Thanks to all who responded to my question.

I went back and took a closer look at other editions of the Handbook,
and at Solid State Design for the Radio Amateur (SSDRA).

The Handbooks all seem to suggest that the shunt feedback networks
using a resistor and a cap in series will somehow result in more
negative feedback at lower frequencies. Given that capacitive
reactance varies inversely with freq, I still can't see how this
happens. Tim's message seems to provide one possible explanation, but
I'm kind of surprised that I haven't seen mention of this in the ham
literature.

When SSDRA discusses shunt feedback, (Chapter 8, page 188) they have a
resistor, a blocking cap, and an inductor all in series between the
collector and the base in a common emmiter amp. "The inductor has the
effect of decreasing the feedback at high frequencies, while the 470
ohm resistor is the dominant element at low frequencies." That's easy
to see. But without the inductor (no inductor in the Handbook
presentations) it is hard to see how this works.

Thanks again to all,

73 Bill M0HBR CU2JL N2CQR
http://www.qsl.net/n2cqr

tim gorman wrote:
wrote:



I'm having trouble understanding how the typical shunt feedback
networks used in RF (solid state) amps work. I'm looking at the 1993
ARRL Handbook. Typical common base broadband amp. For the shunt
feedback (from collector to base) they have two resistors: 560 ohms in
series with 3300 ohms. The 3300 ohm is bypassed by a .01 uf cap.

So far so good. But then the text explains that because you have
rising gain characteristics when the frequency drops you need something
to reduce gain at lower frequencies. That's why the negative feedback
helps.

Here's where I'm having trouble: "As the operating frequency is
decreased the negative feedback increases becasue the network feedback
reactance becomes lower." Huh? Wouldn't that network's reactance
INCREASE as frequency is lowered? The only part of it with reactance
is the .01 cap, correct?

Help! 73!

Bill M0HBR N2CQR CU2JL
http://www.qsl.net/n2cqr

I think to actually figure this out you will need to write the gain transfer
equation for the amplifier. Many of the replies here are talking about an
RF amplifier but are analyzing the feedback at DC.

Remember that the output of a transistor is NOT in phase with the input. As
the base voltage goes UP the collector voltage goes DOWN. (think of a
transistor used as a switch - when the base is biased off the no current
flows and the collector is at power supply potential - when the switch is
biased on then current flows and the collector is driven toward the
potential of the emitter -- i.e. the collector voltage goes down) For RF
the actual phase difference between input and output is dependent upon the
input and output impedances of the transistor as well as the gain transfer
characteristic (things like transit times of the current carriers and
junctions widths and all sorts of stuff figure in here). *THEN* you have to
consider the phase contribution of the RC network in the collector to base
network. At some RF frequency the collector is 180deg out of phase with the
input so a direct feedback link from the collector to the base would be
*negative*. It is quite possible that the phase relationships of this
particular amplifier are such that the negative feedback increases as the
frequency goes down.

You would just have to write the equations and see where they take you.

tim ab0wr

Since your top posting...

The series C has enough reactance to be effective to fairly low
frequencies. Consider that the 0.1uf cap seen has reactance of
10ohms around 150Khz thats insignificant compared to the resistor.
If the resistor is 1000 ohm then R=C around 2khz.

What you have to ask is what frequency does the capacitors
reactance become equal to the series resistor? For even a
..1uF cap that's a fairly low frequency. I suspect not recognizing
that detail may be confusing. As you get to very low frequencies
that cap may appear to be insignificant but those frequencies are
very low.

To appreciate that revisit those circuits in question and calculate
the capacitor Xc for 1khz, 10khz 100khz and 1Mhz.

Allison

On 25 Oct 2005 00:15:04 -0700, wrote:

Thanks to all who responded to my question.

I went back and took a closer look at other editions of the Handbook,
and at Solid State Design for the Radio Amateur (SSDRA).

The Handbooks all seem to suggest that the shunt feedback networks
using a resistor and a cap in series will somehow result in more
negative feedback at lower frequencies. Given that capacitive
reactance varies inversely with freq, I still can't see how this
happens. Tim's message seems to provide one possible explanation, but
I'm kind of surprised that I haven't seen mention of this in the ham
literature.

When SSDRA discusses shunt feedback, (Chapter 8, page 188) they have a
resistor, a blocking cap, and an inductor all in series between the
collector and the base in a common emmiter amp. "The inductor has the
effect of decreasing the feedback at high frequencies, while the 470
ohm resistor is the dominant element at low frequencies." That's easy
to see. But without the inductor (no inductor in the Handbook
presentations) it is hard to see how this works.

Thanks again to all,

73 Bill M0HBR CU2JL N2CQR
http://www.qsl.net/n2cqr

tim gorman wrote:
wrote:



I'm having trouble understanding how the typical shunt feedback
networks used in RF (solid state) amps work. I'm looking at the 1993
ARRL Handbook. Typical common base broadband amp. For the shunt
feedback (from collector to base) they have two resistors: 560 ohms in
series with 3300 ohms. The 3300 ohm is bypassed by a .01 uf cap.

So far so good. But then the text explains that because you have
rising gain characteristics when the frequency drops you need something
to reduce gain at lower frequencies. That's why the negative feedback
helps.

Here's where I'm having trouble: "As the operating frequency is
decreased the negative feedback increases becasue the network feedback
reactance becomes lower." Huh? Wouldn't that network's reactance
INCREASE as frequency is lowered? The only part of it with reactance
is the .01 cap, correct?

Help! 73!

Bill M0HBR N2CQR CU2JL
http://www.qsl.net/n2cqr

I think to actually figure this out you will need to write the gain transfer
equation for the amplifier. Many of the replies here are talking about an
RF amplifier but are analyzing the feedback at DC.

Remember that the output of a transistor is NOT in phase with the input. As
the base voltage goes UP the collector voltage goes DOWN. (think of a
transistor used as a switch - when the base is biased off the no current
flows and the collector is at power supply potential - when the switch is
biased on then current flows and the collector is driven toward the
potential of the emitter -- i.e. the collector voltage goes down) For RF
the actual phase difference between input and output is dependent upon the
input and output impedances of the transistor as well as the gain transfer
characteristic (things like transit times of the current carriers and
junctions widths and all sorts of stuff figure in here). *THEN* you have to
consider the phase contribution of the RC network in the collector to base
network. At some RF frequency the collector is 180deg out of phase with the
input so a direct feedback link from the collector to the base would be
*negative*. It is quite possible that the phase relationships of this
particular amplifier are such that the negative feedback increases as the
frequency goes down.

You would just have to write the equations and see where they take you.

tim ab0wr

Again, the common emitter amplifier gives you 180deg of phase change. If
this were to be applied directly to the base of the transistor through only
a resistor, you would get direct cancellation at a level determnined by the
size of the resistor. The feedback network throws in a phase change
proportional to the tangent of the reactance and the resistance part of the
network. I think the tangent function for the feedback will look something
like:

w(k1) / [k2 + w(k3)]

where w is the radian freq. and k1, k2, and k3 are constants made up of R1,
R2, and C. k1 = C(R2)**2 or something like that.

As the freq goes up the phase change gets larger - i.e. it moves the phase
difference between the collector and the base away from 180deg., which
means less negative feedback. The maximum phase change contribution from
the RC network would be 45deg as w gets very, very large (arctan 1 = 45).
As the frequency goes down, the phase change contribution from the RC
circuit gets to be smaller and smaller which means the feedback moves
closer and closer to being 180deg - which means more negative feedback. (as
w approaches zero arctan 0 = 0)

This is probably only good for RF frequencies. Say, above 1Mhz. I would have
to actually write out the transfer characteristic and graph it to see
exactly what happens. Conceptually, it makes sense though.

tim ab0wr



wrote:



Thanks to all who responded to my question.

I went back and took a closer look at other editions of the Handbook,
and at Solid State Design for the Radio Amateur (SSDRA).

The Handbooks all seem to suggest that the shunt feedback networks
using a resistor and a cap in series will somehow result in more
negative feedback at lower frequencies. Given that capacitive
reactance varies inversely with freq, I still can't see how this
happens. Tim's message seems to provide one possible explanation, but
I'm kind of surprised that I haven't seen mention of this in the ham
literature.

When SSDRA discusses shunt feedback, (Chapter 8, page 188) they have a
resistor, a blocking cap, and an inductor all in series between the
collector and the base in a common emmiter amp. "The inductor has the
effect of decreasing the feedback at high frequencies, while the 470
ohm resistor is the dominant element at low frequencies." That's easy
to see. But without the inductor (no inductor in the Handbook
presentations) it is hard to see how this works.

Thanks again to all,

73 Bill M0HBR CU2JL N2CQR
http://www.qsl.net/n2cqr

tim gorman wrote:
wrote:



I'm having trouble understanding how the typical shunt feedback
networks used in RF (solid state) amps work. I'm looking at the 1993
ARRL Handbook. Typical common base broadband amp. For the shunt
feedback (from collector to base) they have two resistors: 560 ohms in
series with 3300 ohms. The 3300 ohm is bypassed by a .01 uf cap.

So far so good. But then the text explains that because you have
rising gain characteristics when the frequency drops you need something
to reduce gain at lower frequencies. That's why the negative feedback
helps.

Here's where I'm having trouble: "As the operating frequency is
decreased the negative feedback increases becasue the network feedback
reactance becomes lower." Huh? Wouldn't that network's reactance
INCREASE as frequency is lowered? The only part of it with reactance
is the .01 cap, correct?

Help! 73!

Bill M0HBR N2CQR CU2JL
http://www.qsl.net/n2cqr

I think to actually figure this out you will need to write the gain
transfer equation for the amplifier. Many of the replies here are talking
about an RF amplifier but are analyzing the feedback at DC.

Remember that the output of a transistor is NOT in phase with the input.
As the base voltage goes UP the collector voltage goes DOWN. (think of a
transistor used as a switch - when the base is biased off the no current
flows and the collector is at power supply potential - when the switch is
biased on then current flows and the collector is driven toward the
potential of the emitter -- i.e. the collector voltage goes down) For RF
the actual phase difference between input and output is dependent upon
the input and output impedances of the transistor as well as the gain
transfer characteristic (things like transit times of the current
carriers and junctions widths and all sorts of stuff figure in here).
*THEN* you have to consider the phase contribution of the RC network in
the collector to base network. At some RF frequency the collector is
180deg out of phase with the input so a direct feedback link from the
collector to the base would be *negative*. It is quite possible that the
phase relationships of this particular amplifier are such that the
negative feedback increases as the frequency goes down.

You would just have to write the equations and see where they take you.

tim ab0wr

Tim: Thanks again for your help, and thanks again to all the others
who've replied to my question.

I see what you are saying. I reached back to Terman's description of
these kinds of feedback netwoks in tube circuits. I think he describes
just what you are presenting (but I think your presentation is
clearer!)

Obviously I'm still struggling with some very basic amplifier issues.
Am I correct in thinking that shunt feedback would have a number of
different advantages in a common emitter transistor amp? Looks to me
like the feedback network would give the designer the chance to:
1). Manipulate the input and output impedances
2). Counteract the tendency of the amp to "take off" becasue of the
rising gain characteristic (as frequency is lowered).
3). Reduce any distortion (IMD) generated in the amplifier itself.

I'm most shaky on #3. Am I correct in thinking that if you have some
sort of spur or IMD product generated in the amp itelf (say in the
collector circuit)the negative feedback provided by the shunt tends to
nock some (most? all?) of this distortion down?

Thanks, 73
Bill M0HBR N2CQR CU2JL
http://www.qsl.net/n2cqr



tim gorman wrote:
Again, the common emitter amplifier gives you 180deg of phase change. If
this were to be applied directly to the base of the transistor through only
a resistor, you would get direct cancellation at a level determnined by the
size of the resistor. The feedback network throws in a phase change
proportional to the tangent of the reactance and the resistance part of the
network. I think the tangent function for the feedback will look something
like:

w(k1) / [k2 + w(k3)]

where w is the radian freq. and k1, k2, and k3 are constants made up of R1,
R2, and C. k1 = C(R2)**2 or something like that.

As the freq goes up the phase change gets larger - i.e. it moves the phase
difference between the collector and the base away from 180deg., which
means less negative feedback. The maximum phase change contribution from
the RC network would be 45deg as w gets very, very large (arctan 1 = 45).
As the frequency goes down, the phase change contribution from the RC
circuit gets to be smaller and smaller which means the feedback moves
closer and closer to being 180deg - which means more negative feedback. (as
w approaches zero arctan 0 = 0)

This is probably only good for RF frequencies. Say, above 1Mhz. I would have
to actually write out the transfer characteristic and graph it to see
exactly what happens. Conceptually, it makes sense though.

tim ab0wr



wrote:



Thanks to all who responded to my question.

I went back and took a closer look at other editions of the Handbook,
and at Solid State Design for the Radio Amateur (SSDRA).

The Handbooks all seem to suggest that the shunt feedback networks
using a resistor and a cap in series will somehow result in more
negative feedback at lower frequencies. Given that capacitive
reactance varies inversely with freq, I still can't see how this
happens. Tim's message seems to provide one possible explanation, but
I'm kind of surprised that I haven't seen mention of this in the ham
literature.

When SSDRA discusses shunt feedback, (Chapter 8, page 188) they have a
resistor, a blocking cap, and an inductor all in series between the
collector and the base in a common emmiter amp. "The inductor has the
effect of decreasing the feedback at high frequencies, while the 470
ohm resistor is the dominant element at low frequencies." That's easy
to see. But without the inductor (no inductor in the Handbook
presentations) it is hard to see how this works.

Thanks again to all,

73 Bill M0HBR CU2JL N2CQR
http://www.qsl.net/n2cqr

tim gorman wrote:
wrote:



I'm having trouble understanding how the typical shunt feedback
networks used in RF (solid state) amps work. I'm looking at the 1993
ARRL Handbook. Typical common base broadband amp. For the shunt
feedback (from collector to base) they have two resistors: 560 ohms in
series with 3300 ohms. The 3300 ohm is bypassed by a .01 uf cap.

So far so good. But then the text explains that because you have
rising gain characteristics when the frequency drops you need something
to reduce gain at lower frequencies. That's why the negative feedback
helps.

Here's where I'm having trouble: "As the operating frequency is
decreased the negative feedback increases becasue the network feedback
reactance becomes lower." Huh? Wouldn't that network's reactance
INCREASE as frequency is lowered? The only part of it with reactance
is the .01 cap, correct?

Help! 73!

Bill M0HBR N2CQR CU2JL
http://www.qsl.net/n2cqr

I think to actually figure this out you will need to write the gain
transfer equation for the amplifier. Many of the replies here are talking
about an RF amplifier but are analyzing the feedback at DC.

Remember that the output of a transistor is NOT in phase with the input.
As the base voltage goes UP the collector voltage goes DOWN. (think of a
transistor used as a switch - when the base is biased off the no current
flows and the collector is at power supply potential - when the switch is
biased on then current flows and the collector is driven toward the
potential of the emitter -- i.e. the collector voltage goes down) For RF
the actual phase difference between input and output is dependent upon
the input and output impedances of the transistor as well as the gain
transfer characteristic (things like transit times of the current
carriers and junctions widths and all sorts of stuff figure in here).
*THEN* you have to consider the phase contribution of the RC network in
the collector to base network. At some RF frequency the collector is
180deg out of phase with the input so a direct feedback link from the
collector to the base would be *negative*. It is quite possible that the
phase relationships of this particular amplifier are such that the
negative feedback increases as the frequency goes down.

You would just have to write the equations and see where they take you.

tim ab0wr

On 30 Oct 2005 22:43:07 -0800, wrote:

Tim: Thanks again for your help, and thanks again to all the others
who've replied to my question.

I see what you are saying. I reached back to Terman's description of
these kinds of feedback netwoks in tube circuits. I think he describes
just what you are presenting (but I think your presentation is
clearer!)

Obviously I'm still struggling with some very basic amplifier issues.
Am I correct in thinking that shunt feedback would have a number of
different advantages in a common emitter transistor amp? Looks to me
like the feedback network would give the designer the chance to:
1). Manipulate the input and output impedances
2). Counteract the tendency of the amp to "take off" becasue of the
rising gain characteristic (as frequency is lowered).
3). Reduce any distortion (IMD) generated in the amplifier itself.

I'm most shaky on #3. Am I correct in thinking that if you have some
sort of spur or IMD product generated in the amp itelf (say in the
collector circuit)the negative feedback provided by the shunt tends to
nock some (most? all?) of this distortion down?

#3)
Negative feedback of any kind then to make a circuit with nonlinear
action (however small it may be) more linear. IMD (InterModulation
Distortion) is a mixing action from nolinearity that occurs when two
signals are passing through an amplifier that has distortion, less
distortion, less IMD.

Allison
KB!GMX

wrote:



Tim: Thanks again for your help, and thanks again to all the others
who've replied to my question.

I see what you are saying. I reached back to Terman's description of
these kinds of feedback netwoks in tube circuits. I think he describes
just what you are presenting (but I think your presentation is
clearer!)

Obviously I'm still struggling with some very basic amplifier issues.
Am I correct in thinking that shunt feedback would have a number of
different advantages in a common emitter transistor amp? Looks to me
like the feedback network would give the designer the chance to:
1). Manipulate the input and output impedances
2). Counteract the tendency of the amp to "take off" becasue of the
rising gain characteristic (as frequency is lowered).
3). Reduce any distortion (IMD) generated in the amplifier itself.

I'm most shaky on #3. Am I correct in thinking that if you have some
sort of spur or IMD product generated in the amp itelf (say in the
collector circuit)the negative feedback provided by the shunt tends to
nock some (most? all?) of this distortion down?

Thanks, 73
Bill M0HBR N2CQR CU2JL
http://www.qsl.net/n2cqr




Bill,

Negative feedback can be used to help fix input and output impedances. You
have to be careful with shunt type negative feedback to make sure you don't
have any regions where the phase change becomes 0deg. This can happen when
you have both capacitive and inductive reactances in the circuit. That is
one reason parasitic oscillations happen at very high frequencies. Even
connecting wires on resistors can look like inductors at very high
frequencies.

As far as negative feedback having an impact on distortion, you must be very
careful to identify what kinds of distortion you are speaking of and have a
good feel for what the feedback networks are doing. Negative feedback,
improperly designed, can cause audio amplifiers to not be "flat" across the
audio spectrum. This can be considered to be distortion.

Negative feedback, improperly applied can cause gain compression, e.g. in an
RF Linear Amplifier, which distorts the dynamic range of the applied
modulation causing a distortion which impacts intelligibility.

IMD distortion is typically caused by a non-linear device. Since a
transistor is an inherently non-linear device (it is based on "diode"
junctions) at low voltage levels since it acts like an on-off switch, there
will always be some level of IMD that negative feedback won't help.

Where negative feedback *can* help is in keeping a Class A amplifier
operating in the Class A region. This will minimize IMD. If you get an
amplifier out of the Class A region, there is bound to be some IMD. If you
are running a Class B amplifier in push-pull this is a "psuedo" Class A
amplifier and negative feedback that would keep each amplifer right at
cutoff for the negative part of the input would help minimize IMD
distortion. If you are running a Class C amplifier, negative feedback won't
help much because you are running a non-linear amplifer to begin with.

Hope this helps.

tim ab0wr