I've been reading original academic papers on mixers and now a RF textbook and I have been unable to understand a very basic thing, which is how do I get a RF signal, say -65 dBm, at a high enough voltage level for input into a CMOS-based downconverting mixer? Everything I read shows a LNA blackbox, which is fine and good, but LNAs amplify signals at what, 15-20 dB typically? How on earth is that even remotely close to activating a CMOS NPN transistor in the mixer with a typical threshold voltage of 700mV? +15dB gain on -65 dBm doesn't get it to 700mV+. And chaining multiple LNAs to get it to 700mV+ would destroy the signal with noise, no?

I know I'm missing something extremely obvious here so be kind. I'm just starting out. I know there's a lot more to downconversion (filtering, etc). This question is solely about the initial amplification.

  • 2
    \$\begingroup\$ If you've got an amplifier good enough to amplify a -65 dBm signal by 20 dB without "destroying" it, it's certainly not going to destroy a -45 dBm signal. (meaning: the first amplifier stage is the most critical for adding noise) \$\endgroup\$
    – The Photon
    Commented Aug 17, 2018 at 5:20
  • 1
    \$\begingroup\$ Are you missing the concept of biasing? \$\endgroup\$ Commented Aug 17, 2018 at 5:39
  • \$\begingroup\$ And chaining multiple LNAs to get it to 700mV+ would destroy the signal with noise, no? No as you're amplifying both signal and noise so signal/noise stays the same (roughly). But amplifying the RF isn't needed, the LO signal to the mixers needs to be large, the RF signal can be as small as you like. \$\endgroup\$ Commented Aug 17, 2018 at 7:03
  • \$\begingroup\$ Discriminate between the mixer's RF port and local oscillator port. Which port must your -65 dBm signal drive? I'm guessing it goes to RF port. That one operates in the "linear" realm, and can accept small signals down to the noise level. Local oscillator port often doesn't, and may require a large driving signal. \$\endgroup\$
    – glen_geek
    Commented Aug 17, 2018 at 11:48
  • \$\begingroup\$ @immibis I honestly think that's worth an answer. \$\endgroup\$
    – Shamtam
    Commented Aug 17, 2018 at 13:31

4 Answers 4


Here's a schematic of a typical CMOS mixer circuit:

enter image description here

It's the classical "Gilbert" mixer.

Although not clear from this picture, the bottom NMOS is just for biasing, you can view it as a DC current source. That makes the bottom half of this circuit identical to a standard differential pair.

The inputs of this differential pair are connected to the outputs of the LNA so this differential pair simply converts the RF (voltage) signal into a current signal.

That current (containing the RF signal) is then fed into the upper "business" part of the mixer, the actual switching happens here. For that those 4 NMOS in a row need to be switching on/off properly. Therefore the LO signal needs to be large enough. Do we need 700 mV (as you claim), in the order of Vt for that?

No we don't! As long as the switching transistors switch "enough" then we will get an IF output signal. If the switching NMOS have a large enough W/L then even a 200 mV LO signal could be all we need.

What matters is the difference in Vgs for each pair of NMOS. As long as one NMOS has a larger Vgs than the other NMOS and the difference is such that the current from the RF part below chooses one NMOS over the other, then the mixer will "mix".

This is the same principle as applying a bias to a transistor in an amplifier circuit, by applying a bias we overcome the Vt (700 mV) "dead zone" of the NMOS so it is not an issue.

  • \$\begingroup\$ You say "What matters is the difference in Vgs for each pair of NMOS." Which pair are you referring to here? The LO+/LO+ pair, or the RF+/RF- pair? Also, still not understanding how any current can pass through the RF NMOS transistors if the base voltage of those transistors is at microvolt or "few millivolt" level, unless we apply a DC "bias voltage" to the base voltage of the RF+/RF- pair. If DC biasing the RF signal is what should be done, the diagram isn't showing that RF+/RF- base biasing, correct? \$\endgroup\$
    – acker9
    Commented Aug 19, 2018 at 15:40
  • \$\begingroup\$ Which pair are you referring to here? To the LO pair as they are the "switches". if the base voltage You mean gate voltage. But the gate voltage is not a few mV. What actually matters is Vgs and it needs to be > Vt. In this circuit there is no bias voltage being applied but a biasing current, by the bottom NMOS. It is the output of a current source biasing the whole circuit. I suggest reading up on the "differential pair" circuit. \$\endgroup\$ Commented Aug 19, 2018 at 17:28

The mixer is driven by a Local Oscillator. The level is (say +7dBm).

It is never -65dBm.

The RF being switched is -65dBm. It doesn't do anything - just goes through the mixer switches. In the basic and best type, a mixer is switches.

Think relays. The local oscillator energises the coil and must be 12V. The RF goes through the contacts and can be any tiny level at all.

You can make a mixer using a DPDT relay, it wll be a double balanced mixer. It works very nicely with excellent IP3 and low loss. Poor frequency performance but.


Here is approximately what is inside the NE602 bipolar mixer. Note the use of bipolars to establish the biasing, to achieve headroom


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ Hm, so why post a circuit using only bipolar transistors while OP asks about CMOS? If you would have drawn the same circuit using NMOS instead of NPNs it could work just as well. \$\endgroup\$ Commented Aug 19, 2018 at 17:30

I'm going to accept Bimpelrekkie's answer because the suggestion (in the comment section) was that I read up on "differential pair circuits" and this is what got me going in the right direction. In retrospect, this question was really about how a NMOS amplifier works with weak signals and had nothing to do with a mixer. By adding "mixer" to my question I think complicated the question more than necessary.

The answer to how to amplify a weak signal appears to be "use a differential amplifier." The bottom NMOS pair (i.e., the RF+ / RF- NMOS pair) in the diagram Bimpelrekkie posted is a differential amplifier that feeds the rest of the mixer (the rest of the mixer being the "business part" as Bimpelrekkie called it).

In my quest to read up on differential circuits I came across a differential amplifier example provided in Practical Electronics for Inventors by Paul Scherz and Simon Monk, 3rd ed, p.446.

Scherz/Monk didn't provide all the parameters in their example: the signal parameters and NMOS parameters are not given. However, I drew up the Scherz/Monk example in LTSpice and used 100k resistors as Scherz/Monk did. I don't think these resistor values are exactly right for the NMOS transistor and signal levels I used in the simulation, but nevertheless it still shows amplification. Scherz/Monk go into how to determine the R values, but I didn't quite follow. (I think the bottom line is the transistors are supposed to operate near their "Q point.")

My choice of input signals: A 10mV 1kHz sine (Vsig1) and a -10mV 1kHz sine (Vsig2). 10mV is, of course, below the threshold voltage of the NMOS.


Differential amplifier circuit in LTSpice

Waveforms of the two differential signals, V(vout1) and V(vout2) in LTSpice:

Differential output signal pair

Differential output waveform, V(vout1,vout2) in LTSpice:

Differential output signal

It may be hard to see, but this signal is approximately 6.3V peak to peak, from approximately -3.15V to -3.15V.

  • \$\begingroup\$ Any edits should be in your question, not added as an answer. \$\endgroup\$ Commented Aug 20, 2018 at 7:13

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