Using series P-Channel MOSFETs for high-side switching

Question

I am looking at using two p-ch MOSFETs in series as part of a high-side load switching scheme, and do not understand the following circuit behaviour.

In simulation of the simplified schematic below, no current is driven into the load:

Edit: The MOSFETs being modelled are DMP2008UFG, model supplied by manufacturer

If neither Q1 or Q2 are conducting, Q2 has a floating source voltage and therefore undefined Vgs.

What I expected to happen:

• Q1 to turn on, as its Vgs > Vth.
• As Q1 is on, then Q2's source if no longer floating and is instead defined at near Vcc.
• As Q2's Vgs>Vcc, Q2 conducts
• Current flows into the load

Clearly I am misunderstanding the circuit. Why does no current flow into the load?

Many thanks in advance,

Gerry

Answer 1

Without more information, it's hard to say, but I suggest that you may be using ideal FETs in your simulation. Ideal FETs will have no leakage current, so the junction between the two will be floating, so the top FET, even though turned on, draws no current and its model doesn't work well.

I suggest two possibilities. The first is the most obvious - replace the FETs with actual, real part numbers. These will have leakage currents modeled which will allow more realistic operation.

The second is like unto the first. Place a fairly high resistance, say 1 to 10 megohms, between drain and source of each FET.

Answer 2

FETs with trivial models may work different to real ones.
Spurious, leakage and informal coupling can lead to enough drive to "get things going".

Try a large (10 M+) resistor from the floating node to ground or supply.

If the model has no capacitances and the input is "just there" capacitive effects that may make a difference in the 'real' world are missing.
In real world circuits of this general sort, very small leakage and bias currents and trace capacitive coupling as waveforms change or as DC is applied can cause switching of parts which may stay off in "ideal" situations.