# How to change the frequency of this sawtooth generator circuit?

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) 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.

• 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? – ammar Apr 18 '16 at 19:21
• @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. – WhatRoughBeast Apr 18 '16 at 21:04
• 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? – ammar May 1 '16 at 12:08