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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: Basic Gilbert cell schematic 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:

base bias

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:

tail tran

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.

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    \$\begingroup\$ Are you basically just asking about the MC1496? Because, if so, there are lots and lots of white papers on using them. Doing this in discrete form is going to be a nightmare of compromises, I fear. \$\endgroup\$ – jonk Aug 4 '18 at 19:50
  • \$\begingroup\$ I regretfully already am living in a nightmare, been stuck for about at least 6 months now, couldn't find that or any type of mixers here, the only stuff I have is basic components, (resistance, induc, capa) and the 2N2222a to make a mixer, as I said, not even the toroids are available to make 1:1 transformers.. So I literally have to make the first gilbert cell shematic work somehow, I need to find a way to bias the Q transistors... \$\endgroup\$ – Vito_Scaletta Aug 4 '18 at 20:13
  • \$\begingroup\$ In doubt that. Aside from a few embargoed countries, distributors like digikey and arrow ship basically everywhere. \$\endgroup\$ – Marcus Müller Aug 4 '18 at 22:07
  • \$\begingroup\$ @Vito_Scaletta I use the MC1496 when I need well-matched BJTs and can twist the topology for what I want. They were dirt cheap when I bought myself a lifetime supply and now I've got tens of thousands of the darned things. I find it difficult to imagine not being able to get one or two. Are they really unavailable to you? \$\endgroup\$ – jonk Aug 4 '18 at 23:02
  • \$\begingroup\$ @Vito_Scaletta A google search turned up Home Brew Mixer. Can't vouch for it as I haven't had time to look. But for ideas, perhaps? \$\endgroup\$ – jonk Aug 4 '18 at 23:04
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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:

schematic

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:

schematic

simulate this circuit

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

enter image description here

(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

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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)

schematic

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.

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  • \$\begingroup\$ Thank you so much for your answer, well like I said in my bias attempts I failed to bias the RF inputs, since the transistors are linked to the emitters of the upper pair (LO) rather than +vcc, so I can't fix the Q point?! (like I said I always had problems with biasing transistors since uni days) With your indulgence, could you please, if possible provide an example? Thank you \$\endgroup\$ – Vito_Scaletta Aug 4 '18 at 20:56
  • \$\begingroup\$ Resistor voltage divider between vcc and gnd. As Neil said, couple in the rf voltage via capacitors. \$\endgroup\$ – Marcus Müller Aug 4 '18 at 22:09
  • \$\begingroup\$ The whole thing is current-mode, why are you worried about Q point? \$\endgroup\$ – Neil_UK Aug 5 '18 at 4:16
  • \$\begingroup\$ Because I thought I wouldn't want to disfigure the incoming RF signal. The way I see it there surely will be signal amplification, so the Q point needs looked into when biasing th RF section.. Is that wrong? \$\endgroup\$ – Vito_Scaletta Aug 6 '18 at 12:28
  • \$\begingroup\$ @Vito_Scaletta It's wrong to think about Q point in those terms for this circuit configuration. There is signal amplification, and it's the IF load, 100ohms differential load in my case, divided by the emitter resistances of Q2 and Q6 at whatever mean current they're running at. If you want to linearise the gain to RF, then you either a) work with a very small input signal or b) add linear resistance between Q2 and Q6 emitters to degenerate the gain. Think of the RF amplification as a cascode through the LO switch transistors, so the current that Q2/6 controls ends up in the LO TR loads \$\endgroup\$ – Neil_UK Aug 6 '18 at 12:39
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You've got headroom problems. Consider this

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you so much for your reply! Well I'll be sharing I've done so far, it works in simulation, but still having trouble with theory to a certain extent but calculated gain and actual obtained gaon from simulation are way different.. \$\endgroup\$ – Vito_Scaletta Aug 6 '18 at 12:47

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