# Have I missed something?

Status
Not open for further replies.

#### Electronics4you

##### Member
Hi there,

I'm just reading the book "Microelectronic Circuits Fifth Edition" by Sedra/Smith about frequency response for amplifiers. An example shows how to find the DC-gain of an amplifier. The transfer function is (equation #1), which is on the form (equation #2).

The write then uses K as the DC-gain when s-->0 in the transfer function (equation 3). But K is not the DC-gain, but a constant. The DC-gain is the result from (equation 3), when K is known. Right? And how to find K then? One must be known...?

Thanks

#### eddiebtx

##### New Member
I thought that was a good book, im still using it for my circuits class.

i think what K is used for in that sense is a Load Resistance.
so, for instance a High Pass RL filter is H(s) = S / S + (R/L)
if there is a load resistance placed over the inductur than the transfer function changes so then so must H(S) = (Rl/R + Rl) S / S + (Rl/ R + Rl)( R/l)
lets simplify and make k = ( Rl / R + Rl)
and then finally.

H(s) = KS/ S + Wc : Wc = KR/L the cut off freq

Then K is not a constant so much as it just depends on the circuit.

Last edited:

#### Electronics4you

##### Member
That makes very good sense, but in order to complete the Sedra/Smith example 1.5 on page 37, it seems to me, that they are using K as the DC-gain.

The actual DC-gain is, of course, the voltage division between output impedance and load impedance.

In order to determine the constant K, the DC-output must be calculated first.

Status
Not open for further replies.

Replies
38
Views
5K
Replies
7
Views
1K
Replies
9
Views
1K
Replies
2
Views
975
Replies
13
Views
2K 