Electronic – Zener Diode Current Source

Q1 and Q4 are voltage followers. The function of the 3.5-µA current sources on their emitters is simply to provide their operating current. Keeping the current constant improves their speed.

Q2 and Q3 are a differential pair, also called a "long-tailed pair". Both the differential gain and the CMRR increase with the value of the shared emitter impedance. A current source has very high equivalent impedance (ideally infinite), which is why it is used here.

Q7 is a simple common-emitter amplifier. Again, its gain is directly related to the collector load impedance, so a current source is used here for high gain.

As far as implementation, the IC designer has more flexibility than you'll have with discrete parts. Usually, a bipolar current-mirror circuit is used (similar to the Q5-Q6 pair shown above), with one "master" transistor setting a reference current used by several "slave" devices. The current sources you see are all slave devices, and the different currents are set by designing those transistors with different sizes (emitter areas). The ratio of the slave emitter area to the master emitter area determines the slave output current relative to the reference current.

The voltage across a capacitor is the integral of the current through it. If you feed a constant current to a capacitor, its voltage ramps up linearly, which is exactly what you want for a sawtooth waveform generator.

Yes, you're correct that this cannot continue forever; the complete waveform generator circuit will be discharging the capacitor periodically in order to prevent the constant-current circuit from saturating.

In the circuit you show, D1, D2 and R1 provide a voltage reference, in this case, approximately 1.2V, which is the total forward drop across the two diodes. This establishes the voltage across the B-E junction of Q1 and R2. Since the voltage across the B-E junction is also a diode drop, or 0.6V, this means that the remaining voltage, 0.6V appears across R2.

From Ohm's law, we now know the current through R2, 0.6V/2200Ω = 273µA. This is the current that is charging C1.

The voltage across the capacitor is a function of time: V = I×t/C. Let's rewrite this as V/t = I/C, which means that the rate of change of the voltage is the current divided by the capacitance. In this case, 273µA/0.1µF = 2730 V/s, or equivalently, 2.73 V/ms.