Why does the CMOS inverter operate in triode mode in static operation?

In summary, the inverter circuit has a static problem where it cannot decide between triode and saturation mode. If V_sd > V_ov, then V_o > V_t and the circuit is in triode mode. If V_sd < V_ov, then V_o < V_t and the circuit is in saturation mode.
  • #1
iiJDSii
4
0
Hello,

I have a question about how to analyze the CMOS inverter (this circuit: http://www.cs.umass.edu/~weems/CmpSci635A/Lecture2/image10.gif [Broken]). Just to clarify, the input voltage is connected to both gates, and the PMOS on top has its source connected to Vdd. The NMOS on the bottom has its source connected to GND. The drains of both FETs are connected, and it is at this node which we take our output to be.

Assume for now the transistors are matched such that V_tn = |V_tp| = V_t (equal threshold voltages), and k_n*(W/L)n = k_p*(W/L)p. Also say Vdd is 5v. In this example I'll be considering static operation (V_i is set to some value for all time).

So here's me trying to reason out what may happen. I know the answer, I just don't see why I don't exactly get there. Start with V_i = 0 (input voltage). This turns off the NMOS as V_gs = 0, no current flows in this transistor. Now we for the PMOS we know V_sg = Vdd > V_t so we can have either triode or saturation mode. What I can't properly reason through is why we must have triode mode in static operation. Comparing V_sd to V_ov I can see either case being possible.

V_sd = Vdd - V_o

V_ov = V_sg - V_t = Vdd - V_i - V_t = Vdd - V_t (since V_i = 0)

All I really know here is that Vdd is 5v, and V_t is, say, 1v. As I said before, I know the answer is supposed to be that the PMOS takes on triode mode. But I'm having trouble seeing what is wrong with saturation mode. Assuming V_sd > V_ov, this condition for saturation can still be valid. Working through the inequality, I get:

V_o < V_t (= 1v say.)

But clearly this does not give a logic high. Assuming triode mode (V_sd < V_ov) gives me the condition V_o > V_t, and my textbook furthermore replaces the PMOS with a small resistor with the value V_sd/I_d (which can be described purely in terms of device parameters, if V_sd is assumed so small that V_sd^2 can be neglected). Then the circuit just becomes Vdd connected to a resistor connected to the output node V_o. So then they conclude V_o = Vdd.

Any insights into why the circuit must operate in triode as opposed to saturation? Thanks.
 
Last edited by a moderator:
Engineering news on Phys.org
  • #2
As you stated the P is turned on but because it is in DC, and it has no load, then Vs=Vd=Vdd as Id=0A. Therefore Vov=Vsg-Vt=(5-0)-1 > Vsd=0, thus the P is in triode.
 
  • #3
thanks for the reply es1.

So because the PMOS is on but not connected to any load we must have V_o = Vdd. I guess for the case of a parasitic capacitance from V_o to ground this doesn't affect the DC analysis either, thus keeping it in triode mode.

I think I get it, thanks again.
 

1. What is a CMOS inverter?

A CMOS inverter is a type of logic gate that is commonly used in digital integrated circuits. It consists of two complementary metal-oxide-semiconductor (CMOS) transistors, one p-type and one n-type, connected in series. It is used to convert a logic input signal into its complementary output.

2. What is the purpose of DC analysis in a CMOS inverter?

DC analysis is used to determine the steady-state behavior of a CMOS inverter. It helps to calculate the DC voltage levels at the input and output nodes, as well as the current flowing through the transistors. This information is crucial in designing and optimizing the performance of a CMOS inverter circuit.

3. What are the key parameters in CMOS inverter DC analysis?

The key parameters in CMOS inverter DC analysis include the supply voltage, transistor characteristics such as threshold voltage and transconductance, load capacitance, and input signal voltage. These parameters affect the DC voltage levels and current flow in the circuit.

4. How is the DC transfer characteristic of a CMOS inverter represented?

The DC transfer characteristic of a CMOS inverter is typically represented by a plot of output voltage (Vout) versus input voltage (Vin) for different values of the supply voltage. It is a graphical representation of the inverter's behavior and is used to analyze its performance and determine its operating range.

5. What is the importance of DC noise margins in CMOS inverter design?

DC noise margins are used to determine the robustness of a CMOS inverter to noise and variations in its input voltage levels. They are defined as the difference between the minimum input voltage that can be recognized as a logic 0 and the maximum input voltage that can be recognized as a logic 1. A larger noise margin indicates a more reliable and stable CMOS inverter design.

Similar threads

  • Electrical Engineering
Replies
5
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
0
Views
413
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
Back
Top