How MOSFET leaks

[TinselKoala]

That's right, very low. Most of the inductive effects associated with mosfets come from the circuit they are connected to, like the lead wires and circuit board traces, or in the case of scoping, from the probe's ground clip wire. A few inches of wire, or improperly routed PCB traces, will have more inductance than is found in the mosfet itself. However, the Gate-Source capacitance is significant.

[ayeaye]

Ok, i guess you are right, the inductances in the mosfet are too small to cause any considerable effect.

Then i think the only thing that can cause the mosfet leaking, is the gate-drain capacitance, Cgd. Cgd is the same as Crss in the datasheet http://www.irf.com/product-info/datasheets/data/irf530n.pdf, which means that with the gate-source voltage Vds 25V, it is 19pF. But by the figure 5 there, when Vds is 0, this capacitance is 400pF.

Consider a negative current, that is movement of negative particles. And consider that positive direction of the current in the circuit is positive direction for the mosfet, that is drain to source. It is not because the current necessarily goes through the mosfet at all, it is just about direction, to determine the direction. The diagram below shows the current at the gate (red) and the current in the circuit (blue), the 600 units there is 600 mA.

Now what i think happens. When Cgd charges after the start of the pulse, there is current in the negative direction. When Cgd discharges after the end of the pulse, there is current in the positive direction. The directions are such because when the mosfet has a high resistance, one may think that the current only goes from the ground to Cgd through the circuit, or from Cgd to the ground through the circuit.

But why the curves of charging and discharging in the circuit (blue) are not typical exponential. There are two things that happen. First the resistance of the mosfet changes, the lesser it is, the less is resistance between Gds and the ground, and less current goes through the circuit. Second, the capacitance of Gds changes.

What concerns the current curves, the current rises when the pulse starts and there is voltage on the gate. This does not yet open the mosfet though, mosfet opens when the input capacitances are charged so that the voltage on them rises above a certain level. When that happens, the Miller plateau starts. At the end of the Miller plateau, the mosfet is fully open. One can see on the currents diagram, that the current in the circuit goes to zero at that point, because the resistance of the mosfet is close to zero then, and Gds can by all practical means then be considred to be connected to the ground.

I also tried to illustrate it for you on the first figure below to show how the mosfet leaks, hope that helps.

Does that make sense?

[ayeaye]

Ok, no one said that it doesn't make sense.

Now see this.

I did the simulation as before with ngspice, except the circuit was that in the diagram below. I did transient analysis tran 0.1ns 150us and then plotted two different parts of the switching cycle, at the beginning of the pulse in the second figure below, and after the end of the pulse in the third figure below. The diagrams are as before, red is the current at the gate, blue is the current in the circuit, and units are amperes.

As you see, the current in the circuit goes at the beginning of the pulse to 30 mA, but this lasts only 45ns. As you see, at 25ns the mosfet opens, and the Miller plateau starts. The current in the circuit goes to zero at the end of the Miller plateau, when the resistance of the mosfet is almost zero, and Cgd is almost connected to the ground.

The current has almost a triangle shape. So we can calculate root mean square for the triangular wave, a / sqrt(3), the current by that is 17.32mA. The energy leaked into the circuit is I * I * R * t, and this is 1.3nJ.

As you see, at the end of the pulse the current in the circuit goes to 0.5mA, and stays there for 1us. We can consider that the current is almost constant during that time. The energy is I * I * R * t, and this is 2.5nJ.

So the total leaked energy was 3.8nJ. This is the energy necessary for charging a capacitance 0.3pF to 5V. In spite that Cgd, even when the mosfet is fully open, is 19pF, because the time of the mosfet's opening is very short, much less energy can go to that capacitance while the mosfet is still partly closed.

As you see, the energy leaking from the mosfet is very small, so small that sometimes it may even be completey dismissed when calculating a power gain. Power gain inside the circuit that is, not considering the power necessary for switching the mosfet, calculating this in-circuit power gain is important for theoretical reasons.

The leaked energy depends on the impedance of the circuit though, as with a lower impedance the current is higher, and thus more energy can go to Cgd during the time when mosfet is still open. But the resistance of the tested circuit was quite low, only 100 ohms. So if you add a 100 ohms resistor to your circuit, then likely no more energy can leak to the circuit.

The netlist was as follows.

* Spice netlister for gnetlist
.SUBCKT irf530n 1 2 3
**************************************
*      Model Generated by MODPEX     *
*Copyright(c) Symmetry Design Systems*
*         All Rights Reserved        *
*    UNPUBLISHED LICENSED SOFTWARE   *
*   Contains Proprietary Information *
*      Which is The Property of      *
*     SYMMETRY OR ITS LICENSORS      *
*Commercial Use or Resale Restricted *
*   by Symmetry License Agreement    *
**************************************
* Model generated on Sep 21, 01
* MODEL FORMAT: SPICE3
* Symmetry POWER MOS Model (Version 1.0)
* External Node Designations
* Node 1 -> Drain
* Node 2 -> Gate
* Node 3 -> Source
M1 9 7 8 8 MM L=100u W=100u
.MODEL MM NMOS LEVEL=1 IS=1e-32
+VTO=3.79209 LAMBDA=3.64034 KP=81.097
+CGSO=8.9e-06 CGDO=1e-11
RS 8 3 0.056205
D1 3 1 MD
.MODEL MD D IS=9.06112e-12 RS=0.00982324 N=1.13042 BV=100
+IBV=0.00025 EG=1.2 XTI=3.54513 TT=0.0001
+CJO=7.4e-10 VJ=1.52632 M=0.693198 FC=0.5
RDS 3 1 1e+06
RD 9 1 0.0219755
RG 2 7 4.4648
D2 4 5 MD1
* Default values used in MD1:
*   RS=0 EG=1.11 XTI=3.0 TT=0
*   BV=infinite IBV=1mA
.MODEL MD1 D IS=1e-32 N=50
+CJO=9.44672e-10 VJ=0.5 M=0.9 FC=1e-08
D3 0 5 MD2
* Default values used in MD2:
*   EG=1.11 XTI=3.0 TT=0 CJO=0
*   BV=infinite IBV=1mA
.MODEL MD2 D IS=1e-10 N=0.400249 RS=3e-06
RL 5 10 1
FI2 7 9 VFI2 -1
VFI2 4 0 0
EV16 10 0 9 7 1
CAP 11 10 1.29816e-09
FI1 7 9 VFI1 -1
VFI1 11 6 0
RCAP 6 10 1
D4 0 6 MD3
* Default values used in MD3:
*   EG=1.11 XTI=3.0 TT=0 CJO=0
*   RS=0 BV=infinite IBV=1mA
.MODEL MD3 D IS=1e-10 N=0.400249
.ENDS irf530n
V1 n0 0 dc 0 pulse 0 5 0 25n 25n 100u 1m
X1 n2 n1 0 irf530n
R2 0 n2 100
R1 n0 n1 1
.END

So i think the myth of the great mosfet leaking is now destroyed.

Thank you for your attention, or maybe interest, hope it was useful for someone for some purpose. Sorry if not. Have a good time, and do experiments.

[ayeaye]

In the previous post, the leaking energy after the pulse should have been 25pJ. So the total leaked energy during one switching cycle with the circuit resistance 100 ohms, was 1.3nJ.

I also did the same simulation to find out the leaking energy with the circuit resistance 55k. Then the leaked energy at the beginning of the pulse was 2pJ, and the leaked energy after the end of the pulse was 700pJ. So the total leaked energy during one switching cycle with the circuit resistance 55k, was 0.7nJ.

When there are capacitors and coils in the circuit, then what likely matters is the impedance of the circuit. For calculating the impedance, first find the longest curve in the current diagram, like the longest curve above zero, and find its length in time t. Then the minimum frequency should be f = 1 / (2 * t) . With higher frequency, the impedance of capacitors is very small. For coils, the impedance of a coil in ohms is 2 * pi * f * L , where L is the inductance of the coil in henries. Mosfet leakage with that impedance should be the same as with a resistor with the same resistance, by the resistance equivalence. Maybe some don't agree, but should be so by logic.

So i hope that now by knowing that, many avoid the pain that i had due to the mosfet leaking, all like hanged in the air, there was no way to find the real mosfet leaking to the circuit. No more so for these who do experiments now i think, and they can use mosfets because of their very fast switching, at least for initial evaluation.

The energy necessary for switching the mosfet is by that written above much greater than the leaking energy, but most of that is just wasted and goes to the ground at the end of the pulse. Like for the mosfet irf530n, the input capacitance is 900pF, and thus the switching energy in every switching cycle should be 11uJ, which is a lot. But for a theoretical research, only the energy in the circuit matters, and thus only the leaking energy can be considered, which by that above is much less. Yet even 0.7nJ in every switching cycle may cause some remarkable effects, so always find the mosfet's leaking energy before doing the experiments.

Hope it was useful for some.

[forest]

Helpful would be the proven circuit to switch mosfet at nanoseconds time (below 100ns)