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.

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 $$f = \frac{i}{C\times \frac{15}{3}}$$
• 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? Commented Apr 18, 2016 at 19:21
• 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? Commented May 1, 2016 at 12:08