Input coupling for fixed resistor bias CE transitor amplifie

[jalbers@bsu.edu]

I am studying transistor amplifiers, specifically for the moment
Common Emitter with fixed resistor bias. For example Vcc = 15V, Rb =
200K, RC = 1K, Beta = 100 , capacitor coupled input connected to an AC
signal +/- 0.5V for example.

I think that I understand the load line characteristic curve stuff.
Ic will be centered at 7.15 ma and Vce will be centered around
7.85V . I believe that Vbe will fluctuate between .65V and .75V and
Ib will fluctuate above and below 71.5 uA by some value ?? . I guess
this could be determined by plotting a diode curve.

My problem is with the capacitive input coupling and the current flow
and how Vbe changes from .65V through .75V as the book seems to
indicate. Every book and source that I have kind of waves their hands
and says that the capacitor removes the DC bias and allows AC to
flow. The way that I think is to draw many pictures of the same
circuit and think of instances of time with concrete values for the
voltages, currents, direction of currents all over the circuit.

I can see how a capacitor would allow current to flow back and forth
in a circuit containing a capacitor connected in series with a
resistor. I know that current does not cross the dielectric. But
what if you throw a diode (transistor BE) in series with a resistor
and a capacitor. I don't see how this would allow current to flow
back and forth. I guess I need someone to break it way down so that I
can understand it.

Any help would be greatly appreciated. Thanks.

 

[Tim Wescott]

jalbers@bsu.edu wrote:

I am studying transistor amplifiers, specifically for the moment
Common Emitter with fixed resistor bias. For example Vcc = 15V, Rb =
200K, RC = 1K, Beta = 100 , capacitor coupled input connected to an AC
signal +/- 0.5V for example.

You mean the AC signal is 1V p-p, or did you typo and mean to say that
the AC signal goes from -0.05 to +0.05V?

I think that I understand the load line characteristic curve stuff.
Ic will be centered at 7.15 ma and Vce will be centered around
7.85V .

For a mythical transistor with exactly that beta, yes. You'll get into
real circuits later, for now just keep in mind that the biasing method
is for textbook use only.

I believe that Vbe will fluctuate between .65V and .75V and
Ib will fluctuate above and below 71.5 uA by some value ?? . I guess
this could be determined by plotting a diode curve.

Yes, it could be. You can also approximate it by looking at the slope
of the diode curve at your design center voltages.

My problem is with the capacitive input coupling and the current flow
and how Vbe changes from .65V through .75V as the book seems to
indicate. Every book and source that I have kind of waves their hands
and says that the capacitor removes the DC bias and allows AC to
flow. The way that I think is to draw many pictures of the same
circuit and think of instances of time with concrete values for the
voltages, currents, direction of currents all over the circuit.

"Virtual short at AC". Yes, that can be hard to wrap your brain around.
The full explanation is that the capacitor's current is equal to the
rate of change in the voltage, times the capacitance. So for a
capacitance that's relatively big for the rest of the circuit, it'll
charge up to it's quiescent voltage, then the two ends will follow each
other in AC while not flowing current in DC.

I can see how a capacitor would allow current to flow back and forth
in a circuit containing a capacitor connected in series with a
resistor. I know that current does not cross the dielectric.

Yup.

But
what if you throw a diode (transistor BE) in series with a resistor
and a capacitor. I don't see how this would allow current to flow
back and forth. I guess I need someone to break it way down so that I
can understand it.

Any help would be greatly appreciated. Thanks.

In this case the diode is always on, and always flowing current. For
small excursions around it's quiescent point (and 0.1V p-p is pretty
big) you can think of it as a resistor in series with a voltage source
-- in your case the base resistance should, if I have my head screwed on
straight today, be around 360 ohms.

That "looks like a resistor in series with a voltage" approximation
isn't really very good at all for a 100mV swing. What you'd see in real
life with that big of an input signal would be a _very_ peaky output
waveform, and you'd see the base capacitor get "pumped" down in voltage
a bit, until it's average current was once again equal to zero.

For now it may be best to just play along with teacher, but if you just
can't, try your above analysis with 10mV p-p input instead of 100mV.
Things should be better.

--

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

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[John Popelish]

jalbers@bsu.edu wrote:

I never realized that a forward biased diode could pass current in
both directions.

They can conduct in the reverse direction, briefly, but that
is not what you are being told, here.

Which brings up additional questions:

I have always believed that current can flow in only one direction at
a time in a section of wire. We can think of multiple currents
flowing down a section of wire in different directions like when
applying the superposition theorem but there is only "one" current
flowing in "one" direction which would be the algebraic sum of the
individual currents. Is this way of thinking correct?

Yes. You are being told that if the bias current into the
base is larger than the peak excursion of the AC component,
then when they are summed you get a variable unidirectional
total. The instantaneous diode current does not reverse.
Only one of the superposed components reverses.

If this is true, then how can current flow backwards through the
forward biased diode?

Yes, but that is not happening in this case. Search for the
key words "diode reverse recovery".

For example:
http://www.avtechpulse.com/appnote/techbrief9/

--
Regards,

John Popelish

 

[jalbers@bsu.edu]

On Sep 29, 11:19 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:

jalb...@bsu.edu



I can see how a capacitor would allow current to flow back and forth
in a circuit containing a capacitor connected in series with a
resistor.  I know that current does not cross the dielectric.  But
what if you throw a diode (transistor BE) in series with a resistor
and a capacitor.  I don't see how this would allow current to flow
back and forth.

** It won't, unless the diode is forward biased with DC current like it is
in your circuit  -  then a small AC current ( modulating the bias current
level ) can flow without causing the diode to stop conducting.

Another tip:  think of a coupling cap as a battery with fixed DC voltage
that simply moves an AC voltages from one place to another where different
DC levels exist.

.....   Phil

I never realized that a forward biased diode could pass current in
both directions. Which brings up additional questions:

I have always believed that current can flow in only one direction at
a time in a section of wire. We can think of multiple currents
flowing down a section of wire in different directions like when
applying the superposition theorem but there is only "one" current
flowing in "one" direction which would be the algebraic sum of the
individual currents. Is this way of thinking correct?


If this is true, then how can current flow backwards through the
forward biased diode?

 

[Tim Wescott]

jalbers@bsu.edu wrote:

On Sep 30, 9:01 am, "Phil Allison" <philalli...@tpg.com.au> wrote:
jalb...@bsu.edu
"Phil Allison"







I can see how a capacitor would allow current to flow back and forth
in a circuit containing a capacitor connected in series with a
resistor. I know that current does not cross the dielectric. But
what if you throw a diode (transistor BE) in series with a resistor
and a capacitor. I don't see how this would allow current to flow
back and forth.
** It won't, unless the diode is forward biased with DC current like it is
in your circuit - then a small AC current ( modulating the bias current
level ) can flow without causing the diode to stop conducting.
Another tip: think of a coupling cap as a battery with fixed DC voltage
that simply moves an AC voltages from one place to another where different
DC levels exist.
I never realized that a forward biased diode could pass current in
both directions.

** You still don't - cos it cannot.

I have always believed that current can flow in only one direction at
a time in a section of wire.

** Correct.

We can think of multiple currents
flowing down a section of wire in different directions like when
applying the superposition theorem but there is only "one" current
flowing in "one" direction which would be the algebraic sum of the
individual currents. Is this way of thinking correct?

** At any instant in time, the current flow has one value and one
direction.

If this is true, then how can current flow backwards through the
forward biased diode?

** You need to go look up the word " modulation" in a dictionary.

The current flowing in that base-emitter diode never changes direction -
it only changes amplitude.

There is an average DC value plus a smaller AC value superimposed -
MODULATING the DC value at any instant in time.

Think of a lake full of water - with waves on the surface.

It is THAT simple.

..... Phil- Hide quoted text -

- Show quoted text -

Ok, so the current through BE always flows in the same direction but
fluctuates above and below the 71.5 uA . I believe that the BE
current is at its maximum when the input is in the positive cycle and
at its minimum when the input is in its negative cycle.

To help me figure some stuff out, I have been thinking of the input as
a battery and reversing it back and forth to simulate the AC and
looking at how the capacitor would charge and discharge. Suppose that
the input battery is connected so that the negative terminal is
connected to ground. The capacitor should charge up with the side
connected to the base of the transistor more negative than the other
side. The current through BE should have momentarily increased.

Now switch the polarity of the battery around. The polarity of the
capacitor should eventualy switch also. The current from the negative
side of the battery should flow to the non-base side of the
capacitor. Also the negative charge on the base side of the capacitor
needs to go somewhere. Where do these electrons go to? Do they go
move against the BE current flow?

You're using the wrong mental model, then. Visualizing the whole
battery flip-flopping leads you to trying to reverse the current in the
base -- which doesn't work.

Ultimately this whole superposition thing makes the math easier, but it
can be hard to visualize at first (keep at it, it's useful). If trying
confuses you at first, just visualize a DC voltage equal to 0.7V, with a
0.05 (or 0.005) volt sine wave added to it, so that it varies around
0.7V but never goes below zero.

--

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

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[jalbers@bsu.edu]

On Sep 30, 9:01 am, "Phil Allison" <philalli...@tpg.com.au> wrote:

jalb...@bsu.edu
 "Phil Allison"







I can see how a capacitor would allow current to flow back and forth
in a circuit containing a capacitor connected in series with a
resistor. I know that current does not cross the dielectric. But
what if you throw a diode (transistor BE) in series with a resistor
and a capacitor. I don't see how this would allow current to flow
back and forth.

** It won't, unless the diode is forward biased with DC current like it is
in your circuit - then a small AC current ( modulating the bias current
level ) can flow without causing the diode to stop conducting.

Another tip: think of a coupling cap as a battery with fixed DC voltage
that simply moves an AC voltages from one place to another where different
DC levels exist.

I never realized that a forward biased diode could pass current in
both directions.

**  You still don't -  cos it cannot.

I have always believed that current can flow in only one direction at
a time in a section of wire.

**  Correct.

We can think of multiple currents
flowing down a section of wire in different directions like when
applying the superposition theorem but there is only "one" current
flowing in "one" direction which would be the algebraic sum of the
individual currents.  Is this way of thinking correct?

**  At any instant in time, the current flow has one value and one
direction.

If this is true, then how can current flow backwards through the
forward biased diode?

** You need to go look up the word " modulation" in a dictionary.

The current flowing in that base-emitter diode never changes direction  -
it only changes amplitude.

There is an average DC value plus a smaller AC value superimposed  -
MODULATING  the DC value at any instant in time.

Think of a lake full of water -  with waves on the surface.

It is THAT  simple.

.....   Phil- Hide quoted text -

- Show quoted text -

Ok, so the current through BE always flows in the same direction but
fluctuates above and below the 71.5 uA . I believe that the BE
current is at its maximum when the input is in the positive cycle and
at its minimum when the input is in its negative cycle.

To help me figure some stuff out, I have been thinking of the input as
a battery and reversing it back and forth to simulate the AC and
looking at how the capacitor would charge and discharge. Suppose that
the input battery is connected so that the negative terminal is
connected to ground. The capacitor should charge up with the side
connected to the base of the transistor more negative than the other
side. The current through BE should have momentarily increased.

Now switch the polarity of the battery around. The polarity of the
capacitor should eventualy switch also. The current from the negative
side of the battery should flow to the non-base side of the
capacitor. Also the negative charge on the base side of the capacitor
needs to go somewhere. Where do these electrons go to? Do they go
move against the BE current flow?