Talk:Common base

Untitled

Could someone (preferably the author) pleas ex[plain this satment before it is deleted

.... VHF and UHF amplifiers where the fact that the output circuit only passes through a single PN junction rather than the two junctions of a common emitter amplifier gives a crucial advantage. It actually passes through two junctions according to the diagram.

Light current 22:50, 6 August 2005 (UTC)

Relax, please. Maybe you should read through WP:AGF? - Omegatron 14:16, August 7, 2005 (UTC)

Moved this here cos it is really is not correct and should not be on a live page

where the fact that the output circuit only passes through a single PN junction rather than the two junctions of a common emitter amplifier gives a crucial advantage. (which is...?)' Light current 23:42, 7 August 2005 (UTC)

Hello Light Current! I read your profile and enjoyed your humor. Why do you think that the output circuit passes through two junctions? The dependent current in the collector-base junction is relative to the base-emitter junction. When calculating the output impedance we do not cross the collector-base junction. Please reply.

Last comment by Geek22 (new user)

Parallel notation

Which looks better?

$R_{E}\|{r_{\pi } \over 1+\beta _{0}}.$

$R_{E}\left\|{r_{\pi } \over 1+\beta _{0}}\right.$

Neither, in this case. I suggest delete lines and insert phrase 'in parallel with OR: insert 'and' between the two expressions and put words 'in parallel after. Light current 13:46, 8 August 2005 (UTC)

Disagree. :-) - Omegatron 02:58, August 9, 2005 (UTC)

Knew you would:))Light current 03:03, 9 August 2005 (UTC)

The question is, what will the innocent reader make of it? You and I know it is shorthand for 'in parallel with' but really do you think it should be used in pedia? Light current 03:10, 9 August 2005 (UTC)

R_in

how does R1,R2 appear in the expression for R_in? they are shorted in small signal by C_b. 84.95.124.141 11:11, 20 February 2007 (UTC)

Leandrovr: My opinion: True, R1 & R2 shouldn't appear. They're grounded. In example, at audio frequencies (if the capacitor is big enough), the node at the base remains almost at the same potential all the time, because the current sourced or sinked from it is managed mostly by the capacitor, not by the DC source trough R1 or R2 (thevenin equivalent)(in fact, it charges it at circuit startup, until it reaches quiescent mode). I'll try modifing it. (I'm sorry about my english). Regards --Leandrovr 06:32, 19 April 2007 (UTC)

PD: It was true to add R_1 parallel R_2 to r_pi in case that C_b were not present. That's a valid scheme too, with bigger R_in (in fact, r_pi/beta alone is preety small, maybe a few ohms; i.e. 10ohm). It could be added in the future. Regards --Leandrovr 06:53, 19 April 2007 (UTC)

At last, I corrected the expression for r_in and added the previous expression, claryfiying that it's for the case of C_B not present. I think that's better because in some way that doesn't destroy the work of the author or other wikipedian, and adds more information.--24.232.28.122 09:28, 19 April 2007 (UTC)

Suggested changes

Here's one idea I have for a new table and a new diagram:

Figure: Small-signal model for calculating various parameters

	Definition	Expression	Approximate expression	Conditions
Open circuit Voltage gain	${A_{v}}={\begin{matrix}{v_{\mathrm {o} } \over v_{\mathrm {i} }}\end{matrix}}{\Big |}_{R_{L}=\infty }$	${\begin{matrix}{\frac {(g_{m}r_{\mathrm {O} }+1)R_{C}}{R_{C}+r_{O}}}\end{matrix}}$	${\begin{matrix}g_{m}R_{C}\end{matrix}}$	$r_{O}\gg R_{C}$
Short circuit Current gain	$A_{i}={\begin{matrix}{i_{\mathrm {o} } \over i_{\mathrm {i} }}\end{matrix}}{\Big |}_{R_{L}=0}$	${\begin{matrix}{\frac {r_{\pi }+\beta r_{O}}{r_{\pi }+(\beta +1)r_{O}}}\end{matrix}}{\begin{matrix}\end{matrix}}$	$1$	$\beta \gg 1$
Input resistance	$R_{\mathrm {in} }={\begin{matrix}{\frac {v_{i}}{i_{i}}}\end{matrix}}$	${\begin{matrix}{\frac {(r_{O}+R_{C})r_{E}}{r_{O}+r_{E}+{\frac {R_{C}}{\beta +1}}}}\end{matrix}}$	${\frac {1}{g_{m}}}$	$r_{O}\gg R_{C}$
Output resistance	$R_{\mathrm {out} }={\begin{matrix}{\frac {v_{o}}{-i_{o}}}\end{matrix}}{\Big |}_{v_{s}=0}$	$R_{C}\|\{[1+g_{m}(r_{\pi }\|R_{S})]r_{O}+(r_{\pi }\|R_{S})\}$	$R_{C}$	$R_{S}\gg r_{E}\wedge r_{O}\gg (r_{\pi }\|R_{S})$

* Not valid for Rs = 0!

I don't think we need separate sections and tables for the high impedance (current input) and low impedance (voltage input) cases. Given the equations for Rout and Rin, its simple to apply the voltage divider equations and come up with the modified equations for the loaded cases. Maybe we could explain this or put it in another article to avoid repeating it for each (>6) amplifier configuration.

One thing we should agree on is if we should give the current and voltage gains for the loaded or the unloaded cases. The unloaded cases give more compact expressions, but make applying the formulas a bit more tedious as you almost always will have to worry about loading.

Also the formulas could look a little nicer. The equations are too big by default, but using the matrix hack to make them smaller makes them looks uglier and makes manual editing more difficult.

Ideas? -Roger (talk) 04:58, 13 December 2007 (UTC)

To address your points in backward order:

1. The content of the article should never be decided by how good Wiki is at typesetting formulas. Aesthetics are important, but beyond our control.

2. In my mind one critical issue is that the output resistance varies with the source impedance over a very wide range. This seems to require discussion of two cases for the output resistance.

3. The second critical issue is the different behavior of the voltage gain depending upon R_S >> r_E when it is R_L/R_S or << r_E when it is g_m R_L.

I don't think we have the solution yet. Brews ohare (talk) 16:41, 13 December 2007 (UTC)

1. Sorry that was a general comment. I wasn't suggesting we change the equations just to make them look good. But we should agree on a common format to avoid ugly tables like the one above. Either all HTML, default TeX or TeX in a matrix. I prefer default TeX since it matches with the rest of Wikipedia and is easier to edit.

2. We could give the approximations for the low Rs and high Rs cases in the table.

3. I don't think thats a voltage gain issue, but rather an input impedance issue. I think the best definition for voltage gain is vo/vi, following my diagram above where vi in the actual voltage at the amplifier input.

-Roger (talk) 18:47, 13 December 2007 (UTC)

Hi Roger:

OK on items 1 and 2.

As for format, I'm a novice here. Somewhere on Wiki I read that the preferred approach was to use the R_L method in the text and the $R_{L}$ method in equations. I don't recall any suggestions about tables. Maybe you could provide some education here, as just what method of input leads to: HTML; default TeX; or TeX; is greek to me.

As for Item 3, I'd say that the realization that the source in use switches from current division to voltage division depending on its Thevenin resistance is practical knowledge, and should be dealt with in the article in some fashion. I imagine a person not really paying any attention to this resistance and being baffled by what they see happening. What say you? Brews ohare (talk) 21:04, 13 December 2007 (UTC)

By default TeX I just mean TeX such as $R_{L}\,$ which Wikipedia renders rather large. I'm just saying we should be consistent. Either let everything render large (the default) or use matricies so everything has the same small size.

Yes, I think the equations should be practical, but we shouldn't get caught up in special cases. What do you think about switching to the definition of vo/vi for voltage gain (either with or without a load resistance)? If you include Rs it complicates the expression and is redundant because you already know what the input resistance is. One of the points of knowing the input resistance is to be able to calculate the loading effect correction for the voltage gain. -Roger (talk) 22:14, 13 December 2007 (UTC)

Proposal

Hi Roger:

How about setting up your table with the two cases for output R (large and small R_S) using your diagram, and adding a section patterned after the Summary section, which would include a discussion of the way to treat different driver cases and how to do the current or voltage division??

I'll be out of touch for several weeks. Want to take a shot at it?? Brews ohare (talk) 05:12, 14 December 2007 (UTC)

Hi Brews. Yes, I plan on going over as much as I can in the upcoming weeks.

Do you think its best to make the load for the common base (and the other amplifiers) a resistor/impedance or a current source (i.e. active load)? I think the active load makes the analysis less practical, though more ideal. -Roger (talk) 23:05, 15 December 2007 (UTC)

Hi Roger:

I'm away for a few weeks, so my communication is erratic. My opinion is that the active load is preferable because it brings r_O into the equations. Particularly for MOSFET, where r_O is lousy, it cannot be neglected in practice. The cost in complexity is not very great: often it just means R_L is replaced by R_L//r_O. More than that; in IC's active loads are the rule, so the analysis with r_O is in keeping with what is usually done. Brews ohare (talk) 14:53, 17 December 2007 (UTC)

Yes, I realize the active load is closer to the ideal, but I was thinking about a point you made earlier, that the loaded voltage gain is much different from the unloaded voltage gain. If we decide on the active load case then the final table will end up looking like the one above. -Roger (talk) 15:43, 17 December 2007 (UTC)

I am not unhappy about that: after all, there is a column providing the simplified approximate form for those readers that become distressed over too big a formula. BTW, in my lexicon "ideal" means "with all unessential detail stripped away" and "practical" means "with details needed to be useful included". On that basis, I'd say using an active load is more "practical"?? Brews ohare (talk) 19:00, 17 December 2007 (UTC)

Yeah, but the exact expression for the voltage gain of the actively loaded CB is 1+gm*ro, so the simplication doesn't help. I consider this expression ideal because it gives the intrinsic gain of the amplifier, with the bare minimum of additional components. It isn't practical because the instant you connect a (finite) load you need to start thinking about the output impedance and the loading effect and you end up with the much different expression gm*RL.

Incidentally this does expose the useful formula 1+VA/Vt for the maximum gain of the CB. -Roger (talk) 19:12, 17 December 2007 (UTC)

Problems with page

Hi Roger:

Figure 2 is not adequate to describe the use as a current buffer, and this discussion refers to a Norton source, not shown in Figure 2. Some fixing is needed. Brews ohare (talk) 16:33, 6 January 2008 (UTC) I made some changes that I hope you will agree with. Brews ohare (talk) 16:55, 6 January 2008 (UTC)

Hi Brews

Yes, there are still issues with the article that I have to go over. I've been trying to fix the table but it still comes out too large. I think figure 2 should be okay enough since I labeled the input and output currents. Though for input loading I guess another figure would be needed, but I think loading effects are better covered in another article so we don't repeat them ~6 times for every amplifier. -Roger (talk) 16:58, 6 January 2008 (UTC)

Hi Roger: You are not in agreement with the page as amended I take it. Maybe we can discuss this in deatial? I'm on the road today, so maybe tomorrow?? Brews ohare (talk) 17:07, 6 January 2008 (UTC) Roger: I undid your changes because a resistor R_L attached at the output of Figure 1 leads to a load R_C//R_L and makes the table entries all incorrect. It seems easier to just use R_C as load, which is mathematically OK and works with the table and circuit diagram.. Brews ohare (talk) 17:11, 6 January 2008 (UTC)

There's meant to be a difference between R_C and R_L. R_C is the unloaded collector resistance (it could be the r_o of a current mirror or a choke if we were talking about AC, for example) while R_L is the load resistor.

If I just combined RC and RL then it would complicate the definitions of open/short circuit voltage/current gains.

I didn't think about specifying the condition R_L->infinity in the table for the unloaded voltage gain when I drew the circuit, but since I did I guess I can explicitly add R_L to figure 2. I'll work on that, but in the mean time remember there's supposed to be a difference :) -Roger (talk) 17:21, 6 January 2008 (UTC)

OK, Roger. I'll take a look tomorrow and see what wonders have been wrought. 4.240.72.212 (talk) 17:34, 6 January 2008 (UTC)

Contradiction?

There appears to be a contradiction in this article.

firstly: Applications[edit]This arrangement is not very common in low-frequency circuits, where it is usually employed for amplifiers that require an unusually low input impedance, for example to act as a preamplifier for moving-coil microphones. However, it is popular in high-frequency amplifiers, for example for VHF and UHF, because its input capacitance does not suffer from the Miller effect, which degrades the bandwidth of the common-emitter configuration, and because of the relatively high isolation between the input and output. This high isolation means that there is little feedback from the output back to the input, leading to high stability.

This configuration is also useful as a current buffer since it has a current gain of approximately unity (see formulas below). Often a common base is used in this manner, preceded by a common-emitter stage. The combination of these two form the ***cascade*** configuration, which possesses several of the benefits of each configuration, such as high input impedance and isolation.

"a common base is used in this manner, preceded by a common-emitter stage. The combination of these two form the ***cascade*** configuration"

then later: Overview of characteristics[edit]Several example applications are described in detail below. A brief overview follows.

The amplifier input impedance Rin looking into the emitter node is very low, given approximately by , where VT is the thermal voltage and IE is the DC emitter current. For example, for VT = 26 mV and IE = 10 mA, rather typical values, Rin = 2.6 Ω. If IE is reduced to increase Rin, there are other consequences like lower transconductance, higher output resistance and lower β that also must be considered. A practical solution to this low-input-impedance problem is to place a common-emitter stage at the input to form a ***cascode*** amplifier.

"A practical solution to this low-input-impedance problem is to place a common-emitter stage at the input to form a ***cascode*** amplifier"

I looked up cascade and cascode in Wikipedia:

cascade = common collector -> common base cascode = common emitter -> common base

So there is a problem in the first cited definition of cascade in this article.

Im no expert, so its unclear to me what is needed. I have made no changes to the article.

143.92.1.33 (talk) 00:38, 23 July 2013 (UTC)

I took care of it by reverting the latest edit. Dicklyon (talk) 03:43, 23 July 2013 (UTC)

Simplification is required

I added a simple explanation: so college level readers have a starting point — Preceding unsigned comment added by Grahamatwp (talk • contribs) 12:22, 24 May 2014 (UTC)

I doubt that would be any real help to a college student. Your claim that "As current is sunk from the emitter this provides potential difference so causing the transistor to conduct" was initially completely unreferenced. You have now provided a cite to http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/npncb.html#c1 but that page does not make any statement even remotely like that. SpinningSpark 12:33, 24 May 2014 (UTC)

If you have issue with the wording used, then I suggest you improve it. Reverting changes without discussion will not improve the article. I was going for a Faraday description. — Preceding unsigned comment added by Grahamatwp (talk • contribs) 12:58, 24 May 2014 (UTC)

It cannot be improved. It needs discarding and starting again. SpinningSpark 16:47, 24 May 2014 (UTC)

You are being unhelpful. If you can see errors in the text you need to bring these into the discussion. If you feel it's completely wrong, then propose an alternative. — Preceding unsigned comment added by Grahamatwp (talk • contribs) 17:07, 24 May 2014 (UTC)

Added a link to a page which shows the operation better. — Preceding unsigned comment added by Grahamatwp (talk • contribs) 17:39, 24 May 2014 (UTC)

Explain the term large ?

The article is bound in mathematics, and is unclear I hesitate to say to the majority of readers. I would like to see the term large explained for instance, as large can be large in resistance , or large in current which are opposite of each other. just saying something is large without informing if it is large resistance, or large current explains very little. Could I suggest a schematic be added showing stated voltage, stated value of RC in ohms and approximate voltage gain. The video here is I think far more informative and lets readers know how the circuit works; Underwaterlunch (talk) 08:59, 11 March 2015 (UTC)

References