Re: Op amp newbie
- From: kensmith@xxxxxxxxxxxxxxx (Ken Smith)
- Date: Wed, 29 Jun 2005 14:13:42 +0000 (UTC)
In article <1120051752.650007.244640@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Yvan <kayoux@xxxxxxxxxxx> wrote:
>Hi,
>
>I have a very simple question to ask. Say my required Gain-bandwidth
>product is 1kHz * 1 = 1k, what is the minimum gain-bandwidth product of
>the op amp in order to assume an infinite gain?
The actual gain of the circuit is:
G / (1 + GH)
Where:
G is the gain of the op-amp
H is the gain of the feedback section
If G is very large, the one in the denom doesn't matter much and can be
ignored. If yoy do a little math, you will see that the overal gain ends
up as 1/H in that case.
If G is not very large, we can work out what the G has to be to get the
accuaracy we need. For most op-amp, you assume that the G has a phase
shift of 90 degrees.
Take the case of the non-inverting unity gain amplifier working at 1kHz.
In this case, the H is simply 1. Lets assume that we want the overal gain
to be at least 0.99 for a 1% error.
Remember that the phase angle of result is not know until we calculate it.
We use | X | to say ignoring the phase.
0.99 = | G/(1+GH) | = | G/(1+G) |
0.99|(1+G)| = |G|
Remember that we said "assume G is at 90 degrees"
0.99 * sqrt(1^2 + G^2) = | G |
Square both sides:
0.99^2 * (1 + G^2) = G^2
0.99^2 + 0.99^2 * G^ = G^2
0.99^2 = G^2 - 0.99^2 * G^2
0.99^2 = G^ * (1 - 0.99^2)
0.99^2 / (1 - 0.99^2) = G^2
sqrt(0.99^2 / (1 - 0.99^2)) = G
G = 7.018
Given this we check the data sheets to see that the op-amp we have
selected has a gain of at least 7 at 1kHz.
>In the book "sensors and signal conditioning", they gave an example of
>an application that had a GBW of 30 Khz. It mentioned that a opamp GBW
>of 5 Mhz was large enough to assume infinite gain. 5 Mhz/30khz=167
>times. Is there a rule of thumb would help in figuring if a given
>opamp is suitable for my application in terms of GBW?
A good rule of thumb is that an op-amp for a filter must have a GBP
greater than:
GWP > 10 * Q^2 * Fc
The G/(1 +GH) still works here but for a filter, the H is complex value
instead of siply being one.
--
--
kensmith@xxxxxxxxx forging knowledge
.
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