I have a typical Opamp diode log amplifier. formula for it is.Vout=-Vth*ln(Vin/IsR) The positive side of amp has a resistor and then is tied to ground.How does that extra resistor impact my standard forumla.
The Opamp diode log amplifier you mentioned is a typical configuration where an operational amplifier (Opamp) is used along with a diode to implement a logarithmic function of the input (Vin). The formula you provided for the output voltage (Vout) is Vout = -Vth * ln(Vin / (Is * R)).
In this configuration, the positive side of the Opamp is connected to a resistor (R) which is then tied to ground. This additional resistor affects the transfer function of the amplifier. Let's see how it impacts the standard formula.
The transfer function of an Opamp diode log amplifier is derived based on the input-output relationship of the Opamp and the diode. The diode characteristic equation is given by I = Is * exp(Vd / Vth), where I is the current through the diode, Vd is the voltage across the diode, Is is the reverse saturation current, and Vth is the thermal voltage.
To calculate the output voltage, we consider the virtual ground assumption, which states that the voltage at the inverting input of the Opamp is approximately equal to the voltage at the non-inverting input (ground in this case).
Applying this assumption, we can write:
Vin / R = (Vout - 0) / (R + (Vin / Is))
Simplifying this equation, we get:
(Vout - 0) / Vin = R / (R + (Vin / Is))
Now, let's solve for Vout:
Vout = Vin * (R / (R + (Vin / Is)))
Comparing this equation with the standard formula you provided, we can see that the additional resistor (R) modifies the transfer function. Instead of directly multiplying the logarithm of the ratio (Vin / (Is * R)) with -Vth, we now multiply it with (R / (R + (Vin / Is))).
Therefore, the impact of the extra resistor is to shape the response of the log amplifier by introducing a nonlinearity and affecting the gain of the amplifier. It alters the relationship between the input voltage and the output voltage.