Post by thread:[EE] Calculating minimum BJT impedence

Hey all, I'm trying to charge (let's say) a single Li ion battery (float charge) to 4.0V. The maximum charge current will be controlled with a BJT (tell me if there's a better way) and so I've been doing some simulations to see if I can actually get 4.0V across the battery this way. Screenshot of simulation: http://solarwind.byethost7.com/pic4.png I'm pretending that R2 is the battery, gradually increasing its impedance as it's state of charge increases. The voltage across the battery never seems to hit 4.0V. I'm doing another simulation with R2 from 1-100k and that seems to get closer and closer to 4.0V as the current goes to 0 and the BJT's impedance rises rapidly. I can understand that at the beginning, the BJT needs to decrease impedance as R2 increases in order to maintain the overall impedance so that the set amount of current goes through. I don't understand though, why just before 40 ohms on the x axis, the BJT impedance increases again. Shouldn't the BJT try and lower its impedance as much as possible so that the most current can go through (even though the maximum amount allowed by the base current can no longer go through)? What is this minimum impedance and why is it there

On Fri, Jun 17, 2011 at 8:44 PM, V G <spam_OUTx.solarwind.xTakeThisOuTgmail.com> wrote: {Quote hidden}> Hey all, > > I'm trying to charge (let's say) a single Li ion battery (float charge) to > 4.0V. > > The maximum charge current will be controlled with a BJT (tell me if > there's a better way) and so I've been doing some simulations to see if I > can actually get 4.0V across the battery this way. > > Screenshot of simulation: http://solarwind.byethost7.com/pic4.png > > I'm pretending that R2 is the battery, gradually increasing its impedance > as it's state of charge increases. The voltage across the battery never > seems to hit 4.0V. I'm doing another simulation with R2 from 1-100k and that > seems to get closer and closer to 4.0V as the current goes to 0 and the > BJT's impedance rises rapidly. > > I can understand that at the beginning, the BJT needs to decrease impedance > as R2 increases in order to maintain the overall impedance so that the set > amount of current goes through. I don't understand though, why just before > 40 ohms on the x axis, the BJT impedance increases again. Shouldn't the BJT > try and lower its impedance as much as possible so that the most current can > go through (even though the maximum amount allowed by the base current can > no longer go through)? > > What is this minimum impedance and why is it there? > Nobody

Hi, The whole concept of resistance or impedance assumes a linear circuit behavior (or at least it assumes that you want a linearized version of the real behavior). Metals and a few other elements do exhibit a remarkably linear relationship between voltage and current, so Ohm's law holds very well for them and resistance is a solid concept for these materials. Similarly, when you are dealing with ideal inductors and capacitors, extending resistance to the complex plane (giving impedance) allows you to analyze these linear dynamic systems easily. However, for semiconductors, there is often significant nonlinear behavior within the range of current and voltage which you are concerned about - so asking "what is the impedance of this transistor" may or may not be a very meaningful question. It can be helpful when you are operating the transistor within a small-signal region around a DC operating point, but if the transistor is going in and out of saturation or in and out of cutoff, it isn't such a good approximation and can be very misleading. FETs (both JFETs and MOSFETs) have a region where the drain-source current is fairly linearly related to the drain-source voltage (thus often called the ohmic region). There is no nicely analogous region of operation for BJTs. When a BJT is fully "ON" it is in the saturation region, and it roughly has a constant Vce independent of collector current. When the BJT is partially-on, it has a collector current which depends strongly on the base-emitter voltage and only slightly on the collector-emitter voltage. Thus, it acts like a current sink/source. If you compute the impedance of the collector-emitter pair, you'll get a very high value, even though significant current is flowing and the voltage is moderate (for example, 5V for Vce and 10mA current, but the impedance might be 100K ohms - remember that impedance is dV/dI not V/I - resistance can be either dV/dI or V/I depending on whether it is incremental or absolute resistance) There is also a rough relationship in this region between the base current and the collector current (Beta or Hfe or current gain) - but that depends greatly on temperature, manufacturing variation, and somewhat on Vce. Sean On Sat, Jun 18, 2011 at 11:29 PM, V G <.....x.solarwind.xKILLspam@spam@gmail.com> wrote: {Quote hidden}> On Fri, Jun 17, 2011 at 8:44 PM, V G <x.solarwind.xKILLspamgmail.com> wrote: > >> Hey all, >> >> I'm trying to charge (let's say) a single Li ion battery (float charge) to >> 4.0V. >> >> The maximum charge current will be controlled with a BJT (tell me if >> there's a better way) and so I've been doing some simulations to see if I >> can actually get 4.0V across the battery this way. >> >> Screenshot of simulation: http://solarwind.byethost7.com/pic4.png >> >> I'm pretending that R2 is the battery, gradually increasing its impedance >> as it's state of charge increases. The voltage across the battery never >> seems to hit 4.0V. I'm doing another simulation with R2 from 1-100k and that >> seems to get closer and closer to 4.0V as the current goes to 0 and the >> BJT's impedance rises rapidly. >> >> I can understand that at the beginning, the BJT needs to decrease impedance >> as R2 increases in order to maintain the overall impedance so that the set >> amount of current goes through. I don't understand though, why just before >> 40 ohms on the x axis, the BJT impedance increases again. Shouldn't the BJT >> try and lower its impedance as much as possible so that the most current can >> go through (even though the maximum amount allowed by the base current can >> no longer go through)? >> >> What is this minimum impedance and why is it there? >> > > Nobody? >

On Sun, Jun 19, 2011 at 12:40 AM, Oli Glaser <EraseMEoli.glaserspam_OUTTakeThisOuTtalktalk.net>wrote: > Because of the bipolar saturation voltage. A FET would be better as a > pass element, which is why they are used in LDOs. > It seems you are still seeing the bipolar as a variable resistor. It > doesn't work like that - better to see it as a voltage controlled > current source. > No no, I'm not. I fully understand that it's a voltage controlled current source. I just want to observe how the dynamic impedance of the BJT changes.. > Try your simulation again, but set the base current to 1uA and plot > this: d(V(n002))/d(Ic(q1)) > This will give you the dynamic impedance of the collector. > Is it the derivative d(V)/d(Ic)? If so, what is the idea behind that

On Sun, Jun 19, 2011 at 1:06 AM, Sean Breheny <@spam@shb7KILLspamcornell.edu> wrote: > Hi, > > The whole concept of resistance or impedance assumes a linear circuit > behavior (or at least it assumes that you want a linearized version of > the real behavior). Metals and a few other elements do exhibit a > remarkably linear relationship between voltage and current, so Ohm's > law holds very well for them and resistance is a solid concept for > these materials. Similarly, when you are dealing with ideal inductors > and capacitors, extending resistance to the complex plane (giving > impedance) allows you to analyze these linear dynamic systems easily. > However, for semiconductors, there is often significant nonlinear > behavior within the range of current and voltage which you are > concerned about - so asking "what is the impedance of this transistor" > may or may not be a very meaningful question. It can be helpful when > you are operating the transistor within a small-signal region around a > DC operating point, but if the transistor is going in and out of > saturation or in and out of cutoff, it isn't such a good approximation > and can be very misleading. > I thought I was operating in the DC small signal region. Aren't I? {Quote hidden}> FETs (both JFETs and MOSFETs) have a region where the drain-source > current is fairly linearly related to the drain-source voltage (thus > often called the ohmic region). There is no nicely analogous region of > operation for BJTs. When a BJT is fully "ON" it is in the saturation > region, and it roughly has a constant Vce independent of collector > current. When the BJT is partially-on, it has a collector current > which depends strongly on the base-emitter voltage and only slightly > on the collector-emitter voltage. Thus, it acts like a current > sink/source. If you compute the impedance of the collector-emitter > pair, you'll get a very high value, even though significant current is > flowing and the voltage is moderate (for example, 5V for Vce and 10mA > current, but the impedance might be 100K ohms - > remember that > impedance is dV/dI not V/I - resistance can be either dV/dI or V/I > depending on whether it is incremental or absolute resistance) Could you please explain this? I don't understand this > There > is also a rough relationship in this region between the base current > and the collector current (Beta or Hfe or current gain) - but that > depends greatly on temperature, manufacturing variation, and somewhat > on Vce. >

On 19/06/2011 16:57, V G wrote: > On Sun, Jun 19, 2011 at 12:40 AM, Oli Glaser<RemoveMEoli.glaserTakeThisOuTtalktalk.net>wrote: > >> Because of the bipolar saturation voltage. A FET would be better as a >> pass element, which is why they are used in LDOs. >> It seems you are still seeing the bipolar as a variable resistor. It >> doesn't work like that - better to see it as a voltage controlled >> current source. >> > No no, I'm not. I fully understand that it's a voltage controlled current > source. I just want to observe how the dynamic impedance of the BJT changes. > I see - my mistake, sorry. For the dynamic impedance see below. >> Try your simulation again, but set the base current to 1uA and plot >> this: d(V(n002))/d(Ic(q1)) >> This will give you the dynamic impedance of the collector. >> > Is it the derivative d(V)/d(Ic)? If so, what is the idea behind that? To calculate the impedance of the collector d(Vc)/d(Ic) However (I didn't notice the step simulation) you would have to do it with a transient simulation and ramp the collector voltage. Hard to do it with a step .op simulation. It can probably be done but I haven't time to look at the manual right now. I think Sean explained it well. Dynamic as opposed to static, so on a computer where it is a finite difference based derivative (Vx-(Vx-1))/(Ix-(Ix-1)) rather than V1/I1, I2/I2 and so on. So if two points of a changing voltage and current are: V1 = 2, V2 = 1, I1 = 1.0001A, I2 = 1A (V1-V2)/(I1-I2) = 1/0.0001 = 10Kohm as opposed to: V1/I1 = 2ohm and V2/I2 = ~1ohm I have probably not explained it very well. Give it a go - run the simulation and plot the formula I gave you: set the base current to 1uA, resistor to 1 ohm, ramp the collector voltage from 0 to 10V in a transient simulation, and see what you get. Should give you an idea of the collector impedance. If you have problems let me know and I'll send you an LTSpice file over. The Art of Electronics goes into great detail on transistor models, and explains it all very clearly. There are other books which go into more mathematical detail, but the AOE explanation is sufficient to understand enough to use them in most applications.