transitor: the base pin.

[T3sl4co1l]

Masks have always been expensive because they are engraved, etched, deposited or etc. in a very slow sequential method.  Once made... they're easy to use, because they print a complete layer for an entire design in one shot, and that shot is exposed and stepped over the wafer very quickly.

I'm guessing, masks have always been in high demand so production has always been expensive, but the capital equipment to produce it has always been even more expensive, hence the scarcity of supply as well.  (And the supply of THAT in turn, well... same idea?)

Tim

[photon]

Not sure if cool...but close to sci fi for me! 

Wondering what somebody would say to those that commited the "mistake"? Is it? Can you actually blame them? 

no. it is an oversight that wasn't even caught during design review. shit happens. deal with it. that is life. only when playing with 'refined sand' mistakes become very expensive ... but that is life

Engineers definitely do get blamed for mistakes like this. For a chip with million dollar mask costs, more than half the engineers on the team are doing verification. More than half the engineering cost of the chip is spent to prevent bugs like this from happening. If there were a floating net this would be very embarrassing for the verification team to explain how their test plan missed it.
On the other hand, no chip is 100% bug clean on first silicon. I have heard engineers brag about how their chip was perfect, but I tend not to believe them. I have been on the team of about 20 large asics and none of them were perfect on first silicon.
When the budget of a chip is made, a part is set aside for one metal-only spin. First silicon is a declared to be a success if the number and severity of the bugs is small enough that parts can be given to first customers. You will see the terms used like "the part is now sampling, production shipment in the next quarter."

[free_electron]

Masks have always been expensive because they are engraved, etched, deposited or etc. in a very slow sequential method.  Once made... they're easy to use, because they print a complete layer for an entire design in one shot, and that shot is exposed and stepped over the wafer very quickly.

I'm guessing, masks have always been in high demand so production has always been expensive, but the capital equipment to produce it has always been even more expensive, hence the scarcity of supply as well.  (And the supply of THAT in turn, well... same idea?)

Tim

the mask is an optically flawless quartz plate with chrome sputtered on.

an ion-mill then traces the shapes where UV light has to get through. ion milling is done in a vacuum chamber. they ionize some corrosive material ( i don't know what they use to 'eat' chrome ) and then steer a beam of the ionized gas much like a photoplotter would do. this happens at a scale larger than what is ultimately needed.

only a few companies in the world have the equipment to do that. Dupont is by far the largest maskmaker in the world.

[photon]

... so a better solution would be to switch into the s-domain (Laplace transform).
The same three equations then become:
v_g = v_C1 + v_C2 + R₁*v_C2*C₂*s
v_g = v_C1 + v_C3 + R₂*v_C3*C₃*s
v_C1*C₁*s = v_C2*C₂*s + v_C3*C₃*s
The Heaviside function is quite simple to write in the s-domain:
v_g = 1/s

Solve for desired quantity and perform an inverse Laplace transform, so you get the output function in the time domain.

This is how one would solve the circuit analytically. The process is rather tricky and time consuming, but (contrary to what some in this thread claim) it can be done. Kirchoff's laws apply. Always. It is a good idea to help yourself solve problems like these with programs like Matlab.

We could simplify our life and approach the problem by slapping everything into SPICE and running the simulation. This is what you get, if you choose C₁=2µF, C₂=3µF and C₃=1µF:


Interesting, huh?

Yes, so interesting I calculated the closed solution as learned in EE circuits class using a ti89 calculator.

I wrote current equations rather than voltage equations since the arithmetic is easier. Using KVL and KVC I get:
  (I1 + I2)/sC1 + I2/sC2 + RI2 = 1/s
  (I1 + I2)/sC1 + I3/sC3 + RI3 = 1/s
  I1 = I2 + I3

Substituting:
  I2(833333 + 1000s) +  I3(500000)  = 1
  I2(500000) + I3(1500000 + 1000s) = 1

Solving for I2 and I3:
  I2 = (s + 1000)/(1000s^2 + 2333333s + 9999999500)
  I3 = (s + 333.333)/(100s^2 + 2333333s + 9999999500)

Expanding in partial fractions:
  I2 = .000639/(s + 1767.59) + .000361/(s + 565.471)
  I3 = .001193/(s + 1767.59) + .000193/(s + 565.471)

Finally, inverse Laplace transforms gives the solution:
  i2 = .000639(e^(-1767.59t)) + .000361(e^(-565.471t))
  i3 = .001193(e^(-1767.59t)) - .000193(e^(-565.471t))

Switching over to voltages and doing more of the same gives:
  v1 = .51822(1 - e^(-1767.59t)) + .148478(1 - e^(-565.471t))
  v2 = .120503(1 - e^(-1767.59t)) + .212695(1 - e^(-565.471t))
  v3 = .67493(1 - e^(-1767.59t)) - .341146(1 - e^(-565.471t))
  vn2 = .639e^(-1767.59t) + .361e^(-565.471t)
  vn3 = 1.193e^(-1767.59t) - .193e^(-565.471t)

I attach the plot below to show that they agree with your plots and that the arithmetic is correct.

The good people at Mathworks want about $3000 for a Matlab license. One must have a good business using it in order to justify that kind of expense. Yes, LTSpice is an unbelievably good and free tool. And so easy to use. The above calculation is long and tedious, as you mentioned.

I took the circuits class a long (35 yrs) time ago. Back then it was one of the killer classes you had to pass to get your degree. Is it still so important today? It's old (goes back to Heaviside and about 1890), not very helpful in understanding the circuit and not used much, if any, in practice.

[rsjsouza]

Yes, so interesting I calculated the closed solution as learned in EE circuits class using a ti89 calculator.

That reminded me of how many equations and programs I had in my HP48SX back in university... These are pheonomenal calculators. 

The good people at Mathworks want about $3000 for a Matlab license. One must have a good business using it in order to justify that kind of expense.

Instead of Mathworks' package, there is always Octave.

[Syntax_Error]

Circuits 1 is a common "first class" after digital circuits in many university curricula. It is presented early on, with frequent assertions that this is simple; just wait until you get to E&M field theory later, etc.

It's interesting to hear the different perspective on this, coming from "the industry" rather than academia.

[IanB]

I was taught how to solve differential equations analytically using Laplace transforms in the first months of my engineering degree, but it is not something I have found to have much practical use in the days of powerful computers. Many (most?) systems are non-linear, and for any non-trivial system a numerical solution by computer is quicker and more convenient.

In fact we were taught many "hand calculation oriented" design methods that are no longer in common use by modern engineers. The days of charts, nomographs and tables for detailed design are largely history now, and find a use mainly for estimation and for getting a feel for a solution.

[Syntax_Error]

I was taught how to solve differential equations analytically using Laplace transforms in the first months of my engineering degree,

I did read all of what you wrote, but I wanted to key in on this to ask you a question: How on earth is this taught in the first few months of a college education? Did you learn the prerequisite mathematics (differential and integral calculus, and differential equations) prior to university, in high school/secondary school?

This somewhat blows my mind. Not that this is unlearnable, not at all. But thinking back to my high school education, and what I was capable of as a teenager, oh boy...

[IanB]

I was taught how to solve differential equations analytically using Laplace transforms in the first months of my engineering degree,

I did read all of what you wrote, but I wanted to key in on this to ask you a question: How on earth is this taught in the first few months of a college education? Did you learn the prerequisite mathematics (differential and integral calculus, and differential equations) prior to university, in high school/secondary school?

Yes, I had been introduced to differential and integral calculus and differential equations in secondary school. I had studied A level mathematics, which is a prerequisite for acceptance into an engineering degree program. The A level syllabus in the link is representative of the ground we covered in high school before reaching university: http://www.cie.org.uk/images/92083-2014-syllabus.pdf

In the first year of university we were given a core mathematics course which was both a revision of what we were already presumed to know and the introduction of new material relevant to the study of engineering. The new material went into a lot more depth around differential equations than I had encountered before.

[dannyf]

This somewhat blows my mind.

There are differential equations, and then there are differential equations -> they come in different shape, form and complexity.

The simplest differential equations you would have solved in a typical highschool math class, like dx/dt = C, or the maximization of a quadratic equation, or free fall of an object....

Then the overwhelming majority of them have not been solved, or cannot be solved in closed form.

So I wouldn't take that comment too seriously.

[dannyf]

Too bad Ian didn't decide to stay.

To answer some of the questions that have been rolling around, the BJT is best thought of as a voltage-driven device. In the common-emitter configuration for example, the Vbe causes the B-E junction to be forward biased. This generates a diode-type current of electrons into the base. The "efficiency" of the BJT is determined by how many of those electrons are swept across the base into the reverse-biased C-B junction and are collected in the collector region (hence the names "emitter" and "collector"). The electrons lost in the base come out the base terminal. The measure of the efficiency is the ratio of collector current to base current (ie the beta). This is why the base must be very thin in order to have a BJT - just putting 2 dioides back to back does not create a BJT.

"Controlling the BJT with the base current" is not what happens. Even though it appears to be fed by a current, the appropriate Vbe is developed that will support that base current.

My questions to him were trying get him to apply the same logic to a different (current) driving mechanism.

Ian's argument sounds like "if I can cause collector current to flow by applying a voltage to the b-e junction, then a bjt must be a voltage controlled device". Well, if you apply a current to the b-e junction, you can cause collector current to flow too. By the same logic, a bjt must be a current controlled device.

The fact that applying the same principles to the same device yields diametrically opposing conclusions should cause one to question the validity of such "principles".

[Zero999]

Too bad Ian didn't decide to stay.

To answer some of the questions that have been rolling around, the BJT is best thought of as a voltage-driven device. In the common-emitter configuration for example, the Vbe causes the B-E junction to be forward biased. This generates a diode-type current of electrons into the base. The "efficiency" of the BJT is determined by how many of those electrons are swept across the base into the reverse-biased C-B junction and are collected in the collector region (hence the names "emitter" and "collector"). The electrons lost in the base come out the base terminal. The measure of the efficiency is the ratio of collector current to base current (ie the beta). This is why the base must be very thin in order to have a BJT - just putting 2 dioides back to back does not create a BJT.

"Controlling the BJT with the base current" is not what happens. Even though it appears to be fed by a current, the appropriate Vbe is developed that will support that base current.

My questions to him were trying get him to apply the same logic to a different (current) driving mechanism.

Ian's argument sounds like "if I can cause collector current to flow by applying a voltage to the b-e junction, then a bjt must be a voltage controlled device". Well, if you apply a current to the b-e junction, you can cause collector current to flow too. By the same logic, a bjt must be a current controlled device.

The fact that applying the same principles to the same device yields diametrically opposing conclusions should cause one to question the validity of such "principles".

The argument is repeating itself.

The voltage controlled model stems from the fact that the collector current is more strongly linked to the base-emitter voltage, rather than the base current.

We can argue about this until we're blue in the face but it doesn't change the fact that it's better to design an amplifier circuit based on the assumption that a BJT is voltage controlled. The dependence on base current current can quite easily be eliminated by ensuring there's adequate base drive.

As far as the current controlled model is concerned: it's very good to design circuits assuming the worst case for the Hfe. If you're using a BJT as a saturated switch, it's best to assume it's current controlled but one still needs to be aware of the base charge when turning the transistor off.