This first part of the circuit works as follows:

• Assume C1 is discharged to begin with.

• R1 just lets current flow through D1,D2 (if there was no R1, it would be an open circuit), which ensures a known voltage at the base of Q1, since diodes drop a constant voltage.

• Since we know the voltage at the base of Q1, we also know the voltage at its emitter (it behaves as diode), which means we know the drop at R2. It ends up being the same a a diode's voltage drop. Another way to put is is look at the drops from the two branches: D1-D2 and R2-Q1base. Since the diodes drop approximately the same voltage (including Q1's Vbe), then the voltage dropped by R2 will be the same as a diode's drop.

• Since we know the voltage dropped by the R2, we know its current.

  The current is then: $$ I_{R2} \approx V_{diode}/R2 $$

• This current will charge the capacitor C1, and the voltage described will be a linear ramp, because the voltage in a capacitor is proportional to its charge, and we are charging it a constant rate.

• The capacitor C1 will get charged until its voltage, which is the same as the transistor's collector voltage, gets high enough that Vce is too low and Q1 it is not able to provide any more current (hFE drops to zero).

• In the real circuit, this voltage will trigger a discharge before this point is reached, and the process repeats.

You see then that the saw tooth signal is generated by the ability to generate a constant current that charges the capacitor, which generates a voltage ramp. Another way to put it is that we generated a constant signal (the current), and integrated it (with the capacitor), generating a ramp (the capacitor's voltage).

This first part of the circuit works as follows:

• Assume C1 is discharged to begin with.

• R1 just lets current flow through D1,D2 (serves as bias resistor), which ensures a known voltage at the base of Q1, since diodes drop a constant voltage.

• Since we know the voltage at the base of Q1, we also know the voltage at its emitter (it behaves as diode), which means we know the drop at R2. It ends up being the same a a diode's voltage drop. Another way to put is is look at the drops from the two branches: D1-D2 and R2-Q1base. Since the diodes drop approximately the same voltage (including Q1's Vbe), then the voltage dropped by R2 will be the same as a diode's drop.

• Since we know the voltage dropped by the R2, we know its current.

  The current is then: $$ I_{R2} \approx V_{diode}/R2 $$

• This current will charge the capacitor C1, and the voltage described will be a linear ramp, because the voltage in a capacitor is proportional to its charge, and we are charging it a constant rate.

• The capacitor C1 will get charged until its voltage, which is the same as the transistor's collector voltage, gets high enough that Vce is too low and Q1 it is not able to provide any more current (hFE drops to zero).

• In the complete circuit, this voltage will trigger a discharge before this point is reached, and the process repeats.

You see then that the saw tooth signal is generated by the ability to generate a constant current that charges the capacitor, which generates a voltage ramp. Another way to put it is that we generated a constant signal (the current), and integrated it (with the capacitor), generating a ramp (the capacitor's voltage).

This first part of the circuit works as follows:

• Assume C1 is discharged to begin with.

• R1 just lets current flow through D1,D2 (if there was no R1, it would be an open circuit), which ensures a known voltage at the base of Q1, since diodes drop a constant voltage.

• Since we know the voltage at the base of Q1, we also know the voltage at its emitter (it behaves as diode), which means we know the drop at R2. It ends up being the same a a diode's voltage drop. Another way to put is is look at the drops from the two branches: D1-D2 and R2-Q1base. Since the diodes drop approximately the same voltage (including Q1's Vbe), then the voltage dropped by R2 will be the same as a diode's drop.

• Since we know the voltage dropped by the R2, we know its current.

  The current is then: $$ I_{R2} \approx V_{diode}/R2 $$

• This current will charge the capacitor C1, and the voltage described will be a linear ramp, because the voltage in a capacitor is proportional to its charge, and we are charging it a constant rate.

• The capacitor C1 will get charged until its voltage, which is the same as the transistor's collector voltage, gets high enough that Vce is too low and it is not able to provide any more current.

• In the real circuit, this voltage will trigger a discharge before this point is reached, and the process repeats.

You see then that the saw tooth signal is generated by the ability to generate a constant current that charges the capacitor, which generates a voltage ramp. Another way to put it is that we generated a constant signal (the current), and integrated it (with the capacitor), generating a ramp (the capacitor's voltage).

This first part of the circuit works as follows:

• Assume C1 is discharged to begin with.

• R1 just lets current flow through D1,D2 (if there was no R1, it would be an open circuit), which ensures a known voltage at the base of Q1, since diodes drop a constant voltage.

• Since we know the voltage at the base of Q1, we also know the voltage at its emitter (it behaves as diode), which means we know the drop at R2. It ends up being the same a a diode's voltage drop. Another way to put is is look at the drops from the two branches: D1-D2 and R2-Q1base. Since the diodes drop approximately the same voltage (including Q1's Vbe), then the voltage dropped by R2 will be the same as a diode's drop.

• Since we know the voltage dropped by the R2, we know its current.

  The current is then: $$ I_{R2} \approx V_{diode}/R2 $$

• This current will charge the capacitor C1, and the voltage described will be a linear ramp, because the voltage in a capacitor is proportional to its charge, and we are charging it a constant rate.

• The capacitor C1 will get charged until its voltage, which is the same as the transistor's collector voltage, gets high enough that Vce is too low and Q1 it is not able to provide any more current (hFE drops to zero).

• In the real circuit, this voltage will trigger a discharge before this point is reached, and the process repeats.

You see then that the saw tooth signal is generated by the ability to generate a constant current that charges the capacitor, which generates a voltage ramp. Another way to put it is that we generated a constant signal (the current), and integrated it (with the capacitor), generating a ramp (the capacitor's voltage).