Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

How does a transistor amplify current or voltage?

Status
Not open for further replies.
Very good job Mr. Al. I applaud you for your patience. You have more than I have, that's for sure.

Hi,

Ah what the heck, it only took 418 pages right? :)
 
Claude,

MrAl certainly deserves a thank you, but you neglected to mention Jony130 for his contribution. And do I not also deserve an honorable mention for not walking away, and agreeing with something I did not quite believe just to settle the matter? Unfortunately my unfamiliarity with Ltspice and a bum transistor chart did not help to shorten the discussion either.

MrAl,

I said that I agreed with you about Ic controlling the transistor, especially during saturation. That is certainly true, but I was wondering why you made that point. It was never in doubt, was it? Much of the discussion was whether Ic increased enough to bring the transistor out of saturation. And while in saturation, Ib doesn't have much effect on Vce(sat), but it does on what value Ic brings the transistor out of saturation. So it can be said that Ib also controls the transistor during saturation at the same time. I don't want to start another round of discussion on this, because I believe we each see the other's point of view.

Ratch
 
Claude,

MrAl certainly deserves a thank you, but you neglected to mention Jony130 for his contribution. And do I not also deserve an honorable mention for not walking away, and agreeing with something I did not quite believe just to settle the matter? Unfortunately my unfamiliarity with Ltspice and a bum transistor chart did not help to shorten the discussion either.

MrAl,

I said that I agreed with you about Ic controlling the transistor, especially during saturation. That is certainly true, but I was wondering why you made that point. It was never in doubt, was it? Much of the discussion was whether Ic increased enough to bring the transistor out of saturation. And while in saturation, Ib doesn't have much effect on Vce(sat), but it does on what value Ic brings the transistor out of saturation. So it can be said that Ib also controls the transistor during saturation at the same time. I don't want to start another round of discussion on this, because I believe we each see the other's point of view.

Ratch

By thanking Mr. Al, I did not mean to exclude you or Jony130. It's just that Mr. Al has been on this thread since the beginning & has spent much time & energy doing simulation, & presented his findings in an organized manner. That is why I thanked him, because his methods are how science advances. Brownout has also brought good info to the table which we all benefitted from. But now that you mention it, I thank you for agreeing w/ Mr. Al.

One salient point I'd like to make is that earlier on, you mentioned that the I-V curve of a bjt cease to be exponential when the device goes into saturation. Although that is true if we view only Ic vs. Vbe, keep in mind that there are 2 things going on. Ebers & Moll pointed this out in their 1954 paper. In saturation, the b-c jcn is forward biased. The base current has 2 components, base to emitter, & base to collector. In the base region these 2 components add together. But the collector current Ic, is made up of the component due to the b-e jcn forward bias, & the other component, due to the b-c jcn forward bias. The 2nd component subtracts from the 1st. If we visualize an upside down bjt, the collector becomes the emitter & vice-versa.

The relation between Ic & Vbe is still logarithmic, even in saturation, but Ic/Ie consists of 2 parts in opposition.

Ic = alpha_n*Ies*(exp(Vbe/Vt)-1) - Ics*(exp(Vbc/Vt)-1),

Ie = Ies*(exp(Vbe/Vt)-1) - alpha_i*Ics*(exp(Vbc/Vt)-1),

where alpha_n/alpha_i are the normal/inverse common base current gains resp., & Ies/Ics are the reverse saturation scale currents for the b-e/b-c jcns resp.

In the active region, the 2nd term involving Ics/Vbc is small since the b-c jcn is reverse biased, i.e. leakage current. Hence 1 term is all that is needed, & it exhibits exponential I-V. In saturation, 2 terms are involved, & the exponential nature of the junctions is not immediately apparent.
 
Claude,

One salient point I'd like to make is that earlier on, you mentioned that the I-V curve of a bjt cease to be exponential when the device goes into saturation. Although that is true if we view only Ic vs. Vbe, keep in mind that there are 2 things going on. Ebers & Moll pointed this out in their 1954 paper. In saturation, the b-c jcn is forward biased. The base current has 2 components, base to emitter, & base to collector. In the base region these 2 components add together. But the collector current Ic, is made up of the component due to the b-e jcn forward bias, & the other component, due to the b-c jcn forward bias. The 2nd component subtracts from the 1st. If we visualize an upside down bjt, the collector becomes the emitter & vice-versa.

The relation between Ic & Vbe is still logarithmic, even in saturation, but Ic/Ie consists of 2 parts in opposition.

Ic = alpha_n*Ies*(exp(Vbe/Vt)-1) - Ics*(exp(Vbc/Vt)-1),

Ie = Ies*(exp(Vbe/Vt)-1) - alpha_i*Ics*(exp(Vbc/Vt)-1),

where alpha_n/alpha_i are the normal/inverse common base current gains resp., & Ies/Ics are the reverse saturation scale currents for the b-e/b-c jcns resp.

In the active region, the 2nd term involving Ics/Vbc is small since the b-c jcn is reverse biased, i.e. leakage current. Hence 1 term is all that is needed, & it exhibits exponential I-V. In saturation, 2 terms are involved, & the exponential nature of the junctions is not immediately apparent.

I understand and concur.

Ratch
 
Amen. That was one heck of a thread.

Thankfully it all worked out OK.

Cheers
 
voice activated switch.

anyone can explain the circuit with very good working.
http://www.buildcircuit.com/clap-switch/[/URL]
 
It's a very badly-designed circuit.
You cannot connect an electret micrphone to the base of a transistor.
You don't need a BD139 to drive a relay.
 
I spent many years also wondering how a transistor or even its predecessor, a vacuum tube, can amplify. I felt it was creating energy out of nowhere. I was further concerned that the equivalent Thevenin circuit showed a little amplifier. It finally came clear to me when I realized that a transistor or vacuum tube does not amplify anything in a technical sense. All they do is change resistance. A small current from the base to emitter causes an amplified resistance change from the collector to emitter. A small signal change to the grid in a vacuum tube causes a large equivalent resistance change in the anode to cathode. The only ampliebdion going on is the large change in resistance of the main circuit caused by a small change in the grid or base. The rest is all Ohm's Law. Thus, a transistor can't do anything more than a motorized variable resistor (putting aside frequency response). A small change in signal causes a large change in the resistance value. Equivalent circuits showing a little amplifier just confuse the issue. Classically trained electrical engineers will condemn these statements because they are so into the amplification factors and equivalent circuits. But, it's just a variable resistor.
 
rt of as,

"I spent many years also wondering how a transistor or even its predecessor, a vacuum tube, can amplify. I felt it was creating energy out of nowhere. I was further concerned that the equivalent Thevenin circuit showed a little amplifier."

Wonder no more. The energy of the amplified signal comes from the DC bias sources.

" It finally came clear to me when I realized that a transistor or vacuum tube does not amplify anything in a technical sense. All they do is change resistance."

No, they change the current by a control voltage. They do not use a dissipative resistor to raise and lower the output current. Let me explain. Suppose you energize an ideal capacitor from 0 to 100 volts. The current will be large at first and nothing at 100 volts when fully energized. You can simulate the current by adjusting a potientiometer, but that will dissipate heat. An ideal capacitor will not dissipate any heat, so it does not act like a variable resisistor. Even though you can use the resistance formula and get an E/I reading at each point, it is not dissipative resistance. The reduction of the current in the capacitor is caused by the back-voltage resulting from the accumulated charges on the plates, not resistance.

Similarly, the emitter-base bias voltage of a BJT brings the charge carriers up to the base region where most of them are swept across to the collector by the base-collector voltage. Again, the basic control of a BJT is the emitter-base voltage. This voltage controls how much charge is presented to the base region. Increasing the collector voltage will not increase the collector current while operating in the active region. That is why a BJT is a voltage controlled current source (transconductance device), and not a current controlled current source. There is a direct relationship between the base current and collector current, but the base current does not control the collector current. Just as an ammeter in a series circuit not control the current although there is a direct relationship to their values.

A vacuum tube is also a transconductive device. A voltage on the grid repels charge comming from the cathode. No resistance is involved in signal control.

"All they do is change resistance. A small current from the base to emitter causes an amplified resistance change from the collector to emitter. A small signal change to the grid in a vacuum tube causes a large equivalent resistance change in the anode to cathode. The only ampliebdion going on is the large change in resistance of the main circuit caused by a small change in the grid or base. The rest is all Ohm's Law. Thus, a transistor can't do anything more than a motorized variable resistor (putting aside frequency response). A small change in signal causes a large change in the resistance value. Equivalent circuits showing a little amplifier just confuse the issue. Classically trained electrical engineers will condemn these statements because they are so into the amplification factors and equivalent circuits. But, it's just a variable resistor."

I hope I convinced you that both the transistor and vacuum tube to not use resistance to modulate a signal.

Ratch
 
rt of as,
I hope I convinced you that both the transistor and vacuum tube to not use resistance to modulate a signal.

Ratch

You are right at the fundamental level that some people seem to not have a clue about. There is a equivalent resistance that can be modeled and as you say it's not dissipative in the same way the 'resistive' impedance of free space is not. That's just the product of the EM energy propagation medium (charge as electrons/holes/ions for current) in the transistor junction structure or the tube grid/plate field being modified by the applied fields.
 
Last edited:
nsaspook,

I think that perhaps nonresistive immittance is the best way to describe it. Resistance connotes energy transfer to me. Ratch
 
Hello,

It sounds almost like you are saying that the transistor does not dissipate energy. Surely that cant be right, no matter how you choose to analyze it. If we have a constant 1 amp through collector to emitter and 2v drop collector to emitter then we have 2 watts dissipated in the transistor (not including the base emitter dissipation). That's 2 watts that can never be recovered in normal use. This makes it look like a dissipating resistor of value 2 Ohms. If the drive changes then sure the apparent resistance changes, so a higher drive might make it look like only 1 Ohm. That's where the controlled resistance interpretation comes from. Im sure you are aware of this.
 
Mr. Al,

You are right. I was concentrating too much on explaining the control of the BJT and tube. As anyone who has used a linear power supply knows, high current and low output puts a stress on the pass transistors in the form of heat. It takes energy to force a current through these devices even though no physical resistor is present. That energy is dissipated as heat.

Ratch
 
I think if new people to electronics stop thinking of transistors as amps and more like switches there would be a lot less confusion.
 
Hello,

It sounds almost like you are saying that the transistor does not dissipate energy.
...
That's where the controlled resistance interpretation comes from. Im sure you are aware of this.

Sure. My point was the concept of amplification is more complex than just control of resistance. A parametric amplifier like a TWT with velocity modulation or even in a simple Varactor pumped amplifier reactance is varied.
 
I think if new people to electronics stop thinking of transistors as amps and more like switches there would be a lot less confusion.

Hi,

I have to agree. It takes a good hour to convince someone new to transistors that the base current increase turns the transistor "on" and decreasing it to zero makes it turn "off". That is hard enough for them to grasp sometimes so i like to start with that and then progress to "variable switch through variable resistance" after they comprehend that the transistor can act like a switch (common emitter configuration). It is not surprising though to hear them talk about how they wonder how the base drive can create a higher voltage at the collector, or "amplify" that signal to a higher level. They get the impression from the word "amplify" that the signal on the base itself somehow manages to undergo some sort of magical transformation where it can get higher in level all by itself. They think of the signal at the collector as being the SAME thing that went into the transistor, only bigger. It takes a while to convince them it is controlled, not increased itself.
 
I've also tried to teach new engineers what a transistor basically does and I feel that the approach described by 4pyros is the best.
 
Status
Not open for further replies.

New Articles From Microcontroller Tips

Back
Top