Re: Not one single person has answered my MC1530 posting correctly

It looked too much like work, for way too little money.

I used to play chess, too, until I realized that one should always be paid for thinking.

John

Izzat when you joined Mensa ?:-)

...Jim Thompson

No, that's when I joined this newsgroup.

John

LOL ;-)

...Jim Thompson

Hmmph. I posted the correct answer to this on Saturday, but nobody seemed to notice.

I guess that slipped by me since it's numeric rather than a general Algebraic solution. But it looks correct.

...Jim Thompson

That was from the Auton-Bell-Rick consortium of multiple personality- purely blind arithmetic with no insight whatsoever- a worthless response.

Yes.

I was hoping to convey some analytic thinking, but to no avail :-(

...Jim Thompson

Oh yeah- and this "YD" character also posts under the name ChaosMaster- aren't we lucky to have him. YD means Yellow Dirtball or Yankee Detester , take your pick.

I am not exactly sure- but in the case of subcircuits like that Q9/Q10, you may be relying on emitter area ratios to set some of the bias currents. Then Q3-Q4-Q5 is another transcendental. The circuit is tough.

notice.

Q9/Q10 relied on 1:1 area. One year into production without problem, the device was moved into the new Motorola process building "P". Yield went to hell on a handbasket. The problem resolved to offset issues between Q9/Q10. I simply removed 100ohms from R10 and placed

Q3-Q4-Q5 run at essentially the same current, so tracking error is second order.

And this was the good ol' days... this was all breadboarded first.

I don't know that the circuit is "tough", it is simple-minded, but subtle; everything about IC's is transcendental ;-)

...Jim Thompson

I read in sci.electronics.design that Fred Bloggs wrote (in ) about 'Not one single person has answered my MC1530 posting correctly', on Wed, 21 Jan 2004:

That's what I like about you, Fred. You're all heart.

God how awful - the object of the excercise was to determine the output voltage of a relatively simple op-amp, and that's exactly what I - and only I - did. Didn't spot your insightful response anywhere.

Jim - did you not look at my post - it should be fairly evident that I MUST have understood and analysed the circuit in order to be able to provide this numerical answer. If you'd wanted purely algebraic responses, then why include the component values?

Q9/Q10 are, you dimwit, a current mirror. There are newborns who know that the currents in a current mirror scale with the ratio of the emitter areas.

It is neither transcental nor tough.

Q4/Q5/R5/R9 set the voltage on the base of Q3, which defines its emitter voltage and hence the current through R3. This current is the bias current for the long-tail pair. Was that sufficiently non-numerical for you?

Still not enough? Let's go through it piecemeal:

1. Determine current through Q4/Q5: I = (Vee - 2*Vbe) / (R5/R9)
2. Determine voltage on base of Q3: Vb = I*R5 + 2*Vbe
3. Determine voltage on emitter of Q3: Ve = Vb - Vbe
4. Determine current through R3 and hence Q3: Ibias = Ve / R3 = (R5 * (Vee - 2*Vbe)/(R9 + R5) + Vbe) / R3

Of course you did, I just prefer an Algebraic solution... numerically doesn't help the silent lurkers with understanding.

I'll leave the values out next time, and make it an exercise for the student to optimize the circuit performance ;-)

...Jim Thompson

Good idea. Just thought I'd add that to some of us, the dissection of an op-amp at the discrete level is new stuff. 3 or more pairs of current mirrors, biased and matched is an interesting concept, and all the math in the world is not going to discribe the logic behind how it works or what it is doing. There is some info on how these work at the bottom of

Tom Frederiksen's book is a very good tutorial (if you can find a copy)...

"Intuitive IC Op Amps" National's Semiconductor Technology Series © 1984

No ISBN

...Jim Thompson

Mark - the newborn stuff was just a swipe at Mr Bloggs who clearly didn't understand the circuit - or even its bias, for crying out loud - but who still felt the need to denigrate me, even though I was the only one to demonstrate the correct answer.