Draw an equivalent circuit. The purpose of voltage clamp is to set membrane pote

Question

Draw an equivalent circuit. The purpose of voltage clamp is to set membrane potential Vm to clamped value Vc, i.e. making Vm as close to Vc as possible. Assuming the system in steady-state (not transient) so that you can ignore membrane capacitance Cm, derive relationship between Vm and Vc. Now consider transient response. Consider Vc has a abrupt change (e.g. voltage step) thus you have to consider Cm. Derive a differential equation for output Vm using Vc as input. Is the transfer function high pass or low pass? What is the time constant? Ignore the class slide on voltage clamp, which is an illustration but not accurate. You can define K =

Explanation / Answer

1. buffer is nothing but an opamp with unit gain. if we assume the next opamp has gain Av then the equivalent circuit will be, an opamp with the inverting input is fedback from the output through resistances Ro,Rc,Rm and a capacitor Cm to ground.

2.now if the voltage at the inverting port is Vx. then, we can write,

Vout = Av*(Vc - Vx) ------------------------1, where Av is the gain of the second opamp.

so,

Icl = (Vout - Vm)/(Ro + Rm)------------------------------2

again,

Vx = Vm - Icl*Rc------------------------------------3

from the above three equation we get,

Icl*(Ro + Rm - Av*Rc) + Vm*(1+Av) = Av*Vc-----------------------------ans

3.

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