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.

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?

2013-12-22 15:41