I've found the circuit below in "The Art of Electronics" book.

This circuit generates sawtooth signal. This is some of what is written about it:

By using a current source to charge the timing capacitor, you can make a ramp (or "sawtooth-wave") generator. Figure 5.35 shows how, using a simple pnp current source. The ramp charges to (2/3)Vcc, then discharges rapidly (through the 555's npn discharge transistor, pin 7) to (1/3)Vcc, beginning the ramp cycle anew. Note that the ramp waveform appears on the capacitor terminal and must be buffered with an op- amp since it is at high impedance.

I have simulated the circuit and it works well giving a sawtooth output between (1/3)Vcc and (2/3)Vcc.

I have two important questions about this circuit:

  1. How to control the frequency of this circuit? What is the formula that can be used for this?
  2. Is there a way to make the sawtooth signal starts from zero (or somewhere else) rather than (1/3)Vcc?

1 Answer 1


1) The easiest way is by varying the 39k emitter resistor. Calling this resistor Rt, the approximate formula is $$i=\frac {15 -\frac{27k\times 15}{27k + 120k}-0.7}{R_t} $$ and $$f = \frac{i}{C\times \frac{15}{3}} $$

2) No. Or, at least, not easily. If you run the 555 from +10 and -5 volts instead of +15 and ground, the bottom of the sawtooth will be approximately at 0 volts. Or, you can use an operational amplifier to shift your voltage levels by 5 volts, but this will also require using a negative power supply.

  • \$\begingroup\$ 1.a) Thank you @WhatRoughBeast :). But the first formula for i didn't give correct results comparing to simulation results. I tried the following formula and it gave approximately correct results: $$i=\frac{15-V_E}{R_t}$$ where $$V_E=V_B+0.7$$ because it's a PNP transistor. $$V_B=15 \frac{120k}{120k+27k}$$ This assumes that $$I_E \approx I_C$$ and $$I_B \approx 0$$ which means that the transistor in the active region. What do you think? 1.b) Can you please explain how did you get the frequency equation, and what does 15/3 represents? \$\endgroup\$
    – ammar
    Commented Apr 18, 2016 at 19:21
  • 1
    \$\begingroup\$ @ammarx - Oops, forgot the 15 in the current equation. Sorry. I've edited. Your version is correct. Voltage on the capacitor changes as dv/dt = i/C = 5000 v/sec, or 5 volt/msec. The trigger points on a 555 are at 1/3 and 2/3 of Vcc, for as span of 15/3 volts. So i/C =5000, divided by 5 gives 1000 Hz. \$\endgroup\$ Commented Apr 18, 2016 at 21:04
  • \$\begingroup\$ I think it should be $$i=\frac {15 -\frac{120k\times 15}{27k + 120k}-0.7}{R_t}$$ 120k should be placed in the numerator instead of 27k. Right? \$\endgroup\$
    – ammar
    Commented May 1, 2016 at 12:08

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