The MOSFET as an Amp and Switch

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Juniper Byrd
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1 11/1/004 The MSFET as an Amp and Switchdoc 1/14 The MSFET as an Amp and Switch Consider this simple MSFET circuit: Q: h, goody you re going to waste my time with another of these pointless academic problems Why can t you discuss a circuit that actually does something? v 50V R D =1K v K=075 ma/v A: Actually, this circuit is a fundamental electronic device! To see what this circuit does, we need to v = f v determine its transfer function Q: Transfer function! How can we determine the transfer function of a MSFET circuit!? A: Same as with junction diodes we determine the output v for each device mode, and then determine when (ie, for what values of v ) the device is in that mode!

v 50V R D =1K v K=075 ma/v A: Actually, this circuit is a fundamental electronic device!

2 11/1/004 The MSFET as an Amp and Switchdoc /14 First, note that regardless of the MSFET mode: v = v 00 = v GS i D 50V R D =1K and: v = v 00 = v DS K=075 ma/v + v From KVL, we can likewise conclude that: v = v = 50 i R DS D D v + v GS - v DS - Now let s ASSUME that the MSFET is in cutoff, thus ENFRCNG i D =0 Therefore: v = 50 i R D D = = 50V Now, we know that MSFET is in cutoff when: v = v < V = 10 GS t Thus, we conclude that: v = 50V when v < 10 V

v GS - v DS - Now let s ASSUME that the MSFET is in cutoff, thus ENFRCNG i D =0 Therefore: v = 50 i R D D = 50 0 1

3 11/1/004 The MSFET as an Amp and Switchdoc 3/14 Now, let s ASSUME that the MSFET is in saturation, thus ENFRCE: D i = K v V GS t ( V ) = K v And thus the output voltage is: v = 50 i R t = 075 v 10 D D = v 10 1 = v 10 Now, we know that MSFET is in saturation when: and when: v = v > V = 10 GS t v = v > v V = v 10 DS GS t This second inequality means: v > v v 10 > v 10 0> 075 v 10 + v Solving this quadratic, we find that the only consistent solution is: v 10 < 0 v < 30

MSFET is in saturation when: and when: v = v > V = 10 GS t v = v > v V = v 10 DS GS t This second inequality means: v > v 10

4 11/1/004 The MSFET as an Amp and Switchdoc 4/14 Meaning that the MSFET is in saturation when v > 10 and v < 30 Logically, this is same thing as saying the MSFET is in saturation when 10 < v < 30 Thus we conclude: v = v 10 when 10 < v < 30 V Finally, let s ASSUME that the MSFET is in triode mode, thus we ENFRCE: id = K ( vgs Vt ) vds v DS And thus the output voltage is: = 075 v 10 v v v = 50 i R D D = v 10 v v 1 = v 10 v v Rearranging this equation, we get the quadratic form: The solutions of which are: 075 v 15 v 05 v + 50 = 0 v = 15 v 05 ± 15 v

in triode mode, thus we ENFRCE: id = K ( vgs Vt ) vds v DS And thus the output voltage is: = 075 v 10 v v v = 50 i R D D = 50 075 v 10 v v 1 = 50

5 11/1/004 The MSFET as an Amp and Switchdoc 5/14 Note because of the ±, there are two possible solutions However, to be in triode region, the MSFET must not be in pinchoff, ie: v = v < v V = v 10 DS GS t This condition is satisfied with the smaller of the two solutions (ie, the solution with the minus sign!): v = 15 v v So, the above expression provides us with the output voltage if the MSFET is in triode mode The question remaining is thus when (ie, for what values of V ) is the MSFET in triode mode? We could do a lot more math to find this answer, but this answer is actually quite obvious! Recall that we have already determined that: a) The MSFET is in cutoff when v < 10V b) The MSFET is in saturation when 10 < v < 30V Since there are only three modes of a MSFET device, and since the transfer function must well be a function, we can conclude (correctly) that the MSFET will be in triode region when v is the value of the only region that is left--v > 3 0!

): v = 15 v 05 15 v 05 150 15 So, the above expression provides us with the output voltage if the MSFET is in triode mode The question remaining is thus when (ie, for what values of V ) is the MSFET

6 11/1/004 The MSFET as an Amp and Switchdoc 6/14 Thus we can conclude that: 15 v v v = when v > 3 0 V 15 We now have determined the complete, continuous transfer function of this circuit! 0 when v < 1 0 V v = ( v 1 0 ) when 1 0 < v < 3 0 V ( 15 v 05 ) ( 15 v 05 ) 150 when v > 3 0 V v NMS in cutoff NMS in saturation NMS in triode v

7 11/1/004 The MSFET as an Amp and Switchdoc 7/14 Q: thought you said this circuit did something t appears to be just as pointless as all the others! A: To see how this circuit is useful, consider what happens when the input voltage v is 0 V and 5V From the transfer function, we find that if v = 0, the output voltage will be v = 50 Likewise, if the input voltage is v = 50, the output voltage will be small, specifically v = 078V 50 v 078 v Let s summarize these results in a table:

A: To see how this circuit is useful, consider what happens when the input voltage v is 0 V and 5V From the transfer

8 11/1/004 The MSFET as an Amp and Switchdoc 8/14 v v Mode Cutoff 50 small Triode (078) Why, this device is not useless at all! t is clearly a: This circuit provides a simple example of one of the primary applications of MSFET devices digital circuit design We can use MSFETs to make digital devices such as logic gates (AND, R, NR, etc), flip-flops, and digital memory We typically find that, just like this circuit, when a MSFET digital circuit is in either of its two binary states (ie, 0 or 1 ), the MSFETs in the circuit will either be in cutoff (i D =0) or in triode (v DS small) modes Cutoff and Triode are the MSFET modes associated with digital circuits and applications!

devices such as logic gates (AND, R, NR, etc), flip-flops, and digital memory We typically find that, just like this circuit, when a MSFET digital circuit is in either of

9 11/1/004 The MSFET as an Amp and Switchdoc 9/14 Q: So, just what good is the MSFET Saturation Mode?? Sir, it appears to me that the Saturation region is just a useless MSFET mode between cutoff and triode! A: Actually, we will find the MSFET saturation mode to be extremely useful! To see why, take the derivative of the above circuit s transfer function (ie, ): v v

? Sir, it appears to me that the Saturation region is just a useless MSFET mode between

10 11/1/004 The MSFET as an Amp and Switchdoc 10/14 We note that in cutoff and triode: dv dv 0 while in the saturation mode: dv 1 dv >> Q: h goody The slope of the transfer function is large when the MSFET is in saturation Am supposed to be impressed by that?! How are these results even remotely important!? A: Since in cutoff and triode = 0, a small change in input voltage v will result in almost no change in output voltage v Contrast this with the saturation region, where >> 1 This means that a small change in input voltage v results in a large change in the output voltage v! To see how this is important, consider the case where the input signal has both a DC and a small-signal (AC) component: v ( t) = V + v ( t) i

? A: Since in cutoff and triode = 0, a small change in input voltage v will result in almost no change in output voltage v Contrast this with the saturation region, where >> 1 This

11 11/1/004 The MSFET as an Amp and Switchdoc 11/14 As a result, the output voltage likewise has both a DC and smallsignal component: v t = V + v t o Now, let s consider only the DC components We can select the DC input V such that the MSFET is placed in saturation The value V, along with the resulting DC output V, sets a DC bias point for this circuit By selecting the right value of V we could set this DC bias point to where the transfer function slope is the greatest: 50 v 50V DC Bias Point R D =1K V V V v K=075 ma/v V 30 Now, say we add a small-signal v i to this input DC voltage (ie, v ( t) = V + vi ( t) ) This small signal simply represents a small change in the input voltage from its average (ie, DC) value The result is of course as small change in the output voltage the small-signal output voltage v o (t)!

set this DC bias point to where the transfer function slope is the greatest: 50 v 50V DC Bias Point R D =1K V V V v K=075 ma/v 00 10 V 30 Now, say we add a small-signal v i to this input DC voltage

12 11/1/004 The MSFET as an Amp and Switchdoc 1/14 Now for the interesting part ( bet you were wondering when would get around to it)! The small change in the output voltage will have a much larger magnitude than the small change in the input! For example, if the input voltage changes by 1 mv (ie, v i =1mV), the output might change by, say, 5 mv (ie, v o = 5 mv) Q: Goodness! By how much would the output change in our example circuit? How can we determine the smallsignal output v o?? Determining how much the output voltage of our circuit will change when we change the input voltage by a small amount is very straightforward we simply take the derivative of the output voltage v with respect to input voltage v! By taking the derivative of v with respect to v (when the MSFET is in saturation, we find: ( ( 10) ) d v = = 150 v 1 0 for 1 0 < v < 3 0 This expression describes the slope of our circuit s transfer function (for 10 < v < 30) Note the slope with the largest magnitude occurs when v =30, providing a slope of -30 mv/mv

For example, if the input voltage changes by 1 mv (ie, v i =1mV), the output might change by, say, 5 mv (ie, v o = 5 mv) Q: Goodness! By how much would the output change in our example circuit?

13 11/1/004 The MSFET as an Amp and Switchdoc 13/14 Thus, if we DC bias this circuit with V = 30 V (resulting in V = 0 V), we find that the small signal output will be 3 times the small signal input! For example, say that the input to our circuit is: v = cos ωt V (ie, V = 3 0 V and v = 001 cos ωt ) We would find that the output voltage would approximately be: i v = cos ωt V (ie, V = 0 V and v = 0 03 cos ωt ) Note then that: o vo = v = 30 = 30 v = 003 cos ωt i v i n other words, the magnitude of the small-signal output has a magnitude three times larger than the input magnitude We say then that our signal provides small-signal gain our circuit is also a small-signal amplifier!

For example, say that the input to our circuit is: v = 30 + 001 cos ωt V (ie, V = 3 0 V and v = 001 cos ωt ) We would find that the output voltage would approximately be: i

14 11/1/004 The MSFET as an Amp and Switchdoc 14/14 see A small voltage change results in a big voltage change it s voltage gain! The MSFET saturation mode turns out to be excellent Even the simple circuit of this example is sufficient demonstrates to demonstrate the two primary applications of MSFET transistors--digital circuits and signal amplification Whereas the important MSFET regions for digital devices are triode and cutoff, MSFETs in amplifier circuits are typically biased into the saturation mode!

demonstrate the two primary applications of MSFET transistors--digital circuits and signal amplification Whereas the important

Lecture 30: Biasing MOSFET Amplifiers. MOSFET Current Mirrors.

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