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'[EE] Calculating minimum BJT impedence'
2011\06\17@204450 by

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 <x.solarwind.xgmail.com> wrote:

{Quote hidden}

Nobody

On Jun 18, 2011, at 8:29 PM, V G wrote:

>> What is this minimum impedance and why is it there?
>
I wasn't really understanding the question, but: Vcesat ?

BillW

On 18/06/2011 01:44, V G wrote:
> 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.

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.
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.

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

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.xgmail.com> wrote:
{Quote hidden}

>

At 08:44 PM 6/17/2011, you wrote:
>Hey all,
>
>What is this minimum impedance and why is it there?

The transistor saturates and Vce become more-or-less constant, decreasing
slowly with decreasing collector current, given the constant base current.
That's what's happening.

I don't think that viewing it as an "impedance" (ratio of v/i) is
particularly helpful to gaining understanding in this kind of situation.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speffinterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

On Sun, Jun 19, 2011 at 12:40 AM, Oli Glaser <oli.glasertalktalk.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

2011\06\19@115812 by
On Sun, Jun 19, 2011 at 7:35 AM, Spehro Pefhany <speffinterlog.com> wrote:

> The transistor saturates and Vce become more-or-less constant, decreasing
> slowly with decreasing collector current, given the constant base current..
> That's what's happening.
>

Thanks. Now I understand the Vce saturation. I didn't realize it was there
before.

> I don't think that viewing it as an "impedance" (ratio of v/i) is
> particularly helpful to gaining understanding in this kind of situation.
On Sun, Jun 19, 2011 at 1:06 AM, Sean Breheny <shb7cornell.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}

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:
>> >  The transistor saturates and Vce become more-or-less constant, decreasing
>> >  slowly with decreasing collector current, given the constant base current.
>> >  That's what's happening.
>> >
> Thanks. Now I understand the Vce saturation. I didn't realize it was there
> before.
>
>

You were told several times.
At "saturation" a BJT looks like a constant voltage. Nothing like a resistor.

At "saturation" an Enhancement FET is a bit like a voltage controlled resistor. (VGS sets resistance).

If not saturated then a depletion jFET or BJT is a little like a constant current source
On 19/06/2011 17:17, V G wrote:
>
>
>> >  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
>
>

Impedance can be complex.

"Perfect" Resistance is purely a straight line through origin with Volts and Current perfectly in phase and is same for DC and AC.

On Sun, Jun 19, 2011 at 12:33 PM, Michael Watterson <mikeradioway.org>wrote:

>  You were told several times.
>

My humble apologies.

> At "saturation" a BJT looks like a constant voltage. Nothing like a
> resistor.

I understand this. Thank you.

At "saturation" an Enhancement FET is a bit like a voltage controlled
> resistor. (VGS sets resistance).
>
> If not saturated then a depletion jFET or BJT is a little like a
> constant current source.
>
On 19/06/2011 16:57, V G wrote:
> On Sun, Jun 19, 2011 at 12:40 AM, Oli Glaser<oli.glasertalktalk.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.

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