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Is this the way this circuit works ?

circuit

  1. If vref = 5v on the pin #5, then Vthreshold = 5v, and Vtrigger = 2.5v

  2. With Vthreshold = 5v, if voltage on pin #6 Vp6 > 5v, then Vp3 is Vo = 0

  3. With Vtrigger = 2.5v, if voltage on pin #2 Vp2 < 2.5v, then Vp3 is Vo = Vcc

  4. The transistor works as a current source, giving:

    Ic = 15,089mA

Does the capacitor start to charge when the trigger activates in t = T/2, meaning that Vc = 0 until Vp2 < 2.5v ?

While the capacitor is charging, Vo = Vcc ?

And then when the voltage on the capacitor hits the Vc = Vthreshold = 5v, discharges and Vo = 0 ?

Does it work like this ?

NOTE: The max voltage of Vo is VCC - 1.5v, not 5v like in the picture.

Voltages

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  • \$\begingroup\$ Do you really think that circuit, with those resistor values, produces a constant current of over 15 amps? I get 4 mA. \$\endgroup\$
    – AnalogKid
    Oct 26, 2020 at 0:05
  • \$\begingroup\$ This is how i did it: VbaseQ1 = (15*5.6k)/(5.6k + 3.6k), then VbaseQ1 = 9.1304v \$\endgroup\$
    – Samu R
    Oct 26, 2020 at 12:36
  • \$\begingroup\$ With V330 = 15 - 0.7 - VbaseQ, V330 = 5.1696v \$\endgroup\$
    – Samu R
    Oct 26, 2020 at 12:38
  • \$\begingroup\$ If Ie = V330 / 330, Ic = Ie = 15.6654 mA \$\endgroup\$
    – Samu R
    Oct 26, 2020 at 12:39
  • \$\begingroup\$ You do not state the value of Vcc in your text. The highest voltage mentioned in your text is 5 V, so I used that for Vcc. At 15 V and Vbe = 0.6 V, it comes to 16 mA. \$\endgroup\$
    – AnalogKid
    Oct 26, 2020 at 14:00

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