# Circuit Simulator - Emitter-Coupled LC Oscillator

#### Why should I learn to use the circuit simulator to design Emitter-Coupled LC Oscillator circuits?

Learn how to use circuit simulator software to design your own Emitter-Coupled LC Oscillator circuits.

#### Where can I get a Emitter-Coupled LC Oscillator circuit diagram with an explanation?

IndiaBIX provides numerous Emitter-Coupled LC Oscillator circuit diagrams with detailed explanations and working principles.

#### How do I design a Emitter-Coupled LC Oscillator circuit with this circuit simulator?

You can easily design Emitter-Coupled LC Oscillator circuit diagrams by practising with the given circuit simulator. With this online circuit simulator, you can design and simulate your own electronic circuits.

Emitter-Coupled LC Oscillator

Circuit Description:

This is an emitter-coupled oscillator, which uses an LC circuit combined with a transistor for feedback.

When the oscillator starts up, Q2 is conducting; the current comes from the capacitor, charging it until the voltage across it is large enough to get a current across the inductor. As the current across the inductor peaks, the output voltage rises (the inductor's voltage difference is reduced), which causes the current across Q2 to slow down.

But the inductor isn't done providing current, so the voltage across it (and the output voltage) rises as it charges the capacitor. Once the voltage turns positive, Q1 starts conducting, which raises the voltage of the two coupled emitters and prevents Q2 from conducting. This keeps Q2 from sinking any current, which causes the voltage across the inductor to rise faster, since Q1 is not sinking much base current.

Once the output voltage is at about 690mV, Q1 can draw all the current from the inductor, so the voltage across the capacitor (and inductor) peaks. As the current across the inductor peaks, the voltage drops, which causes the Q1 base current to slow down. Eventually the output voltage goes negative, which shuts off Q1 and turns on Q2. Q2 doesn't draw much base current, though, until the output voltage is at about -690 mV. At this point, the cycle begins again.

When the oscillator starts up, Q2 is conducting; the current comes from the capacitor, charging it until the voltage across it is large enough to get a current across the inductor. As the current across the inductor peaks, the output voltage rises (the inductor's voltage difference is reduced), which causes the current across Q2 to slow down.

But the inductor isn't done providing current, so the voltage across it (and the output voltage) rises as it charges the capacitor. Once the voltage turns positive, Q1 starts conducting, which raises the voltage of the two coupled emitters and prevents Q2 from conducting. This keeps Q2 from sinking any current, which causes the voltage across the inductor to rise faster, since Q1 is not sinking much base current.

Once the output voltage is at about 690mV, Q1 can draw all the current from the inductor, so the voltage across the capacitor (and inductor) peaks. As the current across the inductor peaks, the voltage drops, which causes the Q1 base current to slow down. Eventually the output voltage goes negative, which shuts off Q1 and turns on Q2. Q2 doesn't draw much base current, though, until the output voltage is at about -690 mV. At this point, the cycle begins again.

Discussion:

Be the first person to comment on this question !
Post your comments here:

Quick links

Quantitative Aptitude

Verbal (English)

Reasoning

Programming

Interview

Placement Papers