# How can i get the Q operating point of the differential amplifier?

I'm working with this type of configuration: Differential amplifier, and I need to find the current for this circuit so I can have maximum symetrical swing, and for that, I need to find the operating point. But I've searched everywhere and I'm not able to figure how can I get that point in this type of circuit. Any kind of help would be useful, thank you so much.

here you can see the configuration

• Welcome to SE EE. Please copy-paste that image into the question. There is really no need to hide it under a link. Suppose I told you to use a tail current of 10 mA for that circuit, would you then be able to figure out what that "symmetrical swing" would be? Note that the large signal transfer curve of an NPN diffpair (that's what we're dealing with here) is sort of a tanh(Vin) function. So more swing means more distortion. There is no "one value" that will give you maximum swing. Jun 22, 2019 at 15:00
• Sorry! I'm new at this website and I don't know how to do that :( Jun 22, 2019 at 15:03
• I did that and that's how putted the picture. Jun 22, 2019 at 15:05
• Anyways, thanks for the fast answer! Jun 22, 2019 at 15:05
• I'm not a native speaker, so I don't know how do you call the symetrical swing (I searched, and i found that expression) but in spanish at least is called "Máxima excursión simétrica" and it's when you have a circuit and the operating point it's on the most symetrical position Jun 22, 2019 at 15:06

If you substitute the equation for gm in the collector resistors and make Vid=0 you get the QOP dc.

Since current gain drops rapidly to 10% max at Vce(sat) as defined in datasheets with Ic/Ib=10, for low currents keep Vce > 1V and >2V for high currents for max low swing.

For max high swing when voltage drop across collector means current has reduced in that side, this also affects current gain when it starts to get < 10% of I which offers a practical distortion limit for V max unless there is high gain with external negative feedback.

Thus make any reasoned assumptions for Vce-max, min on each side , choose R to be mid scale of those limits from above with Vid=0 for a given I.

Then each side will have max swing with reasoned low distortion from sensitivity to current gain or gm with current.

By reasoned I mean “Rule of Thumb.”

Distortion or asymmetrical gain may be computed, but for DC operating point start with Vce min =1V and for AC use Ice min = 10%I which might also be 1V drop min across Rc thus max (fairly ) linearity is Vc max = Vcc -1 and Vce min =1.

Thus the ideal QOP, Quiescent Operating Point = Vc on both sides. Is the midpoint of min & max.

• But for theoretical academic use, ignore all the above and just focus on the first sentence and use V(Rc)= Vcc/2= I*Rc, however if I current source is at Vee-100V say then you must include Vcc-Vee for max swing which the author neglected to include. so we assume positive outputs only in this case. Jun 22, 2019 at 15:48
• Hahahh yes! It's just for theorical academic use, I didn't understood very well what you wrote abouve, but with your comment I think that I got it, thanks! :) Jun 22, 2019 at 15:57
• So let me see if I understood you everything, for find the QOP I have to use (Vcc-Vee)/2=I*RC? (I didn't notice that the image didn't had the Vee source) Jun 22, 2019 at 16:07