NPN transistor has gain of one in common emitter mode?

[treez]

Hello,

Can you confirm that the gain of Q1 (below) is about one?
(I require it to be about one)

Its a 10V to 5V linear regulator, ltspice sim also attached.

 

[Analog Ground]

Actually, it is about -1 at the gate of the MOSFET. The gain is approximately given by -(R3/R7) or -1.0. Could someone explain the purpose of C4 and R10?

 

[Audioguru]

The voltage gain of a transistor is RC/(RE + Re). Re is internal emitter resistance.
The collector resistors add to make 11k for RC.
RE is 10k so the DC voltage gain is 11k/10k= 1.1. But Re reduces the gain a little.

 

[treez]

C4 and R10 are just to reduce the gain of the BJT dynamically....they basically stop the output oscillating.

 

[Audioguru]

Yes, the voltage gain of the Mosfet adds to the loop gain then the loop becomes unstable and causes oscillation.

 

[LvW]

The voltage gain of a transistor is RC/(RE + Re). Re is internal emitter resistance.

Audioguru, according to my experience it is rather confusing to use a capital symbol Re for the internal effect you call "internal emitter resistance".
May I explain?
1.) Using this symbol Re many beginners believe that this really is an ohmical resistance (like resistance of the base-emitter path). However, that is NOT true.
Rather, it is a DYNAMIC (Differential) resistance which is NOT constant (but depends on Ic) and, therefore, should be called re.
I think, it is absolutely necessary to discriminate between static and dynamic resistances to avoid confusion and to allow a good understanding of electronic circuits (based on physical laws).

2.) But why use a resistance symbol at all? In reality, it is no resistance! It is a parameter which connects the input signal with the ouput signal:
Namely the transconductance gm, which clearly describes the input-output relationship - and it is nothing else than the slope of the Ic=f(Vbe) curve.
Isn`t it logical to use this physically existent parameter in all equations describing the gain properties (instead of re=1/gm)?

 

[Audioguru]

I am sorry that I used a capital letter "R" in my discussion about the internal emitter resistance "re" of a transistor.

I was taught and many other people (Rod Elliot of Elliot sound Products for example) were taught that re is simply calculated by "0.026/emitter current".
I have used this calculation many times and it works.

 

[treez]

..this version of the linear regulator uses a common base stage instead...I presume that this has a gain of one?

Do you also know why this common base version is far more easy to make stable than the common emitter version in the top post?

 

[LvW]

I am sorry that I used a capital letter "R" in my discussion about the internal emitter resistance "re" of a transistor.
I was taught and many other people (Rod Elliot of Elliot sound Products for example) were taught that re is simply calculated by "0.026/emitter current".
I have used this calculation many times and it works.

Audioguru, did I say it wouldn`t work to use "0.026/emitter current" ?
That`s not the question. Perhaps I did not express myself clear enough.

As you know - an OTA is a voltage controlled current source, which is characterized using the transconductance gm to define the output-to-input ratio.
And the same applies to a transistor, which also has a transfer curve Id=f(Vbe) with a slope that is identical to the transconductance gm=Ic/Vt (Vt=26 mV).
My only point is: Is there any good reason to use the inverse form re=1/gm in the gain expression?
Isn´t it logical and straight-forward to use the transconductance of the transfer curve to write for the small signals (ac) ic=gm*vbe ?
Why do you prefer ic=vbe/re (with re=1/gm) ? Does this provide more insight into the transistors operation? I have some doubts.
(By the way: I do not use every form of an equation I was taught many, many years ago).

EDIT: Here comes another point of view:

A resistance is defined as the ratio between a voltage across a conducting device and the current through this device.
However, as we all know - this does NOT apply to the base-emitter path and the collector current.
Thus, it is even not correct to use the resistor symbol re for the (differential) ratio vbe/ic.
Therefore, we must use the term " transconductance" gm=ic/vbe which reflects the physical truth. There is no "internal resistance" re !!!

 

[treez]

Actually this is all very enlightening to myself and others I hope.....I studied this years ago and have gone rusty on it.
In fact, the only reason I used a BJT at all is because the blessed op-amp won't let its output go up to the positive rail, and so it might not be able to turn off the pfet.
(also in some cases, the opamp might not even be connected to the input rail,but to a lesser voltage rail)
The bjt is handy but gives the potential problem of introducing gain into the feedback network....so its gain needs to be one.
i'm actually wondering if I should have used a fet instead of the bjt....I suspect that under certain conditions the base of the common-base bjt may plummet in voltage if the bjt saturates

 

[LvW]

Actually this is all very enlightening to myself and others I hope.....I studied this years ago and have gone rusty on it.
..........

It sounds good.
In this context, I like to emphasize that - for a good understanding of electronic working principles - it is helpful and necessary to know about the physical background of the various parameters describing the properties of electronic parts (in this case: BJT).
And therefore: Why should I use a resistance symbl re (or even Re) for such a "mystery" effect (which, in reality, is NOT resistive) which is not constant but depends on the selected Ic value - when we have a simple and physically true explanation based on the slope gm of the characteristic Ic=f(Vbe), which describes the relation between input and output ?
(Of course, "it works" (as claimed by audioguru) - but is this the only justification?).

 

[treez]

Thanks,
LvW you obviously have great expertise in Transconductance (amongst many other things). I have a longstanding transconductance question which I can't answer, and nobody else can. -that is .. Can the "COMP" pin of the LM3421 PWM controller IC be shorted to ground safely with say a BJT?
LM3421 DATASHEET
https://www.ti.com/lit/ds/symlink/lm3421.pdf

...I need to effect a soft-start in this IC every time it flashs back on in a flash beacon circuit..discharging the "COMP" pin capacitor will effect a soft start, but is it safe to do this?, or will such action give overcurrent to the internal transconductance amplifier output?
(I conjectured that it might be ok to short it to ground as surely a transconductance amplifier always has a high output resistance?, so will have no problem in running into a short)?

 

[FvM]

Can the "COMP" pin of the LM3421 PWM controller IC be shorted to ground safely with say a BJT?

Datasheets have an OTA sink/source current specification talking about this point.

 

[treez]

thanks, yes I saw that, it says 200uA.....but that doesn't seem to make sense to me....I mean , that is an extremely low current, surely it cant damage anything?
200uA is so low that surely that cannot be the limit?

I mean, if a big cap is placed on the comp Pin then the initial charging current into that cap could be more than 200uA, and the datasheet doesn't say you cant have big caps on the comp pin.

I am just wondering if the datasheet means that 200uA is the maximum current that its possible to draw out of the transconductance amplifier?


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I mean, page 33 of the LTC1871 datasheet shows that transconductance amplifier inside the LTC1871 being shorted to ground, so why cant it be done with the one in the LM3421?
LTC1871
http://cds.linear.com/docs/en/datasheet/1871fe.pdf

 

[FvM]

The maximum error amplifier output current inside the supply voltage range will be much lower, typical 30 µA. 200 µA is a maximum rating if you drive the pin beyond the supply limits. So shorting the output to ground isn't a problem.

30 µA is also the current that provides a startup ramp shown in the datasheet.

 

[treez]

Thankyou for your answer FvM.
I asked this on the ti.com forum, and a ti.com member of staff could not answer it.
I wonder why he could not answer it?
I wondered if it were possible to elaborate on where in the datasheet the ~30uA comes from in the datasheet, whence the "comp" pin is shorted to ground?

http://e2e.ti.com/support/power_management/led_driverslcd_bias/f/192/t/207440.aspx

 

[FvM]

Didn't you find it in the error amplifier paragraph?

 

[treez]

The paragraph on page 6 also says transconductance is 100uA/V...so when CSH pin = 0V, then that would mean 123.5uA flowing out of the error amplifier, I would have thought?

(1.235V * 100uA/V = 123.5uA)

 

[FvM]

The output will saturate at +/- 30 µA. That's a typical OTA behaviour, the output can't sink or source more current than than a specific value set by the biasing current source.

 

[treez]

ok thanks, but the datasheet, as you know, suggests that 200uA is possible, if not adviseable.