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I do not understand the basic of a Gilbert cell.

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I see that as its output we consider the output differential current Io. But I do not understand it. Is the load connected between the points 1 an 2? Or is there VCC? Moreover, if the load is connected between them, why is the current flowing on it equal to the difference of Io1 and Io2?

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  • \$\begingroup\$ The circuit is incomplete, omitting the bias arrangements at 1 and 2. Often two large resistors will go to VCC. Often, 1 and 2 will go to a current mirror fed from VCC to get a single-ended bipolar current output. Sometimes 1 and 2 will go to the two ends of a centre-tapped transformer, the centre tap going to VCC. \$\endgroup\$ – Neil_UK May 9 at 18:42
  • \$\begingroup\$ You should check out the MC1496 because there are so many papers and examples around for the IC and the datasheets for it tend to be more comprehensive than most, as well. \$\endgroup\$ – jonk May 9 at 18:59
  • \$\begingroup\$ tinyurl.com/y6xktfcv io=Io1-Io2 is the output just as it says \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 9 at 19:03
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Is the load connected between the points 1 an 2?

Something must be connected to points 1 and 2 that will provide enough operating voltage, and allow \$i_O\$ to be calculated from \$i_{O1}\$ and \$i_{O2}\$.

Or is there VCC?

It is the obligation of the circuit designer using the Gilbert cell to provide this.

Moreover, if the load is connected between them, why is the current flowing on it equal to the difference of Io1 and Io2?

That depends on the load provided by the circuit designer. It is the obligation of the circuit designer to provide circuitry that calculates \$i_O\$ from \$i_{O1}\$ and \$i_{O2}\$.

Check the documentation for the NE612 to see how it's done in an RF circuit (note that the NE612 has circuitry to generate the \$i_{X1}\$ and \$i_{X2}\$ from a single external pin).

Analog Devices also sells a number of analog multipliers that are based on the Gilbert cell.

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  • \$\begingroup\$ You can view that set of 4 transistors as the 2nd stage of an analog EXOR. \$\endgroup\$ – analogsystemsrf May 10 at 4:01
  • \$\begingroup\$ @analogsystemsrf I prefer to think of an EXOR gate as a Gilbert cell mixer that dropped out of Analog School. \$\endgroup\$ – TimWescott May 10 at 4:32

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