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What is the difference between those two circuits, when only output node is changed?

Is there any changes in the output resistance, voltage gain, or somewhere else?

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

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    \$\begingroup\$ What you should read more into would be the "Difference between high-side and low-side MOSFET Drivers". I believe you will find most of your answers there. \$\endgroup\$ – 12Lappie Mar 31 '17 at 13:45
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    \$\begingroup\$ Is there any changes in the output resistance, voltage gain, or somewhere else? Yes, do the small signal analysis of both and you will know. \$\endgroup\$ – Bimpelrekkie Mar 31 '17 at 14:01
  • \$\begingroup\$ Neither of your circuits has an input indicated, so I don't see how either one can be considered an amplifier. \$\endgroup\$ – The Photon Mar 31 '17 at 17:42
  • \$\begingroup\$ @ThePhoton: You know what I meant. I know you could probably and most possibly quess where I am aiming with this question. \$\endgroup\$ – Keno Mar 31 '17 at 21:08
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Yes, there is a huge difference between those two circuits. When the output voltage is taken from the source terminal we have a Source Follower (common-drain amplifier). The Output voltage is Vgs lower than the voltage at the MOSFET Gate. The voltage gain is less than one (\$A_V = \frac{R_4}{\frac{1}{gm} + R_4}\$) and Rout is 1/gm (low).

The second circuit is a classic Common Source with Source Degeneration resistor (R4). The voltage gain is equal to \$A_V = -\frac{R3}{R4+ \frac{1}{gm}}\$

The voltage gain is negative but usually, we ignore this "minus" sign. Because this only informs us about 180-degree phase shift between Vin and Vout. When the voltage at the MOSFET Gate increases, the drain current also increases. The current in the drain resistor (R3) increases (ID = IR3) which increases the voltage drop across it (across R3, VR3 = IdR3) so the drain voltage decreases (Vd = Vdd - IdR3) which is 180 degrees to the change in gate voltage. And the output resistance is Rout = R3.

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  • \$\begingroup\$ Is it true that with "Rs" (by making negative feedback to mosfet) in the circuit "Q" is to be stabilized? I mean if dc bias conditions are applied for "Q" to be in the saturation region and gate voltage for some reason increases, "Q" stays unchanged - because of the negative feedback. Is this true? \$\endgroup\$ – Keno Mar 31 '17 at 21:26
  • \$\begingroup\$ Or this applies if temperature changes - thermal stability? \$\endgroup\$ – Keno Mar 31 '17 at 21:44
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This is the classic 0/180 balanced output reversed-phase amplifier. You use it in certain audio amplifiers. If the active-device is biased for high gm (transconductance), the bottom voltage swing is nearly identical to the top voltage swing.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Which "this" are you looking at? OP showed two entirely different amplifier circuits. \$\endgroup\$ – The Photon Mar 31 '17 at 17:41
  • \$\begingroup\$ Could you maybe explain this in more detailed manner? I know what you want to explain but I would be greateful for more detailed answer. \$\endgroup\$ – Keno Mar 31 '17 at 21:12
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Is there any changes in the output resistance, voltage gain, or somewhere else?

as there is no input signal, those questions are not answerable.

after that, you will need to know about the characteristics of the input signal to answer your questions with confidence.

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  • \$\begingroup\$ How many times do I have to tell you this is only a simbolic circuit. Of course there is an input signal! \$\endgroup\$ – Keno Apr 1 '17 at 16:17

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