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enter image description here The second figure is a modified version of the first figure.

In the second figure, how is the folded cascode stage formed by pnp transistors Q3 and Q4 used to raise the output resistance looking into the collector of Q4 to β4*ro4?

I have trouble understanding how β4*ro4 was derived.

The overall output resistance of the amplifier is: Ro = [β4ro4 || β5ro5/2]

I am referencing "Microelectronic Circuits" by Sedra and Smith, 7th edition page 654.

Egg
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  • There are two folded cascodes (Q1/Q3 and Q2/Q4). Derivation of analytic expression for the common base amp output resistance is a tricky experience. Start with Thevenin/Norton non-ideal current/voltage sources and only in resulting equations make your approximations like those leading to Rout=β*ro. For derivation of the cascode output resistance see, for example, https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-012-microelectronic-devices-and-circuits-fall-2009/lecture-notes/MIT6_012F09_lec21_cascode.pdf page 3. – V.V.T Aug 02 '20 at 08:13
  • Thank you for the link. I checked out the diagram on page 2 first and didn't understand why gpi2, gpi1, go2, and go1 weren't respectively rpi2, rpi1, ro2, and ro1. I also didn't understand why test current "it" didn't include the second current source. As a result, I couldn't gain much clarity. – Egg Aug 06 '20 at 01:53
  • Test current *it* is a sum of currents through the current source *gm1* and the output conductance *go1* (KCL), see the rightmost node in the Page 2 figure. The current source *gm2* is not connected to this node – V.V.T Aug 06 '20 at 06:30
  • Another point of confusion may be that the voltage controlled current source (VCCS) of transistor model is denoted by its transconductance *gm* or *gpi* (VCCS generates a current *i* = *gm·v*, where *v* is a voltage drop across controlling terminals). Here a component name starts with g, but it is not a resistor, it is a current source. – V.V.T Aug 06 '20 at 14:22
  • A starting g or G character (g for small-signal, G for DC) is a convention for naming of conductances, which are inverses of resistances: go1 = 1/ro1 etc., in general, R_x = 1/G_x, r_x = 1/g_x. For a series connection, Rtot = R1 + R2, Gtot = G1*G2/(G1+G2). For a parallel connection, Rtot = R1||R2 = R1*R2/(R1+R2), Gtot = G1 + G2. For serially connected resistors you are comfortable with resistances, for parallel, with conductances. – V.V.T Aug 06 '20 at 14:24

2 Answers2

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In this circuit Q4 working as a common-base amplifier thus, we can use this super simplified small-signal equivalent circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The output resistance is equal to:

$$R_O = \frac{V_X}{I_X}$$

Thus the \$I_X\$ current is equal to:

$$I_X = I_{ro} - I_C = \frac{V_X}{r_o} - \frac{\frac{V_X}{r_o}}{\beta +1}*\beta = \frac{V_X}{r_o} - \frac{V_X \beta}{(\beta +1) r_o} = V_X \left(\frac{1}{r_o} - \frac{ \beta}{(\beta +1) r_o}\right)=$$

$$I_X = V_X \left(\frac{\beta +1}{(\beta +1) r_o} - \frac{ \beta}{(\beta +1) r_o}\right) = V_X \left(\frac{1}{(\beta +1) r_o}\right)$$

Therefore

$$R_O = \frac{V_X}{I_X} = (\beta +1)r_o \approx \beta r_o$$

You can also look here at this example:

finding the output resistance of bjt folded cascode

G36
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  • @V.V.T But I added a link to the "correct " solution as well. – G36 Aug 01 '20 at 12:41
  • No, you can't ("use this super simplified small-signal equivalent circuit"), because this "super simplified" circuit does not include a source current path (not mentioning you are careless with plus/minus signs). First make sure that a source impedance overwhelms the input impedance, R_S >> r_E. If this is the case, the common base amplifier is a current follower and the output resistance can be obtained from a general expression Rout = (1 + gm(r_π||R_S))ro + r_π||R_S. If R_S >> r_E, then Rout = (1 + gm*r_π)*ro) + r_π; From r_π = β0/gm follows that Rout = (1 + β0)*ro . – V.V.T Aug 02 '20 at 08:08
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CB's Rout

I understand my own question now. I read about this on pg 557 of the textbook.

β4*ro4 is the maximum bipolar output impedance.

First I find the general Ro equation of a CB circuit.

  • Ro = ro + (Re||rpi) + (Re||rpi)gmro
  • Ro = ro + (1 + gmro)(Re||rpi)
  • Ro ~= ro + (gmro)(Re||rpi) for gmro>>1
  • Ro,max = ro + gmro*rpi = (1+B)ro
  • Ro,max ~= B*ro

Then I apply this to the figure in question.

Egg
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