P-channel MOSFET gating circuit that passes voltages above 17 V

Question

I'm designing a desktop PC motherboard with AMD Ryzen™ 9 7940HS.

We have an extension board similar to this one that performs negotiations with USB-PD charger and provides 20 V constant voltage to the platform. This board is designed by outsource company, and we don't want to interfere with it.

The 20 V is not guaranteed to be supplied after certain threshold, and we want to do the gating on the platform itself.

To be accurate, we want a circuit that passes voltage and current after the voltage reaches 17 V or more. Below 17 V, we don't want to deliver voltage to the system.

NOTE: The platform takes the 20 V and uses DC-DC to create 3.3 V and 5 V and other voltages for the CPU.

The main consideration of making sure only high voltage (17 V and above) is delivered to the system, is the fact that allowing low voltage at startup can allow high current to the CPU, and we don't want this risk. This idea is based on the fact the USB-PD Charger doesn't provide 20 V immediately like normal PSU, but it starts with lower voltages (5 V/3 A) and then goes up based on negotiations.

I found a gating circuit concept on the internet and tried to simulate it, and I wanted to know how much it fits my goal:

• V1 goes from 0 to 20 V (x-axis)
• We can see that around 16.8 V the FDC638P P-channel MOSFET is open.

Here is the LT-SPICE simulation:

I have few questions regarding this circuit:

1. What is the need of using NPN bipolar junction transistor and not regular MOSFET? Is it for power considerations?
2. Is 5 Ω load realistic to simulate for high current situations for CPU? (4 A and above)?

Note 1: I'm trying to make a simulation with a CPU load that has 35-54 W default TDP

Note 2: In this simulation the load seems to consume -4 A * 20 V =~ 80 W

3. Is FDC638P suitable for situations where current can reach 5 A and 6 A spikes?
4. Are there recommended chips that has all this circuit integrated? where I can configure it to pass voltage only when VIN >=17 V? (maybe more cost effective than my own circuit)

Update 1: Considering the devices suggested in the solution, I decided to use MIC2754 with a voltage divider that gives 2.93 V when VIN_ALW = 17 V:

I want to make sure that the transistor I picked: BSL307SP is suitable for my application. and that I picked the right place for the 17V_VOUT.

There is something that always makes me confused, which is regarding the following graph:

Assuming that threshold is met, and considering VOL(HV#)_MAX = 0.8 V, the P-channel MOSFET VGS = 0.8 V - 17 V = -16.2 V.

Looks like at such VGS at this graph passes really high current, and that contradicts the fact that the system can use up to 4-5 A.

Note 1: According to my previous simulation, seems like when the switching is done, the VDS on the transistor falls back to low voltage (about 320 mV). With such VDS, any VGS above threshold seems to be good for passing current which makes it a good switch. But what explains the VDS falling to such low value?

Note 2: this reminds me of what I learned back in university 5 years ago, that PMOS is good for transferring logic "1", and NMOS is good is good for transferring logic "0", where here we relate to analog voltages, and according to this we expect that the drain voltage be very close to 17 V, am I right?

Note 3: I still didn't understand the purpose of MCR708A SCR and the 1k and 22 Ω resistors, but I'm reading the datasheet of MIC2754 to hopefully understand

Update 2: I added a follow-up question regarding the solution that suggests using TL431.

Update 3: Regarding the solution that suggest using LM393 Compactor, I interested in replacing the M1 (IRF9530) since it has a big package (TO-220AB). I want an SMT P-MOSFET and also give it an option of cooling (for example I can take 5mm x 6mm as maximum size). I got stuck at what Power Dissipation parameter I need to choose regarding my application.

With the suggested solution that uses IRF9530, when VIN=20V, the VDS reaches about -485 mV at -3.903 A (Rds_On = 124.2 mOhm).

P = I^2 * R = (3.903)^2 * 124.2 mOhm = 1.89 W

I have been grabbing my head around the fact about how this P-MOSFET is providing 20V to a platform that has a CPU with TDP of 54 Watt, and the MOSFET itself dissipates only 1.89W, what I'm missing here?

Is the idea here that the P-Mosfet is just a gate with small resistance, and the USB-PD charger is the one responsible for providing the 54 Watt and the rest of power consumption by the platform?

Reminder: The platform takes the 20 V and uses DC-DC to create 3.3 V and 5 V and other voltages for the CPU.

Note 1: The application is an Industrial PC Desktop that should work at a 45 °C Ambient temperature, our thermal design makes sure that the enclosure doesn't go up beyond 70 °C, and we want ICs that still function perfectly with maximal junction temperature 100 °C. Note 2: Eventually I decided to place AOD4189 in my design.

Answer 1

1. The NPN bipolar transistor Q1 is being used as a comparator, and an inverter. When the input voltage rises to 17V, the base of this transistor is at:
   $$ \begin{aligned} V_B &= 17 \times \frac{R_1}{R_1+R_2} \\ \\ &= 17 \times \frac{680}{680+18000} \\ \\ &= 620\text{ mV} \end{aligned} $$
   This is the potential that will just begin to switch on most BJTs. Using a MOSFET in this role is ill advised, since threshold \$V_{GS}\$ is not as predictable as \$V_{BE}\$ for a BJT.

2. 5Ω across 20V is 4A, so it's OK to model a constant load of 4A, but it's borderline. It's not adequate, though, to emulate the rather large capacitance across the supply, due to decoupling caps, or the changing demands of a real life system. That's going to draw a lot more than 4A for a short while after power-on. To get a better idea, use a lower resistance, and include a few tens or hundreds of microfarads of capacitance across it.

3. Worst case on-resistance for this MOSFET is 72mΩ, and with an average of 4A flowing through it, it would dissipate:
   $$ \begin{aligned} P &= I^2 \times R \\ \\ &= 4^2 \times 0.072 \\ \\ &= 1.2\text{ W} \end{aligned} $$
   That's a lot for this package, and I think you'll need heatsinking, even though OK according to the datasheet, it can handle 1.6W continuously.
   The voltage it drops between drain and source at 4A would be: $$ \begin{aligned} V &= I \times R \\ \\ &= 4 \times 0.072 \\ \\ &= 0.29\text{ V} \end{aligned} $$
   I think that's OK too.
   The biggest concern is that this device is rated for an absolute maximum drain-source potential difference of 20V, which leaves you no margin for error.
   Also, the largest gate-source voltage \$V_{GS}\$ is 8V, meaning that the potential divider formed by R3 and R4 is inadequate. You would need to choose values to keep \$V_{GS}\$ 30% or so, of the supply potential.
   In short, I think there are better options, but you'll have to go hunting.

4. Sure, there are ICs that do this kind of thing, but you'll still need an external pass MOSFET. Sorry to say, I don't have any example in mind, but others here might have suggestions.
   Edit: Micrel Inc. has a bunch of such devices, and you can start your search at Digikey here.

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Regarding your Update

The SCR is used as a crowbar, to short-circuit the power supply and blow the fuse in the event of an overvoltage. That's not the kind of behaviour your question is about, and not something we are discussing here.

You're using the MOSFET to both enable the supply, and trigger the crowbar, which doesn't make any sense. You will not need the crowbar SCR or associated resistors in your own application.

Otherwise, you use the HV signal (which is active low) to switch on the MOSFET, which as you've implemented here seems correct to me, but you misunderstand how \$V_{OL(HV)}\$ will influence this behaviour. The value of 0.8V is the highest potential (above ground) that this output will be, when active (low).

Considering that when the input is +17V, any gate potential below about +15V will switch this transistor on, +0.8V at the gate is not a constraint, it's what you want, maybe even too close to zero. The maximum \$V_{GS}\$ for that MOSFET is 20V, and you'd probably even want to raise this minimum somewhat with a resistor potential divider, to keep a safe distance away from \$V_{GS}=20V\$.

As far as that graph is concerned, what it's showing you is how much voltage will be dropped across the drain-source channel given some current \$I_D\$ through it. We need this information to calculate the power that the transistor will dissipate for a certain channel current.

We have some flexibility with gate potential, and we can easily achieve \$V_{GS}=10V\$ when on. You can see from that graph that \$V_{DS}\$ is under 1V for drain currents well over 30A. The values in this graph are not to be interpreted as maximums in any way, except in the sense that you can use them to determine if your device is going to overheat.

Look at the plot of \$V_{GS}=10V\$. I see that voltage across the channel will be about 0.2V at 10A of current through it, for a power dissipation of \$P = I\times V = 10 \times 0.2 = 2\text{ W}\$. That's just OK for this device, but you'll need to research how to get heat away from it, because that's significant power for a tiny thing like that. Since you are operating at around 5A average, you can expect power dissipation of about 1W, when \$V_{GS}=10V\$, which is much less problematic.

This transistor seems to be a better fit for your application.

The MIC2754 datasheet says that the device operates from a supply of 100μA, so you should take care that the divider resistors for \$V_{IN}\$ can supply that current, while producing the correct voltage at \$V_{IN}\$.

It seems to me that you could use the RES output to disable output during an undervoltage condition (which would be the logical choice), or you could use the HV output to enable output during an overvoltage condition. However, since the RES output is low during undervoltage, it would need inverting before reaching the MOSFET gate. So I guess we are going with HV, unless you want another transistor in the mix.

That means you should be looking at the overvoltage thresholds in the datasheet, not undervoltage. Assuming you have the 3.08V threhsold device, you would need 17V input to provide 3.08V at \$V_{IN}\$ while supplying 100μA into that pin. I would use resistors much smaller than the ones you chose, to minimise \$V_{IN}\$ changes as the IC's current demands change. I'm too lazy to to the math, so here's a simulation of something that might work:

^{simulate this circuit – Schematic created using CircuitLab}

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I got hooked, here's an implementation using a comparator, the LM393:

^{simulate this circuit}

R1 and R2 provide a fraction of the input voltage to the comparator, which compares that to a more-or-less fixed 5.6V reference, provided by R3 and D1.

R4 gives some flexibility to that potential, so that it may be modulated somewhat by the comparator's output, via feedback resistor R5. This provides positive feedback for hysteresis, which I'll show you in the graph below.

D2 "adds" 5V or so to the comparator's output, before it's applied to the gate, which serves two purposes. Firstly it ensures that \$V_{GS}\$ never gets close to the maximum of 20V for the MOSFET. Secondly, the behaviour of the comparator is not guaranteed for \$V_{IN}<2V\$, and D2 ensures that the transistor cannot be switched on accidentally, before \$V_{IN}\$ reaches a level where we can be sure that the comparator will behave itself.

Here's a plot of \$V_{IN}\$, an input that sweeps up and down through 17V, and the resulting \$V_{OUT}\$:

The main features here are the very clean transistions of \$V_{OUT}\$, due to positive feedback, and the two different transistion thresholds, or hysteresis.

This hysteresis is probably necessary, because as the input rises through 17V, the sudden additional load can cause it to drop again, a recipe for oscillation. Change the ratio of R1 and R2 to set the midpoint of the switching thresholds, and raise/lower R5 to decrease/increase hysteresis "gap" respectively.

Answer 2

TL431 would be the most rudimentary part to use as a comparator with suitable precision, with widest availability. You can probably scrounge one from a dead switching power supply - a lot of them use that part :)

I decided to use MIC2754

That part is obsolete and is only available on the "secondary" market. The only source for it I'd trust is Rochester Electronics. But why bother? TL431 in its various packages has ample availability, is cheaper in quantity, and will probably remain a supported part for decades to come.

what explains the VDS falling to such low value?

When the MOSFET fully turns on, the channel (drain to source) acts like a low-value resistor, i.e. an "almost" short circuit. VDS is the voltage across this "short circuit", and it better be low when the MOSFET is on, or else it'll heat up rapidly and get destroyed.

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The circuit below should do what you intend.

TL431 becomes a "short" when the REF input goes above about 2.5 V. The R13, R14 divider provides the reference voltage to TL431. When input voltage goes above 17 V, the reference voltage rises above 2.5 V, and TL431 starts conducting hard. This conduction current causes a large voltage drop across R12, and discharges M1's gate in parallel with C1 via R12+R17. M1 turns on gradually, as its absolute gate voltage drops, keeping inrush current manageable.

At the same time, Q12 turns on as well, and shifts the threshold voltage down to 16 V. This is the hysteresis necessary to avoid potential instability during the turn-on and turn-off transient. Increasing R16 decreases the hysteresis voltage*. If the hysteresis is undesired (and it almost never is!), remove Q12, R15 and R16.

When the input voltage drops below 16 V (with hysteresis, or 17 V without), TL431's conduction current drops to idle (<0.5 mA), and C1 in parallel with M1's gate recharges via R12 and D1. This causes the load to be turned off about 10x as quickly as it was turned on.

Since CircuitLab doesn't have a built-in TL431 macro/model, I've added a low-level model of TL431 inside the dashed box. This model is only needed for simulation: in the physical circuit you'd use an actual TL431 part.

^{simulate this circuit – Schematic created using CircuitLab}

The voltage across the load and the absolute V_GS on the mosfet are plotted below vs. the input (USB) voltage.

The transient response on turn-on is plotted below:

*The threshold voltage under hysteresis can be calculated as follows:

$$\begin{aligned} R_p &= 1/(1/14.5k + 1/200k) = 13.5k \\ V_{t'} &= 2.5{\,\rm V} \cdot \frac{2.5k + 13.5k}{2.5k} = 16.0{\,\rm V} \\ \end{aligned}$$

Same in Octave/Matlab:

>> par = @(varargin) 1/sum(1./[varargin{:}]);
>> Vtt = 2.5*((par(14.5e3,200e3)+2.5e3)/2.5e3)
Vtt = 16.020