Formula for gain?

[little billy]

How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

 

[Geir Klemetsen]

"little billy" <littlebilly@btinternet.com> skrev i melding
news:24fcdb4a.0408050623.10732a3a@posting.google.com...

How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

The LM117 and the R2 together, is a current source provides the same current
whatever voltage on the "output".
Idealy, the gain is infinite, because you shouldn't be able to make an
input current so precise so that the outpot voltage stays stable (even for a
short time) between maximum and minimum. (not considering rise/fall time).

 

[Active8]

On Fri, 6 Aug 2004 03:20:42 +0200, Geir Klemetsen wrote:

"little billy" <littlebilly@btinternet.com> skrev i melding
news:24fcdb4a.0408050623.10732a3a@posting.google.com...
How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

The LM117 and the R2 together, is a current source provides the same current
whatever voltage on the "output".
Idealy, the gain is infinite, because you shouldn't be able to make an
input current so precise so that the outpot voltage stays stable (even for a
short time) between maximum and minimum. (not considering rise/fall time).

I was working on a proper derivation but ended up with an extra term
and couldn't divide through by Vin to get Vout/Vin.

The above approach has errors and beta, which I wouldn't use isn't
specified for that transistor, though there is a g_m curve.

For a current source load, gain is limied by the early voltage. I
think the proper approach is to select a current for the 117 and use

Ic = g_m*Vin - Vin being Vbe and

I_R2 = Ic + I_L

to get the load current and thus the voltage. It's an approximation
IMO, but It's quick.
--
Best Regards,
Mike

 

[John Popelish]

Active8 wrote:

On Fri, 6 Aug 2004 03:20:42 +0200, Geir Klemetsen wrote:

"little billy" <littlebilly@btinternet.com> skrev i melding
news:24fcdb4a.0408050623.10732a3a@posting.google.com...
How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

The LM117 and the R2 together, is a current source provides the same current
whatever voltage on the "output".
Idealy, the gain is infinite, because you shouldn't be able to make an
input current so precise so that the outpot voltage stays stable (even for a
short time) between maximum and minimum. (not considering rise/fall time).

I was working on a proper derivation but ended up with an extra term
and couldn't divide through by Vin to get Vout/Vin.

The above approach has errors and beta, which I wouldn't use isn't
specified for that transistor, though there is a g_m curve.

For a current source load, gain is limied by the early voltage. I
think the proper approach is to select a current for the 117 and use

Ic = g_m*Vin - Vin being Vbe and

I_R2 = Ic + I_L

to get the load current and thus the voltage. It's an approximation
IMO, but It's quick.
--
Best Regards,
Mike

Keep in mind that the 'transistor' is an integrated circuit, LM195.

--
John Popelish

 

[Active8]

On Thu, 05 Aug 2004 22:07:41 -0400, John Popelish wrote:

Active8 wrote:

On Fri, 6 Aug 2004 03:20:42 +0200, Geir Klemetsen wrote:

"little billy" <littlebilly@btinternet.com> skrev i melding
news:24fcdb4a.0408050623.10732a3a@posting.google.com...
How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

The LM117 and the R2 together, is a current source provides the same current
whatever voltage on the "output".
Idealy, the gain is infinite, because you shouldn't be able to make an
input current so precise so that the outpot voltage stays stable (even for a
short time) between maximum and minimum. (not considering rise/fall time).

I was working on a proper derivation but ended up with an extra term
and couldn't divide through by Vin to get Vout/Vin.

The above approach has errors and beta, which I wouldn't use isn't
specified for that transistor, though there is a g_m curve.

For a current source load, gain is limied by the early voltage. I
think the proper approach is to select a current for the 117 and use

Ic = g_m*Vin - Vin being Vbe and

I_R2 = Ic + I_L

to get the load current and thus the voltage. It's an approximation
IMO, but It's quick.
--
Best Regards,
Mike

Keep in mind that the 'transistor' is an integrated circuit, LM195.

Right. THat's why the transconductance isn't 40Ie and oops, I said
Vbe which isn't right.

I hit the dead end when I got to:

1.25/R2 = g_m*Vin + Vout/R_L
--
Best Regards,
Mike

 

[little billy]

Active8 <reply2group@ndbbm.net> wrote in message news:<xkeokuu9j7o9.dlg@news.individual.net>...

On Thu, 05 Aug 2004 22:07:41 -0400, John Popelish wrote:

Active8 wrote:

On Fri, 6 Aug 2004 03:20:42 +0200, Geir Klemetsen wrote:

"little billy" <littlebilly@btinternet.com> skrev i melding
news:24fcdb4a.0408050623.10732a3a@posting.google.com...
How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

The LM117 and the R2 together, is a current source provides the same current
whatever voltage on the "output".
Idealy, the gain is infinite, because you shouldn't be able to make an
input current so precise so that the outpot voltage stays stable (even for a
short time) between maximum and minimum. (not considering rise/fall time).

I was working on a proper derivation but ended up with an extra term
and couldn't divide through by Vin to get Vout/Vin.

The above approach has errors and beta, which I wouldn't use isn't
specified for that transistor, though there is a g_m curve.

For a current source load, gain is limied by the early voltage. I
think the proper approach is to select a current for the 117 and use

Ic = g_m*Vin - Vin being Vbe and

I_R2 = Ic + I_L

to get the load current and thus the voltage. It's an approximation
IMO, but It's quick.
--
Best Regards,
Mike

Keep in mind that the 'transistor' is an integrated circuit, LM195.

Right. THat's why the transconductance isn't 40Ie and oops, I said
Vbe which isn't right.

I hit the dead end when I got to:

1.25/R2 = g_m*Vin + Vout/R_L

We can't isolate Vo/Vi so there is no gain because
gain is defined as Vo/Vi. Not all circuits have
gain, this one doesn't, which is very strange
considering the title.

 

[Active8]

On Fri, 06 Aug 2004 12:26:15 GMT, colin wrote:

"little billy" <littlebilly@btinternet.com> wrote in message
news:24fcdb4a.0408060141.4eeb75ce@posting.google.com...
Active8 <reply2group@ndbbm.net> wrote in message
news:<xkeokuu9j7o9.dlg@news.individual.net>...
On Thu, 05 Aug 2004 22:07:41 -0400, John Popelish wrote:

Active8 wrote:

On Fri, 6 Aug 2004 03:20:42 +0200, Geir Klemetsen wrote:

"little billy" <littlebilly@btinternet.com> skrev i melding
news:24fcdb4a.0408050623.10732a3a@posting.google.com...
How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

The LM117 and the R2 together, is a current source provides the same
current
whatever voltage on the "output".
Idealy, the gain is infinite, because you shouldn't be able to make
an
input current so precise so that the outpot voltage stays stable
(even for a
short time) between maximum and minimum. (not considering rise/fall
time).

I was working on a proper derivation but ended up with an extra term
and couldn't divide through by Vin to get Vout/Vin.

The above approach has errors and beta, which I wouldn't use isn't
specified for that transistor, though there is a g_m curve.

For a current source load, gain is limied by the early voltage. I
think the proper approach is to select a current for the 117 and use

Ic = g_m*Vin - Vin being Vbe and

I_R2 = Ic + I_L

to get the load current and thus the voltage. It's an approximation
IMO, but It's quick.
--
Best Regards,
Mike

Keep in mind that the 'transistor' is an integrated circuit, LM195.

Right. THat's why the transconductance isn't 40Ie and oops, I said
Vbe which isn't right.

I hit the dead end when I got to:

1.25/R2 = g_m*Vin + Vout/R_L

We can't isolate Vo/Vi so there is no gain because
gain is defined as Vo/Vi. Not all circuits have
gain, this one doesn't, which is very strange
considering the title.

Vout = Vin/Rin*hfe*RL (asuming rin >>rb)

Colin =^.^=

Too bad thr spec doesn't list hfe :-0
--
Best Regards,
Mike

 

[colin]

"little billy" <littlebilly@btinternet.com> wrote in message
news:24fcdb4a.0408060141.4eeb75ce@posting.google.com...

Active8 <reply2group@ndbbm.net> wrote in message
news:<xkeokuu9j7o9.dlg@news.individual.net>...
On Thu, 05 Aug 2004 22:07:41 -0400, John Popelish wrote:

Active8 wrote:

On Fri, 6 Aug 2004 03:20:42 +0200, Geir Klemetsen wrote:

"little billy" <littlebilly@btinternet.com> skrev i melding
news:24fcdb4a.0408050623.10732a3a@posting.google.com...
How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

The LM117 and the R2 together, is a current source provides the same
current
whatever voltage on the "output".
Idealy, the gain is infinite, because you shouldn't be able to make
an
input current so precise so that the outpot voltage stays stable
(even for a
short time) between maximum and minimum. (not considering rise/fall
time).

I was working on a proper derivation but ended up with an extra term
and couldn't divide through by Vin to get Vout/Vin.

The above approach has errors and beta, which I wouldn't use isn't
specified for that transistor, though there is a g_m curve.

For a current source load, gain is limied by the early voltage. I
think the proper approach is to select a current for the 117 and use

Ic = g_m*Vin - Vin being Vbe and

I_R2 = Ic + I_L

to get the load current and thus the voltage. It's an approximation
IMO, but It's quick.
--
Best Regards,
Mike

Keep in mind that the 'transistor' is an integrated circuit, LM195.

Right. THat's why the transconductance isn't 40Ie and oops, I said
Vbe which isn't right.

I hit the dead end when I got to:

1.25/R2 = g_m*Vin + Vout/R_L

We can't isolate Vo/Vi so there is no gain because
gain is defined as Vo/Vi. Not all circuits have
gain, this one doesn't, which is very strange
considering the title.

Vout = Vin/Rin*hfe*RL (asuming rin >>rb)

Colin =^.^=

 

[little billy]

littlebilly@btinternet.com (little billy) wrote in message news:<24fcdb4a.0408060141.4eeb75ce@posting.google.com>...

Active8 <reply2group@ndbbm.net> wrote in message news:<xkeokuu9j7o9.dlg@news.individual.net>...
On Thu, 05 Aug 2004 22:07:41 -0400, John Popelish wrote:

Active8 wrote:

On Fri, 6 Aug 2004 03:20:42 +0200, Geir Klemetsen wrote:

"little billy" <littlebilly@btinternet.com> skrev i melding
news:24fcdb4a.0408050623.10732a3a@posting.google.com...
How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

The LM117 and the R2 together, is a current source provides the same current
whatever voltage on the "output".
Idealy, the gain is infinite, because you shouldn't be able to make an
input current so precise so that the outpot voltage stays stable (even for a
short time) between maximum and minimum. (not considering rise/fall time).

I was working on a proper derivation but ended up with an extra term
and couldn't divide through by Vin to get Vout/Vin.

The above approach has errors and beta, which I wouldn't use isn't
specified for that transistor, though there is a g_m curve.

For a current source load, gain is limied by the early voltage. I
think the proper approach is to select a current for the 117 and use

Ic = g_m*Vin - Vin being Vbe and

I_R2 = Ic + I_L

to get the load current and thus the voltage. It's an approximation
IMO, but It's quick.
--
Best Regards,
Mike

Keep in mind that the 'transistor' is an integrated circuit, LM195.

Right. THat's why the transconductance isn't 40Ie and oops, I said
Vbe which isn't right.

I hit the dead end when I got to:

1.25/R2 = g_m*Vin + Vout/R_L

We can't isolate Vo/Vi so there is no gain because
gain is defined as Vo/Vi. Not all circuits have
gain, this one doesn't, which is very strange
considering the title.

o.k. guys, thanks for the response.

Gain is gm * RL and from the graph we see that gm
is about 3, so 3RL, so for an 8R speaker we get
Av=24. For a 64R speaker, about 200.

l.b.

 

[Active8]

On 6 Aug 2004 06:44:22 -0700, little billy wrote:

littlebilly@btinternet.com (little billy) wrote in message news:<24fcdb4a.0408060141.4eeb75ce@posting.google.com>...
Active8 <reply2group@ndbbm.net> wrote in message news:<xkeokuu9j7o9.dlg@news.individual.net>...
On Thu, 05 Aug 2004 22:07:41 -0400, John Popelish wrote:

Active8 wrote:

On Fri, 6 Aug 2004 03:20:42 +0200, Geir Klemetsen wrote:

"little billy" <littlebilly@btinternet.com> skrev i melding
news:24fcdb4a.0408050623.10732a3a@posting.google.com...
How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

The LM117 and the R2 together, is a current source provides the same current
whatever voltage on the "output".
Idealy, the gain is infinite, because you shouldn't be able to make an
input current so precise so that the outpot voltage stays stable (even for a
short time) between maximum and minimum. (not considering rise/fall time).

I was working on a proper derivation but ended up with an extra term
and couldn't divide through by Vin to get Vout/Vin.

The above approach has errors and beta, which I wouldn't use isn't
specified for that transistor, though there is a g_m curve.

For a current source load, gain is limied by the early voltage. I
think the proper approach is to select a current for the 117 and use

Ic = g_m*Vin - Vin being Vbe and

I_R2 = Ic + I_L

to get the load current and thus the voltage. It's an approximation
IMO, but It's quick.
--
Best Regards,
Mike

Keep in mind that the 'transistor' is an integrated circuit, LM195.

Right. THat's why the transconductance isn't 40Ie and oops, I said
Vbe which isn't right.

I hit the dead end when I got to:

1.25/R2 = g_m*Vin + Vout/R_L

We can't isolate Vo/Vi so there is no gain because
gain is defined as Vo/Vi. Not all circuits have
gain, this one doesn't, which is very strange
considering the title.

o.k. guys, thanks for the response.

Gain is gm * RL

Can we make this simplification with a current source load? i.e.
treat Rc as infinity (theoretical R of current source) in parallel
with R_L? Even with R2 going to Vcc?

[sotto voce] I guess.

and from the graph we see that gm
is about 3, so 3RL, so for an 8R speaker we get
Av=24. For a 64R speaker, about 200.

And a ton of distortion. Don't fergitcher output cap
--
Best Regards,
Mike

 

[colin]

"Active8" <reply2group@ndbbm.net> wrote in message
news:3mmd8aaipj2n.dlg@news.individual.net...

On Fri, 06 Aug 2004 12:26:15 GMT, colin wrote:

"little billy" <littlebilly@btinternet.com> wrote in message
news:24fcdb4a.0408060141.4eeb75ce@posting.google.com...
Active8 <reply2group@ndbbm.net> wrote in message
news:<xkeokuu9j7o9.dlg@news.individual.net>...
On Thu, 05 Aug 2004 22:07:41 -0400, John Popelish wrote:

Active8 wrote:

On Fri, 6 Aug 2004 03:20:42 +0200, Geir Klemetsen wrote:

"little billy" <littlebilly@btinternet.com> skrev i melding
news:24fcdb4a.0408050623.10732a3a@posting.google.com...
How do I find the gain for the High Gain Amplifier
in the manual of the lm317 page 16

http://www.national.com/ds.cgi/LM/LM117.pdf

Assuming we have a load RL from the output to ground
can we say the following:

IL = Vo/RL where Vo = output current, IL = load current

Denote the constant current of the regulatror as C

C = Ic + IL , where Ic = collector current

Ic = Ib * beta hence Ic = Vi/R1 * beta where Vi = input voltage

all of this gives us

(Vo/RL) + (Vi/R1)*beta = C

what is the gain ?

The LM117 and the R2 together, is a current source provides the
same
current
whatever voltage on the "output".
Idealy, the gain is infinite, because you shouldn't be able to
make
an
input current so precise so that the outpot voltage stays stable
(even for a
short time) between maximum and minimum. (not considering
rise/fall
time).

I was working on a proper derivation but ended up with an extra
term
and couldn't divide through by Vin to get Vout/Vin.

The above approach has errors and beta, which I wouldn't use isn't
specified for that transistor, though there is a g_m curve.

For a current source load, gain is limied by the early voltage. I
think the proper approach is to select a current for the 117 and
use

Ic = g_m*Vin - Vin being Vbe and

I_R2 = Ic + I_L

to get the load current and thus the voltage. It's an approximation
IMO, but It's quick.
--
Best Regards,
Mike

Keep in mind that the 'transistor' is an integrated circuit, LM195.

Right. THat's why the transconductance isn't 40Ie and oops, I said
Vbe which isn't right.

I hit the dead end when I got to:

1.25/R2 = g_m*Vin + Vout/R_L

We can't isolate Vo/Vi so there is no gain because
gain is defined as Vo/Vi. Not all circuits have
gain, this one doesn't, which is very strange
considering the title.

Vout = Vin/Rin*hfe*RL (asuming rin >>rb)

Colin =^.^=

Too bad thr spec doesn't list hfe :-0
--
Best Regards,
Mike

true, it doesnt, it does show a graph of the relationship between ib and ic
tho and its not like an ordinary transistor at all, like i asumed it was.
from the graph delta hfe goes to infinity and then goes negative. hardly
much like a bipolar transistor at all its not realy a transistor anyway, its
even got a pnp input stage.

i think the circuit would be highly unstable. im not sure i cld think of any
use for it that actualy needed such high gain but could cope with the
instability.

although thinking about it again as bias curent is so low the 10k input
resistor hardly has any efect. so my original asumption i stated doesnt
hold.

however it does show colector curent versus colector voltage for various
base voltages..
eg, at IC = 0.825 Amp, vb = 0.825 @ vc 12.5 and vb =0.75 @33

therfore gain at that Ic seems to be 280, asuming infinite colector
reistance.

Colin =^.^=