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I need to calculate the time for C1 to discharge below 60 V using the following constant current sink if it is initially charged to 565 V (MOSFET's VGS(th): 4 V).

enter image description here

At first, I looked at the MOSFET as an ideal switch, but that gave me the wrong time, so what I wanted to do is to have the RDS(on) of the MOSFET, then proceed with Rtotal, and then be left with a simple RC circuit, but I found out it is more difficult then that, because RDS(on) isn't constant, and I'm not sure if in this configuration it is even a linear.

The final answer is 2.23 seconds.

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  • \$\begingroup\$ The approaches you took are not applicable to a "constant current sink", as given in your problem statement. Knowing that it's a constant current sink, do you have any ideas as to how you would calculate this constant current, given that you know Vth, the gate voltage, and the source resistance? Once you have that constant current, it should be more straightforward to estimate the capacitor discharge time. (Please edit your post to include related info, rather than responding in a comment) \$\endgroup\$
    – nanofarad
    Jan 22, 2023 at 21:29
  • \$\begingroup\$ no sir, this is how I received the question: "How long will it take to discharge C1 below 60V using the following constant current sink if it is initially charged at 565V? (MOSFET's VGSth: 4V)" \$\endgroup\$
    – David
    Jan 22, 2023 at 21:34
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    \$\begingroup\$ Why don't you run a simulation? That would seem to be the easiest way to answer your question. \$\endgroup\$ Jan 22, 2023 at 21:43
  • \$\begingroup\$ @ElliotAlderson I can't run simulation because I don't know what MOSFET model should I use \$\endgroup\$
    – David
    Jan 22, 2023 at 22:27
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    \$\begingroup\$ How are you going to do a calculation if you don't know what MOSFET you are using? Is this a homework question? \$\endgroup\$ Jan 22, 2023 at 23:31

3 Answers 3

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The \$\mathrm{R_{DS_{(on)}}}\$ of a MOSFET is a measure of how it behaves when \$\mathrm{V_{DS}}\$, the voltage from source to drain, is very small.

In your circuit, this isn't the case. In your circuit, the MOSFET is controlled, via the gate voltage and R1, such that it pulls a more or less constant drain current.

You need to thumb back through your lecture notes to see how a common-source MOSFET amplifier works. Calculate the voltage across R1 when there's 12V on the gate, work out what the current is through R1 at that voltage, and then use the characteristics of a MOSFET to determine how much of the current through R1 must go through the FET drain.

Then, knowing that current, work out how fast the capacitor will discharge.

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Generic approach:

$$C=\frac{Q}{U} \Leftrightarrow C\cdot U=Q \Leftrightarrow C\cdot\frac{du_c(t)}{dt} = i_c(t)$$

Then you can use $$u_c(t) = i_c(t) \cdot ( R_{fet} + R_1) \Leftrightarrow u_c(t) = ( R_{fet} + R_1) \cdot C\cdot\frac{du_c(t)}{dt}$$

And now you can solve to $$t_{u_c=60\ \mathrm{V}} = [u_c(t) = 60\ \mathrm{V}]^{-1}$$

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  • \$\begingroup\$ but how do I calculate the Rfet? \$\endgroup\$
    – David
    Jan 22, 2023 at 22:25
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The MOSFET operates as a source follower, due to R1 permitting source potential to rise above 0V. With +12V at the gate, the source will be \$V_{GS(TH)}=4V\$ below that at, +8V. This places a permanent 8V across R1, and drain current \$I\$ will be:

$$ \begin{aligned} I &= \frac{8\rm{\ V}}{47\rm{\ \Omega}} \\ \\ &= 170\rm{\ mA} \\ \\ \end{aligned} $$

This current also flows through capacitor C1, and by passive sign convention, is negative from the perspective of C1. Voltage \$V\$ across C1 will change with time according to:

$$ \begin{aligned} I &= C\frac{dV}{dt} \\ \\ \frac{dV}{dt} &= \frac{-170\rm{\ mA}}{1\rm{\ mF}} \\ \\ &= -170 \rm{\ Vs^{-1}} \end{aligned} $$

The required change in voltage is:

$$ \begin{aligned} \Delta V &= 60\rm{\ V} - 565\rm{\ V} \\ \\ &= -505\rm{\ V} \\ \\ \end{aligned} $$

The time \$T\$ that it will take for capacitor voltage to change by \$\Delta V\$:

$$ \begin{aligned} T &= \frac{\Delta V}{\left( \frac{dV}{dt} \right)} \\ \\ &= \frac{-505\rm{\ V}}{-170\rm{\ Vs^{-1}}} \\ \\ &= 2.97\rm{\ s} \end{aligned} $$

\$V_{GS(TH)}\$ is the point where the transistor only begins to conduct, meaning that in practice \$V_{GS}\$ will likely be hundreds of millivolts greater than 4V. This will result in lower source potential, and lower current in R1. The actual time taken is likely to be longer than the calculated 2.97s.

It certainly won't be 2.23s, as you suggest, unless \$V_{GS(TH)}\$ is significantly less than 4V.

The parameter \$R_{DS(ON)}\$ does not feature in the above calculations, because drain current is determined exclusively by gate potential \$V_G\$, the transistor's \$V_{GS(TH)}\$ and resistance \$R_1\$.

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