mixer headroom setting

[zxcv2201]

I want to get headroom from mixer using Gilbert structure.

Is it correct to calculate the headroom in cadence as follows?

For 1,2 tr, vds-vdsat (=Vsat_marg) is obtained, and this value is subtracted from VDD

VDD-(Vsat_marg(about M1))-(Vsat_marg(about M2))

 

[zxcv2201]

I thought about the points to consider in the mixer's headroom as follows.

1. I think that the transconductance stage is actually involved in amplification, so I think that sufficient headroom of the transconductance stage is important.
2. Since the amplified signal passes to the switching stage, headroom similar to that of the transconductance stage is required.

I thought the above factors should be considered. I'd like to know if this is the right thing to consider. Are there any additional things to consider?

 

[BradtheRad]

I suppose the topmost components are resistive (since they're labelled 'loads' in internet articles about the Gilbert mixer)?

Then I think you wish to turn on the mosfet networks to a degree such that their quiescent ohm value is same as the loads.
In this manner the loads operate around 1/2 supply voltage (average).

Moreover you must carefully adjust bias signals at all mosfets so you obtain your desired voltage swings.

 

[zxcv2201]

I suppose the topmost components are resistive (since they're labelled 'loads' in internet articles about the Gilbert mixer)?

Then I think you wish to turn on the mosfet networks to a degree such that their quiescent ohm value is same as the loads.
In this manner the loads operate around 1/2 supply voltage (average).

Moreover you must carefully adjust bias signals at all mosfets so you obtain your desired voltage swings.

Thank you for answer. But I didn't understand. What I'm trying to get is a formula for calculating the headroom in a Gilbert cell.

From what you said, I realized that I forgot the resistance.

VDD-(RL*I)-(Vdsat, switching stage)-(Vdsat, transcondcutance stage)

Is it correct to get the value like this?

 

[BradtheRad]

* Remember that turning on a transistor causes reduced drain-source voltage across it.
Simultaneously it sends greater current through neighboring components thus raising voltage across them.

* The Gilbert cell multiplies two signals. These need to ride a DC component in order to bias the transistors properly (including level 1, level 2, and the bottommost tail transistor).

* It's a complex circuit. I imagine a big challenge is to find a good operating point.
You may find it's crucial to operate all transistors in a linear region. Or instead you may need to operate certain transistors near their threshold of turn-On & turn-Off.

* Headroom seems to be only a minor factor to consider. There's numerous adjustments needed before the circuit works properly. Eventually it might come out that each level operates within a range of voltage, or within a percentage of supply voltage. Can these be determined ahead of time? I believe the circuit is too complex to predict ahead of time.