MOSFET Usage
in [Design]
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From: ian field on 26 May 2010 16:43 "Ron M." <strmbrgr2(a)hotmail.com> wrote in message news:71783a02-e737-4dc5-9584-577b09c02084(a)q8g2000vbm.googlegroups.com... On May 26, 3:11 pm, "ian field" <gangprobing.al...(a)ntlworld.com> wrote: > <op...(a)hotmail.com> wrote in message > > news:8885382a-f931-45f8-977b-4cbb931c0864(a)q8g2000vbm.googlegroups.com... > > > Once you have your 9V battery on top of your 12V rail (21V gate pull-up > > supply) you connect the rail via a pull-up resistor > > Geez! Using a P-Channel MOSFET like the IRF5305 would be so much > simpler. Why has he latched onto the use of a N-Channel MOSFET? > > *************** > > Sounds like he already has the N-channel devices, and is using a > significant > number, so changing route halfway and buying up a batch of P-channel > devices > might be too much over budget. > > P-channel devices are more expensive and less efficient than N-channel and > the OP might not have been able to find a reasonably priced P device that > he > liked for the current rating he wanted. Ok Ian. If I only need a max of 6 amps and am driving it with 12 on gate would that be sufficient for the channel to open enough? I think the easiest way to do this is just build a test circuit and try it. All it will cost me is $1.80. It would be simpler and then I would know for sure. I'm still just having difficulty grasping the extra rail concept. I see the specs but can't figure out why I have to go so much higher than the source for it to conduct enough for my needs. I think a remedial class on MOSFET operation is definitely in order for me. ************* To turn the MOSFET on, the gate voltage has to be more than the source voltage by, at the very least VGSthr (more voltage than that is required to achieve the headline RDSon). To minimise dissipation in the MOSFET you want the source to pull up fully to the drain voltage - which is also your original high current rail that fires the igniters. The very simplest of algebra will show that therefore the gate voltage needs to be capable of exceeding the drain voltage by anything between about 8V and the 20V breakdown limit of the gate oxide layer.
From: ehsjr on 27 May 2010 02:12 Ron M. wrote: > On May 26, 3:11 pm, "ian field" <gangprobing.al...(a)ntlworld.com> > wrote: > >><op...(a)hotmail.com> wrote in message >> >>news:8885382a-f931-45f8-977b-4cbb931c0864(a)q8g2000vbm.googlegroups.com... >> >> >>>Once you have your 9V battery on top of your 12V rail (21V gate pull-up >>>supply) you connect the rail via a pull-up resistor >> >>Geez! Using a P-Channel MOSFET like the IRF5305 would be so much >>simpler. Why has he latched onto the use of a N-Channel MOSFET? >> >>*************** >> >>Sounds like he already has the N-channel devices, and is using a significant >>number, so changing route halfway and buying up a batch of P-channel devices >>might be too much over budget. >> >>P-channel devices are more expensive and less efficient than N-channel and >>the OP might not have been able to find a reasonably priced P device that he >>liked for the current rating he wanted. > > > Ok Ian. If I only need a max of 6 amps and am driving it with 12 on > gate would that be sufficient for the channel to open enough? I think > the easiest way to do this is just build a test circuit and try it. > All it will cost me is $1.80. It would be simpler and then I would > know for sure. I'm still just having difficulty grasping the extra > rail concept. I see the specs but can't figure out why I have to go so > much higher than the source for it to conduct enough for my needs. I > think a remedial class on MOSFET operation is definitely in order for > me. Essentially, you are using the mosfet as a switch, so the voltage at the source pin will be almost equal to the voltage at the drain pin when the mosfet conducts. +12 -------------+ +12 ----+ | | ___|D | || o ||<. ~ \ G_____||_|_ ~ o |S | [Load] [Load] | | Gnd -------------+ Gnd ----+ To make your mosfet conduct (in effect, closing the switch) the voltage at g must be raised *above* the voltage at s by *at least* the "gate threshold voltage". And good design is to raise it more than that minimum, to ensure that the mosfet is turned on hard. In your case, with your mosfet, a 12 volt supply and a 6 amp load current, the gate should be raised to about 20 volts. That makes the gate to source voltage about 8 volts, and ensures that the mosfet is turned on completely. Ed
From: Ron M. on 27 May 2010 15:04 On May 27, 2:12 am, ehsjr <eh...(a)nospamverizon.net> wrote: > Ron M. wrote: > > On May 26, 3:11 pm, "ian field" <gangprobing.al...(a)ntlworld.com> > > wrote: > > >><op...(a)hotmail.com> wrote in message > > >>news:8885382a-f931-45f8-977b-4cbb931c0864(a)q8g2000vbm.googlegroups.com.... > > >>>Once you have your 9V battery on top of your 12V rail (21V gate pull-up > >>>supply) you connect the rail via a pull-up resistor > > >>Geez! Using a P-Channel MOSFET like the IRF5305 would be so much > >>simpler. Why has he latched onto the use of a N-Channel MOSFET? > > >>*************** > > >>Sounds like he already has the N-channel devices, and is using a significant > >>number, so changing route halfway and buying up a batch of P-channel devices > >>might be too much over budget. > > >>P-channel devices are more expensive and less efficient than N-channel and > >>the OP might not have been able to find a reasonably priced P device that he > >>liked for the current rating he wanted. > > > Ok Ian. If I only need a max of 6 amps and am driving it with 12 on > > gate would that be sufficient for the channel to open enough? I think > > the easiest way to do this is just build a test circuit and try it. > > All it will cost me is $1.80. It would be simpler and then I would > > know for sure. I'm still just having difficulty grasping the extra > > rail concept. I see the specs but can't figure out why I have to go so > > much higher than the source for it to conduct enough for my needs. I > > think a remedial class on MOSFET operation is definitely in order for > > me. > > Essentially, you are using the mosfet as a switch, so the > voltage at the source pin will be almost equal to the > voltage at the drain pin when the mosfet conducts. > > +12 -------------+ +12 ----+ > | | > ___|D | > || o > ||<. ~ \ > G_____||_|_ ~ o > |S | > [Load] [Load] > | | > Gnd -------------+ Gnd ----+ > > To make your mosfet conduct (in effect, closing the switch) > the voltage at g must be raised *above* the voltage at s > by *at least* the "gate threshold voltage". And good design > is to raise it more than that minimum, to ensure that the > mosfet is turned on hard. > > In your case, with your mosfet, a 12 volt supply and a 6 amp > load current, the gate should be raised to about 20 volts. > That makes the gate to source voltage about 8 volts, and ensures > that the mosfet is turned on completely. > > Ed- Hide quoted text - > > - Show quoted text - Just had a thought. Went back and looked at the requirements of the actual e-match and what a bonehead I have been. They require 1/2 amp guaranteed fire current. So with that in mind maybe if I pad the Drain down some with a resistor maybe I can still get within the full turn on voltage. Just throwing things out there now but it might work.
From: JosephKK on 28 May 2010 07:27 On Thu, 27 May 2010 12:04:50 -0700 (PDT), "Ron M." <strmbrgr2(a)hotmail.com> wrote: >On May 27, 2:12 am, ehsjr <eh...(a)nospamverizon.net> wrote: >> Ron M. wrote: >> > On May 26, 3:11 pm, "ian field" <gangprobing.al...(a)ntlworld.com> >> > wrote: >> >> >><op...(a)hotmail.com> wrote in message >> >> >>news:8885382a-f931-45f8-977b-4cbb931c0864(a)q8g2000vbm.googlegroups.com.... >> >> >>>Once you have your 9V battery on top of your 12V rail (21V gate pull-up >> >>>supply) you connect the rail via a pull-up resistor >> >> >>Geez! Using a P-Channel MOSFET like the IRF5305 would be so much >> >>simpler. Why has he latched onto the use of a N-Channel MOSFET? >> >> >>*************** >> >> >>Sounds like he already has the N-channel devices, and is using a significant >> >>number, so changing route halfway and buying up a batch of P-channel devices >> >>might be too much over budget. >> >> >>P-channel devices are more expensive and less efficient than N-channel and >> >>the OP might not have been able to find a reasonably priced P device that he >> >>liked for the current rating he wanted. >> >> > Ok Ian. If I only need a max of 6 amps and am driving it with 12 on >> > gate would that be sufficient for the channel to open enough? I think >> > the easiest way to do this is just build a test circuit and try it. >> > All it will cost me is $1.80. It would be simpler and then I would >> > know for sure. I'm still just having difficulty grasping the extra >> > rail concept. I see the specs but can't figure out why I have to go so >> > much higher than the source for it to conduct enough for my needs. I >> > think a remedial class on MOSFET operation is definitely in order for >> > me. >> >> Essentially, you are using the mosfet as a switch, so the >> voltage at the source pin will be almost equal to the >> voltage at the drain pin when the mosfet conducts. >> >> +12 -------------+ +12 ----+ >> | | >> ___|D | >> || o >> ||<. ~ \ >> G_____||_|_ ~ o >> |S | >> [Load] [Load] >> | | >> Gnd -------------+ Gnd ----+ >> >> To make your mosfet conduct (in effect, closing the switch) >> the voltage at g must be raised *above* the voltage at s >> by *at least* the "gate threshold voltage". And good design >> is to raise it more than that minimum, to ensure that the >> mosfet is turned on hard. >> >> In your case, with your mosfet, a 12 volt supply and a 6 amp >> load current, the gate should be raised to about 20 volts. >> That makes the gate to source voltage about 8 volts, and ensures >> that the mosfet is turned on completely. >> >> Ed- Hide quoted text - >> >> - Show quoted text - > >Just had a thought. Went back and looked at the requirements of the >actual e-match and what a bonehead I have been. They require 1/2 amp >guaranteed fire current. So with that in mind maybe if I pad the Drain >down some with a resistor maybe I can still get within the full turn >on voltage. Just throwing things out there now but it might work. If you can handle the change, use NFET to ground switching and put the relays on the positive rail. Alternatively, use PFET to switch the positive rail, the reduction in complexity can buy a worthy change in FET price.
From: ian field on 28 May 2010 09:50
"JosephKK" <quiettechblue(a)yahoo.com> wrote in message news:hv9vv51dda7rsa0g5srjc2bgncd1362vd5(a)4ax.com... On Thu, 27 May 2010 12:04:50 -0700 (PDT), "Ron M." <strmbrgr2(a)hotmail.com> wrote: >On May 27, 2:12 am, ehsjr <eh...(a)nospamverizon.net> wrote: >> Ron M. wrote: >> > On May 26, 3:11 pm, "ian field" <gangprobing.al...(a)ntlworld.com> >> > wrote: >> >> >><op...(a)hotmail.com> wrote in message >> >> >>news:8885382a-f931-45f8-977b-4cbb931c0864(a)q8g2000vbm.googlegroups.com... >> >> >>>Once you have your 9V battery on top of your 12V rail (21V gate >> >>>pull-up >> >>>supply) you connect the rail via a pull-up resistor >> >> >>Geez! Using a P-Channel MOSFET like the IRF5305 would be so much >> >>simpler. Why has he latched onto the use of a N-Channel MOSFET? >> >> >>*************** >> >> >>Sounds like he already has the N-channel devices, and is using a >> >>significant >> >>number, so changing route halfway and buying up a batch of P-channel >> >>devices >> >>might be too much over budget. >> >> >>P-channel devices are more expensive and less efficient than N-channel >> >>and >> >>the OP might not have been able to find a reasonably priced P device >> >>that he >> >>liked for the current rating he wanted. >> >> > Ok Ian. If I only need a max of 6 amps and am driving it with 12 on >> > gate would that be sufficient for the channel to open enough? I think >> > the easiest way to do this is just build a test circuit and try it. >> > All it will cost me is $1.80. It would be simpler and then I would >> > know for sure. I'm still just having difficulty grasping the extra >> > rail concept. I see the specs but can't figure out why I have to go so >> > much higher than the source for it to conduct enough for my needs. I >> > think a remedial class on MOSFET operation is definitely in order for >> > me. >> >> Essentially, you are using the mosfet as a switch, so the >> voltage at the source pin will be almost equal to the >> voltage at the drain pin when the mosfet conducts. >> >> +12 -------------+ +12 ----+ >> | | >> ___|D | >> || o >> ||<. ~ \ >> G_____||_|_ ~ o >> |S | >> [Load] [Load] >> | | >> Gnd -------------+ Gnd ----+ >> >> To make your mosfet conduct (in effect, closing the switch) >> the voltage at g must be raised *above* the voltage at s >> by *at least* the "gate threshold voltage". And good design >> is to raise it more than that minimum, to ensure that the >> mosfet is turned on hard. >> >> In your case, with your mosfet, a 12 volt supply and a 6 amp >> load current, the gate should be raised to about 20 volts. >> That makes the gate to source voltage about 8 volts, and ensures >> that the mosfet is turned on completely. >> >> Ed- Hide quoted text - >> >> - Show quoted text - > >Just had a thought. Went back and looked at the requirements of the >actual e-match and what a bonehead I have been. They require 1/2 amp >guaranteed fire current. So with that in mind maybe if I pad the Drain >down some with a resistor maybe I can still get within the full turn >on voltage. Just throwing things out there now but it might work. If you can handle the change, use NFET to ground switching and put the relays on the positive rail. Alternatively, use PFET to switch the positive rail, the reduction in complexity can buy a worthy change in FET price. I'm starting to get the impression this guy won't take sound advice! |