# Formula for gain?

Discussion in 'Electronic Basics' started by little billy, Aug 5, 2004.

1. ### little billyGuest

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 ?

2. ### Geir KlemetsenGuest

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).

3. ### Active8Guest

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.

4. ### John PopelishGuest

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

5. ### Active8Guest

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

6. ### little billyGuest

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.

7. ### Active8Guest

Too bad thr spec doesn't list hfe :-0

8. ### colinGuest

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

Colin =^.^=

9. ### little billyGuest

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.

10. ### Active8Guest

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 a ton of distortion. Don't fergitcher output cap 11. ### colinGuest

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 =^.^=  