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

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

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

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and R_2 be two kiloohms.

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

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What we know from our previous work,

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is that V_out for an non-inverting amplifier is equal

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to V_s times the ratio of those two resistors,

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

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

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

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

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to six and know that gain is a unitless term.

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

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

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

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Now, let's look at issues of

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saturation by just assuming that we have two voltage sources.

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

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a negative 15-volt source.

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We now ask ourselves,

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what is the range on V_s,

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

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

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Remember what the limitation is.

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

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

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

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than negative 15 volts and less than positive 15 volts.

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Now, substituting V_out for six V_s,

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

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

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so that we get V_s by itself rather,

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

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

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

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

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V_s here, V_out along here.

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

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and negative 15 volts

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there and here's negative 2.5 volts.

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There's positive 2.5 volts.

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Just mark it up there,

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I mark it down here.

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

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saturated at its negative supply voltage of negative 15 volts.

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

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the output is going to equal six times V_s,

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

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hence, it's called the linear region.

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

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

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

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

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it's a nice linear relationship.

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Where then the slope of this line is

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