This is a circuit I have designed to work as an oscillator. But it doesn't work.

enter image description here

The way I supposed this works:

When power is ON, the capacitor(C1) starts charging through 1Kohm resistor. Once it reaches 0.7 volt, it trigger the Q1 which will almost short the point A with ground. This makes the C1 discharged which will switch Off Q1. The operation repeats. But This doesn't work and I am always getting the base voltage as 0.7v.

What is the mistake I am making?

Note: I have not connected any LED or speaker to this circuit. I just used a DMM to check the base voltage to see whether it is fluctuating.

  • \$\begingroup\$ "I just used a DMM to check the base voltage..." That's one mistake. \$\endgroup\$ – Ignacio Vazquez-Abrams May 13 '16 at 5:15
  • \$\begingroup\$ @IgnacioVazquez-Abrams: So this circuit has no problem in design? \$\endgroup\$ – InQusitive May 13 '16 at 5:16
  • \$\begingroup\$ I'm pretty sure it does. After all, the collector is tied directly to the base. But you won't be able to see fluctuations with a DMM unless they're very, very slow. \$\endgroup\$ – Ignacio Vazquez-Abrams May 13 '16 at 5:21
  • \$\begingroup\$ @IgnacioVazquez-Abrams: But what about the Hz measurement of multimeter. It is also giving zero. Atleast this has to work? \$\endgroup\$ – InQusitive May 13 '16 at 5:23


simulate this circuit – Schematic created using CircuitLab

By shorting out the collector-base junction you have converted the transistor into a diode. Diodes can't amplify current. The circuit will not oscillate.

  • \$\begingroup\$ Got it. Is it just for collector base? What if I short Emitter-Base? Is that also a diode equivalent?(In general, not in this circuit) \$\endgroup\$ – InQusitive May 13 '16 at 6:19
  • \$\begingroup\$ Yes. If you test a transistor with your multimeter on "diode" test range you should get a diode reading from b-e and b-c and no reading when leads reversed. It's like two diodes back to back. This is useful for identifying the base and doing a quick "dead or alive" check on the device. \$\endgroup\$ – Transistor May 13 '16 at 6:22

Your misconception about this circuit is that you assume that the transistor "suddenly" does something when its Vbe voltage reaches a certain value.

This does not happen ! What does happen is a much more gradual process. So the transistor will start to conduct a little current when the voltage is around 0.5 V and will increase drastically from there. It will follow a diode IV curve because you shorted Base and Collector making the transistor behave as a diode.

This is a diode IV curve: enter image description here

Note how the curve in the upper right corner has a gradual voltage-current relation and not an abrupt off/on behavior.

So your capacitor will charge to the voltage that is determined by the transistor behaving as a diode. That will be about 0.7 V. Then nothing more will happen.


when the transistor is on the capacitor wont discharge. the current will come from the voltage supply. you can remove the capacitor and nothing will change. for building an oscillator you need an inductor parallel to a capacitor and negative resistor for the oscillation.

  • \$\begingroup\$ Why the current goes to capacitor? If the transistor is ON , current has got least resistance path through Collector-Emitter, so even the capacitor should discharge through this path. Wrong? \$\endgroup\$ – InQusitive May 13 '16 at 6:01
  • \$\begingroup\$ at first the base is 0v. current flows into the cap and charges it until it reaches 0.7 volt and will turn on the transistor. after this the current will go to base and collector. dont \$\endgroup\$ – shayan360 May 13 '16 at 6:35

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