I don't understand how was this long-tailed pair was biased...? (AoE confusion)

[comfortcube]

So I'm reading through the AoE section on differential amplifiers in Ch2 - BJTs, and when I get to the following:

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The red underlined part is my confusion. Without any base voltages, how can we know what any of the currents will be? So far, with the degenerated common-emitter amplifier, the order kind of goes:

• Choose the emitter resistor Re to be >> re (the internal emitter resistance from Eber-Moll) to reduce sensitivity to the thermal changes in Vbe, as well as to put adequate input impedance into the CE amp. For example, let's just say Re = 1k.
• Now that you've got Re, Ve sets the quiescent current. Let's say 1mA is desired. So /w Re = 1k, Ve = 1V. That means Vb is ~0.6V higher, and you can use a resistor divider to set that off of Vcc.
• Now choose Rc to make Vc at 0.5Vcc to maximize output swing. Also -Rc/(Re + re) sets the gain.
• Kind of play around and balance Re and Rc to get the gain you want while working within constraints of input impedance, temperature sensitivity, etc... Maybe add in a bypass emitter cap so that at signal frequencies, we get the gain we want.

But here now, the only assumption is the normal-mode (differential) voltage is 0. If the common-mode voltage is 10V, vs 100V, vs 1V, all that changes the current through everything! I have tried this on LTspice and indeed this is the case! It seems I only get the stated currents when the common-mode voltage is also at 0V. Also, when has Rc ever set the quiescent current? That's always been Re because the emitter voltage is one Vbe below the set base voltage.

Can someone help me out here?

[Ian.M]

Its assumed that the common mode voltage range is small compared to its difference from the negative supply.  If not, you need something better than a simple tail resistor, e.g an active current sink.

Read "Collector resistor Rc is chosen for a quiescent current of 100uA." as "Collector resistor Rc is chosen for use with a quiescent current of 100uA.", i.e. if you change the current you must also change Rc.

[comfortcube]

Ok, I think that makes sense, and I figured it's the only way the circuit works. I was just confused cuz I was like, what am I missing? They would mention an assumption like that, but I guess sometimes not all assumptions are clearly stated. And yeah, I've read on towards the following sections where a current sink is used, and seems to be far superior in pretty much any scenario unless component count mattered (I think...).

[TimFox]

Yes:  start with the assumption that the two bases are at 0V.
Then, the two emitters will be at -0.7V.
The junction of the two emitter resistors Re will be even more negative: 0.1 mA through each 1 k will drop the voltage by 0.1 V to -0.8 V.
That leaves 4.2 V across the common emitter resistor, which bears 0.2 mA current, requiring 4.2/0.0002 = 21 k, or the next standard value of 22 k.

[exe]

I've seen a discrete opamp discussed on youtube. Adding current sink made so much better. If I remember correctly, it was this video: . What I like about it is that the guy on video actually built the circuit and showed voltages, and discussed performance. He also have other tutorials which I found very useful.

[comfortcube]

Awesome! Thanks for clarifying!

[comfortcube]

Haha I watched this video right after I posted! Really really great video! Unfortunately it didn't immediately answer my question of how exactly the AoE example produced the stated 100uA and 200uA of current, but after going over it some more and seeing the previous answers, I got it now. I love w2aew!

[David Hess]

The tail current does vary with common mode voltage which limits the common mode range.  But in a feedback application, the inverting input follows the non-inverting input so the differential voltage is zero, which leaves the common mode voltage following the non-inverting input.

Inverting amplifiers will have the non-inverting input fixed at a constant voltage, usually ground, so the common mode voltage is fixed.  Non-inverting amplifiers will have only a limited input range for the reason you identified, but a current sink can be used for the tail current to correct this, or a much larger supply voltage and higher tail resistance to minimize the variation in tail current.  Before integrated operational amplifiers became less expensive, it was common to have a 50 volt supply just for the tail current of the differential pairs to increase their common mode rang to useful levels.  Note that this does not require high voltage transistors because this supply is applied to the forward biased emitters.

[T3sl4co1l]

Yes, and note that output voltage range is similarly limited by input common mode range.  The diff pair does weird things when you violate that -- namely, one transistor saturates, the other transistor takes up the balance of the tail current, and the emitter and collector voltages track with the input -- this may cause phase inversion, putting a notch or "nipple" on your previously smooth sine wave.

So it's no accident that the two most common improvements to this circuit, are to extend these ranges: a CCS for the tail, and holding the collector load voltage near zero with a current mirror or the base of a common-emitter stage.

Before integrated operational amplifiers became less expensive, it was common to have a 50 volt supply just for the tail current of the differential pairs to increase their common mode rang to useful levels.  Note that this does not require high voltage transistors because this supply is applied to the forward biased emitters.

Yup, same works for the collector load too, of course -- AoE has a problem in one of those sections, asking why the gain of an "ideal" design common-emitter stage (resistor biased) is a constant.  Well, it's ~20 when Vc ~ (V+) / 2, but we don't need to choose this, and indeed we're left with enough headroom for the application, we can set it arbitrarily close to saturation, more than doubling that figure.

Another way to think of it, consider the "barn roof" response of a common-emitter stage with no degeneration: the slope is steepest (instantaneous gain highest) just before saturation.

Tim