Transistor current control with positive feedback differential amplifier

Hi Folks,

I want to analyse the functionality of following OpAmp configuration which seems to be employed as voltage - current conversion. Positive feedback (This is not a Failure and is intentionally done so !) from current sense resistor 1 Ohm is being fed back to Op Amp and compared with reference voltage +15V while being added to \$V_{in}\$. The error signal is amplified by open loop gain of OpAmp to drive the transistor emitter current. The transistor emitter current is almost equal to transistor collector current.

I would appreciate if someone can add further to my analysis.

$$V_{out} = A \cdot (V_+ - V_-)$$

$$Ie = \dfrac{V_{out} - V_{be}}{R_e}$$

\$I_c\$ is approximately equal to \$I_e\$, voltage at feedback node is:

$$V_c = 15 - I_c$$

\$V_+\$ is again depending on output current \$I_c\$, which can be solved using simultaneous equations.

The problem is, when this circuit is simulated in LtSpice the gain of OpAmp which is mentioned to be 85dB in data sheet as typical value doesn't comply exactly with the output voltage from simulation but only for a small range of input voltages. For different range of input values \$V_{in}\$, only certain values of calculated output voltages \$V_{out}\$ come out to be same as from simulation and few even exceed \$+V_{cc}\$ which is not possible in simulation as it saturates at \$+V_{cc}\$.

It looks as if either the gain of OpAmp varies with different input voltages or my mathematical modelling has errors.

I would like to share your experience generally about the configuration, and any hints about mathematical transfer function from \$V_{in}\$ to output emitter current of NPN transistor.

Answer

This circuit is negative feedback due to the inversion provided by the common emitter transistor amplifier.

The collector voltage is: \$V_c=15-I_c\$.

Now, the op amp summing junction must be at 15 V, therefore the relationship for the voltage divider formed by \$R_2\$ and \$R_f\$ is:

$$\frac{V_c-15}{R_f}=\frac{15-V_{in}}{R_2}$$

Substituting \$V_c=15-I_c\$ gives: $$I_c=\frac{R_f}{R_2}(V_{in}-15) $$

Which is a voltage controlled current source.