Potentiometric Mixers

The schematic of a potentiometric mixer is shown in Figure 1.

Potentiometric Mixer

Figure 1. Schematic of a potentiometric mixer

Transistors M1-M4 operate in triode region, and act as resistors.

V_x and V_y form a virtual short through capacitor CV for high frequency signals.

The RF signals v_{RF^+} and v_{RF^-} control the drain resistance of respective transistor.

Drain to source voltage across each transistor is determined by v_{LO}^+ and v_{LO}^-.

Drain currents in the MOS transistor are given by,

(1)   \begin{eqnarray*} i_1 &=& \beta \left((v_{RF^-} - v_T)v_{LO^+} - {{v_{LO^+}^2 \over 2} \right) \\ i_2 &=& \beta \left((v_{RF^+} - v_T)v_{LO^+} - {{v_{LO^+}^2 \over 2} \right) \\ i_3 &=& -\beta \left((v_{RF^+} - v_{LO^-} - v_T)v_{LO^-} - {{v_{LO^-}^2 \over 2} \right) \\ i_4 &=& -\beta \left((v_{RF^-} - v_{LO^-} - v_T)v_{LO^-} - {{v_{LO^-}^2 \over 2} \right) \\ \end{eqnarray*}

The differential output current is given by i_o = i_{op} - i_{om} =(i_1 - i_3) - (i_2 - i_4).

After rearranging above equation and substituting Eq-(1),  it can simplified to

(2)   \begin{eqnarray*} i_o &=& = (i_1 - i_2) - (i_3 - i_4)\\ &=& \beta \left( (v_{RF^-} - v_{RF^+})v_{LO^+} + (v_{RF^+} - v_{RF^-})v_{LO^-} \right)\\ &=& -\beta (v_{RF^+} - v_{RF^-})(v_{LO^+}-v_{LO^-}) \\ \end{eqnarray*}

The differential output voltage is given by

(3)   \begin{equation*} v_o = v_{op}-v_{om}=-(i_{op}-i_{om})R_L = \beta R_L \underbrace{(v_{RF^+} - v_{RF^-})(v_{LO^+}-v_{LO^-})}_{\omega_{LO}\pm\omega_{RF} \text{components}} \end{equation*}

In the absence of mismatch between different paths, the output contains only the sum and different frequency components.  \omega_{LO} + \omega_{RF} is filtered out, whereas the opamp should have the bandwidth  to handle \omega_{LO} - \omega_{RF} .

Leave a Comment