3
\$\begingroup\$

enter image description here

The output voltage of an astable multivibrator is not a perfect square wave; the voltage when viewed from an oscilloscope tends to resemble a curve. I'm told that this is because the voltage across the capacitors cannot change immediately and therefore the R1 capacitor current somehow disrupts the voltage waveform at the transistor collector. Apparently this can be fixed by isolating the capacitor from the collector using a diode:

enter image description here

Now C1 charges through its own resistor, R2. But I don't understand how the R1 current in the first picture disrupts the collector voltage. When Q1 turns off it's off and therefore above ground. How does the capacitor alter this? And when the diodes are included, how is the situation different? To me it seems that the collector is simply around 0.7 volts lower than the capacitors R2 plate.

\$\endgroup\$

1 Answer 1

Reset to default
2
\$\begingroup\$

The point of the diodes is that they stop conducting when the voltage on the collector rises when the transistor switches off.

In the first schematic, when Q1 opens the voltage on the collector of Q1 cannot rise immediately to V+ as C1 is charged to about -0.6 V when Q1 was conducting. So when Q1 switches off C1 wants to keep the collector of Q1 low. A current needs to flow through R1 to charge C1. This charge current causes a voltage drop across R1 and that prevents the voltage on the collector of Q1 to rise immediately to V+.

In the second schematic this works differently. As soon as TR1 switches off R1 can pull the collector to +9V. Now D1 prevents a charge current flowing through R1, instead all the charge current will need to come through R2. As no current will flow through R1 the voltage on the collector of TR1 can rise to +9V immediately.

\$\endgroup\$
7
  • \$\begingroup\$ Rather than immediately you should say much more quickly. \$\endgroup\$
    – Michael Karas
    Jul 6, 2018 at 10:42
  • \$\begingroup\$ A problem can exist with the circuit with the diodes in that it may not startup reliably. To circumvent this problem it is often necessary to put a small value capacitor from one of the transistor bases to GND to imbalance the circuit at power up. This can be an even more critical problem in simulation where the two transistors and the two diodes have identical characteristics. \$\endgroup\$
    – Michael Karas
    Jul 6, 2018 at 10:52
  • \$\begingroup\$ @MichaelKaras Interesting, I have never heard of putting a capacitor between GND and transistor base. Could you tell me more about this, how does it unbalance the circuit? \$\endgroup\$
    – S. Rotos
    Jul 10, 2018 at 7:59
  • \$\begingroup\$ If both sides of the circuit are identical (note how it is symmetrical) then at startup both sides compete to be "first". Then noise and random events determine which side will be first and the oscillation starts. It is possible that this takes some time and that could prevent a quick startup. That can be fixed by deliberately slowing down one of the sides, for example to slow down the TR1 side, add a small capacitor (10 pF or 100 pF) across the base-emitter of TR1. Then TR2 will always "win" and startup becomes more reliable. \$\endgroup\$
    – Bimpelrekkie
    Jul 10, 2018 at 8:06
  • \$\begingroup\$ @S.Rotos - Like Bimpel has said the capacitor causes a small delay to the turn-on of the transistor to which it is attached. The delay is due to the extra time it takes for the base of that transistor to reach its V<sub>BE</sub> threshold before the other transistor reaches that point. The small capacitor will have a minimal effect on the operating frequency or duty cycle of the astable multivibrator. \$\endgroup\$
    – Michael Karas
    Jul 10, 2018 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.