# Please explain the gain function for given circuit

Discussion in 'The Projects Forum' started by pratapkollu, Jul 25, 2007.

1. ### pratapkollu Thread Starter Active Member

Dec 13, 2006
42
0

Dear Members,
For the circuit shown in figure above,
input - 0.5V (p-p),100 kHz sine wave
required output - 10 V (p-p), 100 kHz sine wave
op amp used is TL082
Result: I cannot get 10 V, I can only obtain 5 V (p-p) signal

I changed C7 = 103, R15 = 420 Ohms, Near J3 = 10K(feed back resistor)
Now I can get 12 V(p-p)signal, please explain what is the design equation thats is governing the gain here.

Thanks ans Regards
Pratap

2. ### nomurphy AAC Fanatic!

Aug 8, 2005
567
13
You need to be using +/-12V supplies or greater. Perhaps you should put a meter on these at the opamp and check?

What is the load impedance? Is it driving too much of a load?

The 1K input and 20K feedback should provide the required gain of Av = 20. You're schematic looks okay, provided the feedback isn't getting messed up because of the J3 arrangement.

The output will be 180deg out (inverted) from the input. Although the Unity Gain Bandwidth is only 3MHz, it shouldn't be that bad at 100KHz. Are you sure the input frequency is 100KHz? Try it at 10KHz.

Can the source adequately drive the 1K input? Try changing the input to 10K and the feedback to 200K (the 0.1uF input cap is fine, Fc = ~1.5KHz as shown). Do the actual componet values match those on the schematic?

3. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
$\frac{e_o}{e_i}=\left(\frac{-R_J_3}{R_1_5}\right)\left(\frac{s}{s+\frac{1}{R_1_5C_7}}\f{ }{ }\right)$

Here is the expression for the gain in the s-domain
where $e_i$
- represents the input signal
and $e_o$
- represents the output signal.

and $s = jw$

hgmjr