Predicting Output Voltage Using Small Signal Diode Model

[zak8000]

small signal diode modelling i understand what it is but when i get asked a question like:

if vin(t)=2.5+cos(2*pi*100*t) for 0<t<30

use the small signal model of the diode to predict the output voltage of the circuit i have added. assume the opamp is ideal R=2000 Is=10e-16

so i know that the small signal equation is

id=(ID/VT)*vd where VT=0.0252

now since the opamp is ideal zero current is drawn in and since the configuration is in the inverting configuration then Vout=-vD

now i know i have to somehow dervive and expression for Vout using the small signal diode model equation which is

id(ac current)=(ID(dc current)/0.0252))*vd(ac voltage)

i also know that ID=Is*e(vD/VT)
=(10e-16)*e(vD/0.0252)

subbing into the small signal diode model equation

id=((10e-16)*e(vD/0.0252)/0.0252)*vd

and now i get lost and stuck because i don't know what to do or if i am doing the right thing in order to derive and expression for vout using the small signal diode model

please help!

 

[dlgoff]

For a very small signal, the diode behaves more like a resistor. How could you use this fact in the op-amp circuit?

 

[marcusl]

For a very small signal, the diode behaves more like a resistor. How could you use this fact in the op-amp circuit?

I'm not so sure. The input voltage is +/-2.5V, so the diode quickly becomes fully forward biased on each positive half cycle with usual 0.6 to 0.7V drop, assuming silicon. Output waveform is asymmetric.
EDIT: Sorry, my mistake. I misread it as 2.5 * cos() instead of 2.5 + cos()

 

[waht]

I'm not so sure. The input voltage is +/-2.5V, so the diode quickly becomes fully forward biased on each half cycle with usual 0.6 to 0.7V drop, assuming silicon. Output waveform is asymmetric.

I think we are dealing with an inverting amp.

For DC analysis, one can calculate the diode bias current which is related to R,

and then for AC analysis, replace the diode with a dynamic resistance. Then it looks easy.

 

[zak8000]

would it be correct to say iD=(VD0-Vout)/rd for the circuit where VD0 is the voltage across the diode

 

[waht]

would it be correct to say iD=(VD0-Vout)/rd for the circuit where VD0 is the voltage across the diode

Yes, but for AC analysis.

For simplicity, start with doing DC analysis first. If

v_in = 2.5V,

can you find I_d?

Hint: the negative input of the op-amp is a ground.

 

[zak8000]

well for DC we could use the diode equation and say iD=vin/R and since the diode equation is iD=Ise(vD/VT) we could rearrage it to

Vout= -VT*ln(vin/R*Is) for DC

and then you have stated replace the diode with the dynamic resistance are you referring to rd the resistance across the diode?