I have a situation where I need to use a pretty stiff pull-up resistor on a mosfet gate because I'm going to use a comparator to turn it off, and I plan to limit the current the comparator sinks to about 5 mA. The N-mosfet source will be sitting at 15 volts and the gate drive comes from a voltage doubler. That's a bit less than 30 volts so the pull-up resistor needs to be in the 6k ballpark. I'm going to use the NTP90N02:
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Speed doesn't matter -- this is basically an on-off switch, so I have only to consider the dc behavior of the mosfet. The mosfet has to conduct ten amps. My question is, how much resistance can you put in front of the gate before it gets too weak to turn the mosfet on hard? Mosfet models don't address this, so far as I can tell. I haven't ordered the NTP90N02 yet so I can't simply experiment on it, but I remember other mosfets having problems turning on when the gate resistance gets into the tens of kilohms.
Attention: Keep in mind that the abs max for Vgs is usually no more than
+/-20V, including the FET you have selected.
It can be very high but two things can cause some grief:
a. Slow turn-on, on account of a high Cgs and Cdg. In your case Cgs is over 2000pF so a 6K would slow that down to 10usec or more. During this time the FET transitions through its linear range, briefly dissipating lots of power and possibly going kablouie. Then there is Cgd which works "against" turning it on, further slowing it down.
b. Depending on what is connected Cgd might cause a "dent" in the turn-on phase and possibly oscillation around that point. With a stiff load the latter is usually accompanied by a bang, molten solder splattering about and so on.
--
Since there\'s no DC path through the gate, the only limit to the
value of the pullup will be how long it takes the gate to charge up
through it (and, therefore how much power the MOSFET will be
dissipating) until the MOSFET turns on hard.
You're joking. Right? Back in (what?) the early '70s, EIA decided that instead of using the logical term "Watts, continuous" they would call it "Watts RMS" for audio gear.
That is is similar to *root mean squared* makes this confusing and is a sign of bad judgement on their part.
It is indeed possible to calculate RMS power. However, the result is a mathematical curiosity and is not useful. If you calculate RMS power, the result will be 1.225 X the value you get by multiplying (RMS voltage) X (RMS Current). See the article in JAES (Journal of the Audio Engineering Society) "RMS Power: Fact or Fancy" by Eargle and Locanth for the details. The latest IHF Standard for Amplifiers was released in the late 60's. I does not make reference to "RMS power". In its table of minimum amplifier specifications, it refers to "Continuous output in watts per channel". Partially in response to audio equipment manufacturers' abuse of power ratings ("RMS Power", "Peak Power", "Music Power", etc), the FTC published its "Electronic Code of Federal Regulations". Part 432 of this regulation specifies, among other things, that power output shall be disclosed as "rated minimum sine wave continuous average power output". In fact, the only use of the term "RMS" in this regulation is with respect to the power line voltage and current. "RMS Power" makes absolutely no sense. Regards, Jon
Hey John, I'll explain what I'm doing here and why I think I need a high-side drive.
In the interest of economy the sketch below leaves out most of the circuit I'm building, but it shows the part involving `the mosfet I asked about in the original post. It's a kind of synchronous recitifier that turns on and off at relatively lengthy intervals when the generator revs up or turns off. With the generator turned off or idled, the mosfet acts as a blocking rectifier to keep the battery from discharging into the generator windings.
The rationale: using a mosfet avoids expensive, bulky heatsinking a conventional rectifier would require. The mosfet is also cheaper than a rectifier for the same current. Less than $2 in small quantities for the NTP90N02 from Digikey. It needs a charge pump for the drive, but having the charge pump there also allows me to use a cheap jellybean comparator on the high side. Without the charge pump, to run 10 amps wihout a heatsink, using a p- channel like an STP80PF55 would require two in parallel at $2.50 each, and a more expensive rail-to-rail comparator like the LT1716, which only has one comparator in it and costs $2.50 (in small quantities). But I need four comparators altogether in this circuit, three of them on the high side, and with an LM339 I get it all for about 50 cents. So I come out ahead even though I have to spend an extra buck or two on a charge pump. Two STP80PF55's and three LT1716's comes to more than ten bucks.
Your mosfet has an internal schottky diode from source to drain, which, in your picture, will conduct when the generator is on, regardless of whether you have gate drive. It also enables you to get rid of the external diode.
However, the reason the mosfet is able to deal with 10A is that it has a very low Rds(on). A diode won't have that, and so 10A will cause it to dissipate more heat than you might expect.
Also, I'm not sure what the rest of the circuit does, so I'm not sure what the mosfet drive is going to do. Are you going to charge the battery with it? If so, you probably want the Source to be next to the battery. However, that makes the substrate diode conduct when the generator is off, which you say you don't want.
I wonder if you can go with a simpler circuit? Here is one, assuming I understand your requirements:
generator--->|----(battery +)---->|---load
When the generator is on, it'll charge the battery and power the load. When the generator is off, the battery will drain through the load. You may want to control the charging current somehow, since dumping a huge current into a lead acid battery is a mistake (it'll overheat and vent). For the cost of another a power resistor, you can do that easily:
Here is a schottky diode that is about a buck in single quantity from digikey:
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Here is a suitable resistor:
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It is only 5W, but you'll need to evaludate that, and may need a beefier one.
Total circuit cost is about $2.50.
Regards, Bob Monsen
PS: Vista mail doesn't like this message, so it wouldn't put the appropriate '> ' in front of the replied text. Some formatting thing is screwing up its parser. Please forgive the top posting.
Hey John, I'll explain what I'm doing here and why I think I need a high-side drive.
In the interest of economy the sketch below leaves out most of the circuit I'm building, but it shows the part involving `the mosfet I asked about in the original post. It's a kind of synchronous recitifier that turns on and off at relatively lengthy intervals when the generator revs up or turns off. With the generator turned off or idled, the mosfet acts as a blocking rectifier to keep the battery from discharging into the generator windings.
The rationale: using a mosfet avoids expensive, bulky heatsinking a conventional rectifier would require. The mosfet is also cheaper than a rectifier for the same current. Less than $2 in small quantities for the NTP90N02 from Digikey. It needs a charge pump for the drive, but having the charge pump there also allows me to use a cheap jellybean comparator on the high side. Without the charge pump, to run 10 amps wihout a heatsink, using a p- channel like an STP80PF55 would require two in parallel at $2.50 each, and a more expensive rail-to-rail comparator like the LT1716, which only has one comparator in it and costs $2.50 (in small quantities). But I need four comparators altogether in this circuit, three of them on the high side, and with an LM339 I get it all for about 50 cents. So I come out ahead even though I have to spend an extra buck or two on a charge pump. Two STP80PF55's and three LT1716's comes to more than ten bucks.
** There is no an external diode, that's a diagrammatic representation of the mosfet's internal body diode. Yes, the mosfet conducts anytime the generator comes on. That's what I want. It's not there to block charging current. It's there to block the battery from _dis_charging into the generator.
(snip)
**Yes!
**No!
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**Right, that's why the mosfet is oriented the way it is. So that it won't conduct when the generator is off.
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=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
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=A0 =A0 =A0 =A0 =A0 =A0 |
=A0 =A0 =A0 =A0 =A0 =A0 =A0 .-.
ery =A0 =A0 =A0 =A0 =A0| |
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**I appreciate the effort you went to, Bob. The diagram I posted is only a detail of a larger circuit, which includes voltage regulation. But what I posted ought to be enough to understand what the mosfet is doing. It's a blocking recitifier. At first blush, you might think the mosfet is backwards, but it's not. The body diode acts as a blocking diode (when the generator is turned off). Then, when the generator turns on and the voltage at the source rises above the drain voltage, the comparator detects it and relaxes its hold on the mosfet gate. The voltage doubler then turns on the mosfet through the pullup resistor, allowing it to conduct with much less loss than a mere diode would. If you take a closer look at the circuit, you'll see how it works.
There are ORing-controller ICs that drive FETs of your choice for the power-supply ORing job. For example, the Intersil isl6144, TI tps2410 and 11, and LTC LT4351 for "+" supply lines, and IRF IR5001s for "-" supply lines. All of these feature built-in charge pumps, and have comparators, MOSFET drivers, and other good stuff. The LTC4412 uses a p-channel MOSFET.
Whether any of these ICs interest you or not, studying their datasheets and app notes could turn up aspects you hadn't considered.
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