# How does a transistor amplify current or voltage?

Discussion in 'General Electronics Chat' started by jac4b, Jul 23, 2010.

1. ### MikebitsWell-Known Member

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Here is the exact verbiage that accompanied the transistor man image. Taken from Horowitz and Hill Art of Electronics 2nd ed. pg. 64

2. ### RatchitWell-Known Member

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Mikebits,

That is an absolutely correct statement. But it does not give the reason why that is so. Both the Ib and Ic are exponentially related to Vbe, as I showed in my previous post. Therefore both Ic and Ib will be linearly related to each other, as everyone who works with BJTs knows. But, Ib is a waste current that cannot be avoided, but has to be taken into account. Using β in calculations does that. It still does not take away from the fact that physically speaking, Vbe is what is really controlling the Ic in a BJT. And Ib is a useful indicator, not a controlling factor of Ic.

Ratch

3. ### jangalNew Member

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transfer resistor

hi
transistor means transfer resistor.
when a small current pass a big resistor there will be a high voltage over the transistor.
and the power for amplifying comes from the dc power supply.
and when there is a low voltage over a very low resistor there will be a big current.

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5. ### crutschowWell-Known MemberMost Helpful Member

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The input impedance of a BJT is also very low and that doesn't prevent you from saying that the BJT is voltage controlled. There only difference is that the BJT input resistance is nonlinear and the current amp is not.

My water analogy was just a continuation of BrownOut's. And whether a hydraulic engineer uses electrical analogies is rather irrelevant to this discussion.

You keep referring to the base current as waste current. It is not. It is fundamental to the operation of a BJT. No base current, no collector current (neglecting leakages).

When you design low frequency BJT circuits you seldom worry about the incremental impedance of the transistor and never use a pure voltage to drive the base. You always use a resistor to define the base current (of course you allow for the nominal diode drop at the normal base current, but that's not the incremental impedance you are referring to). So for practical design considerations, a BJT is a current-controlled device.

But as stated by several, if you want to view BJT as voltage operated (which they are in a theoretical sense, since any current also involves a voltage), you're welcome to your opinion. But you are unlikely to convince anyone else of your viewpoint.

Last edited: Jul 24, 2010
6. ### RatchitWell-Known Member

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crutschow,

A low input impedance makes the device amenable to a current source. It does not change whether it is voltage or current controlled. Current controls the magnetic amp, and voltage controls the BJT.

So why does a BJT operate better and better when the β becomes higher and higher and Ib become less and less? Ultimately, when the β becomes super high, the base current becomes insignificant and the BJT behaves like a FET. It does so anyway for very low base currents. You still have not explained the relationship of the equation that relates the Vbe to Ic.

I never said that folks drive a BJT with a voltage source in the active region. That would be folly due to its very nonlinear exponential voltage-current characteristics. Everyone adds external circuitry to linearize the output. That does not take away the fact, as demonstrated by the Vbe to Ic equation, that the basic BJT is a voltage to current responding unit, which mimics a current amplifier.

That is not the reason a BJT is a voltage controlled device. It is because voltage, not current, controls the collector current, as demonstrated by the aforementioned equation. By the way, there is no significant voltage involved with superconductor currents.

Perhaps, but I can be convincing. But can you or anyone else defend your position?

Ratch

7. ### crutschowWell-Known MemberMost Helpful Member

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We have been defending, but you aren't listening. The base current is fundamental to the operation of a BJT, it is not incidental or "waste". It is the recombination of the carriers in the base region that generates the collector current. (See Bipolar Junction Transistors). True a transistor with a higher beta is better, but you can never have a transistor with infinite beta. It requires some finite (however small) base current.

The opposite is not true of a voltage controlled device such as a FET. In that case the gate current is indeed incidental (leakage) and is not necessary for the transistor operation.

The equation relating base voltage to collector current is interesting from a theoretical point of view, but it really has little practical use in the design of most transistor circuits. When you bias a FET you bias it with voltage at the gate, and when you bias a BJT you bias it with current at the base.

Perhaps the difference it that you are talking theoretical aspects of the transistor and I am talking about the practical aspects of circuit design with a transistor. And from the practical aspect, the BJT is definitely a current operated device.

Let us know when you convince someone else, beside yourself, otherwise.

8. ### colin55Well-Known Member

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It just happens that you need a higher voltage to get more current into the base of a transistor. But if you could get more current into a transistor without increasing the voltage, the transistor would produce more current in the collector-emitter circuit. The fact that you need a higher voltage to deliver more current into the base is suggesting to the OP that voltage is the determining factor. This is where he is being mis-lead.
Once you start designing circuits you soon realise you have sufficient voltage to drive an output stage but current is lacking.
That's when you suddenly realise CURRENT IS THE ANSWER.

Last edited: Jul 24, 2010
9. ### RatchitWell-Known Member

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crutschow,

Actually, I thought it was the other way around.

Fundamental with respect to its unavoidable existence.

That unavoidable recombination does not generate the Ic. It is the Vbe that controls the Ic.

Ib is not required, but it is unavoidable. Vbe controls Ic.

As I said before, a BJT at low Ib does behave like a FET.

I said as much many times already. That is why I don't think you are comprehending what I say very well. I said that a BJT is a voltage controlled device. I did not say anyone should design a circuit that way. If fact, I already stated that no one would drive a transistor with a voltage source due to its nonlinear response. I also said that one should use the Ib for calculation and design because it is an indicator of Ic, but not a control. Furthermore, I said that external circuitry is added to linearize the response.

I will do the next best thing. I will show you a couple of folks that I did not convince, but who believe what I do nevertheless.

Look at the second box of this link Transistor Operating Details . Notice the beginning sentence, "The base-emitter voltage can be considered to be the controlling variable in determining transistor action."

On the second to the last paragraph of this link SCIENCE HOBBYIST: how transistors work, an alternate viewpoint.
" It just so happens that the tiny leakage current in the Base connection is proportional to the transistor's main Emitter/Collector current. (This makes perfect sense, since both the Base leakage current and the main Emitter/Collector current are controlled by the insulator thickness, which is set by the Base/Emitter voltage.) ...so it SEEMS as though the Collector current is directly controlled by the Base current. You can even simplify things by pretending that this is true. But in reality, Base current communicates with Collector current via changes in Base-Emitter voltage. It's the Base/Emitter voltage which runs things. You'll never understand "transistor action" and the simple physics inside the box if you think that Base current directly controls the Collector current. It doesn't. The control is there, of course, but it's one stage separated. "

Ratch

10. ### RatchitWell-Known Member

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colin55,

You got that bass-ackwards. A higher voltage on the base results in a higher Vbe, which in turn causes a unavoidable higher Ib to occur.

You would get more Ic because of the higher Vbe. The increase of Ib is unavoidable and not related to control of Ic, but is a good linear indicator of Ic instead.

You do need more Vbe to increase the Ic, and more Ib will ensue. I never said that one should attempt to control Vbe in design. I said many times that Ib should be used as an indicator of what the Ic will be. So where is the confusion?

The answer to what? What is the problem?

Ratch

11. ### ericgibbsWell-Known MemberMost Helpful Member

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We seem to have another member of the Flat Earth society, who just wants to argue about a subject he obviously knows little about.

This thread has gone the same way as the 'speed of sound', 'ohms law' and HHO threads and IMHO should be closed.

12. ### MikebitsWell-Known Member

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Sadly the OP is the one that suffers from such a derailed thread...

13. ### RatchitWell-Known Member

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ericgibbs,

Many people, who receive messages or information that goes against what they believe, will disparage the messenger when they cannot successfully dispel or refute what the message contains. Remember Galileo?

Ratch

14. ### RatchitWell-Known Member

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Mikebits,

Why? He got his questioned answered, didn't he? I hear this type of rejoinder many times from politicians who justify their actions by saying "Its best/good/ for the people".

Ratch

15. ### Claude AbrahamMember

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First of all, Ib is NOT a "waste" current. Without Ib & Ie both, Vbe can never be established. How does Vbe go from 0.00V to 0.65V, for example? The answer is that charge must be transported through the base-emitter region. Upon crossing the junction, the carriers which were formerly majority type become minority type since the material is different. Holes from the base are majority carriers for an npn bjt, & electrons from the emitter are majority in the emitter. But they cross the junction & holes enter the emitter, while electrons enter the base, & they are now minority carriers.

The associated electric field due to the stored minority carriers tends to oppose forward current in the b-e junction. When unbiased, thermal energy generates electron-hole pairs & they cross the junction resulting in an electric field which opposes forward current. The integral of said field over distance is labeled "thermal voltage" aka "built in potential". At room temp it is roughly 25.7mV.

When forward biased, many more charges flow & the junction potential is around 0.65V, give or take w/ temp, & current value. In order for Vbe to be non-zero, a base current & emitter current must first take place. Vbe comes after Ib & Ie. This component of base current is called displacement current. It is fundamental & a bjt cannot operate w/o it. It is like the gate current in a FET. The FET gate current is absolutely indispensable & fundamental for FET operation.

With bjt & FET, a transition frequency, ft, is defined on every data sheet I've seen. This ft parameter is by definition the frequency where the current gain has fallen to unity (0 dB). A device with current gain less than 1 is not very useful as an amplifying device, so ft is stated for that reason. Displacement component of base current is all important & a bjt cannot operate w/o it. Ib is not a residue or "waste" current.

The holes injected from base to emitter do not contribute to collector current. But we can minimize this injection component of current by doping the base much lighter than the emitter. Also, the transport component of base current must be discussed. Electrons emitted from the emitter towards the base diffuse & are drawn into the collector by its local E field. A few electrons recombine in the base never reaching the collector. So the same number of electrons must exit the base to preserve charge neutrality. Otherwise the base region would build up an unbalance of charge & bjt action ceases. But the transport component of Ib is generally much less than the injection component. A super thin base region minimizes said transport component.

So the transport component of Ib is very small, but fundamental to operation, as is the displacement part of Ib. W/o these base current components, there is no bjt action. Vbe is also indispensable as well. But Vbe can also be viewed as a "waste" quantity.

Ideally, a forward biased p-n junction, such as a b-e jcn, carries all current & zero voltage. The notion that a bjt should have positive non-zero Vbe with zero Ib is ludicrous. Such a device is not even BI-polar, but unipolar such as a FET.

Anyone who has ever worked with amplifier bjt power output stages has learned about cross-over distortion 1st hand. In the +/- 0.7V signal region, the device has to cross over. Many papers, patents, articles, books, networks, etc. have been coined to address this issue. There is the class A, AB, B, etc. approach. The objective comes down to 1 thing.

If the Vbe was what it ideally should be, namely zero, all this trouble would not exist. I've designed many networks where the 0.7V Vbe drop was an obnoxious limitation, & wished that a germanium device with 0.3V drop was available. Vbe can be just as much a nuisance as is Ib. Also, with op amps, there is an output error due to Ib mismatch, and one due to Vbe mismatch. If Ib was zero, there is no offset current error. But likewise, if Vbe is zero, there is no voltage offset error.

This notion that Ib is just a "waste" product, is propagated by wannabes, hackers, semi-literates. pretenders, & know-nothings. In the ideal realm, Ib & Vbe both vanish, leaving Ie as the control parameter for Ic. But in the real world, we are stuck with Ib & Vbe whether we like them or not. Still, it is Ie that controls Ic.

The Ebers-Moll equations, published in Dec 1954, describe a bjt as 2 current controlled current sources. Although you can derive Ic as a function of Vbe, it is also true that Ic is related to Ib, as well as Ie.

1) Ic = beta*Ib

2) Ic = alpha*Ies*(exp((Vbe/Vt) - 1)

3) Ic = alpha*Ie.

In equation 1), forcing a specific value of Ib results in Ic per 1). But this approach is avoided most of the time due to beta dependency. It is hard to predict beta with differing speciman & temp. In 2), Ies varies with speciman, hugely with temp, & a bjt driven w/ a voltage source across Vbe is thermally unstable. A Vbe connected to a source, results in Ic, some power, a temp rise, then Ies increases due to elevated temp, Ic increases, temp increases, etc. Thermal runaway means that you can never "voltage control" a bjt.

In eqn 3), alpha is very stable in value. Controlling Ie results in a very predictable Ic. It is thermally stable & predictable. Ic is ultimately controlled by Ie, not Vbe.

To see this, take a bjt w/ a very thick base region. A Vbe of 0.7V is set up, & Ie = 10 mA, which also equals Ib. Because the base is very thick, alpha=0, & all electrons emitted are recombined in the base, so Ic = 0. "Transistor action" does not occur until electrons emitted from the emitter are "collected" by the collector. Although Vbe=0.7V, there is no Ic.

What controls the number of electrons in the collector is the number emitted from the emitter. If 1010 electrons are emitted in 1.0 picosecond, then the Ic value can never exceed 1010 e-/psec, regardless of Vbe. The emitter emission literally controls the collector collection.

A bjt that is "voltage driven" is thermally unstable, & unpredictable. A bjt that is "base current driven" is thermally stable, but beta dependent. Still we use base current control for saturated bjt switches. By forcing a base current larger than Ic/beta_min, we assure saturation, & beta dependency is mitigated. A bjt that is "emitter current driven" is thermally stable & predictable. The value of Ic is influenced by temp, Ib, Vbe, doping, illumination, Vbc (Early effect), but Ie is the parameter that chiefly controls Ic.

Again, the number of electrons per picosecond "collected" by the collector is proportional to the number "emitted" by the emitter. The emitter & collector leads are so named for a very good reason. Questions/comments are welcome.

Claude

Last edited: Jul 25, 2010
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16. ### RatchitWell-Known Member

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Claude Abraham,

OK, let's see how you defend that position.

Ib is a consequence of Vbe, not a requirement for control. In other situations, it is possible to have a voltage without a current, but not in the base circuit of a BJT.

That is the reason why a base current exists and is inevitable, but although the base current tracks the collector current, it is still Vbe that controls Ic.

I think you are trying to say that the uncovered charges oppose the diffusion voltage. Let's see where that leads us.

Yes, Vbe beats down the back voltage caused by the uncovered charges, and allows the charges to pass through the base into the collector. Vbe, Ib, and Ic occur concurrently, but Vbe is what causes things to happen.

No need to get involved with the high frequency characteristics of a BJT in this discussion. I still think you are confusing what is essential with what is a consequence.

Now you are describing what goes into manufacturing a BJT, and why Ib is inevitable. It still does not take away from the fact that Vbe controls Ic.

Again, you are confusing requirement with consequence. If Vbe is a waste quantity, why is it the primary term in the diode equation?

I already pointed out that at a very low Ib, a BJT behaves like a FET. Voltage without current is possible, as in a perfect energized capacitor. But it is not possible in a BJT due to its inevitable leakage into the base circuit.

All the problems caused by Ib and Vbe are vexing, but they do not disprove the contention that Vbe controls Ic.

It is also promoted by the equation I referenced in the textbook by Sedra & Smith.

Models show what a device does. They do not show how or why it works that way.

Yes, I said already that no one would drive a BJT by a voltage source. The above paragraph does not address what really controls Ic. Vbe controls Ic and thereby Ie, too.

That is well known and obvious, but not germane to the question of what really controls Ic.

Ratch

17. ### Claude AbrahamMember

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You insist that Ib is a consequence of Vbe, but cannot prove it. It just seems logical to you, so it must be true. Ib takes place ahead of Vbe chronologically. Take a bjt excited at a small signal frequency of 1.0 MHz, or better. Examine ib & vbe, the ac components, & it is obvious that ib leads vbe. The separated charges act like a capacitor. According to Eli the ice man, current leads voltage in a cap. This is so well known, elaboration should not be needed. Ib is not a consequence of Vbe. That is just a silly prejudice. It sounds crazy to you, but Ib & Vbe cannot exist independently.

As far as Vbe being the "primary term" in the diode equation, let me remind you that in a diode, just as we can write

Id = Is*((exp(Vd/Vt)-1), we can also write

Vd = Vt*ln((Id/Is)+1).

Just as I is a function of V, so it is that V is a function of I. Neither "causes" the other.

Before we discuss transistor action, we must get to the root of your contrarian position. Every semiconductor OEM classifies bjt as current controlled. But it is apparent that you regard voltage as the driver of current in all general circuits. I won't put words in your mouth, so please answer the following in your own words.

In Ohm's Law, V = I*R, which of the following is true.

a) The voltage is what causes the current.

b) The current is what causes the voltage.

c) It's a circular relation, akin to chickens & eggs.

It is my belief that you choose answer a). I choose answer c).

That seems to be the crux of the debate. All who insist that the current control model is wrong, also insist that current is just a consequence of voltage. If that were true, every electrical device would be "voltage controlled" & nobody would bother mentioning it.

The equation you referenced is not the complete equation. The alpha factor in my equation 2) cannot be ignored when comprehending bjt action. Vbe is related to Ib & Ie as the b-e jcn is a forward biased p-n junction, & hence is described by a diode relationship. But Ic exists due to alpha*Ie. Vbe does not control Ic. Ic is due to transistor action, which is merely emitted charges transported through & past the base & into the collector.

As far as high beta goes, remember that Ie controls Ic, so that the parameter relating the two is alpha, not beta. But regarding beta, sure higher beta is desirable. That does not diminish the importance of Ib. With FETs, the gain is "gm", the transconductance. The higher the gm the better. One disadvantage w/ FETs vs. bjt, is that a FET has a lower gm. When amplifying current, it is desirable to have the highest current gain possible. Hence high beta is better than low beta. But the same holds true with voltage amplification. A higher gm is better than a lower gm. Hence it is desirable to obtain the highest output voltage with the smallest input voltage.

If "beta were infinite", the bjt is still a current controlled device, because Ie still emits carriers & that determines the collection of carriers in the collector. Remember an ideal p-n jcn has a zero forward voltage drop. An ideal bjt would have a b-e junction with no forward Vbe. Thus any signal at the base appears at the emitter. The emitter resistance determines emitter current. Then Ic = alpha*Ie. But alpha=1 for an ideal bjt. Again, Ie controls Ic.

To say that Ib is ideally zero flies in the face of science. If Vbe is non-zero & positive, an E field exists from base to emitter due to charge separation. This E field exerts a force on all free charges in the base region, emitter, & depletion. The base is p type, so its holes feel a force in the direction towards the emitter. The electrons in the n type region feel a force towards the base.

The hole motion from base to emitter is the dominant component of Ib at low frequencies. How does this Ib "ideally" vanish. To reduce it we dope the p type base lightly. Even undoped, the silicon still has carriers, & Ib results. To reduce Ib to zero requires that the base be an insulator. But then the device is no longer "bipolar". "Bi" means 2, & as long as there are 2 types of material there has to be base current. Those who neglect the importance of Ib seem to overlook the bi in bipolar. In addition to the injection part of Ib, as I stated earlier, the displacement part of Ib is critical to establishing Vbe.

Once again, your answer to Ohm's law question is the reason you are at odds with semicon OEMs. If you insist that current is a consequence of voltage, then debating any device, bjt or otherwise, is pointless. That seems to be your mental block you must overcome.

If your answer to Ohm's law is NOT a), then we can have a more productive discussion. So please answer my Ohm question, & we'll take it from there.

Claude

Last edited: Jul 25, 2010
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18. ### BrownOutBanned

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He can learn alot if he reads posts by crutschow, colin55 and Claude Abraham, who have posted correct informaton about current control in BJT's. Others should read, understand and learn from these members.

19. ### RatchitWell-Known Member

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Claude Abraham,

I have reasons for my beliefs other than what you think they are. I will concede that I did not explain thoroughly enough why Vbe is the cause of Ic. Although the equations are correct in showing that a Ib will always exist in a BJT operating in the active region, they do not show why voltage is the motivating factor. I will attempt to do that below. Let's not get into the AC characteristics of the transistor. The DC characteristics will suffice. I agree that Ib and Ic cannot exist independently in a BJT. I said so already, and the diode equation shows that.

I understand what you are saying, it just verifies that Ib and Vbe are related to each other. I already said as much. It still begs the question of whether Ib or Vbe controls Ic. I will explain that below. I know of no basic device that is a current amplifier except a magnetic amplifier.

Again you are referencing a model. As I said before, a model is good for explaining what a device does, but not how or why it does it.

I disagree. An ideal diode would not have any bulk resistance. It would still have a forward voltage due to the physics of the material. Ic is the majority part of Ie, so Ic does not control Ic, except in a model perhaps.

It flies in the face of the physics of the BJT. One cannot have zero Ib. I said that before. I also said that at very small values of Ib, the BJT behaves like a FET.

I never said that Ib can vanish. I said that Ib was inevitable, but not a controlling factor in Ic. I also said it was a indicator of what Ic would be, but did not control Ic. Ib is important for being a linear indicator of Ic.

Just to digress, I do not believe Ohm's law is V=IR. What you believe is Ohm's law, I call the resistance formula, and Ohm's law relates to the linearity of the material. In other words, it is a property of a material. My arguments to that are posted at the end of this thread. http://www.electro-tech-online.com/content/222-amp-hour-fallacy-ohm-s-law-noobies.html. You can contribute to it if you wish.

Of course the resistance formula can be written three different ways. But that does not mean that Ib or Vbe are both correct in what controls Ic. My further explanation follows.

In a NPN BJT, if the N-type material of the emitter is put in contact with a P-type material of the base, the electrons if the N-type material will migrate over into the P-type material material and vice versa. Eventually this will stop happening, because the N-type material will have positive ions along the boundary due to the absence of electrons that departed. These positive ions will keep the holes of the base from coming over. The same thing happens in the base in reverse. The electrons coming from the emitter neutralize the holes in the base and make negative ions which oppose further migration of the electrons from the emitter. Eventually a equilibrium is reached. These repulsive forces form a region along the NP boundary called the depletion layer. By applying a voltage Vbe, one can effectively make the boundary layer thinner and allow the charge carriers to move again. Or a opposite Vbe can increase the boundary layer and restrict the charge carriers. This is why I and others consider a BJT to be a voltage controlled device. Only a voltage on the base-emitter junction can do what needs to be done, that is, control the thickness of the depletion layer. Some electrons will get into the base circuit, the amount depending on the exponential rate of the diode equation. The Ic will also vary at a exponential rate. Therefore Ic will appear to be controlled by Ib in a linear manner. But a BJT is only mimicking a current amplifier. Nevertheless, this is a useful property that should always be used for design and analysis. But the operation of the transistor is dependent on the value of Vbe. Any bias arrangements or signal applied to the base will change Vbe, to a first approximation, by a few millivolts or microvolts. All the active region of the BJT is controlled exponentially by a Vbe of between zero and 1 volt. Therefore no one tries to control a BJT directly by a voltage source unless they are building a logarithmic amplifier. So, take it from there.

Ratch

Last edited: Jul 25, 2010
20. ### RatchitWell-Known Member

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BrownOut,

Yes, he can learn a lot from all of us. We all posted correct information. It is the conclusions we arrived at that are different. Everyone should read, evaluate, learn and arrive at their own conclusions.

Ratch