When there are three amplifiers that are cascaded to produce higher gain:

  • The source resistance should be lower than input impedance of first amplifier
  • The output impedance of first amplifier should be lower than input impedance of second amplifier
  • The output impedance of second amplifier should be lower than input impedance of the last amplifier
  • Then the output impedance of last amplifier should be matched/equal to load resistance for max power transfer

In short,am I correct to conclude that: On the way to the load resistance, we maximize voltage transferred but when in front of the load resistance, we maximize power transferred by matching the impedances, but this means the output voltage will then be half of what the last amplifier should have gained.

I also wonder why we don't match impedance between two amplifiers(just like how we match impedance of the last amplifier to load resistance) to also transfer max power?

  • 1
    \$\begingroup\$ I believe you should specify the nature of the amplifier stages. Are they all voltage amplifiers? Are they current amplifiers? Is the intermediate stage put there to 'mediate' between the first and last stage? Each amplifier has different needs in terms of the best input and output impedance. Also, maximum power transfer does not necessarily require matched impedances: it's only when you are stuck with a given source output impedance that you can't do better than matching its impedance. But if you can design the source, you can choose the best mismatch that maximize power transfer. \$\endgroup\$ Commented Jun 16, 2020 at 4:57
  • \$\begingroup\$ They are all voltage amplifiers and each amplifiers are use to increase gain. But the last one is a buffer amp. \$\endgroup\$
    – hontou_
    Commented Jun 16, 2020 at 5:12
  • \$\begingroup\$ It's late and I need to sleep so I will only add this as a hint: if you are given a fixed Rout and you can choose the load RL, by choosing RL=Rout you maximize the power transfer on RL. But if you are stuck with a load RL and you are free to choose Rout (because you can shop for different amplifiers or you are designing it) it is not wise to choose Rout = RL because in that case you will maximize the power transfer on Rout, not RL. And it's not the same thing. (this should explain why you do not match impedances in between). As for the rest... a buffer is not a voltage amplifier. Why is there? \$\endgroup\$ Commented Jun 16, 2020 at 5:44
  • \$\begingroup\$ Bullet point 4 is usually wrong. There are specific circumstances where Max Power Transfer is required - at either end OR in the internal stages - but that is application dependent. Specific counter-example : nobody would design an audio power amplifier with 8 ohm output impedance. \$\endgroup\$
    – user16324
    Commented Jun 16, 2020 at 10:39
  • \$\begingroup\$ An ingeniously simple question that deserves an ingeniously simple answer... Congratulations! \$\endgroup\$ Commented Jun 17, 2020 at 16:48

4 Answers 4


[I discuss Noise Voltage versus Noise Figure at end of this answer.]

simply stated

  • matching will cost you 6dB per interface on the voltage levels

  • I once lead a team doing RF design on silicon; we concluded there was no need to match over our 500 micron distances on the silicon

  • I guided the team (all coming from past PCB work, where matching WAS needed), to view the silicon design as broadband opamps where you can use an emitter follower to achieve low Rout, and use diffpairs (bipolar or FET; we have biCMOS process) for input circuit, thus HIGH_RIN, to the next signal_processing circuit

  • we learned, in our simulations, the matching made no sense after building a precision gain/phase circuit at substantial power consumption and THEN to throw away 6dB voltage level


At the time of this design team's learning of RFIC methods, a big topic at technical conferences was Noise Figure versus Noise Voltage.

simply put:

  • Noise Figure requires a given noise density at the signal source

  • a "noise density" seems to require an Output Resistor

  • we don't want to insert lossy resistors, just to add noise

  • so we went with the OpAmp_as_broadband_amplifier for our mindset; we did no matching; we used Noise Voltage as our UHF (300MHz to 3,000MHz) design goal


For practical purposes, the answer depends very much on the frequency band that you're working with, and on the length of your transmission lines = whether TML properties need to be considered at all, at those frequencies.

My first experiences as a DIY has been with audio circuitry. The classic is a phono preamp, followed by some line-level corrections and level adjustment stages, followed by a power output stage. In this range, cca 20 Hz to 20 kHz, your transmission lines would need to be in the range of kilometers, to matter in terms of RF impedance, at the upper end of the band. For many purposes, the audio amps treat their signal as DC, and the load is also considered "pretty much DC" - at least in terms of transmission line properties. (You only start to border on AC when working with highpass/lowpass filters in the signal path, power decoupling, feedback stability and the possible capacitive component to the output load in the power stage.) These amps are the voltage-oriented classic, where the current is considered a mere pesky sideways property of your load, that you need to cater for. To answer your question for this partial topic, here you typically construct the output of an amplifier stage as "fairly low impedance", and the input of the next stage as "fairly high impedance". The motivation is to minimize distortion. The preceding stage should have no problem to drive the load presented by the follow-up stage. Which might lead you to prefer FET-based differential inputs as the ultimate holy grail of amplifier design - because they have so little input current, right? Well not necessarily. There are prophets in audio amp design, who would tell you to use BJT-based diff inputs, and design their quiescent current relatively high (input impedance low, maybe in the kiloOhms range), as that is good for noise performance. Your op-amp-based preamp stages have an output impedance in units of ohms, so what. Also, where you start moving op-amps closer to RF, i.e. you need extreme slew rates or Gain*Bandwidth product, that's where current-coupled design principles start to rear their head.

Proper RF design is a whole different game. You need to consider transmission lines and impedance-matching at output and input - as a way of preventing reflections at the "handover points". Reflected energy is not your friend, as it can overload the output of a power amp stage, can distort the shape of your signal in the time domain, can "ring" (resonate) in the frequency domain, even to the extent that it can make amp stages self-oscillate etc.

You have areas where a high-speed op-amp-based signal chain design meets RF and transmission lines - one classic I can think of is analog video or maybe DSL modem transmissions.

Then you have areas where digital stuff meets RF and transmission lines - pretty much any kind of digital communication. From RS485 at the low end (hundreds of meters of cabling, kbps transfer rate) to modern computer busses over distances of just a few inches - think of the several digital display standards (TMDS and DP) or PCI-e or USB3. The PCB routing of their symmetrical pairs is all in the RF territory.


You wrote:

On the way to the load resistance, we maximize voltage transferred but when in front of the load resistance, we maximize power transferred by matching the impedances

and this seems to imply that you believe that the maximum power on the load will be transferred when the impedances are matched. Well, this is not necessarily the case. In fact it is in general not true. Let me take a level of complexity out of the equation by considering resistances and not impedances, to make my case.

Impedance matching does NOT imply the load gets the highest power

Now, suppose you are given a 1 ohm load and you want to power it with a 10 volts battery, hopefully getting it to dissipate a power close to V^2/R = 100W. You go to the mall and find three 10V batteries each with a different internal resistance (it's a strange shop, not only the sell batteries with nonstandard voltages, but they also state their internal resistance...). The first battery has an internal resistance of 0.1 ohms, the second 1 ohm, the third a whooping 10 ohms (for some reason they are in discount).

Which battery will maximize the power transfer to your 1 ohn load? Certainly not the one with the humongous internal resistance. But not even the one with the 'matched' resistance of 1 ohm. In fact, unsurprisingly, it's the 0.1 ohm battery that will deliver the highest power to RL.

  • V = 10 V, Rs = 0.1 ohm, RL = 1 ohm

  • PRL = 82.6 W

  • V = 10 V, Rs = 1 ohm, RL = 1 ohm

  • PRL = 25 W

  • V = 10 V, Rs = 10 ohm, RL = 1 ohm

  • PRL = 0.82 W

So, what is this maximum power transfer theorem about then? Well, consider the battery alone, what is the maximum power it can develop (not deliver, develop)? Short circuit it and you will see all its power going into heat, dissipated by its internal resistance (representing complex internal processes we are not interested in). For the three batteries above, the short circuit powers are

  • Rs = 0.1 ohm, PCC = 1000 W
  • Rs = 1 ohm, PCC = 100 W
  • Rs = 10 ohm, PCC = 10 W

the problem is that in the best of cases (yes, when there is matching) only a fourth of this power can be delivered to the load. So, in the first case you can get at most 250W on a 0.1 ohm load, in the second 25W on a 1 ohm load, and in the third 2.5W on a 10 ohm load.

If you choose to buy a battery that matches your load, i.e. the one with an internal resistance of 1 ohm, well congratulations, you are satisfying the maximum power transfer theorem but instead of the ideally maximum 100W on your load you are getting only 25W. Yes, 25W is the maximum you can get for that kind of battery, but this is a meager satisfaction since your load is underpowered.

A few formulas

To see what is going on graphically, let's consider the expressions for the power dissipated on the load and on the internal resistance in a circuit constituted by a voltage generator with internal resistance Rs that creates a voltage divider with the load RL. By choosing rms values we can pretend we are still in a DC condition and the formulae are:

enter image description here enter image description here

If we focus on the power dissipated by the load we see that it has two different functional forms, depending on which parameter we consider it function of. If we see it as a function of RL it has a bell shaped form with a peak for RL = Rs (yes, it's actually a bell if you plot it on log scale),

enter image description here

but if we see it a function of Rs, it is a monotonically decreasing function (for Rs>=0) that has a finite maximum in Rs=0.

enter image description here

So, the choice RL = Rs maximize power on RL when Rs is given and you are seeing the power as a function of RL, but if you are given RL and you can choose Rs, then the value that maximize power on RL is Rs=0, and NOT Rs=RL.

Incidentally, the power dissipated by Rs has a dual functional form and when you see PRS as a function of RL you will see that the choice Rs = RL is what maximize the power lost by Rs.

Here are the plots (sorry, here Rs is called Rout) for the powers dissipated by RL (in blue) and Rs (in red) as functions of RL

enter image description here

and here are the same powers seen as functions of Rs (called Rout)

enter image description here

Finally, here is the power dissipated by RL as a function of both RL and Rs (it is still called Rout, and the numerical values used for this graph are different from the two-dimensional ones - just for aesthetics)

enter image description here

Impedance matching runs along the RL=Rs line in the horizontal plane but it's not necessarily the choice that will deliver the highest power on RL.

So, why doing impedance matching in the first place?

Well, from the point of view of power, you want to match the impedances when you want to extract all the juice you can from the source. Getting back at the battery example: the best of the three choices that gives the higher power to the load is not the matched impedance one, but the one with Rs=0.1 ohm, which delivers 82.6 W on RL. Even if this is the closest to the nominal 100W you would expect applying 10V on a 1 ohm load, this is just a tiny fraction of the maximum deliverable power of 250W for a battery of that kind. But even if I'm just extracting a fraction of what I could, if my goal is to get the nominal power to my load (irregardless of how efficiently I am squeezing juice out of it) that's the best choice. If I wanted to suck all I can from that battery, I'd have to choose a different load, one that matches its internal impedance, i.e. a 0.1 ohm load; in that case I would be able to suck all the deliverable 250W.

  • So, if the source was a solar cell, I'd do my best to get all the juice out of it by design the first stage of my solar battery charger in such a way as to match its impedance - not to let it just sit there in the sun.
  • In the battery example above, I'd go for impedance matching if I were to do a youtube video on melting metals with a car battery: too low a load and I'll just boil the battery, too high the load and nothing spectacular happens (even if the load is drawing almost all of its nominal power), match RL = Rs,and maximum danger and fireworks ensue.

If you are working in AC and have to juggle impedances, matching has the added bonus of compensating the reactances of Zs and ZL.

If you are into RF, matching has a completely new reason to exists since it eliminates the reflections that happen when there is an impedance mismatch, thus reducing or removing signal integrity problems, and inefficient power transmission.

And sometimes matching is required by sheer compliance to standards. Imagine a world where function generators and electronic instrumentation had random impedances. Even if you did not reach frequencies high enough where reflections could be a problem, you would still have the divider problem: your scope has 13 ohm input resistance, your generator has 74 ohm output impedance... what voltage will you see? And when you use another function generator with 123 ohms output impedance? Madness. Let's get a reasonable standard value - or a limited set of such values - and everything is easier.

In amplifiers, well, usually you have to know what you want to amplify. If it's voltage, you want the highest input impedance possible while if you want to amplify current you will look for the lowest input impedance. Yes, power drawn will be negligible, but you can count on the last stage to deliver the right amount of power to your load. That's what the last buffer is for: you amplify voltage along the chain, then 'add' current. Sometimes you add in a stage whose sole purpose is to translate a low impedance into a high impedance (or vice versa). The power supplied to the stage will provide the extra current or voltage required to avoid losing dBs. Of course, in RF amplifiers you might want to impedance-match each stage, but the threshold frequency above which this make sense depends on the scale of integration (as described in another answer).

  • \$\begingroup\$ "the problem is that in the best of cases (yes, when there is matching) only a fourth of this power can be delivered to the load", is that a general assumption? Or was it derived from an equation? \$\endgroup\$
    – hontou_
    Commented Jun 21, 2020 at 11:05
  • \$\begingroup\$ Sorry I just got it now, that's a stupid question I guess \$\endgroup\$
    – hontou_
    Commented Jun 21, 2020 at 12:24
  • \$\begingroup\$ @IwataniNaofumi I made it explicit in the pictures now. \$\endgroup\$ Commented Jun 21, 2020 at 16:49
  • \$\begingroup\$ Actually I can see two facets of this topic where the "quarter power rule" applies: 1) as Sredni Vashtar implies, given a non-zero inherent series resistance in the source, if you attach an impedance-matched load, the voltage is one half, the current is one half the short-circuit current, hence the power delivered to load is a quarter. 2) as analogsystemsrf mentioned earlier, if you consider a source with a 0 internal resistance, and a load with a particular impedance, then matching the source resistance to your load will result in -6 dB = 1/4 power to the load. Thumbs up to Sredni Vashtar :-) \$\endgroup\$
    – frr
    Commented Jun 21, 2020 at 20:11
  • \$\begingroup\$ @frr exaclty, point 2) means choosing, in the second graph, the point on the blue curve PRL where the red curve PRout has a maximum. That is, it means maximizing the power transfer to the source resistance (at the expense of the power delivered to the load). Something you would not want to do, unless forced to by other reasons (reflection, compliance to standard, noise...) \$\endgroup\$ Commented Jun 21, 2020 at 22:45

I have to admit that I felt a (noble) envy when I saw the OP's answer. I realized that I have known this way of cascading amplifier stages since I was a student but I have not seriously thought about why it is. And now I have seriously thought... and I think I have found "an ingeniously simple answer to this ingeniously simple question"…

The reason for such a separation of the amplifier stages into two types - voltage amplifiers followed by a "current amplifier", is that we can not make a full-fledged power amplifier that combines both. Let's see why…

Amplifier structure. In essence, transistor amplifier stages are "voltage dividers" - a circuit of two elements E1 and E2 in series. By controlling the resistance of the one, the other or both, the input voltage controls the output voltage ... and we call this "amplification." The problem here is how to connect (incorporate) the load here - in a potentiometer or rheostat manner...

"Potentiometer". To obtain a huge voltage gain by this humble network, we apply various clever tricks to control dynamically the resistance of the elements (the so-called dynamic load)... and it acts like an "electronic potentiometer". We take the voltage drop across the one of elements (usually, the grounded transistor) as output voltage and apply it as input voltage to the input of the next stage. It is obvious that the next input should not load the "amplifier divider" output; so it has to have high input resistance (ideally, an open circuit).

"Rheostat." To control the high current through the final load, the "potentiometer arrangement" is inappropriate… we need the "rheostat connection"... i.e., to connect the load in series to the transistor... to replace the one of elements E with the load. The most appropriate place is between the emitter and ground (emitter follower aka common collector). Thus the same voltage will be applied to the load (no voltage gain) but the current will be as high as the load wants (the load can be as low as it wants). We call this "current amplification", although this configuration is not exactly a current amplifier (this remark was for pedants). Of course, this stage has some low output resistance... and if you want, you can match the load resistance to it.

By the way, the common-emitter stage can also act as a "rheostat" if we connect the load in the collector (instead Rc)... and it will amplify both voltage and current... but there are two problems. First, the load will be driven by constant current; so the voltage across it will depend on its resistance (this can cause nonlinear distortions). Second, we cannot apply clever "dynamic load" techniques to obtain extremely high gain.

In conclusion, we build power amplifiers by cascading a few "potentiometric voltage amplifiers" and an output "rheostat current amplifier" (simply speaking, the input stages are "potentiometers" and the output stage is a "rheostat"). That is why the input voltage amplifier stages are matched accordibg to the max voltage divider requirements while the output current amplifier stage obeys the max power matching rule.


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