OP-AMP output voltage swing

[mattko]

Hello!

In this document: http://www.ti.com/lit/an/sloa011/sloa011.pdf
Page 29. It is stated that maximum output voltage is VO <= +VCC – VR1 – VBEQ6 – VSATQ6.
I don't understand why it is not just VO <= +VCC – VR1 – VBEQ6  OR  VO <= +VCC – VR1 – VSATQ6.
Why do they count the base-emitter drop and Vce sat drop?
The same goes for minimum output voltage.

Thanks

[Doctorandus_P]

I also do not understand this.
I also think that the current source for the base of the transistors would need some voltage drop to work properly. Stating that Vbq6 equals Vcc seems ... incomplete.

But on a practical term it would not matter much.  opamps with this bipolar output stage can not output voltages which is within 1V5 to 2V of either supply rails and this is usually given in the datasheets.

Picture is screenshot of page 29 of sloa011.pdf

[RobK_NL]

Stating that Vbq6 equals Vcc seems ... incomplete.

But that is not what they say, is it?

You are most likely right that Vbq6 will be lower than Vcc, but any which way you turn it, Vbq6 can never be higher than Vcc.

For the transistor to conduct, Vbq6 must be at least V_BEQ6 higher than the voltage at the emitter of Q6 (Veq6 for short). So it follows that Veq6 cannot be higher than Vcc-V_BEQ6.
With Q6 fully saturated, there will be a voltage drop of V_CEsat across it, which also detracts from the maximum output. This then gives us VO <= +VCC - VBEQ6 – VSATQ6

[mattko]

But that is not what they say, is it?

They say "The maximum value that VBQ6 can be is +VCC", which is close.


Lets analyze the circuit based on Kichoff 2. law: (see attachment)

Loop 1: Vbe6 + Vr1 + Vo = Vcc -> Vo=Vcc-Vr1-Vbe6
Loop 2: Vce(sat)+Vr1+Vo=Vcc -> Vo=Vcc-Vr1-Vce(sat)
Loop 3: Vbe6=Vce(sat)

(we assume that current source has no voltage drop, which I think is reasonable)

From this equations, we can see that TI has it wrong. No?

[David Hess]

I don't understand why it is not just VO <= +VCC – VR1 – VBEQ6  OR  VO <= +VCC – VR1 – VSATQ6.

I do not know either; lately Texas Instruments has been producing some pretty poor documentation.

Vce saturation is less than Vbe forward so the later is all that matters.  The PNP current source has its own Vce saturation and that does add to the output but it is not shown in detail in the schematic.  So the output is Vcc - Vce saturation of PNP - Vbe forward of NPN - VR1.