On 08/19/2013 12:26 PM, John Larkin wrote:> On Mon, 19 Aug 2013 12:07:44 -0400, Phil Hobbs > <pcdhSpamMeSenseless@electrooptical.net> wrote: > >> On 08/19/2013 11:38 AM, Jan Panteltje wrote: >>> On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs >>> <pcdhSpamMeSenseless@electrooptical.net> wrote in >>> <kuta6k$3gg$1@dont-email.me>: >>> >>>> The base current is really a loss mechanism--an ideal BJT has zero base >>>> current, because all of the emitter current makes it to the collector >>>> and none recombines in the base region. (Recombination is where the base >>>> current comes from.) >>> >>> Well, eh, I disagree, seems you want infinite beta, I do not want a BJT with infinite beta. >>> Changes in beta, and its dependence on Ic for example, can be used for gain control, >>> many other circuits with BJTs use beta in some way or the other, >>> say for example current limiting comes to mind. >>> >>> A BJT is NOT a voltage amplifier, it is a CURRENT amplifier. >>> You want a MOSFET if you want a voltage amplifier. >>> >> >> I didn't say what I wanted. The base current comes from recombination, >> which in a transistor is a loss mechanism. That's why beta is not a >> good design parameter in most cases. >>> >>> >>> >>>> The fundamental control mechanism of a BJT is the base-emitter voltage, >>>> which provides pretty tight voltage feedback in an emitter follower. >>> >>> Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload). >>> Asymmetric circuit really. >> >> In a voltage regulator, one hopes that the base voltage only goes down >> much when power is removed. And the input impedance when the voltage is >> going down is much larger than Zload until the base-emitter junction zeners. >> >> Cheers >> >> Phil Hobbs > > I "design" around beta a lot more often than around small-signal input impedance > or transconductance. But then, I mostly use bipolars for switching and emitter > followers and simple things, not the sort of thing that you do with laser noise > cancellers and such. I think most people use opamps for precise things nowadays, > and transistors to do dumb stuff. > > But I did design the analog multiplier safe-operating-area computer recently, > and it works! That involved most of the low-frequency transistor effects. > > https://dl.dropboxusercontent.com/u/53724080/Circuits/Power/SOAR_4.asc > > >I agree that most discrete circuits are simple things, and that one normally thinks about beta more than about delta-Vbe. I still say that the transconductance mechanism is more fundamental, because you can explain the current amplification by the transconductance (with recombination), but you can't explain the transconductance by the current amplification. The reason that beta varies all over the place is that it's the reciprocal of a small loss term that depends on a lot of fine processing details. The OP's confusion seemed to be that he didn't understand the voltage feedback mechanism in an emitter follower, though, and seemed to think that current amplification alone was an adequate basis for understand a BJT. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Zener regulator : load and zener current
Started by ●August 18, 2013
Reply by ●August 19, 20132013-08-19
Reply by ●August 19, 20132013-08-19
On Sun, 18 Aug 2013 05:12:11 -0700 (PDT), sridhar09.cherukuri@gmail.com wrote:>In the simple zener regulators like the one below, If the load current is 100ma and zener current is 10ma we calculate the resistor to allow 110ma. > >If Vin is above Vz and the load draws 50ma, the remaining 60ma flows through zener. What makes it that the remaining current after the load current go through the zener? Whay can't it be the otherway? i.e after the zener current the remaining to the load. Is it due to the load impedance being less than the zener impedance? > >Vin------/\/\/\/\--------|-----------| > | | > | | > | | > | / RL > /--- | > /\ | > / \ | > ------ ----- > | --- > | - > | > ----- > --- > ---- http://www.onsemi.com/pub_link/Collateral/HBD854-D.PDF -- JF
Reply by ●August 19, 20132013-08-19
On a sunny day (Mon, 19 Aug 2013 12:07:44 -0400) it happened Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> wrote in <5uKdnXf4NsxN34_PnZ2dnUVZ_tOdnZ2d@supernews.com>:>On 08/19/2013 11:38 AM, Jan Panteltje wrote: >> On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs >> <pcdhSpamMeSenseless@electrooptical.net> wrote in >> <kuta6k$3gg$1@dont-email.me>: >> >>> The base current is really a loss mechanism--an ideal BJT has zero base >>> current, because all of the emitter current makes it to the collector >>> and none recombines in the base region. (Recombination is where the base >>> current comes from.) >> >> Well, eh, I disagree, seems you want infinite beta, I do not want a BJT with infinite beta. >> Changes in beta, and its dependence on Ic for example, can be used for gain control, >> many other circuits with BJTs use beta in some way or the other, >> say for example current limiting comes to mind. >> >> A BJT is NOT a voltage amplifier, it is a CURRENT amplifier. >> You want a MOSFET if you want a voltage amplifier. >> > >I didn't say what I wanted. The base current comes from recombination, >which in a transistor is a loss mechanism. That's why beta is not a >good design parameter in most cases.After I posted that (sure eat your popocorn Spehro) I thought it is a bit of chicken and egg, or religious thing too. What was first, the voltage or the current? We are talking electons.... little balls.... that move (never mind the holes and golfing). Beta is a variable, depending on many things, but it is an essential design parameter. The Vbe is _also_ depending on many things. But if you want any kind of linearity, you need to drive the transistor (look at the transistor) as current amplifier. Driving a sine voltage into the base emitter junction will not produce a sine wave form in the collector current. Modulation Ib with a sine _will_ much more result in sine in Ic.> >> >> >>> The fundamental control mechanism of a BJT is the base-emitter voltage, >>> which provides pretty tight voltage feedback in an emitter follower. >> >> Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload). >> Asymmetric circuit really. > >In a voltage regulator, one hopes that the base voltage only goes down >much when power is removed. And the input impedance when the voltage is >going down is much larger than Zload until the base-emitter junction zeners.Yes, yes. A classical one is the emitter follower cable driver... This always bothered me: + | c video -------- b NPN +1 to +2V e | [ ] 75 75 Ohm coax 10 meter |----------==============================================-----------||--- amp | | | | [ ] 1k /// /// [ ] 75 | | /// ///
Reply by ●August 19, 20132013-08-19
On 08/19/2013 01:56 PM, Jan Panteltje wrote:> On a sunny day (Mon, 19 Aug 2013 12:07:44 -0400) it happened Phil Hobbs > <pcdhSpamMeSenseless@electrooptical.net> wrote in > <5uKdnXf4NsxN34_PnZ2dnUVZ_tOdnZ2d@supernews.com>: > >> On 08/19/2013 11:38 AM, Jan Panteltje wrote: >>> On a sunny day (Mon, 19 Aug 2013 10:30:43 -0400) it happened Phil Hobbs >>> <pcdhSpamMeSenseless@electrooptical.net> wrote in >>> <kuta6k$3gg$1@dont-email.me>: >>> >>>> The base current is really a loss mechanism--an ideal BJT has zero base >>>> current, because all of the emitter current makes it to the collector >>>> and none recombines in the base region. (Recombination is where the base >>>> current comes from.) >>> >>> Well, eh, I disagree, seems you want infinite beta, I do not want a BJT with infinite beta. >>> Changes in beta, and its dependence on Ic for example, can be used for gain control, >>> many other circuits with BJTs use beta in some way or the other, >>> say for example current limiting comes to mind. >>> >>> A BJT is NOT a voltage amplifier, it is a CURRENT amplifier. >>> You want a MOSFET if you want a voltage amplifier. >>> >> >> I didn't say what I wanted. The base current comes from recombination, >> which in a transistor is a loss mechanism. That's why beta is not a >> good design parameter in most cases. > > After I posted that (sure eat your popocorn Spehro) I thought it is a bit > of chicken and egg, or religious thing too. > What was first, the voltage or the current? > We are talking electons.... little balls.... that move (never mind the holes and golfing). > > Beta is a variable, depending on many things, but it is an essential design parameter. > The Vbe is _also_ depending on many things.I don't agree at all. Apart from I_S, which is a more-or-less constant device parameter, V_BE depends only on fundamental constants and the junction temperature, and at lowish currents, it follows the Ebers-Moll curve to absurd accuracy. Beta depends on everything including what the silicon crystal-puller had for lunch. You can explain the current gain by the transconductance mechanism (electrons being emitted over top of the potential barrier between base and emitter, a few recombining in the base), but you can't explain the transconductance from the current gain.> But if you want any kind of linearity, you need to drive the transistor (look at the transistor) > as current amplifier.That's one approach, but there are lots of others, e.g. current mirrors, negative feedback, and translinear things like using junctions for collector loads.> Driving a sine voltage into the base emitter junction will not produce a sine wave form in the collector current. > Modulation Ib with a sine _will_ much more result in sine in Ic.For the same output swing, transconductance works fine. If you want better performance, put in an emitter resistor. An emitter-degenerated transconductance amp is far more predictable than a beta-dependent circuit, for one thing, and for another, the beta linearity of a BJT isn't necessarily anything nice. To do a good job, you're going to need feedback components either way.>>>> The fundamental control mechanism of a BJT is the base-emitter voltage, >>>> which provides pretty tight voltage feedback in an emitter follower. >>> >>> Zi is beta * Zload in an emitter follower (and that only when going 'up', else it is Zload). >>> Asymmetric circuit really. >> >> In a voltage regulator, one hopes that the base voltage only goes down >> much when power is removed. And the input impedance when the voltage is >> going down is much larger than Zload until the base-emitter junction zeners. > > Yes, yes. > A classical one is the emitter follower cable driver... > > This always bothered me: > + > | > c > video -------- b NPN > +1 to +2V e > | > [ ] 75 75 Ohm coax 10 meter > |----------==============================================-----------||--- amp > | | | | > [ ] 1k /// /// [ ] 75 > | | > /// /// > >You don't even need the resistors on the BJT side--it'll drive the coax fine with all of the DC load at the far end. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Reply by ●August 19, 20132013-08-19
On a sunny day (Mon, 19 Aug 2013 14:40:52 -0400) it happened Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> wrote in <FcGdnWTRmPIp-4_PnZ2dnUVZ_hCdnZ2d@supernews.com>:>> What was first, the voltage or the current? >> We are talking electons.... little balls.... that move (never mind the holes and golfing). >> >> Beta is a variable, depending on many things, but it is an essential design parameter. >> The Vbe is _also_ depending on many things. > >I don't agree at all. Apart from I_S, which is a more-or-less constant >device parameter, V_BE depends only on fundamental constants and the >junction temperature, and at lowish currents, it follows the Ebers-Moll >curve to absurd accuracy. Beta depends on everything including what the >silicon crystal-puller had for lunch.I though: "Now he will come up with bandgap reference..." What you say is correct, I look at it maybe from an aplication POV.>You can explain the current gain by the transconductance mechanism >(electrons being emitted over top of the potential barrier between base >and emitter, a few recombining in the base), but you can't explain the >transconductance from the current gain.And you more from the scientific POV.>> But if you want any kind of linearity, you need to drive the transistor (look at the transistor) >> as current amplifier. > >That's one approach, but there are lots of others, e.g. current mirrors, >negative feedback, and translinear things like using junctions for >collector loads.Yes, but junctions for collector load is also driving current, Zc is [almost] infinite impedance.>> Driving a sine voltage into the base emitter junction will not produce a sine wave form in the collector current. >> Modulation Ib with a sine _will_ much more result in sine in Ic. > >For the same output swing, transconductance works fine. If you want >better performance, put in an emitter resistor. An emitter-degenerated >transconductance amp is far more predictable than a beta-dependent >circuit, for one thing,I am not convinced about that, first putting in an emitter series resistor to get better linearity is IMNSHO not a very nice solution, but it will work of course. In the same way you can put a collector-base resistor as feedback (I do that a lot to get the DC working point, correct Vc).>and for another, the beta linearity of a BJT >isn't necessarily anything nice. To do a good job, you're going to need >feedback components either way.That depends, take an RF amp, AGC done by changing Ic, signals are small, no feedback. I will tell you a story, told it before here years ago, I was designing (1968 or so) a pre-amp for a vidicon camera. This is interesting, as you have the giga Ohm fetish, so maybe it will make you think. The usual way they do this (industrial espionage) is make a very high impedance amplifier, and then use feedback. A vidicon target electrode is a very high impedance, really (so voltage drive). I tried that, but had endles problems with interference pickup from the scan coils.. Experimenting I accidently shorted the vidicon target to the base of the second amplifier stage, a normal BJT, and saw a perfect picture for a moment. Curious I just removed the high impedance input stage and used the target electrode as currect feed into the base of that transistor. Absolute gain is not that important (light level changes constantly), but the low impedance presented by the base shorted, got rid of, all the interference, current drive! No feedback! Amazing picture!, Shorted the target and wiring capacitance out too, good resolution. I left it that way... For those who have a GOhm fetish...>> + >> | >> c >> video -------- b NPN >> +1 to +2V e >> | >> [ ] 75 75 Ohm coax 10 meter >> |----------==============================================-----------||--- amp >> | | | | >> [ ] 1k /// /// [ ] 75 >> | | >> /// /// >> >> > >You don't even need the resistors on the BJT side--it'll drive the coax >fine with all of the DC load at the far end.The 75 Ohm on the left,.. somehow you need to match the cable impedance.
Reply by ●August 19, 20132013-08-19
On 08/19/2013 03:21 PM, Jan Panteltje wrote:> On a sunny day (Mon, 19 Aug 2013 14:40:52 -0400) it happened Phil > Hobbs <pcdhSpamMeSenseless@electrooptical.net> wrote in > <FcGdnWTRmPIp-4_PnZ2dnUVZ_hCdnZ2d@supernews.com>: > >>> What was first, the voltage or the current? We are talking >>> electons.... little balls.... that move (never mind the holes and >>> golfing). >>> >>> Beta is a variable, depending on many things, but it is an >>> essential design parameter. The Vbe is _also_ depending on many >>> things. >> >> I don't agree at all. Apart from I_S, which is a more-or-less >> constant device parameter, V_BE depends only on fundamental >> constants and the junction temperature, and at lowish currents, it >> follows the Ebers-Moll curve to absurd accuracy. Beta depends on >> everything including what the silicon crystal-puller had for >> lunch. > > I though: "Now he will come up with bandgap reference..." > > What you say is correct, I look at it maybe from an aplication POV. > > >> You can explain the current gain by the transconductance mechanism >> (electrons being emitted over top of the potential barrier between >> base and emitter, a few recombining in the base), but you can't >> explain the transconductance from the current gain. > > And you more from the scientific POV. > > > >>> But if you want any kind of linearity, you need to drive the >>> transistor (look at the transistor) as current amplifier. >> >> That's one approach, but there are lots of others, e.g. current >> mirrors, negative feedback, and translinear things like using >> junctions for collector loads. > > Yes, but junctions for collector load is also driving current, Zc is > [almost] infinite impedance. > > >>> Driving a sine voltage into the base emitter junction will not >>> produce a sine wave form in the collector current. Modulation Ib >>> with a sine _will_ much more result in sine in Ic. >> >> For the same output swing, transconductance works fine. If you >> want better performance, put in an emitter resistor. An >> emitter-degenerated transconductance amp is far more predictable >> than a beta-dependent circuit, for one thing, > > I am not convinced about that, first putting in an emitter series > resistor to get better linearity is IMNSHO not a very nice solution, > but it will work of course. > > In the same way you can put a collector-base resistor as feedback (I > do that a lot to get the DC working point, correct Vc).>> and for another, the beta linearity of a BJT isn't necessarily >> anything nice. To do a good job, you're going to need feedback >> components either way. > > That depends, take an RF amp, AGC done by changing Ic, signals are > small, no feedback.But that's a transimpedance effect, not a current-gain effect. If beta is constant, changing I_C doesn't change the gain of a current-mode amplifier. A base resistor functions a lot like emitter degeneration, except it isn't as predictable because beta varies orders of magnitude more than g_M for a given collector current.> I will tell you a story, told it before here years ago, I was > designing (1968 or so) a pre-amp for a vidicon camera. This is > interesting, as you have the giga Ohm fetish, so maybe it will make > you think. > > The usual way they do this (industrial espionage) is make a very high > impedance amplifier, and then use feedback. A vidicon target > electrode is a very high impedance, really (so voltage drive). > > I tried that, but had endles problems with interference pickup from > the scan coils.. Experimenting I accidently shorted the vidicon > target to the base of the second amplifier stage, a normal BJT, and > saw a perfect picture for a moment. Curious I just removed the high > impedance input stage and used the target electrode as currect feed > into the base of that transistor. Absolute gain is not that important > (light level changes constantly), but the low impedance presented by > the base shorted, got rid of, all the interference, current drive! No > feedback! Amazing picture!, Shorted the target and wiring capacitance > out too, good resolution. I left it that way...Good news.> > For those who have a GOhm fetish...I don't like using huge resistors like that, for all the reasons you give, plus the fact that they make everything slow. Even using all sorts of bootstrap-driven shields and stuff, it's still slow. The only reason to do it is if there's no other way to get the signal-to-noise ratio you need. Even with picoamp things, I try to use noninverting bootstraps rather than giga- or teraohm resistors wherever possible. With photodiodes, for instance, if you make a sufficiently good bootstrap, you often don't need a TIA at all--just take the output from the bootstrap. (We can argue over a beer sometime whether that's actually any different from a plain-Jane TIA. I claim that it is, but the other side is also arguable.) Of course I did recently build a transistor tester box with a 1-Tohm resistor in it (and two each of 100G, 10G, 1G, etc, down to 100 ohms), so you had room for reasonable doubt about the gigohm enthusiasm. (And besides, your average physicist or chemist thinks nothing of putting a 1G resistor on the end of a cable.) ;) Cheers Phil Hobbs> >>> + | c video -------- b NPN +1 to +2V e | [ ] 75 >>> 75 Ohm coax 10 meter >>> |----------==============================================-----------||--- >>> amp | | | >>> | [ ] 1k /// /// [ ] >>> 75 | | >>> /// /// >>> >>> >> >> You don't even need the resistors on the BJT side--it'll drive the >> coax fine with all of the DC load at the far end. > > The 75 Ohm on the left,.. somehow you need to match the cable > impedance. >Nah, as long as the far end is properly terminated, the near end looks like a 75-ohm resistor anyway. It's the dual of series-termination--short circuit on one end, Z0 on the other. Not a lot of circuit protection, of course. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Reply by ●August 19, 20132013-08-19
"John Fields" <jfields@austininstruments.com> wrote in message news:8gm419tl4nns6p0jd3jfgrd0ibkl4oth8c@4ax.com...> On Sun, 18 Aug 2013 05:12:11 -0700 (PDT), > sridhar09.cherukuri@gmail.com wrote: > >>In the simple zener regulators like the one below, If the load current is >>100ma and zener current is 10ma we calculate the resistor to allow 110ma. >> >>If Vin is above Vz and the load draws 50ma, the remaining 60ma flows >>through zener. What makes it that the remaining current after the load >>current go through the zener? Whay can't it be the otherway? i.e after the >>zener current the remaining to the load. Is it due to the load impedance >>being less than the zener impedance? >> >>Vin------/\/\/\/\--------|-----------| >> | | >> | | >> | | >> | / RL >> /--- | >> /\ | >> / \ | >> ------ ----- >> | --- >> | - >> | >> ----- >> --- >> - > --- > http://www.onsemi.com/pub_link/Collateral/HBD854-D.PDF > > -- > JFWow, great link, John. Thanks. -Tom
Reply by ●August 19, 20132013-08-19
On Monday, August 19, 2013 10:30:43 AM UTC-4, Phil Hobbs wrote:> > The base current is really a loss mechanism--an ideal BJT has zero base > > current, because all of the emitter current makes it to the collector > > and none recombines in the base region. (Recombination is where the base > > current comes from.)That's not quite true. Base current is also due to minority carrier diffusion as well as recombination.> > > > The fundamental control mechanism of a BJT is the base-emitter voltage, > > which provides pretty tight voltage feedback in an emitter follower.Maybe, but you don't drive transistor ports with ideal voltage sources.
Reply by ●August 19, 20132013-08-19
On 8/19/2013 10:09 PM, bloggs.fredbloggs.fred@gmail.com wrote:> On Monday, August 19, 2013 10:30:43 AM UTC-4, Phil Hobbs wrote: > >> >> The base current is really a loss mechanism--an ideal BJT has zero >> base current, because all of the emitter current makes it to the >> collector and none recombines in the base region. (Recombination is >> where the base current comes from.) > > That's not quite true. Base current is also due to minority carrier > diffusion as well as recombination.True. I think that's a small effect in practical devices, though, since it wouldn't vary much from device to device whereas beta is all over the map.> >> >> The fundamental control mechanism of a BJT is the base-emitter >> voltage, which provides pretty tight voltage feedback in an >> emitter follower. > > Maybe, but you don't drive transistor ports with ideal voltage > sources.I'm not sure I'm getting your point here. The OP was thinking of the BJT solely as a current-gain device. That idea doesn't in itself constrain the base-emitter voltage at all, whereas thinking about it in terms of transconductance does. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 USA +1 845 480 2058 hobbs at electrooptical dot net http://electrooptical.net
Reply by ●August 20, 20132013-08-20
On Monday, August 19, 2013 10:20:26 PM UTC-4, Phil Hobbs wrote:> On 8/19/2013 10:09 PM, bloggs.fredbloggs.fred@gmail.com wrote: >=20 > > On Monday, August 19, 2013 10:30:43 AM UTC-4, Phil Hobbs wrote: >=20 > > >=20 > >> >=20 > >> The base current is really a loss mechanism--an ideal BJT has zero >=20 > >> base current, because all of the emitter current makes it to the >=20 > >> collector and none recombines in the base region. (Recombination is >=20 > >> where the base current comes from.) >=20 > > >=20 > > That's not quite true. Base current is also due to minority carrier >=20 > > diffusion as well as recombination. >=20 >=20 >=20 > True. I think that's a small effect in practical devices, though, since==20>=20 > it wouldn't vary much from device to device whereas beta is all over the==20>=20 > map. >=20 >=20 >=20 > > >=20 > >> >=20 > >> The fundamental control mechanism of a BJT is the base-emitter >=20 > >> voltage, which provides pretty tight voltage feedback in an >=20 > >> emitter follower. >=20 > > >=20 > > Maybe, but you don't drive transistor ports with ideal voltage >=20 > > sources. >=20 >=20 >=20 > I'm not sure I'm getting your point here. The OP was thinking of the=20 >=20 > BJT solely as a current-gain device. That idea doesn't in itself=20 >=20 > constrain the base-emitter voltage at all, whereas thinking about it in==20>=20 > terms of transconductance does. >=20 >=20 >=20 > Cheers >=20 >=20 >=20 > Phil Hobbs >=20 >=20 >=20 > --=20 >=20 > Dr Philip C D Hobbs >=20 > Principal Consultant >=20 > ElectroOptical Innovations LLC >=20 > Optics, Electro-optics, Photonics, Analog Electronics >=20 >=20 >=20 > 160 North State Road #203 >=20 > Briarcliff Manor NY 10510 USA >=20 > +1 845 480 2058 >=20 >=20 >=20 > hobbs at electrooptical dot net >=20 > http://electrooptical.netI don't know how gm helps you in this case. For purposes of determining the= transistor loading on the zener, most people agree it will be (beta +1) x = Rload. If this estimate is all over the map because of beta, there's nothin= g you can do to improve on it by using gm, the loading will be still be all= over the map. Manufacturers help by providing beta variation with Ic and m= in and max values, things that help bound the variation.
