It's only a theoretical question; no particular circuit needed.

Some backstory: Suppose there are, say, two NMOS transistors with different W/L ratios in series. Further, suppose that both of them have their gate voltages equal, and their corresponding drain and source voltages. Neglect body and channel length modulation effects. Assume both are in saturation voltages.

The current in first NMOS: Id1= (W1/L1)* kn' *(Vgs - Vt)^2

The current in second NMOS: Id2= (W2/L2)* kn' * (Vgs - Vt)^2

All the terms in both the equations are equal except the W/L ratios.

My Question: When individually calculated, the drain currents of both the transistors are different. More specifically, the drain current is higher in the transistor with higher W/L ratio, and smaller in the other one. But, when both of them are in series, what would be the actual/total drain current in the circuit? Will the current be equal to Id1 or Id2 or something else?

I'm still posting a link of a random figure(ignore the PMOS circuit) for a better idea: http://ece-research.unm.edu/jimp/vlsi/misc/IMG00002.GIF


2 Answers 2


You would not be able to control both series source-drain voltages simultaneously. Try to draw out this circuit, with the controlling voltage sources in place. You would need to control both FET's to have the same Vgs and Vds. Once you attempt to draw it out you will realize that it cannot happen (falstad circuit simulator is nice for this, you can probe V and I, in real time).

If you attempt to do this (set both transistors Vgs and Vds simultaneously), you will not have a series circuit, since the voltage supplies would be in parallel. In this situation, the power supplies would source/sink whatever currents were needed to make the FET's Vds equal, but equal currents would not flow through each FET.

In series, there will be more voltage drop on whichever FET is weaker. This provides equal current to the FET which is stronger (with lesser Vsd drop). If one (or both) FET must go into the linear region for this to occur, it will.

Two transistors in series with different W/L ratios combine exactly like parallel resistors. This makes sense because the W/L ratios can be considered as a conductance.

As two numerical examples, consider two similar FET's with the same W/L = 2

If these FET's are placed in series, the 'equivalent' single FET W/L is ~2//2 = 1.

Now, consider two similar FET's with W/L of 1 and 2. The equivalent W/L, if these devices are placed in series and driven with the same gate signal, is ~1//2 = 0.66

So, adding a similarly driven strong FET in series with a weaker FET just makes the chain weaker overall.

Of course there is the body effect (if these are built on-die, which is implied since you can vary the W/L), which increases the threshold voltage of whichever FET's do not have their source at the most negative/positive voltage, for NMOS and PMOS respectively. This can be thought of as reducing the W/L ratio. This occurs if you have two or more of either type in series (2+ NMOS or 2+ PMOS). A CMOS inverter does not suffer the body effect since both NMOS and PMOS have their sources at the respective supplies.

  • \$\begingroup\$ how would you not have equal flux? that is plainly incorrect if you do not lose charge \$\endgroup\$
    – b degnan
    Aug 16, 2016 at 6:47
  • \$\begingroup\$ @bdegnan: You would lose charge, as I said in my answer. It would go into the voltage sources you are trying to set both Vds voltages with. You can check this very easily using Falstad circuit simulator ... \$\endgroup\$
    – jbord39
    Aug 16, 2016 at 14:34
  • \$\begingroup\$ For simulation, I agree, but for a real device, you would have conservation of flux of that set by the device that is constraining. Your simulator has created energy. \$\endgroup\$
    – b degnan
    Aug 16, 2016 at 15:33
  • \$\begingroup\$ @bdegnan: Then there would be voltage division and Vds would not be equal for two series FET's with different W/L (creating equal current) ... as i said in my answer ... Instead of not comprehending the situation, and asking many questions, why not read the answer? \$\endgroup\$
    – jbord39
    Aug 16, 2016 at 15:50
  • \$\begingroup\$ @bdegnan: but for a real device, you would have conservation of flux of that set by the device that is constraining. How exactly are you going to simultaneously set both Vds and Vgs of both FET's at the same time?? If you would draw this out you would realize that in a real device, or in a simulation, you cannot do this. When you do, you necessarily parallelize whichever node you are trying to set equal Vds with. Or, if the FET's are truly in series, the Vds would not be equal, but the current would. \$\endgroup\$
    – jbord39
    Aug 16, 2016 at 15:56

Due to conservation of flux in the system, the current will be set by the constraining device. This means that \$V_{ds}\$ and \$V_{gs}\$ for the controlling device will dominate. In the following image,

enter image description here

I show two devices in saturation for unknown conditions. The voltage at the drain of the first device \$V_{d1}\$ is equal to \$V_{S2}\$, and because flux is conserved, the current will be dominated by the device with the greatest "slope" between the source and the drain. Therefore, the minimum \$V_{gs}\$ sets the current through the system.

Whichever has the smaller W/L ratio will set the total current through the series stack for the same \$V_{gs}\$ in saturation.


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