Gyrator questions

Discussion in 'General Electronics Chat' started by Piggins, Jun 28, 2014.

  1. Piggins

    Thread Starter New Member

    May 5, 2014
    It wouldnt let me reply to the older threads about gyrators.

    Okay, I was reading a tutorial about op amps and their basic use in circuits and I came upon something called a gyrator. From what I gathered from these forums it is an op amp connection that reverses the effect of the capacitor. So instead of causing a sudden current spike and gradual voltage rise/current fall we get the reverse effects.

    I found a simulator about the gyrator too.

    I did make some notes. I am referring to the falstad circuit in the text below.

    Since the op amp has a feedback it tries to hold the inputs at the same level and since the output is connected straight to one of the inputs it shares their voltage aswell. So all the three op amp connectors share the same voltage at all times.

    When the voltage changes polarity the cap and 20kohm resistor share the voltage to some extent. The voltage on the noninverting input is the same that is over the 20kohm resistor. As the cap charges that voltage is becomes lower and lower. That also means that the other input and the output will have that same voltage. Starting from a higher value at the rising and falling edge and falling to near zero volts.

    But the thing I dont understand is why the current rises as the voltage falls? The 1kohm resistor the current has to go through does not change.
    So how does this happen?

    I can understand that the max amperage comes from I = U/R and that is 5v/1000ohms tops, 5milliamps that is. But at the time the op amp outputs that 5 milliamps the voltage it outputs is at the lowest.

    Oh! I just might have gotten it now as I type this. The op amp output has to push against the input voltage that is of the same polarity at the 1kohm resistor. So, the current equals (input wave)-(op amp output) / 1kohm ?

    In the beginning when the cap has not yet charged this will be lower and going towards 5volts load the amperage will increase towards the maximum possible. This is limited by the op amp current sinking/ sourcing capabilities?
  2. dougc314


    Dec 20, 2013
    Hers my explanation of that simulation.
    In the gyrator circuit...
    First off the capacitor will not allow sudden voltage changes across it, so when the square wave input step changes, the voltage at Vpos changes exactly the same amount, because the capacitor passes the entire step. The drop across the Vneg resistor does not change, because the closed loop nature of the circuit requires the Vneg input to be equal to Vpos so it jumps the same amount as well. That means there is no instantaneous voltage change across the resistor, and no change in current. Now of course the capacitor starts to change its current to the new condition current, and its Vpos end returns to the ground state. This means the Vpos voltage slowly changes, and the Vneg tracks it, and so the current across the Vneg resistor slowly changes to Vin/R.

    In the bottom circuit, the inductor doesn't allow sudden changes in current, so every time the voltage changes the entire voltage change appears across the inductor, and the current through the resistor (and inductor) is the same (as just before the step change). Then, of course the current through the inductor starts to change slowly, to Vin/R.

    Another thing to realize (fairly obvious) that the output voltage of the opamp is indeed jumping by the change of the square wave, just as the voltage across the inductor. (It then sags towards 0)
  3. crutschow


    Mar 14, 2008
    I believe you have it. When the square-wave goes positive, the two op amp inputs are at the same voltage as the signal and the output is the same voltage as the input so no current flows. As the capacitor voltage starts to drop, due to the 20kΩ resistor, the output voltage also drops, causing the input current through the 1kΩ resistor to increase with time, similar to applying the same input voltage to an inductor.

    Yes, the maximum current is limited by the op amp maximum output current capabilities.

    This "inductance" can be used in AC filtering applications but, since it doesn't actually store energy as a real inductor, cannot be used where that characteristic is needed.