L4 2 2Noninverting ExampleChap 4

Derivation of the gain of a Non-inverting Op Amp Example

>> All right. Let's take a look at

an example of a non-inverting amplifier with some real numbers.

Let's let R_1, the feedback resistor be 10 kiloohms,

and R_2 be two kiloohms.

Let's calculate what the output voltage will be in terms of V_s in these ratios.

What we know from our previous work,

is that V_out for an non-inverting amplifier is equal

to V_s times the ratio of those two resistors,

R1 over R2, or 10 kiloohms divided by two kiloohms plus one.

Well, 10 divided by 2 is 5 plus 1 is 6.

So, V_out then is equal to six times V_s. In this case, then,

we would say that the gain of the amplifier G is equal

to six and know that gain is a unitless term.

It's equal to the ratio of V_out divided by V_s

or we say then the gain is equal to the ratio of V_out divided by V_s,

the volts cancel, and gain is again a unitless term.

Now, let's look at issues of

saturation by just assuming that we have two voltage sources.

The positive supply is going to be a plus 15 volts and let's make the negative supply,

a negative 15-volt source.

We now ask ourselves,

what is the range on V_s,

so that the amplifier will stay in it's linear range,

and the output voltage will be six times whatever V_s is.

Remember what the limitation is.

The output voltage can't be any greater than the supply voltages.

It can't be any greater than 15 volts and it can't be any less than negative 15 volts,

or to put it in terms of inequalities V_out has got to be greater

than negative 15 volts and less than positive 15 volts.

Now, substituting V_out for six V_s,

we have then six V_s must be greater than a negative 15 volts and less than

a positive 15 volts dividing both this side and this side by the six,

so that we get V_s by itself rather,

we get V_s must be greater than a negative 15 over six and it must be

less than a positive 15 over six or V_s that is constrained to

being greater than a negative 2.5 volts and less than a positive 2.5 volts.

Let's graph this by looking at V_out as a function of V_s. So,

V_s here, V_out along here.

Let's go ahead and mark our positive 15 volts there

and negative 15 volts

there and here's negative 2.5 volts.

There's positive 2.5 volts.

Just mark it up there,

I mark it down here.

So, for V_s less than 2.5 volts the amplifier is going to be

saturated at its negative supply voltage of negative 15 volts.

As V_s gets larger than negative 2.5 volts and less than positive 2.5 volts,

the output is going to equal six times V_s,

or a nice sort of linear, you get the idea, relationship in that linear region,

hence, it's called the linear region.

When V_s is greater than 2.5 volts, it saturates at the upper or

the positive supply voltage and while the graph may not be beautiful, you get the idea.

Saturation means that the output can't be any less than negative 15 volts,

can't be any greater than positive 15 volts and in between the constraints,

it's a nice linear relationship.

Where then the slope of this line is

15 divided by 2.5 which is six and that's just our gain term.