Linear Current Sources

Op Amp Current Sources

Because in the feedback path of inverting and non-inverting amplifier the current is proportional to the incoming voltage these two circuits may be used as linear voltage controlled current sources.

$\begin{circuitikz} \draw (5,2) node[op amp] (opamp) {} (0.5,2.5) to [short, o-, i^>=$I_{in}$] (2,2.5) to [R, l=$R_{in}$] (opamp.-) (0.5,2.5) node[above] {$V_{in}$} (3.8,1) node [ground] {} to [short] (opamp.+) (3.8,3.5) to [R, l=$R_{load}$] (6.2,3.5) (3.8,3.5) to [short, -*] (opamp.-) (6.2,3.5) to [short, -*, i^>=$I_{load}$] (opamp.out) (opamp.out) to [short, -o] (7, 2);\end{circuitikz}$

The current through $R_{load}$ is given by $I_{load} = \frac{V_{in}}{R_{in}}$ . The main drawback of this design is that the load connected instead of $R_{load}$ must not be grounded (or generally speaking: connected to a fixed potential) because this would cause a short circuit of the op amp's output.

TODO: op amp current source for grounded loads, derivations, input+output impedance

Transistor Current Sources

Simple NPN current sink

$\begin{circuitikz} \draw (3,2) node[npn] (npn) {} (1,2) node[above] {$V_B$} to [short, o-] (npn.base) (3,5) node[above] {$V_{cc}$} to [short, o-, i^>=$I_{C}$] (3,4) (3,4) to [R, l=$R_{load}$] (npn.collector) (npn.emitter) to [R, l=$R_E$] (3,0) to [short, i^>=$I_E$] (3,-.5) (3,-.5) node[ground] {} ;\end{circuitikz}$

First, to make the transistor conduct, we choose a control voltage $V_B > 0.6$ . Thus $V_E = V_B - 0.6V$ . The emitter current is

$I_E = \frac{V_E}{R_E} = \frac{V_B-0.6V}{R_E}$

If the transistor is not saturated

$I_C \approx I_E = \frac{V_B-0.6V}{R_E}$

which gives the current that the NPN transistor sinks.

Simple PNP current source

$\begin{circuitikz} \draw (3,2) node[pnp, yscale=-1] (pnp) {} (3,4) to [short, o-, i^<=$I_{C}$] (pnp.collector) (pnp.emitter) to [R, l=$R_E$] (3,0) (pnp.base) to [short] (1,2) to [short] (1,1) (1,1) node[ground] {} (1,-1) to [short, o-] (3,-1) to [short, i^>=$I_E$] (3,0) (1,-1) node[above] () {$V_E$} ;\end{circuitikz}$

The derivation for a simple PNP transistor current source is similar to that for the NPN case. This time we have the current flowing into the emitter. If $V_E > 0.6$ (breakpoint voltage of a typical BJT) the current $I_C$ sourced at the collector is given by

$I_C \approx I_E = \frac{V_E-0.6V}{R_E}$

This configuration is useful for instance to drive the control current input of an OTA.

simple FET current source
precision current sources with transistor in feedback path of op amp

Current Mirror

A current mirror is a special variety of a current controlled current source. The incoming current controls an outgoing current of opposite sign, hence the term current mirror. The simplest realisation is a voltage controlled current source in which the incoming control current is converted to a control voltage using a resistor according to Ohm's law. To create current mirrors with different input impedances one may instead connect a voltage controlled current source in series with a current to voltage converter (cf. OpAmpCircuits).

TODO: derivations, circuits, is this interesting because of OTAs or expendable?