has a specified rise time about 3x slower than the fall time, (78ns rise time and 25ns fall time with 1.6ohm gate resistor)
and this 75V n-channel mosfet:
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?name=IRFB3077PBF-ND
has a specified fall time about equal to the rise time. (87ns rise time and 95ns fall time with 2.1ohm gate resistor)
What are some of the factors inside an n-mosfet that effect the mosfet rise/fall times? Or are these results just because of the test setup used? ie. perhaps schottkys parallel with the gate resistors were used in the case of fast fall times.
Also for an H-bridge, can switching losses be minimized by taking these relative switching speeds into account? For example by putting different mosfets with faster fall times on either the upper or lower legs of the Hbridge, assuming the same total switching times for the upper and lower fets used.
My mental model of a mosfet is an infinitely fast chip that has parasitic capacitance on-chip and wirebond inductance out to the leads. In the practical case, most mosfets will switch much faster than the datasheets suggest if you can drive the gates hard enough and keep external circuit parasitics down. The rise/fall difference on the datasheets is mostly an artifact of the test circuit: turn-on impedance (Rds-on) is a lot lower than turn-off (whatever else is out there.)
This is a 50-volt pulse into a 50 ohm load, transformer coupled to boot:
What is Vgs(th)? I can imagine that, if V(low) isn't as far below Vgs(th) as V(high) is above it, that will directly affect how fast the miller capacitance charges through the gate resistor (and whatever source impedance there is).
Re: high speed stuff- parallel stuff should come as no surprise (although I am a little surprised 2N7002's are a bit faster than
2N3904s :) ). That's exactly what they do for RF stuff- you can get chips that consist of myriad little junctions (or FET channels), either strategically wired together with metallization or tediously wired, individually, with bond wires! Think of a wafer of 2N3904 dies, without cutting them apart, and lashing them all together:
If a bipolar transistor has an Ft of 300 MHz, it has a beta of 1 at
300 MHz. It's going to take a monstrous amount of base drive to get the collector to move fast. Even 45 GHz SiGe transistors are pig-slow switches compared to phemts, and the phemts are far easier to drive.
Bipolars are OK in tuned circuits, but fets are much better wideband/fast switches.
The photo cleverly doesn't show the entire waveform, so you can't tell if it's a positive or negative-going pulse, or what the duty cycle is.
Assuming the gates are driven from a low impedance, that probably means a pulse generator with 50 ohm source impedance, perhaps driving an attenuator to further reduce the Thevenin impedance at the gates. The case of the generator is at ground, so the 2N7002 sources would be at ground also.
So the only thing left to wiggle is the drains. And that is a problem.
The attenuator is a 40 dB 18 GHz attenuator, so it has an input impedance of 50 ohms.
There doesn't seem to be a coupling cap between the drain and the attenuator, since the waveform shown goes from zero to +0.5V.
The attenuator case is grounded, so you are probably not connecting the center pin of the attenuator to +50V and driving the outside shield with the drain of the 2N7002. That would mean lifting the scope off ground, which is difficult to do at 1GHz.
If you connected the input pin directly to the drains of the 2N7002, you would need to connect a 50 ohm resistor to a 100V supply in order to get a 50V swing at the drain.
But that would burn out the attenuator if the 2N7002's were left off.
But with the 2N7002's turned on, the 50 ohm drain resistor would dissipate 200 watts.
That would mean you are using a very good high frequency resistor, since there is little or no overshoot in the waveform. That would be very expensive.
You might be cheating a bit by switching the circuit on briefly to capture the waveform, then turning it off to save the resistor. But that seems unlikely.
The 2N7002's are only rated for 60V VDSS, so there's not much room for other approaches that might involve higher supply voltages.
So it's not definite how you are getting the waveform, but it is possible to eliminate many of the options due to bandwidth, cost, or power considerations.
That leaves few remaining alternatives, so your circuit has to be in one of them.
Use your imagination: the flats stay flat between pulses. The next pulse looks just like this pulse. The little "ground" symbol represents ground. The duty cycle is obviously low.
Why is that a problem? It is, as I've said, transformer coupled.
See above. It's actually 50 volts into the 100:1 attenuator.
Everything that's normally grounded is still grounded. You don't float an 11801. One person can barely lift an 11801.
This is getting too weird.
Like I said: two 2N7002's driving a transmission-line isolation transformer, with brutal gate drives. The point being that mosfets can be made to switch a lot faster than their datasheets usually suggest.
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Next time, we're going to use a GaN fet. That should really scream.
No, the attenuator pad is a symmetrical pi; it doesn't matter which way you put it in the circuit. And there's only one ground, the one in the sampling head.
--
Fine; then the sampling head input looks like a high Z and it gets its
signal from one of the pad\'s shunt resistors, like this:
SAMPLING
40dB PAD HEAD
+------------------+ +--------+
| | | |
+-->>--|---+--[4950]--+---|-->>--|IN |
P S | | | | | |
R E | [50] [50] | | |
I C | | | | | |
+-->>--|---+----------+---|-->>--|GND |
| | | |
+------------------+ +--------+
JF
There is distributed resistance in the gate. That is, the gate is poly, but picked up well with "metal". [Metal is often a mixture of stuff on chips, hence the quotes.]
I wasn't 100% sure of this since I've only done lateral flow devices, but found a good paper on power semiconductors in general:
John, a series resistor of 2499.75 ohms might be impractical for this attenuator.
A stray capacitance of only 3.53E-15 F. between the ends would have the same reactance as the series resistor at 18Ghz. This would probably put the unit out of spec.
Also, a small amount of capacitance from the center of a long high value resistor to the outside shield might have serious effects on the bandwidth. For example, see Win's description of "High voltage probe frequency response", at
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However, if the attenuator were split into three cascaded 10dB sections, the resistors would be 96.25, 71.15, and 96.25 ohms. Ref:
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This would dramatically reduce the effects of stray capacitance.
Are you sure you are considering possibilities?
I find it stimulating, probably because I like to consider possibilities:)
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