On 8/13/19 7:42 PM, George Herold wrote:> On Tuesday, August 13, 2019 at 2:42:54 PM UTC-4, John Larkin wrote: >> On Mon, 12 Aug 2019 10:45:47 -0700 (PDT), George Herold >> <gherold@teachspin.com> wrote: >> >>> On Monday, August 12, 2019 at 1:33:01 PM UTC-4, John Larkin wrote: >>>> On Mon, 12 Aug 2019 09:37:28 -0700 (PDT), George Herold >>>> <gherold@teachspin.com> wrote: >>>> >>>>> On Sunday, August 11, 2019 at 5:53:07 AM UTC-4, amal banerjee wrote: >>>>>> On Saturday, August 10, 2019 at 8:16:01 PM UTC+5:30, jurb...@gmail.com wrote: >>>>>>> People did this years ago without PCs or even calculators in some cases. I just happen to have a little time before people start bugging me. >>>>>>> >>>>>>> Transistors have a gain curve. Generally people want linear amplification so bias it near the apex of the gain curve. >>>>>>> >>>>>>> You need an emitter resistor, almost everybody uses them. As long as the transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, consider it a ratio. Wherever you bias it, after looking at the spec sheet, the emitter is going to have a certain voltage at the operating point YOU SET. >>>>>>> >>>>>>> The collector resistor should usually be dropping about half the Vcc. The emitter resistor much less, being a lower value of course. Now comes the bias. You know where you want it so you can figure what the emitter voltage should be. to the base you make a resistive divider providing about 0.6 volts above your intended emitter voltage. >>>>>>> >>>>>>> If you got a transistor with a gain of 100 it doesn't much matter the base current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe. >>>>>>> >>>>>>> Anyway, you choose the collector resistor to give you the output impedance (or lower ) you want. then you go to the transistors and figure out one that is at the apex of its hfe curve at the current YOU WANT. >>>>>>> >>>>>>> An emitter resistor around a tenth of the collector resistor will almost always be fine. And it will make it thermally stable IF the base is biased from a voltage divider. If you just take a like 560K from the collector to the base it will bias it, but it will not be stable thermally. >>>>>>> >>>>>>> That is as simplistic as it can really get if you really want to build anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead of like 470K and 47K you might want 47K and 4.7K. Lower input impedance but more bandwidth, and the same voltage gain pretty much. >>>>>>> >>>>>>> And then there's bootstrap. Took me some time to figure that out. I had the concept when I saw it but the math was very elusive. But using bootstrap you can make a transistor twice s good as it really is. These days the transistors are so good there isn't much need for it, but a few applications do still exist. >>>>>>> >>>>>>> You want to design kid ? (I can call you that because I am old) Put the mouse down and step away from the PC. Get some graph paper and a bunch of mechanical pencils. I recommend the kind that screws the lead out rather than the kind where you push the edge. Or you can just buy some pencils and a pencil sharpener. Get one of those old programming flow charts, that's what I use. It has lines all over it and is easy to align to the graph paper. Then your design at least looks neat and that helps to understand it. >>>>>>> >>>>>>> So you can know the characteristics of the device, a transistor. A wonderful thing invented before we were born. >>>>>>> >>>>>>> And you can have all the info, if Digikey doesn't have a spec sheet that's any good someone does. Most manufacturers make them available, after all they are trying to sell them. >>>>>>> >>>>>>> If you need a certain lowness of output impedance and you want to bias the transistor at the apex of hfe and find that volts times amps add up to 1,450 mW and the rating of the transistor is 700mW, you have to find a different transistor. >>>>>>> >>>>>>> That's life. >>>>>> >>>>>> I use the standard design rules and equations. The >>>>>> starting point is of course the transistor datasheet - specifically the Ic vs. Vce curves, and the listed >>>>>> absolute maximum values for Ic etc. The target Ic >>>>>> value is checked to ensure that it would be less >>>>>> than the absolute maximum value for Ic. >>>>>> >>>>>> Vc is selected at ~ 0.5Vcc, and the Ve ~ 0.1Vcc. >>>>>> Then the value of Vcc is easily calculated from >>>>>> Vcc = 0.6Vcc + Vce >>>>>> The values for Rc, and Re are calculated easily >>>>>> from target Ic value and Vcc. >>>>> >>>>> Hmm OK.. Amal, I think you are using the 'current gain' model of a transistor. >>>>> This is often just fine. >>>> >>>> But not to compute voltage gain. VG is, first order, independent of >>>> beta. >>> OK. (I am but a tadpole when it comes to design/ understanding of transistors.) >>> But don't you need the 'intrinsic' emitter resistance (~25 mV/Ic) >>> to get the gain? >>> (assuming no/small resistance from emitter to ground.) >> >> Exactly. But that's independent of beta. >> >> The transconductance Gm is 1/(.025/Ic) = 40 * Ic, and the voltage gain >> is Gm * Rload. That magically collapses to Vgain = 40 times the DC >> voltage drop in the collector resistor. >> >> Transistors are actually pretty simple. > > Well we are in violent agreement, my only point was that > you need to Ebers-Moll or some other model to figure > out what the intrinsic resistance is. > (and no one wants to run a transistor gain so high that > the intrinsic resistance matters.)Generally not in a vanilla CE amplifier, right--the emitter degeneration provides local feedback, which linearizes the stage. But you can also apply local resistive feedback from the collector to the base, as in your bog-standard bipolar MMIC. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics Briarcliff Manor NY 10510 http://electrooptical.net http://hobbs-eo.com
Amplification factor for common emitter amplifier
Started by ●August 8, 2019
Reply by ●August 13, 20192019-08-13
Reply by ●August 13, 20192019-08-13
On Tue, 13 Aug 2019 15:26:09 -0700, John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:>On Tue, 13 Aug 2019 20:14:05 GMT, Steve Wilson <no@spam.com> wrote: > >>John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote: >> >>> The transconductance Gm is 1/(.025/Ic) = 40 * Ic, and the voltage gain >>> is Gm * Rload. That magically collapses to Vgain = 40 times the DC >>> voltage drop in the collector resistor. >> >>> Transistors are actually pretty simple. >> >>Here are two circuits with identical operating points. Your formula sets >>the gain at 40 * 5 = 40. > >You might ask someone to check that multiply for you. > >> >>Circuit A has a gain of 0.535. Circuit B has a gain of 176.7. >> >>Your equation is wrong. > >George specified > > (assuming no/small resistance from emitter to ground.) > >Why did you snip that? > >176 is close to 200. A little Early effect will drop the gain some >from an ideal transistor. The 0.025 factor isn't exact either. And the >estimate is of course small-signal.And there are some real, ohmic resistances too, wire bonds and contact/spreading resistances in the silicon of a real transistor. All that reduces the gain a bit from the theory.
Reply by ●August 14, 20192019-08-14
John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:> On Tue, 13 Aug 2019 20:14:05 GMT, Steve Wilson <no@spam.com> wrote: > >>John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote: >> >>> The transconductance Gm is 1/(.025/Ic) = 40 * Ic, and the voltage gain >>> is Gm * Rload. That magically collapses to Vgain = 40 times the DC >>> voltage drop in the collector resistor. >> >>> Transistors are actually pretty simple. >> >>Here are two circuits with identical operating points. Your formula sets >>the gain at 40 * 5 = 40. > > You might ask someone to check that multiply for you.Already posted the fix. Also included Legg's formula Vg = Zc/Ze, which you ignored. Here's my post: Steve Wilson <no@spam.com> wrote:> Here are two circuits with identical operating points. Your formula sets > the gain at 40 * 5 = 40.40 * 5 = 200 Legg is right. Vg = Zc / Ze>>Circuit A has a gain of 0.535. Circuit B has a gain of 176.7.>>Your equation is wrong.> George specified> (assuming no/small resistance from emitter to ground.)> Why did you snip that?You have to include the impedance in the emitter.> 176 is close to 200. A little Early effect will drop the gain some > from an ideal transistor. The 0.025 factor isn't exact either. And the > estimate is of course small-signal.Your formula is useless and misleading. You normally never run a transistor with the emitter grounded. You need some way to stabilize the bias. Also, there is usually some load on the collector, which you ignore. You could also have transformer coupling or a resonant circuit in the collector, which would produce a low dc resistance and consequently low voltage drop. This would produce low gain in your formula. Legg's formula includes the impedance on both the collector and emitter, and includes loading due to the following stage and bypassing on the emitter. Legg's formula can also consider the DC case by omitting the load impedance and emitter bypass. So you can get both the DC gain and high frequency gain. Your equation is only for DC gain, and ignores the resistance in the emitter. It does not include the load impedance on the collector or degeneration in the emitter. It is useless for real-world circuits.
Reply by ●August 14, 20192019-08-14
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:> Your formula is useless and misleading. You normally never run a transistor > with the emitter grounded.odd thing to say> You need some way to stabilize the bias. Also,There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits. NT
Reply by ●August 14, 20192019-08-14
On Wed, 14 Aug 2019 10:09:49 GMT, Steve Wilson <no@spam.com> wrote:>John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote: > >> On Tue, 13 Aug 2019 20:14:05 GMT, Steve Wilson <no@spam.com> wrote: >> >>>John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote: >>> >>>> The transconductance Gm is 1/(.025/Ic) = 40 * Ic, and the voltage gain >>>> is Gm * Rload. That magically collapses to Vgain = 40 times the DC >>>> voltage drop in the collector resistor. >>> >>>> Transistors are actually pretty simple. >>> >>>Here are two circuits with identical operating points. Your formula sets >>>the gain at 40 * 5 = 40. >> >> You might ask someone to check that multiply for you. > >Already posted the fix. Also included Legg's formula Vg = Zc/Ze, which you >ignored. Here's my post: > >Steve Wilson <no@spam.com> wrote: > >> Here are two circuits with identical operating points. Your formula sets >> the gain at 40 * 5 = 40. > >40 * 5 = 200 > >Legg is right. > >Vg = Zc / ZeThat's infinite with no emitter resistor, right?> >>>Circuit A has a gain of 0.535. Circuit B has a gain of 176.7. > >>>Your equation is wrong. > >> George specified > >> (assuming no/small resistance from emitter to ground.) > >> Why did you snip that? > >You have to include the impedance in the emitter.That's trivial. If you assume that the active emitter resistance is only the resistor that you add external to the transistor, that's wrong. The emitter diode has a dynamic resistance that matters, and limits the gain to below infinite.> >> 176 is close to 200. A little Early effect will drop the gain some >> from an ideal transistor. The 0.025 factor isn't exact either. And the >> estimate is of course small-signal. > >Your formula is useless and misleading.Merely correct. I can't help what people want to believe. You normally never run a transistor>with the emitter grounded.Why not? It's done all the time.>You need some way to stabilize the bias.Lots of ways to do that. For example, bypass the emitter as you did in your sim. Also,>there is usually some load on the collector, which you ignore.Also trivial. Any real signal source drops when you load it. Do that after you calculate the unloaded gain. You could>also have transformer coupling or a resonant circuit in the collector, >which would produce a low dc resistance and consequently low voltage drop. >This would produce low gain in your formula.There a million ways a smart person can wreck the gain of an amplifier. Actually, a dumb person can do that too.> >Legg's formula includes the impedance on both the collector and emitter, >and includes loading due to the following stage and bypassing on the >emitter. > >Legg's formula can also consider the DC case by omitting the load impedance >and emitter bypass. So you can get both the DC gain and high frequency >gain. > >Your equation is only for DC gain, and ignores the resistance in the >emitter.Dead wrong. It is an estimate of the small-signal (that means AC) gain. The "resistance in the emitter" is the dynamic impedance of the b-e junction. That's fundamental to understanding transistors. It does not include the load impedance on the collector or>degeneration in the emitter. It is useless for real-world circuits. >It's an algebraic result of the actual transistor behavior in one simple stated case. It's merely right.
Reply by ●August 14, 20192019-08-14
On Wed, 14 Aug 2019 04:46:28 -0700 (PDT), tabbypurr@gmail.com wrote:>On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote: > >> Your formula is useless and misleading. You normally never run a transistor >> with the emitter grounded. > >odd thing to say > >> You need some way to stabilize the bias. Also, > >There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits. > > >NTThe same simple concepts can be appied to a differential amplifier biased by a current sink. Suicide bias has poor control over Ic hence poor control over the voltage drop in the collector resistor, hence less predictable voltage gain. For max gain and lowest noise in a single transistor amp, ground the emitter.
Reply by ●August 14, 20192019-08-14
On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:> On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote: > > > Your formula is useless and misleading. You normally never run a transistor > > with the emitter grounded. > > odd thing to say > > > You need some way to stabilize the bias. Also, > > There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits. > > > NTWhat's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^) GH
Reply by ●August 14, 20192019-08-14
On Wed, 14 Aug 2019 08:24:28 -0700 (PDT), George Herold <gherold@teachspin.com> wrote:>On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote: >> On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote: >> >> > Your formula is useless and misleading. You normally never run a transistor >> > with the emitter grounded. >> >> odd thing to say >> >> > You need some way to stabilize the bias. Also, >> >> There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits. >> >> >> NT > >What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^) > >GHA resistor from Vcc to the base, with the emitter grounded. That's basically a constant base bias current, so Ic is proportional to beta. Early transistor circuits tended to do that. As do some cheap products still. I've done it in some special cases. One can now buy beta-graded transistors, like BCX70J, that make the idea less silly. This is cool: https://www.dropbox.com/s/tqw2qe9mi4t8zmz/Suicide_Slicer.JPG?raw=1 I'm doing that as a CML-to-TTL converter.
Reply by ●August 14, 20192019-08-14
On Wednesday, 14 August 2019 15:25:53 UTC+1, jla...@highlandsniptechnology.com wrote:> On Wed, 14 Aug 2019 04:46:28 -0700 (PDT), tabbypurr wrote: > >On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:> >> Your formula is useless and misleading. You normally never run a transistor > >> with the emitter grounded. > > > >odd thing to say > > > >> You need some way to stabilize the bias. Also, > > > >There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits. > > > > > >NT > > The same simple concepts can be appied to a differential amplifier > biased by a current sink. > > Suicide bias has poor control over Ic hence poor control over the > voltage drop in the collector resistor, hence less predictable voltage > gain.V_Rc_quiesc is determined by Ib & beta. Beta varies, but within limits. Sometimes it's good enough. Often not. Suicide never happens if Ic is low, thus tr heating is low.> For max gain and lowest noise in a single transistor amp, ground the > emitter.and max output swing. NT
Reply by ●August 14, 20192019-08-14
On 14/08/2019 4:24 pm, George Herold wrote:> On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote: >> On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote: >> >>> Your formula is useless and misleading. You normally never run a transistor >>> with the emitter grounded. >> >> odd thing to say >> >>> You need some way to stabilize the bias. Also, >> >> There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits. >> >> >> NT > > What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^) > > GH >If want it to be linear but emitter well and truly grounded one can always de-suicide the bias by using feedback resistor to base from collector. For max AC gain decouple it midway with a capacitor to ground. piglet
