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.
Figure 1. Schematic of double balanced switching mixer
The differential current in a single balanced mixer is given by
(1) 
where, 
Due to DC component 
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) 
where, 
The differential output current flowing into the load is, 
.
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*}](https://analog.intgckts.com/wp-content/ql-cache/quicklatex.com-65e7671b4577e14a481d2045550abe38_l3.svg "Rendered by QuickLaTeX.com")
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