# Switch-on time of an active p-channel MOSFET current limiting circuit

I have recently posted a question on the design of an inrush-current limiting circuit:

Current limiting circuit with P-Channel MOSFET

I have decided to post a new question on how to analytically calculate the actual MOSFET opening time when adding an external capacitor C2, which increases the Miller capacitance of the FET.

So the following circuit will limit the inrush current into capacitor Cin when hot plugging the power. The circuit works as expected as can be seen from the following plot: I am now trying to understand how exactly the inrush time - so the time from the FET starting to open when the gate reaches the threshold voltage until fully open - can be calculated.

Calculating the threshold voltage - which is about Vth=-1V - is easy at it is determined by the time constant of the C1*R1||R2 circuit. It however becomes tricky once the threshold voltage is reached. C2 provides negative feedback to the gate. In other words it increases the Miller capacitance and causes the FET to open more slowly. I figured this can be best modelled as a integrating OP Amp:

Initially, the output voltage Vout is 0V and C2 is charged to Vin, which is 48V. C2 discharges over R2 to Gnd and once the Threshold Voltage Vth is it, the output voltage rises slowly. However, how do I bring R1 and C1 into play to match the timing of the voltage rise?

How can the actual inrush time be calculated?

EDIT 1

Based on jonk's answer below, the switch-on time can be estimated by the discharging current of $$\C_2\$$ and $$\C_2\$$ over the voltage divider network $$\R_1\$$ and $$\R_2\$$. The answer has the details in it and I will not cover them in this edit.

I also built a breadboard to compare the simulation results with the real-word circuit. I have to note that the initially selected MOSFET SI7465DP was not available, so I selected the slightly larger IRF9Z34PB P-Channel FET. I also measured all ceramic capacitors that I used for the breadboard and put the exact values into the simulation model. Firstly, here are the updated simulation results: The linear region is about 2.3 ms: The following photo shows the breadboard circuit: And here are the measurement results with a linear region of 3.4 ms: The threshold voltage is at around 2.6 V: As can be seen, in the real circuit the linear time is about 3.4 ms, which is more than one ms longer than the simulation. What is causing this mismatch?

• The circuit example could be better redrawn. But for a later time. First question I have: "How would you work out the delay until the transistor 'opens' if you didn't have C2 there?" I'd like to see the quantitative details of your calculation, including references to the model parameters for your FET. You should be able to make a specific prediction. There seems little point to more nuance, if that's not already in hand.
– jonk
Aug 28, 2021 at 22:44
• The voltage divider of R1 and R2 is set to deliver a gate voltage of about 8.3V. When first applying the source voltage of 48V, C1 will charge up with a time constant of C1*R1||R2, which is 20.7ms. According to the data sheet of Si7465DP, the minimum threshold votlage is -1V. In the Spice model it opens at about -1.4V, which is set after about 3.8ms. Aug 29, 2021 at 6:48
• Excellent. My spice model has it at -2.4 V. But your calculations match my own thoughts, given your Spice model's VTO. It's nice to see we are on the same page, then. You might include that thinking process in your question. (A few hours ago I used $-20.7\:\text{ms}\cdot\ln\left(1-\frac{2.4\:\text{V}}{8.27\:\text{V}}\right)\approx 7.1\:\text{ms}$. But with your modifications I also get your result.)
– jonk
Aug 29, 2021 at 6:56
• Thanks. I will include it in an edit. However, the tricky part is now to get the actual opening time of the FET. In the OP Amp model my first thought was to set the gain to match the timing but I couldn't figure out how. Aug 29, 2021 at 7:33

I think we both see things similarly, without the use of $$\C_2\$$. So that's a great base upon which I can attempt an answer.

To begin, though, I'd like to simplify your schematic. It will help a lot and it's my hope you will agree with the simplifications I'll make here: simulate this circuit – Schematic created using CircuitLab

(My apologies for renaming $$\C_{in}\$$ to $$\C_3\$$. I hope you won't mind.)

## Cutoff Phase

Initially, $$\C_1\$$ charges via $$\R_{\text{TH}}\$$, driving $$\V_{_\text{G}}\$$ towards $$\V_{\text{TH}}\$$ (which must have been designed through the values of the resistor divider, $$\R_1\$$ and $$\R_2\$$, be lower than the threshold voltage of the FET, for obvious reasons.) We both agree that the RC time constant formed by $$\\tau=C_1\cdot R_{\text{TH}}\$$, combined with the FET's threshold voltage, is key in determining how long this initial phase takes and that something new happens, once the threshold voltage is reached.

So, up to this point, I think we both see very similarly. $$\C_3\$$ holds the drain voltage close to ground. And, of course, holds that shared end of $$\C_2\$$ close to ground, as well. $$\C_2\$$ is relatively small, compared to $$\C_1\$$. So as $$\C_1\$$ charges via $$\R_{\text{TH}}\$$ a small current is stolen from $$\C_1\$$'s charging in order to allow $$\C_2\$$ to follow the charging process of $$\C_1\$$. This stolen current will slightly alter the time that would otherwise be computed using $$\\tau\$$. But not by a lot. And at the end of this process, before something new happens, $$\C_2\$$ will charged to approximately $$\V_{_\text{CC}}\$$, less the threshold voltage of the FET.

Now, we come to that something new part.

## Ohmic Phase

Once the FET moves out of its cutoff region, it moves into its linear/Ohmic region. Without $$\C_2\$$, this would be a rather shorter period of time. But with $$\C_2\$$ in place, this period is dragged out.

As the voltage on $$\C_3\$$ rises, it drives the other end (gate-side) of $$\C_2\$$ also upward. This acts in a way to drive the FET gate back towards cutoff, again. Now, it won't succeed in completely doing so, as there is a path for $$\C_2\$$ to continue discharging. But it will be held right at the brink.

So for a period of time, $$\C_2\$$ slowly discharges and slowly allows the voltage on $$\C_3\$$ to increase along a controlled ramp. Eventually, $$\C_3\$$ reaches the output voltage and $$\C_2\$$ no longer constrains the FET in its sub-threshold region. At this point, the whole process of this phase ends, as $$\C_2\$$ has been charged to exactly the threshold voltage of the FET (which has the opposite sign to when this process started) and the circuit enters the next phase, with the gate of the FET slowly declining towards $$\V_{\text{TH}}\$$ along the same curve it otherwise would have followed had $$\C_2\$$ not been present.

All $$\C_2\$$ does here is create a plateau where the FET operates sub-threshold.

Let's look at a simplified schematic for this phase: simulate this circuit

Note that during this phase, the gate voltage remains essentially fixed. It does not change. Again, the reason is simply that any changes at the output voltage caused by a trickle from the FET is immediately applied as a negative feedback to the gate voltage forcing the FET to move back towards cutoff. So the gate voltage during this phase can't really change much. It's stuck.

This leaves the primary discharging current for $$\C_2\$$ as $$\\mid \:I_{discharge}\mid \:=\:\frac{V_{_\text{CC}}-\mid V_{_\text{TO}}\mid-V_{_\text{TH}}}{R_{_\text{TH}}}\$$. All of this current goes to $$\C_2\$$ and not $$\C_1\$$, since the voltage across $$\C_1\$$ isn't changing.

In your case, this current is $$\\mid \:I_{discharge}\mid \:=\:\frac{48\:\text{V}-1.4\:\text{V}-39.7\:\text{V}}{20.7\:\text{k}\Omega}\approx 330\:\mu\text{A}\$$.

You need to completely discharge and then recharge $$\C_2\$$ over the entire $$\48\:\text{V}\$$ range. So you can compute the time required as $$\t=\frac{10\:\text{nF}\:\cdot\: 48\:\text{V}}{330\:\mu\text{A}}\approx 1.5\:\text{ms}\$$.

(Once this phase ends and $$\C_2\$$'s discharge/recharge process is over, the gate voltage can once again continue towards $$\V_{_\text{TH}}\$$ per the prior $$\\tau\$$ curve until it reaches $$\V_{_\text{TH}}\$$.)

You can easily adjust this time period by altering the value of $$\C_2\$$.

Also note that, as this is a relatively fixed $$\C_2\$$ discharge/recharge current, the voltage across $$\C_2\$$ is a linear ramp and therefore the output voltage will follow that same linear ramp.

As you can see, an overly complicated view isn't required and a simplified approach can get credibly close to actual behavior. The time period doesn't (much) depend on FET characteristics, as $$\C_2\$$'s current in this phase is set by the circuit and much less so by the FET.

## Afterwards

With the output now settled, $$\C_2\$$ now provides feedback to the gate. If the output starts to decline, $$\C_2\$$ will pull down on the gate encouraging more current from the FET to compensate. If the output starts to rise, $$\C_2\$$ will drive upward on the gate discouraging current from the FET. So $$\C_2\$$ impacts two important behaviors.

## Summary

So, set $$\\tau_{_1}=R_{_\text{TH}}\cdot C_1\$$ and $$\\tau_{_2}=R_{_\text{TH}}\cdot C_2\$$ and then you can compute the time for the first phase as:

$$t_{_\text{cutoff}}= -\tau_{_1}\cdot\ln\left(1-\frac{\mid V_{_\text{TO}}\mid}{V_{_\text{CC}}-V_{_\text{TH}}}\right)$$

and for the following phase as:

$$t_{_\text{linear}}=\tau_{_2}\cdot \frac{V_{_\text{CC}} }{V_{_\text{CC}}-\mid V_{_\text{TO}}\mid-V_{_\text{TH}}}$$

In your example, you'd find that $$\t_{_\text{cutoff}}\approx 3.8\:\text{ms}\$$ and $$\t_{_\text{linear}}\approx 1.4\:\text{ms}\$$.

(Note that none of this includes your series resistor, $$\R_5\$$. But at $$\1\:\text{k}\Omega\$$ and $$\330\:\mu\text{A}\$$ the voltage drop is $$\330\:\text{mV}\$$ and I don't see how it significantly impacts the current or the timing. You should be able to vary that value somewhat without much impact.)

The most important aspect of this circuit isn't so much the timing of the cutoff phase (which is highly sensitive to specific FET parameter values and isn't so important) but is instead much more about the Ohmic region, which is where the output follows a controllable linear rise time. That's the main point of a circuit like this. And the prediction of that time should be fairly accurate using this simplified view.

Bear in mind that for these calculations, $$\C_2\$$ must be much larger than the FET capacitance. Otherwise, the error increases.

There is an additional time that continues $$\\tau_{_1}\$$'s decay, after $$\t_{_\text{linear}}\$$, but I'm not covering those final details here.

Anyway, I hope that helps.

• First I would like express my appreciation for the time you took to help educate some stranger on the internet. The topic is now much clearer for me! I agree with all expect that the $I_{discharge}$ is only contributed by $C_2$. During the Ohmic Phase, $C_1$ also discharges but only by a little amount due to its size compared to $C_1$ causing $V_{TO}$ further drop by $0.5 V$. In fact $I_{discharge} = I_{C1}+I_{C2} \approx 330 \mu A$ with $I_{C1} \approx I_{C2}$ causing $t_{linear} \approx 3ms$, which is closer to the actual linear region as can be seen in the above graph. Aug 30, 2021 at 7:27
• @F.Heisenberg I tried to keep the analysis as simple as possible. And there is a small voltage change during this Ohmic change so $C_1$ isn't completely isolated. But as an approximation, it's close enough. There is about $100\:\text{mV}$ change during this period, though. So there is something there to account for. Good point. My main goal is in showing that you can roughly approximate the most important aspect of the circuit, which is the linear ramp of the output voltage. The volts per second ramp is key for design. And this can be easily approximated.
– jonk
Aug 30, 2021 at 7:32
• @F.Heisenberg In any case, the basic idea is there. $C_2$ holds the FET in its Ohmic region for a while and you can reasonably compute the time period so long as certain values are far enough away from each other. It's not complex, then. You can adjust $C_1$ to set a delay -- it won't be precise, but it doesn't need to be. The ramp period, on the other hand, is important as many devices specify their power-up ramp rate -- especially in the case of FPGAs, for example. And here, the value of $C_2$ and the Thevenin of the divider determine what's important. It's a nice concept.
– jonk
Aug 30, 2021 at 7:40
• And what about the "knee" region after $V_{TO}$ is reached and before the ohmic phase begins. It seems that this timing is dependent on the FET's characteristics. Aug 30, 2021 at 7:43
• @F.Heisenberg Yeah. I think so. In fact, I think the timing of the combined period of the first two phases is mostly determined by the FET and that the Ohmic region subtracts from this result. Or, put another way, the sum of the two periods is computed by using the first equation with a slightly adjusted VTO value and that the Ohmic region ends at this time, but begins earlier by exactly the timing computed for the Ohmic period. It's different in this way than what I wrote. But the main point remains -- the ramp period, which is important, is easily understandable and computable.
– jonk
Aug 30, 2021 at 7:53