1) How do I interface LC balun with mixer using differential current RF+ and RF- input signals ?

2) How should I deal with the load impedance (RL) for LO+ and LO- which is almost infinite because it is mosfet gate ?

gilbert cell mixer waveform

LC balun calculation


1 Answer 1


Your circuit and simulation does appear to work already (you do get a mixed down signal) so you don't have to change it.

You do not explain why any change is needed but I can guess: you're now feeding the RF signal into the mixer as a current. Normally those currents would come from a pair of NMOS transistors which are biased such that they deliver a DC current. Then the RF is added to the gate of these NMOS so that the DC current contains an RF current as well. I generally do that like this:


simulate this circuit – Schematic created using CircuitLab

Note how the RF transistors (M1, M4) are part of a current mirror with M3 as input. For DC no current flows into a gate so there will be no DC current through R1 and R2 so the current mirror will work normally.

The RF signal is superimposed onto that DC biasing voltage at the gates (of M1 and M4) by capacitors C1 and C2. Note that the RF signal is a differential signal.

I make the differential signal by using two Voltage-controlled-Voltage-Sources, one with a gain = 0.5 and the other gain = -0.5. I use +0.5 and -0.5 as then the amplitude of the signal generated by Vrf stays the same. If Vrf = + 1 V then at C1 we get -0.5 V and at C2 we get +0.5 V.

You can also use an LC balun instead of VCVS1 and VCVS2.

Indeed then the balun will then not be loaded by 50 ohms. Sometime inductors are used to "tune out" the input capacitance of M1 and M4. The real part of the impedance might then be 200 ohms for example, the balun then needs to be designed for that impedance.

  • \$\begingroup\$ Given the LO inputs are capacitive, you can match (using inductors), or just use differential current steering into 2 load resistors that drive the two LO-input gates. \$\endgroup\$ Apr 2, 2019 at 13:43
  • \$\begingroup\$ @analogsystemsrf The purpose of the LC balun is to perform single-ended to differential conversion \$\endgroup\$
    – kevin998x
    Apr 3, 2019 at 3:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.