Compensating transistor biasing - Page 1
Forum section: electronics / beginners

author: David1
author_rank: Contributor
timestamp: February 28, 2019, 08:22:58 pm
content: How to analyze the quiescent conditions of the circuit below assuming the tansistors are on an array such as CA3096? (Fig 5N.21 from Learning the Art of Electronics)Both emitters are at ground, so both bases at about 0.7V.The questions:Is the collector of Q1 also at about 0.7V or is there a voltage drop across the 10k base resistor of Q1 and if so, how to calculate how much? If there's no voltage across the base resistor, why not? (I have measured this with a VM so know the answer in practice but want to understand it.)Assuming collector of Q1 is also at about 0.7V, the current in the collector of Q1 is a bit less than 1mA (19.3V across 20k).The rest I think is fairly straightforward: since the bases of both transistors are at the same voltage and have equal value base resistors, the base currents will be the same and therefore so too the collector currents (give or take unintended component differences). So the collector current of Q2 will also be about 1mA, hence the 10k Q2 collector resistor to set the output at the 10V midpoint.So, the bit I'm struggling to understand is how to analyse what happens across the base resistor of Q1. If I replace both base resistors with 5k, it still seems to work fine.

author: rstofer
author_rank: Super Contributor
timestamp: February 28, 2019, 09:13:47 pm
content: If you want to analyze the circuit, you need to replace the transistor with an equivalent circuit. Like in this video:You set up the base voltage, determine the current and then multiply by Hfe to get the collector current. This is going to be a little hard to handwave because the collector current sets up the maximum base voltage (before it gets dropped in the base resistor).I suspect you have to write mesh or node equations for the entire model. Not a big deal but solving the matrix will not be fun unless you use a solver like MATLAB or Octave.Here's a video on collector feedback biasing:

author: bson
author_rank: Unknown
timestamp: February 28, 2019, 09:59:50 pm
content: Remember,Vce = Vbe + Vbc-and-Ic = Ib + IeDo the loops and nodes.

author: Ian.M
author_rank: Super Contributor
timestamp: February 28, 2019, 10:19:42 pm
content: From a DC bias point of view, short out the base resistors and its a simple two transistor current mirror. Q1 would then be diode connected and the control current of slightly under 1mA flows through it. Adding back the base resistors doesn't stop it acting as a current mirror, it simply increases Q1's collector voltage slightly, such that there is Rb*Ic/hFEacross the base resistors, but Q2's base resistor prevents Q1 shorting out the AC signal. As you have reasoned, Q2's load resistor is half of Q1's current set resistor, so will have half the supply voltage across it (less about 0.3V), hence the nominal 10V bias point of Q2 collector.Of course, hFEis a fiction as a BJT is actually a voltage controlled device with its control input shunted by a diode junction, but as datasheets typically don't provide Gummel–Poon model parameters for transistors, unless you are lucky enough to have a manufacturer provided SPICE model, or are prepared to spend the time to curve trace a statistically significant number of devices, then curve fit the Gummel–Poon model to the gathered data to get the model parameters, the best one can do is roughly estimate the operating point then scale off from the datasheet's hFEvs Ic and Vbe vs Ic graphs (assuming you have a comprehensive datasheet . . .) and iteratively refine it.