Exercise 10.11 in "Microelectronic Circuits" 2nd Ed.

Beta=3D 200 is only 0.5% so that means you can ignore base currents for practical purposes. Then you compute IE1 as [+15V-VI-VBE1]/RE where VBE1=3D VTxLN(IE1/IS), and similarly for IE2. This must be done numerically. Then since Q3 and Q4 are in series, the VBE are the same and therefore must equal 1/2 x (VBE1 + VBE2) from which you obtain IC3,4=3DISxEXP(VBE3,4/VT).

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Ok, which is what I've done so far.

Are you talking for Q1 or for Q3? I already did Q1 & 2. Q1 and Q2 were easy because there was an emitter R=3D5k.

The problem is that R3=3DR4=3D0 for Q3 & 4. To make it worse, there is no RL either, in this problem. So I can't use that approach for Q3 or Q4.

Which is what I've assumed so far but I'm beginning to doubt that.

I do know that Vbb =3D 1.4V.

That's what I thought at first, but, when you calculate that out, this is what you get:

Ic3 =3D Is * exp(0.7V/0.0252V) =3D 3.3E-14 * 1.158 =3D 38.2 mA

Their answer is 1.95 mA.

I got to thinking that Vbe3 and Vbe4 must be different. They must add up to 1.4V of course. But VCE3 =3D 5V, while VCE4 =3D 25V. From this I would expect VBE3 to be a higher than VBE4.

There is a term that involves VCE, but you need the value of Va. I don't have it in front of me but I think it was:

Ic =3D Is * exp(Vbe/vt) * ( 1 + Vce/Va )

Where Va ranges from 50 to 100. But the difference is too small that I ruled out VCE in the calculation.

So the only thing left is Vbe3 vs Vbe4.

Still stumped.

Warren

I am talking about Q1 and Q2, the input buffers.

Actually it is not a problem, RL=3Doo means IC3=3DIC4, no current into a load, the transistors are therefore in series which means their collector currents are the same and therefore their VBE are the same.

You might want to redo that calculation- your answer is way off even with those numbers...

Nope.

You're too sloppy with that VBE1=3DVBE2=3D0.7V. This problem requires more precision than that, there is a nearly 6:1 ratio of IC2:IC1. Use VBE1,2=3DVT x LN( IC1,2/IS) to find the Q1 , Q2 VBE, then take the mean to be equal to VBE3=3DVBE4, as explained above- from which you obtain the IC3, 4 via the exponential formula as explained above.

...

e

Ok, I see what you mean. I've recalculated the following:

Vbe1 =3D 0.78V Ve1 =3D 10 + 0.78 =3D +10.78V

Vbe2 =3D 0.82V Ve2 =3D 10 - 0.82 =3D +9.18V

Vbb =3D 10.78 - 9.18 =3D 1.6V

Vbe3 =3D Vbe4 =3D 1.6V / 2 =3D 0.8V

Assuming the above is correct, then I try:

Iq3 =3D Is * e ^ ( Vbe3 / 25.2 mV ) =3D 3.3E-14 * e ^ ( 0.8 / 0.0252 ) =3D 3.3E-14 * e ^ 37.74603175 =3D 3.3E-14 * 6.125286538E+13 =3D 2.0214 A !!

I must be missing something really stupid here. Arrrg.

Warren

ore

Oops: Is =3D 3.3E-15 (3.3 * 10^-14)

But this still computes: Iq3 =3D 0.2021 A, which is far from their 1.95 mA.

ore

Okay- assuming for IE1 calculation purposes thet VBE1=3D0.7V then VE1=3D VI

+VBE=3D10.7V, so IE1=3D(15-10.7)/5K=3D0.86mA. The back calculating VBE1=3D0.025 x LN( 0.86 E-3/3.3E-14)=3D0.599V Same with IE2=3D (10-.7+15)/5K=3D4.86mA making VBE2=3D 0.025 x LN( 4.86E-3/3.3E-14)=3D0.642V.

Then 0.5( VBE1 + VBE2)=3D0.5 x ( 0.599+0.642)=3D0.620 =3D VBE3=3DVB4, so IC3=3DIC4=3D 3.3E-14 x EXP ( 0.620/0.025)=3D1.95 mA

You need to double check how you're using your calculator or whatever- when in doubt about priorities, use parantheses. Right now, your calculations are wildly off...

more

With Is corrected, I've recalculated the following:

Vbe1=3D0.83V Ve1=3D10+0.83=3D10.83v

Vbe2=3D0.88v Ve2=3D10-.88=3D9.12V

Vbb=3D10.83-9.12=3D1.71V

Vbe3 =3D 1.71V / 2 =3D 0.855V

Iq3 =3D Is * e ^ ( Vbe3 / 25.2 mV ) =3D 3.3E-15 * e ^ ( 0.855 / 0.0252 ) =3D 3.3E-15 * 5.432395119E+14 =3D 1.79269 A (book says 1.95 mA)

Arrg. I need a rum and coke. I've been=20 working on this too long.

..

..

OK, I'm getting tired now (been working on this=20 exercise most of the day). But correct Is is=20 indeed 3.3E-14 which is the same=20 as 3.3 * 10^-14 (duh!)

I was missing the re-computing Vbe1 & 2, which you pointed out. Using what you posted (with VT=3D25.2mV)=20 I got Vbe1=3D0.604v and Vbe2=3D0.648v.

If I use VT=3D25.0 mV, I can repeat your numbers.

Using Vbe3 =3D Vbe4 =3D 0.626mV, I calc Iq3 =3D 2.03 mA,=20 which is close to the 1.95 mA in the book. (Yeah!)

I have no idea why my exponential calculation was=20 not correct tonight. I shouldn't drive the calculator=20 when I get this tired. =20

My good calculator is at work (Sharp I think). This=20 one is a TI-30XA, which I don't like as much. The=20 one thing that really irritates me (about both) is=20 that the EE button don't allow for negative=20 exponents. So you have to divide to get one. Maybe I need a good calculator for home :)

Tomorrow I'll attack the rest of the exercise.=20 I'm very interested in this class AB design at=20 the moment. So I need to get this right.=20

Thanks for your help.

Warren -

Try entering the number then "EE" then the "chs" (or +/-) button then the exponent.

John

Or the other way around; number => [EE] => exponent => [+/-]

ote:

I gave the calculator away (it didn't have a sign change button BTW).

The new calculator does, which I like.

I got the remainder of the exercise worked out, except for (c) for >

The book lists 0.02mA as the answer.

I figured Iq3=3D100mA approx, since vo=3D10V/RL=3D100

Since Iq3 * Iq4 =3D Iq ^2; Where Iq is quiescent current.

So I computed Iq4 =3D 2.87 * 2.87 / 100mA =3D 0.082 mA

But the book lists the answer as 0.02 mA.

What am I missing this time?

Darn good calculator for your computer:

OK, hold the phone- I figured it out.

After I computed:

Vbb =3D Vbe1 + Vbe2 =3D 0.579 + 0.643 =3D 1.222 volts

Based upon Iq3 =3D 100 mA, Vbe3 =3D 0.717 volts, leaving Vbb - Vbe4 =3D

0.504 mA

From Vbe4, Iq4 =3D 0.019 mA (very close to book value of 0.02 mA)

I found my answers agreed best using VT=3D25mV, vs 25.2mV. It would be nice to know what the precision of the calculations were. The 1st edition was published in 1982. Assuming 2 years to publication, the numbers might have been crunched ca 1980. I recall owning a scientific calculator between '76 and '80 so I doubt they would have used book tables. The only thing that I can think of that might change the results is if the values were rounded and then used along the way.

Warren

I don't really like sitting at the computer when I'm working from a book.

I just gave away the problem TI calculator. It was cheap and I don't want to waste any more time with it.

I believe that the trouble with that calculator was that it required the use of a "2nd" key, prior to hitting the ln(x) key. Either it didn't always register or due to keybounce I didn't always get the e^x that I was expecting. Otherwise there was a "fat finger syndrome". No matter the reason, why waste more time on bad results? That calculator's keypad was too plasticky and cheap.

The new calculator does finally offer a sign change key, which also works for exponents (it's about time!) The new calculator is a Casio fx-991MS. Hardly top of the line but I like the way it works including the feel of the keypad.

Warren

I bought my HP45 in 1973, so yes, scientific calculators were available in

1980. They even had a key. ;-)

Excalibur is also a nice calculator. I use both. I'd like one that had a decent binary mode, though. *Very* few do binary fractions.

At 100mA into the load, the base current for Q3 becomes significant. You know that VEB1 and VBE2 are on the order of 0.600V so IE1=3D(15-0.6-10)/5K - 100mA/200=3D0.38mA from which VEB1=3D0.025 x LN( 0.38E-3/3.3E-14)=3D0.579V. For Q2 you have IE2=3D(10-0.6+15)/5K=3D4.88m= A as before for a VBE2=3D0.643V. Using the fact VBE3+VEB4=3DVEB1+VBE2 you have IC3 x IC4=3D IS^2 x EXP((VBE3+VEB4)/VT)=3DIS^2 x EXP ((VEB1+VBE2)/ VT)=3DISx EXP(VEB1/VT) x IS x EXP( VBE2/VT)=3DIC1 x IC2=3D0.38E-3 x

4.88E-3=3D1.85E-6. Now IC3=3D Iload +IC4. Therefore IC3 x IC4=3D(100mA +IC4)xIC4=3DIC4^2 +100E-3xIC4=3D1.85E-6 from which IC4=3D0.5 x( -100E-3 + SQRT((100e-3)^2+4 x 1.85E-6))=3D1.85E-5=3D0.0185mA ~ 0.02mA to two places. Same result as just assuming IC3=3D100mA so that IC4=3D1.85E-6/100E-3=3D0.02mA....

Here's the circuit in ASCII, for anyone trying to follow. The problem asks:

a) For Vi=0 and RL=infinity, find Icq1, Icq2, Icq3, Icq4, and Vo. b) Same as (a) except solve for Vi=+10 and Vi=-10. c) Same as (b) except RL=100.

It's given that for case (a) above, that Vbe=0.7 for Q1 and Q2.

Solutions:

a) Vbq3=+0.7 Vbq4=-0.7 Icq1=(Vbq3-(+15))/5k=-2.86mA, Icq2=(Vbq4-(-15))/5k=+2.86mA, since Icq3=Icq4, it must be that Vbeq3=Vbeq4, so Vo is half way between Vbq3 and Vbq4, or Vo=0V and Vbeq3=Vbeq4=Vbeq1=Vbeq2, so Ic for each must also be the same, Icq3=+2.86mA, Icq4=-2.86mA

b) Vi=+10 Temporarily assume Vbeq1=Vbeq2=0.7, so, Vbq3=Vi+.7=+10.7 Vbq4=Vi-.7=+9.3 Icq1=(Vbq3-(+15))/5k=-860uA, so Vbeq1=.669 Icq2=(Vbq4-(-15))/5k=+4.86mA, so Vbeq2=.714 this corrects the earlier assumption of 0.7, so, Vbq3=Vi+.669=+10.669 Vbq4=Vi-.714=+9.286 Icq1=(Vbq3-(+15))/5k=-866uA Icq2=(Vbq4-(-15))/5k=+4.857mA since Icq3=Icq4, it must be that Vbeq3=Vbeq4, so Vo is half way between Vbq3 and Vbq4, or Vo=((10+.669)+(10-.714))/2 ~= 9.978V Vbeq3=Vbeq4=(.669+.714)/2=.6915, or 8.5mV less than .7 Icq3=Icq4, but since the nominal Vbe is .7@2.86mA (given for case (a)), then Icq3=Icq4=2.86mA/e^(8.5mV/26mV), or Icq3=Icq4=2.06mA (roughly the same logic now for Vi=-10)

c) Vi=+10, RL=100 Temporarily assume IL=10/100=100mA, that comes from Q3, not Q4, so Icq3 estimated at 100mA, at Icq3, Vbeq3=.7+26mV*ln(100mA/2.86mA) or about .792 base drive for Q3 is say about 150 times less, or about Ibq3=100mA/150=666uA assume the following from (b) above, Vbq3=+10.669, Vbq4=+9.286, IR1=866uA, IR2=4.857mA, it's clear that Ibq3 is most of IR1, so Icq1 will be reduced to Icq1=IR1-Ibq3=866uA-666uA=200uA, so Vbeq1=.7-ln(2.86mA/200uA)=.631 (so IR1 now 874uA) Vbq3=10.631 so, Vo=10.631-.792=9.839, IL=98.39mA there's not much base drive for Q4, so it is still, Vbq4=9.286 (taken from case (b)) so Vbeq4=9.839-9.286=.553, so, Icq4=2.86mA/e^((.7-.553)/26mV)=10uA

Anyway, that's what I got from the problems. It's a little different from the link you gave, but close.

Jon

snipped-for-privacy@att.bizzzzzzzzzzzz expounded in news: snipped-for-privacy@4ax.com:

Ya, I don't know why it is not std equipment. It seems they try to squeeze every other function in (those I don't need) rather than the essentials.

I saw many of the newest calculators in the store yesterday have a "(-)" key, which is not actually a change sign key. It is a minus sign, as opposed to subtraction. On mine, this allows entry of a negative exponent when used with the Exp key. Ahhhhh....

Warren