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In Razavi's Design of Analog CMOS Integrated Circuits (page 126 in second edition), he presents two variations of the Gilbert cell, one were the control voltage is applied to tail current sources (left) and one where the control voltage is applied to cascode devices (right):

tail control cascode control

For the version on the left, he says that if \$V_{cont1}-V_{cont2}\$ goes very high, then the current is routed through the diff pair of transistors \$M_1, M_2\$ and the gain is negative. If the differential control voltage goes low, then the tail current is routed through the other diff pair and the gain is positive. This makes sense since it's just two side by side diff pairs whose gains make sense given the definition of the output voltage polarity.

For the gilbert cell on the right, there are the following two situations depending on if the control voltage goes high (left) or low (right):

cascoded

Here, Razavi says that the left version has positive gain and the right version has negative gain. The output voltage polarity isn't explicitly shown, but I would assume that it should be the same as before since nothing special was said, so left branch is the positive terminal, right branch is the negative terminal. If that were the case, then shouldn't the left version actually have negative gain and right version have positive gain?

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    \$\begingroup\$ Swap labels VIN and VCONT and see that both circuits are identical. So wI agree with you. In his favour he didn't specify the output polarity one way or the other on thae second circuit. \$\endgroup\$ Oct 18 '20 at 17:59
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In one example current through M5 connects to the right terminal, in the other it connects to the left terminal.

Just relabel and they become identical.

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  • \$\begingroup\$ Assuming the left branch is the positive output terminal and the right branch is the negative output terminal, then I would expect gain to be negative when M5 routes current from the left and gain to be positive when M5 routes current from the right. Is this correct? Razavi says the opposite but he doesn't label the output voltage polarity so maybe he switched it. \$\endgroup\$
    – knzy
    Oct 18 '20 at 17:51
  • \$\begingroup\$ @knzy - "I would expect gain to be negative when M5 routes current from the left and gain to be positive when M5 routes current from the right. Is this correct?". Do you mean "to the left"? The gain will be positive when M5 current is routed to the right and negative to the left. Vin going positive increases the current through M5 and so causes the voltage at the right output to drop (ie go more negative). \$\endgroup\$ Oct 18 '20 at 18:58
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View these Gilbert_cell circuits as analog versions of an Exclusive Or gate.

Thus there is a polarity interaction within the circuit.

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