Gilbert cell biasing

Question

I'm new here so I'm going to make is as simple as it can get.. How can I bias a Gilbert cell with bjt transistors like this one: What methods can I use to bias the transistors? Is there any example schematic where they show how biasing resistance are put and calculated?

I've tried many methods but I got stuck, because I suck at dealing with transistors since uni days.. For instance, I could bias the diff amps with base resistance method, like this one:

but got stuck in biasing tail transistor since it's its collector is linked to the upper diffs emitters rather than +vcc like in this example:

I may have done my research in the wrong way, but I've looked in many articles, files etc even in different languages, but all I can find is basic gilbert cell theory and that's it, I managed to find only one practical example so far.

Please, I'm getting really desperate, I've been stuck with this for a very long time now, the place where I live, electronic devices are really limited, I can't find mixers in any shape or form, IC mixers, or even toroid to make a diode ring mixer since it's simpler. All I have is only 2N2222A transistors that could operate up to 300 MHz, so they're my only way. I appreciate any type of help, advice or information you can give me.

Answer 1

Well guys thank you all so much for your help, I immensely appreciate it!!

Well while discussing the matter in the comments with jonk (thanks for the link) and Marcus Müller, I had the idea of simply treating the upper diff pair as a simple load that has 20mA through, and made a voltage divider biased current sink, and calculated the resistance values with the same method of biasing a base divider biased common emitter biased amplifier, and with simulation i got mixed signal.

So here's a recap:

i biased the upper pair with simple base resistance biasing method I made the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

Well I wanted my Q point for each transistor to be 6V, 10mA, for that the value R1 and R2 is 600Ohms, and to make sure that the base runs with ic/hfe (with hfe around 200) R3 and R4 have 226kOhms.

Now thanks to the discussion I've had with the guys i thought about treating the whole block as a load itself that requires 20mA, therefore i added the following circuit instead of the constant current source:

^{simulate this circuit}

And what i obtained from the simulation (Multisim) is the following:

(sorry didn't know how to resize it) The expected results are shown, and it's the same situation for the Gilbert made out of this simple balanced mixer.

Only 2 issues bother me, the 226k resistance (very high value) I'm wondering if i can remove it or not.. and inputs need to be very small..

theory behind it as well as schematics can be found on: http://michaelgellis.tripod.com/gilbert.html and the link jonk proposed: https://smallwonderqrp.blogspot.com/2016/02/home-brew-your-own-mixer-ics.html

So how catastrophic does this look?? And what do you guys think? Please be gentle haha

Answer 2

Bias the RF inputs at the same voltage. The RF signal is then applied differentially. Use a voltage that keeps enough headroom on the tail current source.

Bias the LO inputs at the same voltage. The LO signal is then applied differentially. Use a voltage that keeps enough collector headroom on the RF pair, perhaps RF+2.0v.

Bias the IF outputs at the same voltage. The IF signal is then developed differentially. Use a voltage that keeps enough collector headroom on the LO pairs, perhaps LO+2.0v.

To put those words into pictures with numbers, try this (biassing only, I'm not going to draw the whole Gilbert Cell)

^{simulate this circuit – Schematic created using CircuitLab}

The current source is designed to deliver about 20mA. The voltage bias on its base gives about 1.4v across R2. The 20mA splits nominally 10mA per path.

V3 is used to bias Q2 and 6 to give about 2v VCB on I1. By fixing the base voltage, the emitter voltage is fixed at about 0.7v below that. The 100ohm resistors R4 give a resistance across which you can apply the RF input voltage.

You don't need much more than 2v, any extra is just waste heat in the transistors. You can use less, but be careful to use transistors which retain good parameters at low VCB, many will lose beta and fT.

V2 is used to bias Q1,3,4,5 to give 2v VCB on Q2,6. The LO input voltage is applied across resistors R3.

The power supply is chosen to give 2v VCB on Q1,3,4,5, and another 1v peak signal swing. I've shown the R1 output resistors as 51 ohms, which gives a nominal 0.5v voltage drop when the currents are split equally, and 1v output when the current is all in one output.

Answer 3

You've got headroom problems. Consider this

^{simulate this circuit – Schematic created using CircuitLab}