Double Balanced Switching Mixer


The schematic of a double balanced  switching mixer circuit is shown in Figure 1. The double balanced mixer can be realized by connecting the input and output terminal of two single balanced circuits appropriately.

Double-Balanced-Switching-Mixer

Figure 1. Schematic of double balanced switching mixer

The differential current in a single balanced mixer is given by

(1)   \begin{eqnarray*} i_{o,I} &=& i_{o2,I}-i_{o1,I}\\ &=& g_m v_{i,p}(t) \left( -{4\over \pi} \sum\limits_{n=1,3,5}^{\infty}{1\over n}\sin(n\omega_{LO}t)\right) \end{eqnarray*}

where, v_{ip}=V_b + {1\over 2}V_{RF}\cos(\omega_{RF}t)

Due to DC component V_b in the RF signal, the output of single balanced mixer contains LO feedthrough( LO fundamental and it’s harmonics). This LO feedthrough can be suppressed using another single balanced mixer(stage-II) by forming a double balanced mixer.

By flipping the LO signals to the switches, the differential output current from the second single balanced mixer (mixer-II) is given by

(2)   \begin{eqnarray*} i_{o,II} &=& i_{o2,II}-i_{o1,II}\\ &=& g_m v_{i,m}(t) \left({4\over \pi} \sum\limits_{n=1,3,5}^{\infty}{1\over n}\sin(n\omega_{LO}t)\right) \end{eqnarray*}

where, v_{im}=V_b - {1\over 2}V_{RF}\cos(\omega_{RF}t)

The differential output current flowing into the load is, i_o = i_{o,I} + i_{o,II}.
Therefore the differential output voltage of double balanced mixer is given by,

(3)   \begin{eqnarray*} v_o(t) &= & -i_o R_L \\ &=& {4\over \pi} g_m (v_{i,p}(t)-v_{i,m}(t)) \left(\sum\limits_{n=1,3,5}^{\infty}{1\over n}\sin(n\omega_{LO}t)\right) \\ &=& {4\over \pi}g_m R_L (V_{RF}\cos(\omega_{RF}t)) \left(\sum\limits_{n=1,3,5}^{\infty}{1\over n}\sin(n\omega_{LO}t)\right) \\ &=& {2\over \pi}g_m R_L V_{RF} \underbrace{\begin{cases}\left(&[\sin(\omega_{LO}-\omega_{RF})t+\sin(\omega_{LO}+\omega_{RF})t] \\ & + {1\over 3}[\sin(3\omega_{LO}-\omega_{RF})t+\sin(3\omega_{LO}+\omega_{RF})t]+\ldots \right) \end{cases}}_{\text{mixed components}} \end{eqnarray*}

From Eq.(3), we can notice that the output voltage spectrum contain only sum and difference components. LO feed-through and RF feed-through are suppressed.
Disadvantage :
1) Double power consumption for the same power output
2) More area or circuit complexity

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