Uneven differential pair DC current calculation ? - Page 1
Forum section: electronics / beginners

author: MathWizard
author_rank: Super Contributor
timestamp: November 16, 2022, 06:10:39 am
content: I'm making an AM modulator circuit, and it has a BJT diff. pair, like below. And the additional 2.2k base resistor, makes the larger base current ~1.7x the smaller base current. So that's not trivial at all. And it in the DC analysis it doesn't have all simple meshes, since they share that voltage divider.I did some DC/AC analysis on the tail current BJT, so I have that. And I've done a bit on BJT diff pairs before, but with each BJT having it's own base divider (or voltage).So what's the correct way to calculate those currents, I either don't know or don't remember. And I don't think any approximation I would try, would be 1.7x between the base currents.I can't remember all the resistor values in that circuit, but I made it IRL (w/ a Colpitts osc and electret mic audio amp. The diff pair had the 2.2k between the shared divider that was about 39k/47k for sure.

author: wasedadoc
author_rank: Super Contributor
timestamp: November 16, 2022, 07:25:38 am
content: If you want a better balance then:1. Add a 2k2 resistor in series with the base of Q4 and decouple that base to 0 volt rail.2. Use matched transistors for Q3 and Q4.

author: MathWizard
author_rank: Super Contributor
timestamp: November 23, 2022, 05:13:17 pm
content: I'm not trying to make a better/different diff pair, I'm just wondering how people would calculate that setup.Ok so it is just a planer circuit, and if I set some tail current and assume both BJT's are the same and in the linear region, I can calculate V1. Then with a few guesses of the base current splits, and the BJT diode eqn, I can get a pretty close value, such that VBE1=VBE2+R3*Ibase2For instance with fixed Ie=3.8mA, (Ib1+Ib2)=Ie/301=12.62uA, if Ibase1=8uA, then 719.6mV=705.4mV + 3300*4.63uA=720.6mV, pretty closeThat's with Beta=300 and Is=1.96fAWhat's a better way than trial/error to solve something like VBE1=VBE2+R3*Ibase2 ?

author: wasedadoc
author_rank: Super Contributor
timestamp: November 23, 2022, 06:13:47 pm
content: I don't understand where you are trying to go with this. If you want to calculate the relative current in each of the two transistors that will depend greatly on the characteristics of the individual transistors and the temperature. (The latter is why you often see the use of transistors in a shared package or separate transistors thermally joined). I have told you how you can make the currents more equal. You can also add emitter resistors (with a bypass capacitor between the emitters if needed) to reduce the susceptibility.As a side issue, why those weird resistor values around the bottom transistor?

author: MathWizard
author_rank: Super Contributor
timestamp: November 23, 2022, 06:19:49 pm
content: The originals were just what I measured in the real circuit, I just snapped it's pic. I could add wire resistances too.But I'm saying, how would a real EE or mathematician calculate the unknowns in that model ? When I did mesh/node analysis, I still had too many unknown's.With the base currents I got, at least i could make the AC model tho.