0
\$\begingroup\$

Consider a circuit consisting of a variety of linear and/or non-linear elements and independent voltage and/or current sources. Now imagine a dependent current or voltage source is added to an arbitrary part of the circuit, and assume that the voltage or current the source depends on is non-zero. The dependent source would generate a non-zero current or voltage which would affect the rest of the circuit.

It's easy to imagine that the dependent source would affect the current or voltage it depends on, which in turn would change the magnitude of the dependent source, which in turn would affect the current or voltage it depends on again, essentially creating an infinite loop of the dependent source and the voltage/current it depends on increasing each other until both reach infinity.

An analogy might help. A man tells his friend that for everything dollar he has, he will give him another. (Analogous to the dependent source "agreeing" to "contribute" an amount that is a function of what another part of the circuit "has"). His friend starts off with $10, so the man pays him $10. His friend now has $20, so the man now pays him another $20. The man now has $40; this continues on until the man is paying his friend an infinite amount of money.

\$\endgroup\$
  • \$\begingroup\$ It's called positive feedback. \$\endgroup\$ – Andy aka Sep 16 '14 at 7:20
0
\$\begingroup\$

Yes, absolutely it can make a circuit unstable.

A voltage controlled voltage source is nothing more than a gain block, and if you have other circuit elements that cause the open loop phase to be greater than -180 degrees when the open loop gain crosses 0dB you have an oscillator when you close the loop.

Same goes for other dependent sources with appropriate circuit elements.

So you can easily make an unstable circuit with a dependent source, but there's no reason you can't design a stable circuit as well.

\$\endgroup\$
  • \$\begingroup\$ Why are not all circuits containing dependent sources unstable? Wouldn't my analogy be valid for all circuits containing one? If not, what part of the analogy would fail? \$\endgroup\$ – Phidias Sep 15 '14 at 23:57
  • \$\begingroup\$ @FarhadYusufali, if the output of the controlled source can affect the input to the controlled source, there is feedback. If the feedback is negative, the circuit is stable. If the feedback positive, the circuit is unstable. Have you studied feedback? \$\endgroup\$ – Alfred Centauri Sep 16 '14 at 0:10
  • \$\begingroup\$ @A.Centauri:In his answer, JohnD mentioned -180deg phase shift (for loop gain>0 dB) as a valid criterion for instability. Hence, he assumes negative feeedback which is correct. That means: Negative feedback for low frequencies ensures DC stability only! Positive feedback always causes instability, but negative feedback requires additional considerations for stability (For example: Nyquist criterion). \$\endgroup\$ – LvW Sep 16 '14 at 7:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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