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I have the below circuit in which I am trying to perform simulations.

Simulation Tool used - Falstad Simulator

12V Zener Diode

Top MOSFET

Bottom MOSFET

Question 1 :

enter image description here

The Zener diode is 12V Rated.

I want to simulate a current of 12A through both the MOSFETs

So, for my 16V input voltage, I calculated my Resistance to be 1.33Ohms to get a current of 12A. But when both the MOSFETs are ON, my drain current through them is only 91mA. Can someone tell me what is the problem and why am I not getting a drain current of 12A? And I tried to change the Rds(on) MOSFET parameter in the tool, but not able to find and change it.

Next Question :

enter image description here

When I apply a negative voltage of -14V to the gate of the bottom MOSFET as shown below, I get a voltage at the Zener cathode as 2.195 and a voltage between the MOSFET nodes as 2.7V with 6.4mA of current?

Can someone tell me how this 2.195V and 2.7V is appearing? Just want to understand the circuit behaviour in these conditions. Please help.

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  • \$\begingroup\$ Can someone help me with the understanding of the circuit during these conditions? \$\endgroup\$
    – user220456
    Commented Jul 7, 2020 at 12:08
  • \$\begingroup\$ 1- wrong simulation model. 2 - The upper MOSFET (T1) is ON and the lower (T2) could also be ON or be on the edge of turning ON. Because the Zener diode is forward biased (0.6V), thus the voltage at the T1 source (T2 drain) is 5V - Vgs = 2.7V. And the voltage at the T2 gate is 2.7V -0.6V = 2.1V. So this voltage at the T2 gate could turn T2 ON. \$\endgroup\$
    – G36
    Commented Jul 7, 2020 at 15:22
  • \$\begingroup\$ @G36, thank you for the analysis. I am not able to get on why you did - "5V-Vgs = 2.7V" Also, I would like to understand what is wrong with the simulation model in (1). Please explain me how you arrived at 5V-Vgs = 2.7V and how is the 2.7V voltage arrived in the simulation too? Please help in simple terms. Have been stuck at this place for long time. \$\endgroup\$
    – user220456
    Commented Jul 7, 2020 at 15:51
  • \$\begingroup\$ Show us the link to the simulation. \$\endgroup\$
    – G36
    Commented Jul 7, 2020 at 16:01
  • \$\begingroup\$ tinyurl.com/y9r67vsk \$\endgroup\$
    – user220456
    Commented Jul 7, 2020 at 16:01

1 Answer 1

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1- From your simulation, I can see that you decided to use wrong MOSFET parameter.

\$V_T = 1.5V\$ and \$\beta = 20 \text{m}\:A/V^2 \$

And this, for example, will give us the drain current for \$V_{GS} = 4.5V\$ equal to:

$$I_D = \frac{\beta}{2}(V_{GS} - V_T)^2 = 90 \text{mA} $$

Therefore you need to increase the \$\beta\$ value.

We can calculate the needed \$\beta\$ value by using this approximate equation (MOSFET in Triode region)

$$\beta = \frac{1}{R_{DS(ON)}(V_{GS} - V_T)}$$

For example for \$V_T = 1.5V\$, \$V_{GS} = 10V\$,\$ R_{DS(ON)} = 30\text{m}\Omega \$

$$\beta = \frac{1}{R_{DS(ON)}(V_{GS} - V_T)} = \frac{1}{30\text{m}\Omega(10V - 1.5V)} \approx 4 \:A/V^2$$

Thus to fix your simulation you need to increase the \$\beta\$ value.

When I apply a negative voltage of -14V to the gate of the bottom MOSFET as shown below, I get a voltage at the Zener cathode as 2.195 and a voltage between the MOSFET nodes as 2.7V with 6.4mA of current?

Can someone tell me how this 2.195V and 2.7V is appearing? Just want to understand the circuit behaviour in these conditions. Please help.

Hmm...strange circuit.

The upper MOSFET will be ON because the gate of the upper MOSFET is at \$+5V\$. So the current will continue to flow through the forward-biased Zener diode and into the lower MOSFET gate resistor. And if the MOSFET threshold voltage is low (as is in your case) the lower MOSFET will also be ON.

We can apply KVL here and see what we get:

schematic

simulate this circuit – Schematic created using CircuitLab

$$V_G=V_{GS1}+V_D+V_{GS2}$$

And in your simulation, we have perfectly matched MOSFET's (Id1 = Id2 ignoring diode current, thus Vgs1 = Vgs2 )

$$V_{GS} = \frac{V_G - V_D}{2} = \frac{5V - 0.6V}{2} = 2.2V$$

And the voltage at the \$T_2\$ drain will be \$V_{D2} = V_{GS}+V_D = 2.8V\$

\$I_{D2} = \frac{\beta}{2}(V_{GS} - V_T)^2 = \frac{20\text{m}}{2}(2.2V - 1.5V)^2 = 4.9\text{mA}\$

\$I_{D1} = I_{D2} + I_{DZ} = 4.9\text{mA} + \frac{2.2V+14V}{10k\Omega} = 6.5\text{mA}\$

And I must say that I don't like this circuit at all.

Why the Zener diode is connected between the gate and the drain of a lower MOSFET?

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  • \$\begingroup\$ Thank you for the detailed answer. The Zener diode can be connected between the gate and drain, right? Like, if the Lower MOSFET is turned ON, the Zener diode is between the supply and the ground only. That's why \$\endgroup\$
    – user220456
    Commented Jul 9, 2020 at 2:52
  • \$\begingroup\$ But why you want to "protect" the gate and the drain terminals? \$\endgroup\$
    – G36
    Commented Jul 9, 2020 at 13:03
  • \$\begingroup\$ I want to protect them because, the absolute maximum ratings of gate to source voltage of the MOSFET is 20V. So, I am protecting it with a zener diode so that the maximum voltage of the gate to source voltage of the MOSFET does not exceed the Zener breakdown voltage which is not more than 12V \$\endgroup\$
    – user220456
    Commented Jul 10, 2020 at 2:52

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