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I'm reading an basic electronics textbook, the chapter on MOSFETs, and it has started with a simple model of the MOSFET as a switch (the "S model"). It shows a circuit like this:

enter image description here

And says: Here we see the purpose of the load resistor R -- it provides a logical 1 output when the MOSFET is off. Huh? Without that resistor (ie, replacing it with wire) the value at Vout would still be 1/high when the switch was off, because there is an open circuit between drain and source. So why does it need the resistor?

(This is on p.292 of Foundations of Analog and Digital Electronic Circuits, by Agarwal and Lang. I'm trying to follow the MIT open courseware, first course.)

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  • \$\begingroup\$ If it makes more sense for you, you can always put the resistor between S and GND. \$\endgroup\$ – Dave Sep 19 '12 at 20:59
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    \$\begingroup\$ Putting the resistor to ground will only work if you use a PMOS transistor instead of an NMOS. \$\endgroup\$ – Joe Hass Sep 19 '12 at 21:21
  • \$\begingroup\$ You are making a major incorrect conceptual assumption. Removing R does NOT connect Vs to Vout - it separates them with an infinite resistance (= an open circuit.) Without a source of voltage to drive Vout its state is undefined.In an ideal circuit there is no reason for it to be high or low or any other value when the FET is off. In a real wolrd circuit with no other connections it will tend to remain at about its last "on" value for an undefined period as the capacitance will retain its last voltage value until current flow from wherever changes it. \$\endgroup\$ – Russell McMahon Sep 20 '12 at 1:57
  • \$\begingroup\$ Ah, I'll think about editing the question. When I said remove R I meant replacing it with wire. \$\endgroup\$ – Rob N Sep 20 '12 at 2:11
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    \$\begingroup\$ Just a side comment: if you're taking the open courseware 6.002 course, you really should be aware that they started it as a full-on, free online course on edXonline.org September 5th. That'd be a whole lot better than OCW, in my opinion (I took it the first time around). \$\endgroup\$ – exscape Sep 20 '12 at 12:06
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First with the FET switched on. The on-resistance of a FET can be very low, even as low as a few mΩs for high current ones, but let's take an average FET with a 1 Ω on-resistance, and a 10 kΩ pull-up resistor. Let's say \$V_S\$ = 5 V. The FET pulls the output level almost to ground; it forms a resistor divider with R, so that

\$ V_{OUT} = \dfrac{R_{DS(ON)}}{R + R_{DS(ON)}} V_S = \dfrac{1 \Omega}{10000 \Omega + 1 \Omega} 5 V = 0.5 mV \$

So with the FET on we have as good as zero.

Next with the FET off. Then there's no current through R, and since the voltage across R = R \$\times\$ current (Ohm's Law) the voltage is also zero. If \$V_S\$ is 5 V, and there's no voltage difference across R, then \$V_{OUT}\$ also must be 5 V.

Just like the FET isn't a perfect switch when closed it isn't a perfect switch when open either. There's a small leakage current, say up to 1 µA. That will cause (again due to Ohm) a voltage drop of 1 µA \$\times\$ 10 kΩ = 10 mV across the resistor, and the output will be 4.99 V instead of 5 V. The leakage current is the reason why you shouldn't choose R too high. If R would be 1 MΩ then the voltage drop would be 1 V and that may be too much.


So it works with the resistor. What if we omit it? With the FET on the output woill be drawn to ground, but with the FET off the output would be floating if our FET was a perfect switch, so it would be undefined. With the leakage current it might still pull the output low, if the input impedance of the load was very high. So the resistor is needed to define the level when the FET is off.

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  • \$\begingroup\$ Okay, I followed you until the last paragraph. If I remove the resistor, with the FET off, Vout and Vs are the same node, so I thought that I could then say that Vout = Vs = 5V. But you're saying: no, Vout is floating. I'm a beginner and must be missing something basic. Anyway, thanks, I'll try to think about these answers for a while... \$\endgroup\$ – Rob N Sep 19 '12 at 23:50
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    \$\begingroup\$ @RobN - No, by omitting I didn't mean replace it with a wire, I meant no connection between Vs and Vout at all. You can't just connect Vout to Vs, because if you would switch on the FET you would short-circuit the power supply: Vs would be directly connected to ground. That's how the resistor is a solution: it will pass Vs to Vout, but you can still pull Vout to ground without shorting the power supply. \$\endgroup\$ – stevenvh Sep 20 '12 at 6:47
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The resistor is needed; otherwise you'd have no way of getting a logic 0/1; also, in this case, when the MOSFET is on it shorts Vs to ground.

E.g. in the example below, the blue line is the input voltage (Vin) and the green line is the current that the voltage source (Vs) is giving. As you can see, the current reaches extreme values.

Example curves

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But if you turned the switch on, what would your state be? You need the resistor to keep the Vs from shorting to ground and to give you a defined 0V when the transistor is on thus the inverting property.

EDIT: go look up 'ohm's law'. Looking at the image below, for the "LOW" case, where I=Vs/R, change R to 0ohms and tell me what your current is. Or for an empirical experiment, go short your car battery terminals together or stick your finger in a socket.

enter image description here

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  • \$\begingroup\$ That makes sense, but then the author said it backwards. Do you agree it is better said: "the purpose of the resistor is to provide the low voltage (and not too much current) when the MOSFET is on"? Then you'd think it would be called "pull-down" not "pull-up". \$\endgroup\$ – Rob N Sep 19 '12 at 21:05
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    \$\begingroup\$ @RobN It's called a pull-up because it ties the signal to the upper voltage (Vcc, Vdd, Vs). A pull-down resistor ties the signal to ground (0 V) | Pull-up resistor \$\endgroup\$ – m.Alin Sep 19 '12 at 21:10
  • \$\begingroup\$ @RobN No, the author got it just right. The MOSFET provides a (switched) path from the output to ground. The resistor provides a path from the output to Vs. If the resistor value is too small the MOSFET will have hard time pulling the output all the way to ground, but if the resistor is too large then it will take a long time for it to pull the output high. Without the resistor (an open circuit to Vs) the output never goes high. \$\endgroup\$ – Joe Hass Sep 19 '12 at 21:24
  • \$\begingroup\$ -1 I feel that this answer could have been much better. It assumes that the reader is on the same page as him/her. For example, the state depends greatly of on the resistor is removed entirely or the resistor is replaced with wire. \$\endgroup\$ – user3624 Sep 19 '12 at 22:16
  • \$\begingroup\$ If the Resistor was ABSENT but you were still connected high, you'd still get a high when the FET is off. The problems would come when the FET is on, obviously. Seems semantic to me, though. \$\endgroup\$ – Scott Seidman Sep 24 '12 at 23:18
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If the resistor is replaced with a short circuit then turning the transistor on tries to short out the power supply. Either the output will stay high and the transistor will get very hot or the power supply will collapse. Both are bad.

If the resistor is replaced with an open circuit then when the transistor is off there is nothing left connected to the output other than some stray capacitance and so most likely it will stay low.

Which brings us to the root of the question. You have interpreted "without that resistor" as "replace the resistor with a short circuit". The author clearly intended it to be interpreted as "replace the resistor with an open circuit".

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