RCE4 decreases but simultaneously RCE6 increases like the two partial resistances of the potentiometer when moving the slider to right. But the total resistance RCE4 + RCE6 remains constant so the common current flowing through the network remains constant as well and the output voltage VA vigorously changes.

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  • \$\begingroup\$ I imagined that since current is constant, even if Q2 wants to surge more current, It will be blocked since Q2 cannot change the constant current that is produced by Q6 that's why I thought it will be considered having a very large resistance(nearly infinite), am I right at this?And yes, infinite impedance is hard to imagine intuitively \$\endgroup\$ – Iwatani Naofumi Jun 25 at 13:36
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    \$\begingroup\$ Yes, it is hard to imagine intuitively the infinite impedance if it was "static"... but if it was... I hope you can finish the sentence:) Thanks for the +1 that has neutralized the usual -1 from the usual person... \$\endgroup\$ – Circuit fantasist Jun 25 at 13:48
  • \$\begingroup\$ I will tell you something else as another hint... The two transistors are controlled in opposite ways, in a differential manner... but the current through them does not change. What then can change (differentially)? \$\endgroup\$ – Circuit fantasist Jun 25 at 13:55
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    \$\begingroup\$ The resistance, so it needs to be dynamic resistance, then \$\endgroup\$ – Iwatani Naofumi Jun 25 at 14:00
  • \$\begingroup\$ Eureka! We discovered America:) Now all we have to do is visualize it in an attractive way... \$\endgroup\$ – Circuit fantasist Jun 25 at 14:01

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