Editing Common base (section)

== Overview of characteristics ==

Several example applications are described in detail below. A brief overview follows.
* The amplifier input impedance ''R''<sub>in</sub> looking into the emitter node is very low, given approximately by
*: <math> R_\text{in} = r_\text{E} = \frac{V_\text{T}}{I_\text{E}}, </math>
: where ''V''<sub>T</sub> is the [[Boltzmann constant#Thermal voltage|thermal voltage]], and ''I''<sub>E</sub> is the DC emitter current.
: For example, for ''V''<sub>T</sub> = 26&nbsp;mV and ''I''<sub>E</sub> = 10&nbsp;mA, rather typical values, ''R''<sub>in</sub> = 2.6&nbsp;Ω. If ''I''<sub>E</sub> is reduced to increase ''R''<sub>in</sub>, there are other consequences like lower transconductance, higher output resistance and lower ''β'' that also must be considered. A practical solution to this low-input-impedance problem is to place a common-emitter stage at the input to form a [[cascode]] amplifier.
* Because the input impedance is so low, most signal sources have larger source impedance than the common-base amplifier ''R''<sub>in</sub>. The consequence is that the source delivers a ''current'' to the input rather than a voltage, even if it is a voltage source. (According to [[Norton's theorem]], this current is approximately ''i''<sub>in</sub> = ''v''<sub>S</sub> / ''R''<sub>S</sub>). If the output signal also is a current, the amplifier is a current buffer and delivers the same current as is input. If the output is taken as a voltage, the amplifier is a [[transresistance]] amplifier and delivers a voltage dependent on the load impedance, for example ''v''<sub>out</sub> = ''i''<sub>in</sub> ''R''<sub>L</sub> for a resistor load ''R''<sub>L</sub> much smaller in value than the amplifier output resistance ''R''<sub>out</sub>. That is, the voltage gain in this case (explained in more detail below) is
*: <math> v_\text{out} = i_\text{in} R_\text{L} = v_\text{s} \frac{R_\text{L}}{R_\text{S}} \Rightarrow A_v = \frac{v_\text{out}}{v_\text{S}} = \frac{R_\text{L}}{R_\text{S}}. </math>
: Note that for source impedances such that ''R''<sub>S</sub> ≫ ''r''<sub>E</sub> the output impedance approaches ''R''<sub>out</sub> =  ''R''<sub>C</sub> || [''g''<sub>m</sub> (''r''<sub>&pi;</sub> || ''R''<sub>S</sub>) ''r''<sub>O</sub>].
* For the special case of very low-impedance sources, the common-base amplifier does work as a voltage amplifier, one of the examples discussed below. In this case (explained in more detail below), when ''R''<sub>S</sub> ≪ ''r''<sub>E</sub> and ''R''<sub>L</sub> ≪ ''R''<sub>out</sub>, the voltage gain becomes
*: <math>A_\text{v} = \frac{v_\text{out}}{v_\text{S}} = \frac{R_\text{L}}{r_\text{E}} \approx g_\text{m} R_\text{L}, </math>
: where ''g''<sub>m</sub> = ''I''<sub>C</sub> / ''V''<sub>T</sub> is the transconductance. Notice that for low source impedance, ''R''<sub>out</sub> = ''r''<sub>O</sub> || ''R''<sub>C</sub>.
* The inclusion of ''r''<sub>O</sub> in the hybrid-pi model predicts reverse transmission from the amplifiers output to its input, that is the amplifier is '''bilateral'''. One consequence of this is that the input/output [[Electrical impedance|impedance]] is affected by the load/source termination impedance, hence, for example, the output resistance ''R''<sub>out</sub> may vary over the range ''r''<sub>O</sub> || ''R''<sub>C</sub> ≤ ''R''<sub>out</sub> ≤ (''β'' + 1) ''r''<sub>O</sub> || ''R''<sub>C</sub>, depending on the source resistance ''R''<sub>S</sub>. The amplifier can be approximated as unilateral when neglect of ''r''<sub>O</sub> is accurate (valid for low gains and low to moderate load resistances), simplifying the analysis. This approximation often is made in discrete designs, but may be less accurate in RF circuits, and in integrated-circuit designs, where active loads normally are used.

=== Voltage amplifier ===
[[Image:Common-base small signal.svg|thumbnail|300px|Figure 2: Small-signal model for calculating various parameters; Thévenin voltage source as signal]]

For the case when the common-base circuit is used as a voltage amplifier, the circuit is shown in Figure 2.

The output resistance is large, at least ''R''<sub>C</sub> || ''r''<sub>O</sub>, the value which arises with low source impedance (''R''<sub>S</sub> ≪ ''r''<sub>E</sub>). A large output resistance is undesirable in a voltage amplifier, as it leads to poor [[voltage division]] at the output. Nonetheless, the voltage gain is appreciable even for small loads: according to the table, with ''R''<sub>S</sub> = ''r''<sub>E</sub> the gain is ''A''<sub>v</sub> = ''g''<sub>m</sub> ''R''<sub>L</sub> / 2. For larger source impedances, the gain is determined by the resistor ratio ''R''<sub>L</sub> / ''R''<sub>S</sub>, and not by the transistor properties, which can be an advantage where insensitivity to temperature or transistor variations is important.

An alternative to the use of the hybrid-pi model for these calculations is a general technique based upon [[two-port network]]s. For example, in an application like this one where voltage is the output, a g-equivalent two-port could be selected for simplicity, as it uses a voltage amplifier in the output port.

For ''R''<sub>S</sub> values in the vicinity of ''r''<sub>E</sub> the amplifier is transitional between voltage amplifier and current buffer. For ''R''<sub>S</sub> ≫ ''r''<sub>E</sub> the driver representation as a [[Thévenin's theorem|Thévenin source]] should be replaced by representation with a [[Norton's theorem|Norton source]]. The common base circuit stops behaving like a voltage amplifier and behaves like a current follower, as discussed next.

=== Current follower ===
[[Image:Common base with Norton driver.svg|thumbnail|300px|Figure 3: Common base circuit with Norton driver; ''R''<sub>C</sub> is omitted because an [[active load]] is assumed with infinite small-signal output resistance]]

Figure 3 shows the common base amplifier used as a current follower. The circuit signal is provided by an AC [[Norton's theorem|Norton source]] (current ''I''<sub>S</sub>,  Norton resistance ''R''<sub>S</sub>) at the input, and the circuit has a resistor load ''R''<sub>L</sub> at the output.

As mentioned earlier, this amplifier is '''bilateral''' as a consequence of the output resistance ''r''<sub>O</sub>, which connects the output to the input. In this case the output resistance is large even in the worst case (it is at least ''r''<sub>O</sub> || ''R''<sub>C</sub> and can become (''β'' + 1) ''r''<sub>O</sub> || ''R''<sub>C</sub> for large ''R''<sub>S</sub>). Large output resistance is a desirable attribute of a current source because favorable [[current division]] sends most of the current to the load. The current gain is very nearly unity as long as ''R''<sub>S</sub> ≫ ''r''<sub>E</sub>.

An alternative analysis technique is based upon [[two-port network]]s. For example, in an application like this one where current is the output, an h-equivalent two-port is selected because it uses a current amplifier in the output port.