one transistor amplifier help

Jim Thompson · Apr 19, 2017

On Tue, 18 Apr 2017 19:49:21 -0500, amdx <nojunk@knology.net> wrote:

On 4/18/2017 6:58 PM, George Herold wrote:
[snip]
What's the electrical equivalent of the ratchet and pawl?
(I've got my answer.)

Sample and hold?
Mikek

[snip]

amdx · Apr 19, 2017

On 4/15/2017 5:09 PM, Jasen Betts wrote:

On 2017-04-15, amdx <nojunk@knology.net> wrote:
I have the following amp.
It originally had a 12V supply voltage, In an effort to get a larger
swing on the output, I raised that to 24V.
I now get 12.5Vpp on the output, I want to maximize the output signal.
A gain of 3 will work for my circuit.
What changes can I make to maximize the output swing.
The input is at maximum before distortion is seen.

maybe

reduce R9, R10 to 39R
reduce R8 to something like 680 - your aimimg to get the collector to 13V

I came up with 33 ohm resistors for R9 and R 10, and 549 ohms for R8.
I got 20Vpp before distortion, however my gain is about 13.
The value of R8 is critical, small change make big differences in the
collector voltage.

Thanks for the input
Mikek

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Phil Hobbs · Apr 19, 2017

On 04/18/2017 02:37 PM, George Herold wrote:

On Tuesday, April 18, 2017 at 1:54:15 PM UTC-4, Tim Wescott wrote:
On Tue, 18 Apr 2017 10:21:20 -0700, George Herold wrote:

On Tuesday, April 18, 2017 at 12:21:45 PM UTC-4, Tim Wescott wrote:
On Tue, 18 Apr 2017 11:11:03 +1000, David Eather wrote:

On Mon, 17 Apr 2017 07:07:12 +1000, Jim Thompson
To-Email-Use-The-Envelope-Icon@on-my-web-site.com> wrote:

On Sat, 15 Apr 2017 10:18:32 -0500, amdx <nojunk@knology.net> wrote:

I have the following amp.
It originally had a 12V supply voltage, In an effort to get a
larger swing on the output, I raised that to 24V.
I now get 12.5Vpp on the output, I want to maximize the output
signal.
A gain of 3 will work for my circuit.
What changes can I make to maximize the output swing.
The input is at maximum before distortion is seen.

https://www.dropbox.com/s/07jq88cqh8sxr7j/Kliejers%20amp%
20ascii.jpg?
dl=0

I would like to put matching load on the output so my swing will
drop in 1/2.

Fundamental novice error... amplifier outputs should be LOW
impedance for maximum power transfer.

actually for maximum power transfer the amp output impedance should
be equal to next stages input impedance. If the amplifier has a low
impedance it results im maximum efficiency not power transfer

So, I have a 50 ohm load, a +12V supply, and I build an amplifier that
can do 12Vp-p open-circuit, and has a 50 ohm output impedance. Working
into 50 ohms, it has a 6Vp-p output. For a sine-wave signal, that
works out to 90mW.

Contrast that with the same load and p-p limits, but an amplifier with
zero ohms output impedance. That one can put 12Vp-p into the load, or
360mW.

You're saying that 90mW is bigger than 360mW?

I think the argument (problem) usually starts with some output
impedance. (Physics types always start with a battery and the internal
battery resistance.) And then asks for which load is there a maximum of
power transferred from the device to the load. So if you had 50 ohm
Z-out for which load do you get maximum power?
It's a silly argument. (And sets me off, every time I hear it.)

There are times when impedance matching is important -- but for the most
part impedance matching isn't the best thing to do. Even in a radio
receiver, the best SNR is often achieved at some input impedance that
does _not_ match the impedance of the line seen by the receiver.

I'm not sure about radio receivers. Somewhere in there I think the antenna
is doing some sort of impedance matching.

Going out on a limb, I think you need impedance matching when you are
interested in energy (power) and not just the signal (amplitude) and don't
have any gain available.

So; microwave horns, trumpet horns, ultra sound transmitters/receivers,
AR coating on optics... there's probably some mechanical examples...?

Acoustic transducers, shock absorbers,...

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

Phil Hobbs · Apr 19, 2017

On 04/18/2017 09:26 PM, Jim Thompson wrote:

On Tue, 18 Apr 2017 19:49:21 -0500, amdx <nojunk@knology.net> wrote:

On 4/18/2017 6:58 PM, George Herold wrote:
[snip]
What's the electrical equivalent of the ratchet and pawl?
(I've got my answer.)

Sample and hold?
Mikek

[snip]

TRACK and hold.

...Jim Thompson

Nah, that can go up and down. An electrical analogy to a ratchet and
pawl is a diode charge pump.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

Phil Hobbs · Apr 19, 2017

On 04/17/2017 10:15 PM, Jim Thompson wrote:

On Tue, 18 Apr 2017 11:11:03 +1000, "David Eather" <eather@tpg.com.au
wrote:

On Mon, 17 Apr 2017 07:07:12 +1000, Jim Thompson
To-Email-Use-The-Envelope-Icon@on-my-web-site.com> wrote:

On Sat, 15 Apr 2017 10:18:32 -0500, amdx <nojunk@knology.net> wrote:

I have the following amp.
It originally had a 12V supply voltage, In an effort to get a larger
swing on the output, I raised that to 24V.
I now get 12.5Vpp on the output, I want to maximize the output signal.
A gain of 3 will work for my circuit.
What changes can I make to maximize the output swing.
The input is at maximum before distortion is seen.

https://www.dropbox.com/s/07jq88cqh8sxr7j/Kliejers%20amp%20ascii.jpg?dl=0

I would like to put matching load on the output so my swing will drop in
1/2.

Fundamental novice error... amplifier outputs should be LOW impedance
for maximum power transfer.

actually for maximum power transfer the amp output impedance should be
equal to next stages input impedance. If the amplifier has a low impedance
it results im maximum efficiency not power transfer

Really?

Care to demonstrate that ?>:-}

...Jim Thompson

C'mon, Jim, all op amps deliver their maximum power into a 1-milliohm
load at DC.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

George Herold · Apr 19, 2017

On Wednesday, April 19, 2017 at 8:03:48 AM UTC-4, Phil Hobbs wrote:

On 04/18/2017 09:26 PM, Jim Thompson wrote:
On Tue, 18 Apr 2017 19:49:21 -0500, amdx <nojunk@knology.net> wrote:

On 4/18/2017 6:58 PM, George Herold wrote:
[snip]
What's the electrical equivalent of the ratchet and pawl?
(I've got my answer.)

Sample and hold?
Mikek

[snip]

TRACK and hold.

...Jim Thompson

Nah, that can go up and down. An electrical analogy to a ratchet and
pawl is a diode charge pump.

OK I was thinking of a stepper motor. (Or maybe an up/down counter.)

George H.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

Jim Thompson · Apr 19, 2017

On Wed, 19 Apr 2017 08:08:29 -0400, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

On 04/17/2017 10:15 PM, Jim Thompson wrote:
On Tue, 18 Apr 2017 11:11:03 +1000, "David Eather" <eather@tpg.com.au
wrote:

On Mon, 17 Apr 2017 07:07:12 +1000, Jim Thompson
To-Email-Use-The-Envelope-Icon@on-my-web-site.com> wrote:

On Sat, 15 Apr 2017 10:18:32 -0500, amdx <nojunk@knology.net> wrote:

I have the following amp.
It originally had a 12V supply voltage, In an effort to get a larger
swing on the output, I raised that to 24V.
I now get 12.5Vpp on the output, I want to maximize the output signal.
A gain of 3 will work for my circuit.
What changes can I make to maximize the output swing.
The input is at maximum before distortion is seen.

https://www.dropbox.com/s/07jq88cqh8sxr7j/Kliejers%20amp%20ascii.jpg?dl=0

I would like to put matching load on the output so my swing will drop in
1/2.

Fundamental novice error... amplifier outputs should be LOW impedance
for maximum power transfer.

actually for maximum power transfer the amp output impedance should be
equal to next stages input impedance. If the amplifier has a low impedance
it results im maximum efficiency not power transfer

Really?

Care to demonstrate that ?>:-}

...Jim Thompson

C'mon, Jim, all op amps deliver their maximum power into a 1-milliohm
load at DC.

Cheers

Phil Hobbs

Phil Hobbs · Apr 19, 2017

On 04/19/2017 10:57 AM, Jim Thompson wrote:

On Wed, 19 Apr 2017 08:08:29 -0400, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

On 04/17/2017 10:15 PM, Jim Thompson wrote:
On Tue, 18 Apr 2017 11:11:03 +1000, "David Eather" <eather@tpg.com.au
wrote:

On Mon, 17 Apr 2017 07:07:12 +1000, Jim Thompson
To-Email-Use-The-Envelope-Icon@on-my-web-site.com> wrote:

On Sat, 15 Apr 2017 10:18:32 -0500, amdx <nojunk@knology.net> wrote:

I have the following amp.
It originally had a 12V supply voltage, In an effort to get a larger
swing on the output, I raised that to 24V.
I now get 12.5Vpp on the output, I want to maximize the output signal.
A gain of 3 will work for my circuit.
What changes can I make to maximize the output swing.
The input is at maximum before distortion is seen.

https://www.dropbox.com/s/07jq88cqh8sxr7j/Kliejers%20amp%20ascii.jpg?dl=0

I would like to put matching load on the output so my swing will drop in
1/2.

Fundamental novice error... amplifier outputs should be LOW impedance
for maximum power transfer.

actually for maximum power transfer the amp output impedance should be
equal to next stages input impedance. If the amplifier has a low impedance
it results im maximum efficiency not power transfer

Really?

Care to demonstrate that ?>:-}

...Jim Thompson

C'mon, Jim, all op amps deliver their maximum power into a 1-milliohm
load at DC.

Cheers

Phil Hobbs

C'mon Phil, OpAmps with low output impedance deliver more power to a
load than those that have output impedance that equals the load.

...Jim Thompson

'S a joke, son, jes' a joke.

The closed-loop output impedance of your average op amp is in the 1
milliohm ballpark at DC. If you put 1 milliohm on the output, it'll
either current limit or blow up, and will be delivering only microwatts
to the load, whereas it would be delivering many milliwatts into some
sane load like 500 ohms.

It sure won't put out VDD**2/0.001 ohm, which is like 200 kW for 15-volt
supplies.

Cheers

Phil "Senator Claghorn fan" Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

rickman · Apr 29, 2017

On 4/28/2017 7:24 PM, rickman wrote:

On 4/15/2017 11:18 AM, amdx wrote:
I have the following amp.
It originally had a 12V supply voltage, In an effort to get a larger
swing on the output, I raised that to 24V.
I now get 12.5Vpp on the output, I want to maximize the output signal.
A gain of 3 will work for my circuit.
What changes can I make to maximize the output swing.
The input is at maximum before distortion is seen.

https://www.dropbox.com/s/07jq88cqh8sxr7j/Kliejers%20amp%20ascii.jpg?dl=0

I would like to put matching load on the output so my swing will drop in
1/2.

I read part of the thread and it seemed to go a bit off topic and I
don't see where you acknowledge an understanding of this amp. Did you
get what you need?

First let me say that I think people get a bit too compulsive with
matching output and input impedances. If you are looking for maximum
power transfer into a load of fixed impedance, that will happen when you
minimize the output impedance. The impedance matching goal comes in
when you have a fixed output impedance and can only adjust the load
impedance. If you don't care about this here, fine, but in the future
remember that impedance matching only optimizes power transfer when you
are adjusting the load, not the source.

This amp is pretty simple to analyze. It is doing *two* things. It is
providing gain while at the same time it is minimizing the capacitive
load on the previous stage. The bootstrap is reducing the capacitive
loading while taking away some of the output swing. If you really want
to maximize the output swing you need to make this two stages, one a
bootstrap to minimize capacitive loading and another to provide the gain
with maximum headroom.

Is that at all clear? The nearly 1 gain on the emitter reduces the
headroom available for the output swing. Max headroom is approximately

24 - 0.2 * R11 / (R11 + R10) or with the current values of R10 and R11,
20.5 volts.

But you won't want to push the transistor into saturation I expect, so a
little less.

If you split this into two stages or just don't bother with the
bootstrap (I'm not sure it is needed, but maybe, you'd need to analyze
the impact of the capacitive loading on the earlier stage), the max
headroom equation becomes,

24 - 0.2 or 23.8 volts

Which of these is what you want?

You also need to set the bias point to the center of the headroom once
you decide which approach you want. That is set with R7 and R8.

--

Rick C

amdx · Apr 29, 2017

On 4/28/2017 6:24 PM, rickman wrote:

On 4/15/2017 11:18 AM, amdx wrote:
I have the following amp.
It originally had a 12V supply voltage, In an effort to get a larger
swing on the output, I raised that to 24V.
I now get 12.5Vpp on the output, I want to maximize the output signal.
A gain of 3 will work for my circuit.
What changes can I make to maximize the output swing.
The input is at maximum before distortion is seen.

https://www.dropbox.com/s/07jq88cqh8sxr7j/Kliejers%20amp%20ascii.jpg?dl=0

I would like to put matching load on the output so my swing will drop in
1/2.

I read part of the thread and it seemed to go a bit off topic and I
don't see where you acknowledge an understanding of this amp. Did you
get what you need?

First let me say that I think people get a bit too compulsive with
matching output and input impedances. If you are looking for maximum
power transfer into a load of fixed impedance, that will happen when you
minimize the output impedance. The impedance matching goal comes in
when you have a fixed output impedance and can only adjust the load
impedance. If you don't care about this here, fine, but in the future
remember that impedance matching only optimizes power transfer when you
are adjusting the load, not the source.

This amp is pretty simple to analyze. It is doing *two* things. It is
providing gain while at the same time it is minimizing the capacitive
load on the previous stage. The bootstrap is reducing the capacitive
loading while taking away some of the output swing. If you really want
to maximize the output swing you need to make this two stages, one a
bootstrap to minimize capacitive loading and another to provide the gain
with maximum headroom.

Is that at all clear? The nearly 1 gain on the emitter reduces the
headroom available for the output swing. Max headroom is approximately

24 - 0.2 * R11 / (R11 + R10) or with the current values of R10 and R11,
20.5 volts.

But you won't want to push the transistor into saturation I expect, so a
little less.

If you split this into two stages or just don't bother with the
bootstrap (I'm not sure it is needed, but maybe, you'd need to analyze
the impact of the capacitive loading on the earlier stage), the max
headroom equation becomes,

24 - 0.2 or 23.8 volts

Which of these is what you want?

This was an amp I gave up on because it rolled off at 10MHz.
Well, so did the next amp, but as we have discussed, that is a
measurement error. With that in mind, I will go back to this amp
because it did give me very close to my needed output swing.
I may end up trying to figure out how to make it roll off
to a specified degree, because I found my diode detector has more output
as frequency increases.
As far as matching, nah, I found out my Diode detector has about a 6kÎŠ
input impedance, so my 500ÎŠ collector resistor is plenty low, and I have
room to increase it, if I have a reason.
Mikek

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rickman · Apr 29, 2017

On 4/28/2017 9:31 PM, amdx wrote:

On 4/28/2017 6:24 PM, rickman wrote:
On 4/15/2017 11:18 AM, amdx wrote:
I have the following amp.
It originally had a 12V supply voltage, In an effort to get a larger
swing on the output, I raised that to 24V.
I now get 12.5Vpp on the output, I want to maximize the output signal.
A gain of 3 will work for my circuit.
What changes can I make to maximize the output swing.
The input is at maximum before distortion is seen.

https://www.dropbox.com/s/07jq88cqh8sxr7j/Kliejers%20amp%20ascii.jpg?dl=0

I would like to put matching load on the output so my swing will drop in
1/2.

I read part of the thread and it seemed to go a bit off topic and I
don't see where you acknowledge an understanding of this amp. Did you
get what you need?

First let me say that I think people get a bit too compulsive with
matching output and input impedances. If you are looking for maximum
power transfer into a load of fixed impedance, that will happen when
you minimize the output impedance. The impedance matching goal comes
in when you have a fixed output impedance and can only adjust the load
impedance. If you don't care about this here, fine, but in the future
remember that impedance matching only optimizes power transfer when
you are adjusting the load, not the source.

This amp is pretty simple to analyze. It is doing *two* things. It
is providing gain while at the same time it is minimizing the
capacitive load on the previous stage. The bootstrap is reducing the
capacitive loading while taking away some of the output swing. If you
really want to maximize the output swing you need to make this two
stages, one a bootstrap to minimize capacitive loading and another to
provide the gain with maximum headroom.

Is that at all clear? The nearly 1 gain on the emitter reduces the
headroom available for the output swing. Max headroom is approximately

24 - 0.2 * R11 / (R11 + R10) or with the current values of R10 and
R11, 20.5 volts.

But you won't want to push the transistor into saturation I expect, so
a little less.

If you split this into two stages or just don't bother with the
bootstrap (I'm not sure it is needed, but maybe, you'd need to analyze
the impact of the capacitive loading on the earlier stage), the max
headroom equation becomes,

24 - 0.2 or 23.8 volts

Which of these is what you want?

This was an amp I gave up on because it rolled off at 10MHz.
Well, so did the next amp, but as we have discussed, that is a
measurement error. With that in mind, I will go back to this amp
because it did give me very close to my needed output swing.
I may end up trying to figure out how to make it roll off
to a specified degree, because I found my diode detector has more output
as frequency increases.
As far as matching, nah, I found out my Diode detector has about a 6kÎŠ
input impedance, so my 500ÎŠ collector resistor is plenty low, and I have
room to increase it, if I have a reason.

If you want higher frequency response, higher drive will help or at
least won't make it worse. Nothing wrong with a lower impedance on the
output other than using more power supply current.

--

Rick C

Ralph Mowery · Apr 29, 2017

In article <oe18kk$ajd$1@dont-email.me>, gnuarm@gmail.com says...

On 4/28/2017 9:31 PM, amdx wrote:
On 4/28/2017 6:24 PM, rickman wrote:
On 4/15/2017 11:18 AM, amdx wrote:
I have the following amp.
It originally had a 12V supply voltage, In an effort to get a larger
swing on the output, I raised that to 24V.
I now get 12.5Vpp on the output, I want to maximize the output signal.
A gain of 3 will work for my circuit.
What changes can I make to maximize the output swing.
The input is at maximum before distortion is seen.

https://www.dropbox.com/s/07jq88cqh8sxr7j/Kliejers%20amp%20ascii.jpg?dl=0

I would like to put matching load on the output so my swing will drop in
1/2.

I read part of the thread and it seemed to go a bit off topic and I
don't see where you acknowledge an understanding of this amp. Did you
get what you need?

First let me say that I think people get a bit too compulsive with
matching output and input impedances. If you are looking for maximum
power transfer into a load of fixed impedance, that will happen when
you minimize the output impedance. The impedance matching goal comes
in when you have a fixed output impedance and can only adjust the load
impedance. If you don't care about this here, fine, but in the future
remember that impedance matching only optimizes power transfer when
you are adjusting the load, not the source.

This amp is pretty simple to analyze. It is doing *two* things. It
is providing gain while at the same time it is minimizing the
capacitive load on the previous stage. The bootstrap is reducing the
capacitive loading while taking away some of the output swing. If you
really want to maximize the output swing you need to make this two
stages, one a bootstrap to minimize capacitive loading and another to
provide the gain with maximum headroom.

Is that at all clear? The nearly 1 gain on the emitter reduces the
headroom available for the output swing. Max headroom is approximately

24 - 0.2 * R11 / (R11 + R10) or with the current values of R10 and
R11, 20.5 volts.

But you won't want to push the transistor into saturation I expect, so
a little less.

If you split this into two stages or just don't bother with the
bootstrap (I'm not sure it is needed, but maybe, you'd need to analyze
the impact of the capacitive loading on the earlier stage), the max
headroom equation becomes,

24 - 0.2 or 23.8 volts

Which of these is what you want?

This was an amp I gave up on because it rolled off at 10MHz.
Well, so did the next amp, but as we have discussed, that is a
measurement error. With that in mind, I will go back to this amp
because it did give me very close to my needed output swing.
I may end up trying to figure out how to make it roll off
to a specified degree, because I found my diode detector has more output
as frequency increases.
As far as matching, nah, I found out my Diode detector has about a 6kO
input impedance, so my 500O collector resistor is plenty low, and I have
room to increase it, if I have a reason.

Coming into this late,but if the problem is to raise the output of a
signal generator, look at some of the 8 to 12 dollar amps coming in from
China on ebay. They use an IC and can provide about 20 db of gain.
Either low noise amps, or some that will give about 1/2 of a watt
output. If you want, you can buy just the ICs that are on them and
build your own. Just a couple of capacitors and a resistor and maybe a
RF choke.

rickman · Apr 29, 2017

On 4/29/2017 10:06 AM, Ralph Mowery wrote:

In article <oe18kk$ajd$1@dont-email.me>, gnuarm@gmail.com says...

On 4/28/2017 9:31 PM, amdx wrote:
On 4/28/2017 6:24 PM, rickman wrote:
On 4/15/2017 11:18 AM, amdx wrote:
I have the following amp.
It originally had a 12V supply voltage, In an effort to get a larger
swing on the output, I raised that to 24V.
I now get 12.5Vpp on the output, I want to maximize the output signal.
A gain of 3 will work for my circuit.
What changes can I make to maximize the output swing.
The input is at maximum before distortion is seen.

https://www.dropbox.com/s/07jq88cqh8sxr7j/Kliejers%20amp%20ascii.jpg?dl=0

I would like to put matching load on the output so my swing will drop in
1/2.

I read part of the thread and it seemed to go a bit off topic and I
don't see where you acknowledge an understanding of this amp. Did you
get what you need?

First let me say that I think people get a bit too compulsive with
matching output and input impedances. If you are looking for maximum
power transfer into a load of fixed impedance, that will happen when
you minimize the output impedance. The impedance matching goal comes
in when you have a fixed output impedance and can only adjust the load
impedance. If you don't care about this here, fine, but in the future
remember that impedance matching only optimizes power transfer when
you are adjusting the load, not the source.

This amp is pretty simple to analyze. It is doing *two* things. It
is providing gain while at the same time it is minimizing the
capacitive load on the previous stage. The bootstrap is reducing the
capacitive loading while taking away some of the output swing. If you
really want to maximize the output swing you need to make this two
stages, one a bootstrap to minimize capacitive loading and another to
provide the gain with maximum headroom.

Is that at all clear? The nearly 1 gain on the emitter reduces the
headroom available for the output swing. Max headroom is approximately

24 - 0.2 * R11 / (R11 + R10) or with the current values of R10 and
R11, 20.5 volts.

But you won't want to push the transistor into saturation I expect, so
a little less.

If you split this into two stages or just don't bother with the
bootstrap (I'm not sure it is needed, but maybe, you'd need to analyze
the impact of the capacitive loading on the earlier stage), the max
headroom equation becomes,

24 - 0.2 or 23.8 volts

Which of these is what you want?

This was an amp I gave up on because it rolled off at 10MHz.
Well, so did the next amp, but as we have discussed, that is a
measurement error. With that in mind, I will go back to this amp
because it did give me very close to my needed output swing.
I may end up trying to figure out how to make it roll off
to a specified degree, because I found my diode detector has more output
as frequency increases.
As far as matching, nah, I found out my Diode detector has about a 6kO
input impedance, so my 500O collector resistor is plenty low, and I have
room to increase it, if I have a reason.

Coming into this late,but if the problem is to raise the output of a
signal generator, look at some of the 8 to 12 dollar amps coming in from
China on ebay. They use an IC and can provide about 20 db of gain.
Either low noise amps, or some that will give about 1/2 of a watt
output. If you want, you can buy just the ICs that are on them and
build your own. Just a couple of capacitors and a resistor and maybe a
RF choke.

He is designing a high input impedance amp with a very low input
capacitance. The front end is looking good and now he is looking at the
gain stage to drive a diode circuit to measure the signal amplitude.

--

Rick C

one transistor amplifier help

Jim Thompson

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amdx

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Phil Hobbs

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Phil Hobbs

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Phil Hobbs

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George Herold

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Jim Thompson

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Phil Hobbs

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rickman

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amdx

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rickman

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Ralph Mowery

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