Why is there a resistor in a MOSFET-based inverter?

[zenterix]

Homework Statement

I will show below the S circuit model of the MOSFET inverter. It contains a resistor in it. I'd like to know why that resistor is needed.

Relevant Equations

It seems to me that the inverter would work the same without the resistor.

Here is a MOSFET-based inverter combinational gate

which we represent with the symbol

The MOSFET in the first diagram is itself an abstraction for another circuit (the S model of the MOSFET), which is as follows

The MOSFET is a nonlinear circuit but the two sub-circuits above are linear and represent the ON and OFF states of the MOSFET.

Now, if we put this S model into the first diagram above, then we get the "S circuit model of the MOSFET inverter"

My question is about the presence of the resistor.

The book I am reading says

When $v_{IN}$ is high, the MOSFET is in the ON state (assuming that the high voltage level is above the threshold $V_T$), thereby pulling the output voltage to a low value. In contrast, when the input is low, the MOSFET is off, and the output is raised to a high value by $R_L$. Here we see the purpose of the load resistor $R_L$ - it provides a logical 1 output when the MOSFET is off. Futhermore, $R_L$ is chosen to be large so that the current is limited when the MOSFET is on.

In the case of the MOSFET being OFF, wouldn't $v_{OUT}$ be $V_S$?

Is this not higher than when the resistor is present?

 

[phyzguy]

Without the resistor, Vout would be shorted to Vs and would never change, so it would not function as an inverter.

 

[berkeman]

Homework Statement: I will show below the S circuit model of the MOSFET inverter. It contains a resistor in it. I'd like to know why that resistor is needed.
Relevant Equations: It seems to me that the inverter would work the same without the resistor.

Without the pullup resistor, what would pull $v_{OUT}$ high when the pulldown transistor is off?

 

[zenterix]

Without the pullup resistor, what would pull $v_{OUT}$ high when the pulldown transistor is off?

When the pulldown transistor is off (ie, $v_{IN}=\text{Low}$, then without the resistor we would have $v_{OUT}=V_S$.

Now, when the pulldown transistor is on and we have a short circuit, then without the resistor (ie, just the wire without any resistance), then there would be enormous current flow and now that I think of it I am not even sure what the voltage is at any point in the short circuit. I mean one end of the short is on the positive end of a power source and the other is at the negative (ground) end of the power source.

If we have a resistor, then it means that $v_{OUT}$ is at the same voltage as ground.

 

[zenterix]

Without the resistor, Vout would be shorted to Vs and would never change, so it would not function as an inverter.

What is $v_{OUT}$ when the transistor is ON?

 

[berkeman]

When the pulldown transistor is off (ie, $v_{IN}=\text{Low}$, then without the resistor we would have $v_{OUT}=V_S$.

Why? How is the output pulled to the supply voltage if there is no pullup resistor? By magic?

Now, when the pulldown transistor is on and we have a short circuit, then without the resistor (ie, just the wire without any resistance), then there would be enormous current flow

What? Why? The next logic stage is high input impedance, so where is this enormous current coming from?

[EDIT -- Oh, I see now, you are saying that if you replace the resistor with just a wire, that will have high current when the transistor is on. That is correct, but nobody would do that anyway.]

Is this your first time studying logic circuits?

 

[phyzguy]

What is $v_{OUT}$ when the transistor is ON?

If you just have a wire instead of the resistor, the impedance of the wire is much, much less than the impedance of the ON transistor. So even when the transistor is ON, Vout would still be Vs. As you said, there would be a large current flowing through the ON transistor, but even this large current wouldn't pull Vout significantly lower than Vs.

 

[zenterix]

Why? How is the output pulled to the supply voltage if there is no pullup resistor? By magic?What? Why? The next logic stage is high input impedance, so where is this enormous current coming from?

[EDIT -- Oh, I see now, you are saying that if you replace the resistor with just a wire, that will have high current when the transistor is on. That is correct, but nobody would do that anyway.]

Is this your first time studying logic circuits?

It is my first time studying electric circuits and combinational gates, yes.

I can see what "pulldown" means in "pulldown transistor": when the transistor is in the ON state it "pulls" the voltage $v_{OUT}$ down to the same voltage as ground by creating a short circuit between the $v_{OUT}$ node and the ground node.

I can't understand what it means for the resistor to "pull up" the voltage $v_{OUT}$. When the transistor is in the OFF state, we have an open circuit (infinite resistance), and so as far as I can see, $v_{OUT}$ is separated from ground by this infinite resistance.

It seems to me that the main function of the resistor is actually in the transistor ON state to make sure there isn't a huge current flowing from drain to source. In this case, in purely semantic terms, I can see that when we go to the OFF state since the voltage $v_{OUT}$ goes up then it somehow "pulled up". I just can't see why we say that this is due to the resistor.

 

[berkeman]

I can't understand what it means for the resistor to "pull up" the voltage $v_{OUT}$. When the transistor is in the OFF state, we have an open circuit (infinite resistance), and so as far as I can see, $v_{OUT}$ is separated from ground by this infinite resistance.

Yes, so the pullup resistor can pull the output voltage to V_s.

It seems to me that the main function of the resistor is actually in the transistor ON state to make sure there isn't a huge current flowing from drain to source.

Nope.

The alternate logic gate circuit uses an active pullup transistor. Can you say what the function of each of the transistors in this simple TTL inverter is?

[/QUOTE]

 

[Tom.G]

Well, the Vout must control something, otherwise it would not be needed. Usually that something is the input to another logic circuit (gate, transistor).

To make things easier to think about, assume that Vout is connected to a 3V flashlight bulb, with the other end of the light bulb connected to Ground.

If Vs is a rather common 5V supply, when the transistor is Off, the flashlight bulb will turn On (from the current thru R_L).

When the transistor is On, Vout is connected to Ground and there is no voltage to light the bulb.

Of course the equivalent thing happens if Vout is connected to Vin of another transistor.

I can see that when we go to the OFF state since the voltage goes up then it somehow "pulled up". I just can't see why we say that this is due to the resistor.

You already indicated (correctly) that if the resistor is replaced by a wire, the transistor would be destroyed by the large current when On. If the Drain is not connected to anything, then there can not be any voltage on Vout when the transistor is Off

So a compromise is found by using a resistor, R_L, that allows a smaller Drain current to flow when the transistor is On.

Hope this helps!

Cheers,
Tom