I'm given this 555 timer based monostable saw-tooth generator. This is a known arrangement for this part. The difference from other circuits I've seen is that the bias network of the external transistor is composed from a Zener + regular (small signal) diode. Why is the Zener diode been used? What advantage does it have over a simple resistor and why is the other diode needed?

555 saw tooth generator


2 Answers 2


The PNP transistor is part of a constant current source for the capacitor. The current is determined by the voltage across \$R_E\$ (ignoring the relatively small base current).

The voltage across \$ R_E\$ is \$ V_Z - V_{BE} + V_D \$ where \$V_D\$ is the forward voltage across the diode, so we can say that \$I_C \approx\$ \$ (V_Z - V_{BE} + V_D)\over R_E \$

Although the currents through the diode and the base-emitter junction are typically quite different, the voltages are not that far apart, and the magnitude of the temperature coefficients of both will be about 2mV/°C, and they will tend to cancel.

In that case, \$I_C \approx\$ \$ V_Z\over R_E \$

The Zener voltage in this circuit must be less than 5V for this circuit to work (as the capacitor has to charge to 2/3 of 15V). The temperature coefficient of zeners in the 4.3-4.7V range is approximately zero.

(Ref Motorola Semiconductor Data Library 1N47xx series)

enter image description here

The combination of the zener diode and the diode stabilizes the constant current against both temperature changes and changes in the supply voltage, however it actually increases the variation in time with supply voltage since the threshold voltage of the 555 is derived by a divider from the supply voltage.

Edit: In reference to an additional question, I include a "Functional Table" from a version of the 555 datasheet (CMOS type, but the logic is the same).

enter image description here

  • \$\begingroup\$ I only have one question regarding what you wrote here... If the trigger level is held low, the capacitor starts charging (assuming it was discharged beforehand), this goes on until Vc crosses 2/3Vcc (10V) but if trigger is still held low at that point, Vc will only be limited by the transistor right? So it can go over 2/3Vcc. \$\endgroup\$
    – user34920
    Mar 8, 2014 at 11:42
  • \$\begingroup\$ Yes, according to the function table (see edit above), the "S" is dominant over the "R" on the internal flip-flop, so Vc will increase until the transistor saturates (a bit more than \$V_{RE}\$ below 15V). \$\endgroup\$ Mar 8, 2014 at 18:05

From the capacitor charging equation, CV = It

V =( I/C)t

So, if I is made constant , V =kt where k =(I/C) It is a Ramp generator

So, the transistor and zener creates a constant current source . D1 is used to compensate for B-E drop of the transistor as well as for temperature compensation. Hence, current IC =IE = Vz/RE = constant as required. Normal resistor used in the place of zener won't provide constant voltage in the case of line voltage variation or temperature variations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.