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So I am trying to develop a saw tooth generator using a 555 timer however I have a couple of problems regarding the calculations. The saw tooth generator is built in this way:

Saw tooth wave generator using 555 timer

The mathematical way to describe the behavior of this is by the following formula: F = (Vcc-2.7) / (RCVpp)

where Vcc is the voltage from the power supply, R stands for Resistance, C for the capacitor and Vpp for the output voltage of the wave generated.

My problem is that I need the saw tooth wave to last 60 seconds which is 0.016 Hz, a really low frequency, also it will use a 12V power supply. The voltage Vpp I have no idea how to find it and it is exactly the value that is affecting my circuit. According to my research the Vpp value comes directly from the input voltage used, this means the R and C values won't affect it, therefore I can't use the equation to Vpp's value. How can I find the Vpp value?

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    \$\begingroup\$ One way would be to build the circuit and stuff in some values of R and a 100uF C and see what you get. Then adjust accordingly. \$\endgroup\$ – Michael Karas Feb 4 '17 at 2:44
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    \$\begingroup\$ You should study the workings of a 555 chip. The cap C will charge up the upper threshold detected by the pin 2 TRIG input. That will be two thirds of the VCC level. \$\endgroup\$ – Michael Karas Feb 4 '17 at 2:56
  • \$\begingroup\$ In the previous comment I meant the pin 5 THRESHOLD pin. Not pin 2. \$\endgroup\$ – Michael Karas Feb 4 '17 at 3:06
  • \$\begingroup\$ Typo in your title Danny, should be -> frequencies \$\endgroup\$ – crowie Feb 4 '17 at 5:22
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A common alternate starting place is to plug the circuit into simulator and see what you get. I did this with the following similar circuit:

enter image description here

The simulated saw tooth waveform came out as:

enter image description here

Note that in my circuit I replaced the zener diode with two forward biased signal diodes. One diode drop approximates the Vbe drop of the PNP transistor. This means that the R1 resistor has an ~0.65V drop across it which defines the current through the resistor. This is an easy way to make a current source that works with a small drop across the resistor.

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  • \$\begingroup\$ I tried to build this circuit, however the 2.2k resistor keeps on melting, which shouldn't be happening according to the simulation. \$\endgroup\$ – Danny Feb 10 '17 at 0:34
  • \$\begingroup\$ @Danny - Yeah, if that 2.2K resistor is melting you must have something built wrong or shorting out. As the schenatic stands that resistor should have ~10.5V across it. Power dissipated in the resistor should be about 50mW. Even an 0402 SMT resistor should be able to survive that. \$\endgroup\$ – Michael Karas Feb 10 '17 at 1:32
  • \$\begingroup\$ now the interesting thing about this is that I replaced that resistor with a 100K and everything works exactly the same, both in the simulation and in real life. Despite being exactly the same circuit. \$\endgroup\$ – Danny Feb 10 '17 at 4:08
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More than 10 seconds is a problem for the old bipolar 555 and leaky electrolytic capacitors. But using the CMOS LMC555 version you can get a 60 second delay with 10M and 10uF. However this wont be a linear sawtooth. This requires a constant current source using a low input bias current FET Op Amp and a low leakage plastic cap.

If you want a true linear sawtooth, you must pay the price of a precision CC design with a 10 GOhm leakage resistance Cap and picoamp input bias OA for a 60 second cycle.

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  • \$\begingroup\$ I used to generate 1 hour perfect linear sawtooths (or any long term stable slope rate) by S&H mixing using 2 much higher frequencies with beat frequency of 2 OCXO crystal oscillators with one square wave converted to triangle then a sawtooth and the other pulse S&H of the sawtooth. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 4 '17 at 4:12
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For a 555, the value of Vpp will be 2/3 the power supply voltage minus 1/2 the output voltage + a diode drop. Output voltage of a bipolar 555 will be about 100mV, closer to 0 for a CMOS 555. The diode drop will be about 0.7V.

From the TI datasheet:

enter image description here

The three equal-valued resistors R split the supply voltage V into V/3 and 2V/3, so the cap voltage would swing between those two voltages, with a total swing of V/3, but the added diode between pins 3 and 5 means that the control voltage is reduced to Vo + Vf when the output is low (assuming a bipolar 555) so the lower voltage will be about 0.4V. Vpp will thus be 8V-0.4V = 7.6V

The biggest variable in your circuit is likely the 2.7V Zener voltage. You may need to set the Zener current near to the datasheet Iz to get close to 2.7V. Another possibility would be to use a green LED rather than the Zener and adjust the resistor value according to the Vf of the LED. That could allow a lower current.

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  • \$\begingroup\$ Not so....the capacitor is discharged to zero and the comparator at 2V/3 is the termination point for charging. The lower V/3 is used for the trigger comparator. \$\endgroup\$ – Jack Creasey Feb 4 '17 at 4:55
  • \$\begingroup\$ @JackCreasey thanks- I missed the diode which reduces 1/3 V to (Vf + Vo)/2 or about 0.4V. Without that the sawtooth goes from 1/3 to 2/3 because the discharge turns off at that point (trigger). Michael's simulation confirms this result. \$\endgroup\$ – Spehro Pefhany Feb 4 '17 at 7:52

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