Input impedance of a BJT in CE config

[riccardo manfrin]

I'm designing a very simple (dummy I'd say) linear amp with a biasing network like the one shown in this image:

http://pui.chorwong.com/devry/wp-content/uploads/2009/08/bjtbias1.jpg

Ignore the values reported in the image. I'm interested in understanding the input impedance seen at the base by the AC source, using the \pi model of the BJT.

For AC analysis DC sources are shutdown (become grounds), the capacitor near the emitter shortcircuits the parallel resistor, and all the other caps caps are shortcircuited hence letting AC through.

The impedance is the ratio between voltage and current, so in my case,

r_\pi = v_{BE} / i_B

Unfortunaltely v_BE and i_B are related to the exponential relationship between current and voltage of a diode.

Looking around, I found mention to formulas like

r_\pi = V_T / i_B

where V_T= kT/q, that is the thermal voltage of the BE junction which accounts for the so called "ohmic effects" of the diode model.

Unfortunately, if both the above formulas are correct it looks like V_{BE} = V_T. If this is the case, then my input source signal (V_s in the image) is not varying the V_{BE} voltage which is only driven by thermal noise. Is this a correct guess?

Can you explain me the procedure to derive the input impedance in AC, given the knowledge of the voltage V_s and its series resistance R_s.

Thanks,
RM

 

[riccardo manfrin]

VS and its series resistance do not influence the _circuit_ input

impedance.



Does your homework require an accurate calculation of AC input

impedance good up to substantial frequency ?>:-}

NOTATION
upper case: DC current/voltages (bias),
lower case: AC current/voltages (signals),

Well, my goal is to dimension R_C in such a way to ensure that the range of the input signal i_B guarantees exact full swinging of the output signal V_{CE} in the admitted datasheet dynamic.

For the purpose, I need to know the range of i_B.

In order to know the max range of i_B I need to know the source and its series resistance and the input impedance of the transistor, which will define how much current will be injected in the base, and therefore the range of i_B.

Precisely, given a range of frequencies of my input signal v_s, I need to know the minimum BJT base AC input impedance in that range, which will admit the maximum current i_B into the base.

Once I know the range of i_B, I can dimension R_C so to admit exactly that range and guarantee that the maximum i_B value causes V_CE to get near saturation, and the minimum i_B value causes V_CE to reach V_CC.

Am I misunderstanding on this?

Thanks in advance,
R

 

[Jim Thompson]

On Wed, 2 Oct 2013 04:56:59 -0700 (PDT), riccardo manfrin
<riccardomanfrin@gmail.com> wrote:

I'm designing a very simple (dummy I'd say) linear amp with a biasing network like the one shown in this image:

http://pui.chorwong.com/devry/wp-content/uploads/2009/08/bjtbias1.jpg

Ignore the values reported in the image. I'm interested in understanding the input impedance seen at the base by the AC source, using the \pi model of the BJT.

For AC analysis DC sources are shutdown (become grounds), the capacitor near the emitter shortcircuits the parallel resistor, and all the other caps caps are shortcircuited hence letting AC through.

The impedance is the ratio between voltage and current, so in my case,

r_\pi = v_{BE} / i_B

Unfortunaltely v_BE and i_B are related to the exponential relationship between current and voltage of a diode.

Looking around, I found mention to formulas like

r_\pi = V_T / i_B

where V_T= kT/q, that is the thermal voltage of the BE junction which accounts for the so called "ohmic effects" of the diode model.

Unfortunately, if both the above formulas are correct it looks like V_{BE} = V_T. If this is the case, then my input source signal (V_s in the image) is not varying the V_{BE} voltage which is only driven by thermal noise. Is this a correct guess?

Can you explain me the procedure to derive the input impedance in AC, given the knowledge of the voltage V_s and its series resistance R_s.

Thanks,
RM

VS and its series resistance do not influence the _circuit_ input
impedance.

Does your homework require an accurate calculation of AC input
impedance good up to substantial frequency ?>:-}

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Jim Thompson]

On Wed, 2 Oct 2013 08:07:01 -0700 (PDT), riccardo manfrin
<riccardomanfrin@gmail.com> wrote:

VS and its series resistance do not influence the _circuit_ input

impedance.



Does your homework require an accurate calculation of AC input

impedance good up to substantial frequency ?>:-}

NOTATION
upper case: DC current/voltages (bias),
lower case: AC current/voltages (signals),

Well, my goal is to dimension R_C in such a way to ensure that the range of the input signal i_B guarantees exact full swinging of the output signal V_{CE} in the admitted datasheet dynamic.

For the purpose, I need to know the range of i_B.

In order to know the max range of i_B I need to know the source and its series resistance and the input impedance of the transistor, which will define how much current will be injected in the base, and therefore the range of i_B.

Precisely, given a range of frequencies of my input signal v_s, I need to know the minimum BJT base AC input impedance in that range, which will admit the maximum current i_B into the base.

Once I know the range of i_B, I can dimension R_C so to admit exactly that range and guarantee that the maximum i_B value causes V_CE to get near saturation, and the minimum i_B value causes V_CE to reach V_CC.

Am I misunderstanding on this?

Thanks in advance,
R

I posted an answer before...

NNTP-Posting-Date: Thu, 26 Sep 2013 11:41:25 -0500
From: Jim Thompson <To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com>
Newsgroups: alt.binaries.schematics.electronic
Subject: Biasing Question from S.E.D - BiasQuestionManfrinSED.pdf
Date: Thu, 26 Sep 2013 09:41:24 -0700
Message-ID: <nqo849hdre4ahhgo2q8rnisf8gp337994n@4ax.com>

If you can't access A.B.S.E, let me know and I'll post to my website.

NOTE: I'm moving into our new house, so it may be a day before I
respond again.

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[George Herold]

On Wednesday, October 2, 2013 7:56:59 AM UTC-4, riccardo manfrin wrote:

I'm designing a very simple (dummy I'd say) linear amp with a biasing network like the one shown in this image:



http://pui.chorwong.com/devry/wp-content/uploads/2009/08/bjtbias1.jpg



Ignore the values reported in the image. I'm interested in understanding the input impedance seen at the base by the AC source, using the \pi model of the BJT.



For AC analysis DC sources are shutdown (become grounds), the capacitor near the emitter shortcircuits the parallel resistor, and all the other caps caps are shortcircuited hence letting AC through.

You still have to consider the bias resistors.
which look like 30k // 130k (//= in parallel)
Then there's the 1.5k
For just the transistor I think Zin looks like beta times the 1.2k emitter resistor..(?)
But I should check in Art of electronics by H&H to be sure, has your copy arrived yet? They explain all this stuff very well...

The impedance is the ratio between voltage and current, so in my case,



r_\pi = v_{BE} / i_B



Unfortunaltely v_BE and i_B are related to the exponential relationship between current and voltage of a diode.



Looking around, I found mention to formulas like



r_\pi = V_T / i_B



where V_T= kT/q, that is the thermal voltage of the BE junction which accounts for the so called "ohmic effects" of the diode model.



Unfortunately, if both the above formulas are correct it looks like V_{BE} = V_T. If this is the case, then my input source signal (V_s in the image) is not varying the V_{BE} voltage which is only driven by thermal noise. Is this a correct guess?



Can you explain me the procedure to derive the input impedance in AC, given the knowledge of the voltage V_s and its series resistance R_s.



Thanks,

RM

 

[Tim Wescott]

On Wed, 02 Oct 2013 04:56:59 -0700, riccardo manfrin wrote:

I'm designing a very simple (dummy I'd say) linear amp with a biasing
network like the one shown in this image:

http://pui.chorwong.com/devry/wp-content/uploads/2009/08/bjtbias1.jpg

Ignore the values reported in the image. I'm interested in understanding
the input impedance seen at the base by the AC source, using the \pi
model of the BJT.

For AC analysis DC sources are shutdown (become grounds), the capacitor
near the emitter shortcircuits the parallel resistor, and all the other
caps caps are shortcircuited hence letting AC through.

The impedance is the ratio between voltage and current, so in my case,

r_\pi = v_{BE} / i_B

Unfortunaltely v_BE and i_B are related to the exponential relationship
between current and voltage of a diode.

Looking around, I found mention to formulas like

r_\pi = V_T / i_B

where V_T= kT/q, that is the thermal voltage of the BE junction which
accounts for the so called "ohmic effects" of the diode model.

Unfortunately, if both the above formulas are correct it looks like
V_{BE} = V_T. If this is the case, then my input source signal (V_s in
the image) is not varying the V_{BE} voltage which is only driven by
thermal noise. Is this a correct guess?

Can you explain me the procedure to derive the input impedance in AC,
given the knowledge of the voltage V_s and its series resistance R_s.

Thanks,
RM

In comparing those two resistances you're forgetting that it's an AC
model, and letting the DC voltages creep in.

The AC impedance of the base, when the emitter is shorted, is V_t/i_B,
where V_t = kT/q (so note that it changes with temperature!), and i_B is
the --> DC <-- current through the base.

The AC impedance at the emitter of the transistor is (more or less) the
series combination of the base resistance and the un-bypassed resistor
(the 200 ohm resistor in your drawing), times the transistor's current
gain.

The actual impedance seen by the source will be the transistor's emitter
impedance in parallel with the bias network, then that combination in
series with the resistor between "Vi" and the source.

BUT! And this may be a very big but -- that only applies at frequencies
well below the f_t of the transistor. It also ignores the base spreading
resistance of the transistor. If you're working at audio and using a
normal small-signal transistor and values about what are in this
schematic, then you can maybe ignore the base spreading resistance and
f_t.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[Jim Thompson]

On Wed, 02 Oct 2013 19:25:22 -0500, Tim Wescott
<tim@seemywebsite.really> wrote:

On Wed, 02 Oct 2013 04:56:59 -0700, riccardo manfrin wrote:

I'm designing a very simple (dummy I'd say) linear amp with a biasing
network like the one shown in this image:

http://pui.chorwong.com/devry/wp-content/uploads/2009/08/bjtbias1.jpg

Ignore the values reported in the image. I'm interested in understanding
the input impedance seen at the base by the AC source, using the \pi
model of the BJT.

For AC analysis DC sources are shutdown (become grounds), the capacitor
near the emitter shortcircuits the parallel resistor, and all the other
caps caps are shortcircuited hence letting AC through.

The impedance is the ratio between voltage and current, so in my case,

r_\pi = v_{BE} / i_B

Unfortunaltely v_BE and i_B are related to the exponential relationship
between current and voltage of a diode.

Looking around, I found mention to formulas like

r_\pi = V_T / i_B

where V_T= kT/q, that is the thermal voltage of the BE junction which
accounts for the so called "ohmic effects" of the diode model.

Unfortunately, if both the above formulas are correct it looks like
V_{BE} = V_T. If this is the case, then my input source signal (V_s in
the image) is not varying the V_{BE} voltage which is only driven by
thermal noise. Is this a correct guess?

Can you explain me the procedure to derive the input impedance in AC,
given the knowledge of the voltage V_s and its series resistance R_s.

Thanks,
RM

In comparing those two resistances you're forgetting that it's an AC
model, and letting the DC voltages creep in.

The AC impedance of the base, when the emitter is shorted, is V_t/i_B,
where V_t = kT/q (so note that it changes with temperature!), and i_B is
the --> DC <-- current through the base.

The AC impedance at the emitter of the transistor is (more or less) the
series combination of the base resistance and the un-bypassed resistor
(the 200 ohm resistor in your drawing), times the transistor's current
gain.

The actual impedance seen by the source will be the transistor's emitter
impedance in parallel with the bias network, then that combination in
series with the resistor between "Vi" and the source.

BUT! And this may be a very big but -- that only applies at frequencies
well below the f_t of the transistor. It also ignores the base spreading
resistance of the transistor. If you're working at audio and using a
normal small-signal transistor and values about what are in this
schematic, then you can maybe ignore the base spreading resistance and
f_t.

It appears to be homework.

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[riccardo manfrin]

The AC impedance of the base, when the emitter is shorted, is V_t/i_B,

where V_t = kT/q (so note that it changes with temperature!), and i_B is

the --> DC <-- current through the base.

This was definitely something I was missing and I realized only after posting. My fault for not investigating more.

Now that the I realize that we are talking of DC bias current I_B, I still miss the motivation behind the use of V_t instead of using the DC voltage between base and emitter (which would be V_{BE} ~= 0.7V >> V_T ~= 26mV).

I'm going to read more about it in the documentation (btw no I'm still waiting for that book to ship).
R

 

[Julian Grodzicky]

On Thursday, October 3, 2013 6:05:33 PM UTC+10, riccardo manfrin wrote:

I'm going to read more about it in the documentation (btw no I'm still waiting for that book to ship).

R

The book is available in its entirety online.

 

[riccardo manfrin]

> The book is available in its entirety online.

Yes, 2nd ed. is available in a "non-profit" way ..
I'm reading through it right now.. page 76 says

"C is chosen so that all frequencies of
interest are passed by the high-pass filter
it forms in combination with the parallel
resistance of the base biasing resistors (the
impedance looking into the base itself will
usually be much larger because of the way
the base resistors are chosen, and it can be
ignored);"

This is unexpected.. if the emitter is bypassed by C_E, the impedance looking into the base is solely given by the internal impedance of the BJT from the base to the emitter, which is V_T/I_B.
For V_T ~=26mV and I_B in ranges from 0.1 to 1mA isn't the input resistance going to be low enough to be the dominant component in the parallel with the base biasing voltage divider resistors?

I'm reading further to see if something comes out..

 

[Tim Wescott]

On Thu, 03 Oct 2013 01:05:33 -0700, riccardo manfrin wrote:

The AC impedance of the base, when the emitter is shorted, is V_t/i_B,

where V_t = kT/q (so note that it changes with temperature!), and i_B
is

the --> DC <-- current through the base.

This was definitely something I was missing and I realized only after
posting. My fault for not investigating more.

Now that the I realize that we are talking of DC bias current I_B, I
still miss the motivation behind the use of V_t instead of using the DC
voltage between base and emitter (which would be V_{BE} ~= 0.7V >> V_T
~= 26mV).

I'm going to read more about it in the documentation (btw no I'm still
waiting for that book to ship).
R

How much formal education do you have in circuit design?

The AC impedance of the emitter terminal comes directly from the equation
for emitter current, which, if my memory serves, is

i_e = Iss * (e^(v_be/V_t) - 1),

where Iss is the saturation current of the BE junction. If you solve for
Iss at any given combination of v_be and i_e, then take the derivative

1 / R_e = di_e / dv_be

you'll get the whole R_e = V_t / I_e thing.

When I took circuits they practically drilled a hole in your head and
shoved in a note card with this analysis written on it, so they must have
thought it was important.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[Tim Wescott]

On Thu, 03 Oct 2013 06:49:18 -0700, riccardo manfrin wrote:

The book is available in its entirety online.

Yes, 2nd ed. is available in a "non-profit" way ..
I'm reading through it right now.. page 76 says

"C is chosen so that all frequencies of interest are passed by the
high-pass filter it forms in combination with the parallel resistance of
the base biasing resistors (the impedance looking into the base itself
will usually be much larger because of the way the base resistors are
chosen, and it can be ignored);"

This is unexpected.. if the emitter is bypassed by C_E, the impedance
looking into the base is solely given by the internal impedance of the
BJT from the base to the emitter, which is V_T/I_B.
For V_T ~=26mV and I_B in ranges from 0.1 to 1mA isn't the input
resistance going to be low enough to be the dominant component in the
parallel with the base biasing voltage divider resistors?

I'm reading further to see if something comes out..

If you're looking at the same picture I'm looking at, the emitter
resistor is not bypassed. Generally if you choose a set of bias
resistors and emitter resistor that give good performance over a wide
range of part variations and temperature, then his statement will be true.

If you bypass the emitter resistor then yes, the base resistance may
start to matter.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[riccardo manfrin]

If you bypass the emitter resistor then yes, the base resistance may

start to matter.

In the image, the 0.2KOhm is not bypassed, the 1K is, so you are right to say that R_E is not bypassed. Nevertheless, the 200 Ohm R_E is not that much and I assumed that the resistance seen through base to emitter started to matter.

 

[riccardo manfrin]

> How much formal education do you have in circuit design?

Definitely not enough

then take the derivative

1 / R_e = di_e / dv_be



you'll get the whole R_e = V_t / I_e thing.

That is what I was looking for (and actually also found in "the art of electronics" meanwhile).
I was considering resistance as the ratio between voltage and current whereas for a non linear device it is rather the derivative of these two guys around the point of operation. I was missing something so simple as linearization..

 

[Tim Wescott]

On Fri, 04 Oct 2013 00:57:04 -0700, riccardo manfrin wrote:

If you bypass the emitter resistor then yes, the base resistance may

start to matter.

In the image, the 0.2KOhm is not bypassed, the 1K is, so you are right
to say that R_E is not bypassed. Nevertheless, the 200 Ohm R_E is not
that much and I assumed that the resistance seen through base to emitter
started to matter.

Well, do the math. At low frequencies the base resistance due to emitter
resistance is beta * Re, with Re being the resistance both internal and
external.

In that particular schematic, because the base bias network is fairly
high resistance, the base resistance will, indeed, have an effect.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[Tim Wescott]

On Fri, 04 Oct 2013 01:05:50 -0700, riccardo manfrin wrote:

How much formal education do you have in circuit design?

Definitely not enough

then take the derivative

1 / R_e = di_e / dv_be



you'll get the whole R_e = V_t / I_e thing.

That is what I was looking for (and actually also found in "the art of
electronics" meanwhile).
I was considering resistance as the ratio between voltage and current
whereas for a non linear device it is rather the derivative of these two
guys around the point of operation. I was missing something so simple as
linearization..

This may be overly pedantic, but the _AC_ resistance comes from dv/di.
"DC resistance", and by extension "resistance" is fairly meaningless.
It's probably more accurate to talk about "small signal resistance", or
even "differential resistance".

Be careful with this AC resistance stuff, by the way -- I went into
college an electronics hobbyist, so when I hit this small signal analysis
stuff I was overjoyed. I immediately went home and designed myself some
circuits that worked like absolute crap, because my "small" was about 20
times bigger than the book's "small". The distortion was bad enough that
the bias got all messed up, and -- well, it was a disaster.

Doing some checking with circuit simulations or just plain math to see
how much distortion you'll be getting with seemingly-tiny signals is a
Good Thing if you're at the level that I think you are.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[Jim Thompson]

On Wed, 2 Oct 2013 04:56:59 -0700 (PDT), riccardo manfrin
<riccardomanfrin@gmail.com> wrote:

I'm designing a very simple (dummy I'd say) linear amp with a biasing network like the one shown in this image:

http://pui.chorwong.com/devry/wp-content/uploads/2009/08/bjtbias1.jpg

Ignore the values reported in the image. I'm interested in understanding the input impedance seen at the base by the AC source, using the \pi model of the BJT.

For AC analysis DC sources are shutdown (become grounds), the capacitor near the emitter shortcircuits the parallel resistor, and all the other caps caps are shortcircuited hence letting AC through.

The impedance is the ratio between voltage and current, so in my case,

r_\pi = v_{BE} / i_B

Unfortunaltely v_BE and i_B are related to the exponential relationship between current and voltage of a diode.

Looking around, I found mention to formulas like

r_\pi = V_T / i_B

where V_T= kT/q, that is the thermal voltage of the BE junction which accounts for the so called "ohmic effects" of the diode model.

Unfortunately, if both the above formulas are correct it looks like V_{BE} = V_T. If this is the case, then my input source signal (V_s in the image) is not varying the V_{BE} voltage which is only driven by thermal noise. Is this a correct guess?

Can you explain me the procedure to derive the input impedance in AC, given the knowledge of the voltage V_s and its series resistance R_s.

Thanks,
RM

Here's a simple example...

<http://www.analog-innovations.com/SED/Quasi_hybrid-pi_Example.pdf>

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.