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Sci.Electronics.Basics -> mosfet? transistor? hexfet?
There are 12 messages in this thread.
You are currently looking at messages 1 to 12.
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When would one use a transistor? mosfet? hexfet?
Basically looking into PWM switching of medium-sized DC motors (12V,
~5A or so).
Which should I use...? A beefy transistor like a TIP36C? Or a hexfet
like IRLZ34NPBF (98c at Mouser)?
Thanks...
Michael
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Author: Charles SchulerDate: 16:35 21-12-06
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<mrdarrett@gmail.com> wrote in message
news:1166730368.844108.175760@i12g2000cwa.googlegroups.com...
> When would one use a transistor? mosfet? hexfet?
>
> Basically looking into PWM switching of medium-sized DC motors (12V,
> ~5A or so).
>
> Which should I use...? A beefy transistor like a TIP36C? Or a hexfet
> like IRLZ34NPBF (98c at Mouser)?
Generally, you want the lowest on resistance and the fastest switching
times. So, look at the spec sheets and pick the best part within your price
range.
It's more complicated than that, but it is a start. Maximum ratings and
drive requirements are also critical.
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Author: Jonathan KirwanDate: 17:47 21-12-06
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On Thu, 21 Dec 2006 16:35:09 -0500, "Charles Schuler"
<charleschuler@comcast.net> wrote:
><mrdarrett@gmail.com> wrote in message
>news:1166730368.844108.175760@i12g2000cwa.googlegroups.com...
>> When would one use a transistor? mosfet? hexfet?
>>
>> Basically looking into PWM switching of medium-sized DC motors (12V,
>> ~5A or so).
>>
>> Which should I use...? A beefy transistor like a TIP36C? Or a hexfet
>> like IRLZ34NPBF (98c at Mouser)?
>
>Generally, you want the lowest on resistance and the fastest switching
>times. So, look at the spec sheets and pick the best part within your price
>range.
>
>It's more complicated than that, but it is a start. Maximum ratings and
>drive requirements are also critical.
Another part of the answer is to use what you know better about and
can design. Getting something working reasonably well is better, even
if it dissipates more, than having something not dissipate as much but
also not work quite right.
Jon
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Author: Jonathan KirwanDate: 18:00 21-12-06
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On 21 Dec 2006 11:46:08 -0800, mrdarrett@gmail.com wrote:
>When would one use a transistor? mosfet? hexfet?
>
>Basically looking into PWM switching of medium-sized DC motors (12V,
>~5A or so).
>
>Which should I use...? A beefy transistor like a TIP36C? Or a hexfet
>like IRLZ34NPBF (98c at Mouser)?
First off, I should say that I design and use discrete BJTs into
circuits, but don't use Mosfets that much. Partly, because I just
find BJTs more interesting and I enjoy learning and partly because I
don't design anything for others to use. Plus, I suppose, I get BJTs
in groups that work out to about 1-3 cents each, so I don't mind
freely giving away some to interested students without worrying
whether or not they will actually ever get around to using them. I'm
decidedly NOT a designer of motor controls or anything else. Just an
occasional hobbyist. That said....
Mosfet's source-drain connection act resistive (which is why their
on-resistance is commonly specified) and BJTs don't act quite that
way, which is why several of their saturation Vce's are given in a
table, often along with a chart. (There is _some_ resistance there,
though. So it is a mixture of effects.)
Neither STMicro nor OnSemi seems to provide a Vce vs Ib graph for
various Ic on the TIP36C, like I usually find for the 2N3055 or
MJ2955, for example. But the TIP36C table shows 1.8V at 15A for
Vcesat (and 4.0V at 25A.) At 5A, it will be an even better experience
than what 15A will show, but at 15A this works out to 1.8V/15A or 0.12
ohms. Probably, you will get a little better than the extrapolated
0.6V Vcesat, though, at 5A. You will also be driving current into the
base, though, and this will also dissipate. The 2N3055/MJ2955, by
comparison, shows slightly more than 0.5V at 5A, so it seems in the
same ballpark. I'd estimate .1 ohms equivalent. Then take into
account the drive currents working against the Vbe, which may be
around 1.5V itself at say .5A, so not insignificant. Maybe 2.5W +
0.75W, for 3.25W total. Just rough numbers for comparison.
The IRLZ34N shows an on resistance of 0.06 ohms at Id=16A with a Vgs
of 4V. And it will be a little better than that at 5A. So it will
probably dissipate 1.5W at 5A. I don't think the rough 1nF input
capacitance will cause you trouble in switching it fast enough for
your needs and it probably won't dissipate that much during that
transition time. So 1.5W is probably good.
From that, you can probably see that the hexfet looks better in terms
of dissipation. About twice as good. Driving either case is probably
not too hard, but the hexfet drive will probably be a little cheaper
for which to buy parts (supplying .5A will use somewhat larger parts
than supplying say 1mA for a short time to the mosfet requires.)
There are other considerations. Runaway thermal problems are possible
with BJTs, but I seem to recall some discussion about similar effects
with IRF's hexfets, in particular, on sed. I don't know about it,
though.
Why not try both and see what you think about it? That's probably
what I'd try doing in a case like this.
Jon
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Jonathan Kirwan wrote:
> On 21 Dec 2006 11:46:08 -0800, mrdarrett@gmail.com wrote:
>
> >When would one use a transistor? mosfet? hexfet?
> >
> >Basically looking into PWM switching of medium-sized DC motors (12V,
> >~5A or so).
> >
> >Which should I use...? A beefy transistor like a TIP36C? Or a hexfet
> >like IRLZ34NPBF (98c at Mouser)?
>
> First off, I should say that I design and use discrete BJTs into
> circuits, but don't use Mosfets that much. Partly, because I just
> find BJTs more interesting and I enjoy learning and partly because I
> don't design anything for others to use. Plus, I suppose, I get BJTs
> in groups that work out to about 1-3 cents each, so I don't mind
> freely giving away some to interested students without worrying
> whether or not they will actually ever get around to using them. I'm
> decidedly NOT a designer of motor controls or anything else. Just an
> occasional hobbyist. That said....
>
> Mosfet's source-drain connection act resistive (which is why their
> on-resistance is commonly specified) and BJTs don't act quite that
> way, which is why several of their saturation Vce's are given in a
> table, often along with a chart. (There is _some_ resistance there,
> though. So it is a mixture of effects.)
>
> Neither STMicro nor OnSemi seems to provide a Vce vs Ib graph for
> various Ic on the TIP36C, like I usually find for the 2N3055 or
> MJ2955, for example. But the TIP36C table shows 1.8V at 15A for
> Vcesat (and 4.0V at 25A.) At 5A, it will be an even better experience
> than what 15A will show, but at 15A this works out to 1.8V/15A or 0.12
> ohms. Probably, you will get a little better than the extrapolated
> 0.6V Vcesat, though, at 5A. You will also be driving current into the
> base, though, and this will also dissipate. The 2N3055/MJ2955, by
> comparison, shows slightly more than 0.5V at 5A, so it seems in the
> same ballpark. I'd estimate .1 ohms equivalent. Then take into
> account the drive currents working against the Vbe, which may be
> around 1.5V itself at say .5A, so not insignificant. Maybe 2.5W +
> 0.75W, for 3.25W total. Just rough numbers for comparison.
>
> The IRLZ34N shows an on resistance of 0.06 ohms at Id=16A with a Vgs
> of 4V. And it will be a little better than that at 5A. So it will
> probably dissipate 1.5W at 5A. I don't think the rough 1nF input
> capacitance will cause you trouble in switching it fast enough for
> your needs and it probably won't dissipate that much during that
> transition time. So 1.5W is probably good.
>
> From that, you can probably see that the hexfet looks better in terms
> of dissipation. About twice as good. Driving either case is probably
> not too hard, but the hexfet drive will probably be a little cheaper
> for which to buy parts (supplying .5A will use somewhat larger parts
> than supplying say 1mA for a short time to the mosfet requires.)
>
> There are other considerations. Runaway thermal problems are possible
> with BJTs, but I seem to recall some discussion about similar effects
> with IRF's hexfets, in particular, on sed. I don't know about it,
> though.
>
> Why not try both and see what you think about it? That's probably
> what I'd try doing in a case like this.
>
> Jon
Thank you for the very informative reply.
I *think* I've figured out the difference between MOSFETs and bipolar
transistors: the gates of MOSFETs are opened by application of a
*voltage*; the base of a bipolar transistor is activated by application
of *current*. (Right? Wrong?)
So here's a spec sheet of a 2N3055 (what a beauty):
http://www.st.com/stonline/books/pdf/docs/4079.pdf
So, is the 5 mA for I_ebo the minimum current needed to "activate" the
base (fully open the "gate" from collector to emitter)?
So, would applying 1 mA to 5 mA activate the base linearly (for
example, 1 mA at the base = 20% collector-emitter, 2 mA base = 40%
collector-emitter?)
I guess I need help figuring out how to read these spec sheets.
Thanks,
Michael
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Author: Ben JacksonDate: 18:47 21-12-06
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On 2006-12-21, mrdarrett@gmail.com <mrdarrett@gmail.com> wrote:
>
> I *think* I've figured out the difference between MOSFETs and bipolar
> transistors: the gates of MOSFETs are opened by application of a
> *voltage*; the base of a bipolar transistor is activated by application
> of *current*. (Right? Wrong?)
This page explains the relationships between Ice, Vbe and beta for BJTs
and Id and Vgs for FETS: http://en.wikipedia.org/wiki/Transistor
> So here's a spec sheet of a 2N3055 (what a beauty):
> http://www.st.com/stonline/books/pdf/docs/4079.pdf
>
> So, is the 5 mA for I_ebo the minimum current needed to "activate" the
> base (fully open the "gate" from collector to emitter)?
No, not fully. The parameter you really want to look at is the hFE (or
"beta") which is the DC current gain. From the data sheet you can only
count on switching 20x more current than you supply for turn-on. So
if you have a 5A motor you'd need to supply 250mA or more to the base
to ensure the transitor was saturated. And you need to do it *fast* to
avoid having the transistor live in the linear region where it is
dissipating the power (and heat!).
The FET presents a different problem: The gate appears to you as a
capacitance and you have to drive Vgs up to the saturation point quickly
(which will be a higher voltage than the BJT). Once you *get* there,
though, you don't have to supply any more current. The bigger the
device, the higher the Ciss. Then, to turn it off, you have to suck
that charge back out.
--
Ben Jackson AD7GD
<ben@ben.com>
http://www.ben.com/
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Author: JoergDate: 19:11 21-12-06
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Ben Jackson wrote:
> On 2006-12-21, mrdarrett@gmail.com <mrdarrett@gmail.com> wrote:
>
>>I *think* I've figured out the difference between MOSFETs and bipolar
>>transistors: the gates of MOSFETs are opened by application of a
>>*voltage*; the base of a bipolar transistor is activated by application
>>of *current*. (Right? Wrong?)
>
>
> This page explains the relationships between Ice, Vbe and beta for BJTs
> and Id and Vgs for FETS: http://en.wikipedia.org/wiki/Transistor
>
>
>>So here's a spec sheet of a 2N3055 (what a beauty):
>>http://www.st.com/stonline/books/pdf/docs/4079.pdf
>>
>>So, is the 5 mA for I_ebo the minimum current needed to "activate" the
>>base (fully open the "gate" from collector to emitter)?
>
>
> No, not fully. The parameter you really want to look at is the hFE (or
> "beta") which is the DC current gain. From the data sheet you can only
> count on switching 20x more current than you supply for turn-on. So
> if you have a 5A motor you'd need to supply 250mA or more to the base
> to ensure the transitor was saturated. And you need to do it *fast* to
> avoid having the transistor live in the linear region where it is
> dissipating the power (and heat!).
>
> The FET presents a different problem: The gate appears to you as a
> capacitance and you have to drive Vgs up to the saturation point quickly
> (which will be a higher voltage than the BJT). Once you *get* there,
> though, you don't have to supply any more current. The bigger the
> device, the higher the Ciss. Then, to turn it off, you have to suck
> that charge back out.
>
Also, don't forget Cds which works against your efforts here. It may
appear small but there is a multiplier effect. The higher the drain
voltage when open the greater the detrimental effect on gate turn-on.
When looking at Vgs turn-on levels it is prudent not to pay too much
attention to what the front page says. That's often marketing spin. What
matters are the guaranteed values in the tables. There needs to be a
specified maximum Rdson at a given Vgs. Often you'll find 2-3 values
there. Anything else can't really be relied upon IMHO.
--
Regards, Joerg
http://www.analogconsultants.com
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Author: Jonathan KirwanDate: 19:29 21-12-06
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On 21 Dec 2006 15:20:42 -0800, mrdarrett@gmail.com wrote:
>Jonathan Kirwan wrote:
>> On 21 Dec 2006 11:46:08 -0800, mrdarrett@gmail.com wrote:
>>
>> >When would one use a transistor? mosfet? hexfet?
>> >
>> >Basically looking into PWM switching of medium-sized DC motors (12V,
>> >~5A or so).
>> >
>> >Which should I use...? A beefy transistor like a TIP36C? Or a hexfet
>> >like IRLZ34NPBF (98c at Mouser)?
>>
>> First off, I should say that I design and use discrete BJTs into
>> circuits, but don't use Mosfets that much. Partly, because I just
>> find BJTs more interesting and I enjoy learning and partly because I
>> don't design anything for others to use. Plus, I suppose, I get BJTs
>> in groups that work out to about 1-3 cents each, so I don't mind
>> freely giving away some to interested students without worrying
>> whether or not they will actually ever get around to using them. I'm
>> decidedly NOT a designer of motor controls or anything else. Just an
>> occasional hobbyist. That said....
>>
>> Mosfet's source-drain connection act resistive (which is why their
>> on-resistance is commonly specified) and BJTs don't act quite that
>> way, which is why several of their saturation Vce's are given in a
>> table, often along with a chart. (There is _some_ resistance there,
>> though. So it is a mixture of effects.)
>>
>> Neither STMicro nor OnSemi seems to provide a Vce vs Ib graph for
>> various Ic on the TIP36C, like I usually find for the 2N3055 or
>> MJ2955, for example. But the TIP36C table shows 1.8V at 15A for
>> Vcesat (and 4.0V at 25A.) At 5A, it will be an even better experience
>> than what 15A will show, but at 15A this works out to 1.8V/15A or 0.12
>> ohms. Probably, you will get a little better than the extrapolated
>> 0.6V Vcesat, though, at 5A. You will also be driving current into the
>> base, though, and this will also dissipate. The 2N3055/MJ2955, by
>> comparison, shows slightly more than 0.5V at 5A, so it seems in the
>> same ballpark. I'd estimate .1 ohms equivalent. Then take into
>> account the drive currents working against the Vbe, which may be
>> around 1.5V itself at say .5A, so not insignificant. Maybe 2.5W +
>> 0.75W, for 3.25W total. Just rough numbers for comparison.
>>
>> The IRLZ34N shows an on resistance of 0.06 ohms at Id=16A with a Vgs
>> of 4V. And it will be a little better than that at 5A. So it will
>> probably dissipate 1.5W at 5A. I don't think the rough 1nF input
>> capacitance will cause you trouble in switching it fast enough for
>> your needs and it probably won't dissipate that much during that
>> transition time. So 1.5W is probably good.
>>
>> From that, you can probably see that the hexfet looks better in terms
>> of dissipation. About twice as good. Driving either case is probably
>> not too hard, but the hexfet drive will probably be a little cheaper
>> for which to buy parts (supplying .5A will use somewhat larger parts
>> than supplying say 1mA for a short time to the mosfet requires.)
>>
>> There are other considerations. Runaway thermal problems are possible
>> with BJTs, but I seem to recall some discussion about similar effects
>> with IRF's hexfets, in particular, on sed. I don't know about it,
>> though.
>>
>> Why not try both and see what you think about it? That's probably
>> what I'd try doing in a case like this.
>
>Thank you for the very informative reply.
>
>I *think* I've figured out the difference between MOSFETs and bipolar
>transistors: the gates of MOSFETs are opened by application of a
>*voltage*; the base of a bipolar transistor is activated by application
>of *current*. (Right? Wrong?)
Kind of. Actually, the purist will point to the basic Shockley diode
equation (it's a part of the Eber-Moll model for BJTs) and point out
that a voltage determines BJT collector current, as well. But the
voltage determines collector current in a very non-linear way (small
increases leading to large Ic changes, collector load permitting) and
to achieve that voltage, you have to drive current through a diode
sufficiently to create that voltage, and this base-emitter voltage
itself increases only slightly to large increases in that base
current, Ib. Together, these two non-linear behaviors conspire
together to be nearly close enough to each other that it will roughly
appear that the Ic will vary nearly linearly with Ib. Hence, the idea
of 'beta' arises and from that, the idea that the BJT's collector
current is a function of base current (for some range of various
values.)
In the case of switching, rather than amplification, your interest is
mostly in just driving the Vce to a low value. As low as is possible,
consistent without having to go to ridiculous extremes ramming in base
current. Here, the usual beta figure isn't all that meaningful, so a
rule of thumb is to estimate the base current as 1/10th of the
collector current to drive Vce "suffiently down" that you've reached
diminishing returns by then. My experience for smaller BJTs is that
about 1/30th is where diminishing returns are met. But with 5A
devices, like you are talking about, I'd definitely go with the 1/10th
figure.
>So here's a spec sheet of a 2N3055 (what a beauty):
>http://www.st.com/stonline/books/pdf/docs/4079.pdf
No curves to speak of. I think you'll find this better:
http://www.onsemi.com/pub/Collateral/2N3055-D.PDF
Take a look at Figure 6. To read it, just look at the curve for "4.0
A". This is your collector current curve. Now, note the y-axis is
your Vce you can expect to get. Lower is better, here. Then look at
the x-axis. This is your base current. You can see it knee-over
around 200mA (1/20th of collector current) at around Vce=0.55V and
that if you push this to 400mA (1/10th of the collector current) you
get about 0.50V. As you can see, cramming in more base current pushes
the Vce down a little bit (50mV), but that you have to double the base
current to do it. I think that curve will help you think about this.
By the way, take note of the fact that the minimum guaranteed beta for
this device is 20, maximum is 70. For switching purposes you need to
be using an estimate that is substantially less than the minimum
figure. So your base current should be well more than 1/20th of the
collector current. At least 1/10th and if you are being really
conservative about getting the worst of a batch of these parts,
probably even 1/5th (which isn't so beautiful.) Look at Figure 4,
though, and look up from the Ic=5A line. You can see that we are
talking about a typical beta of 30, or so. As a hobbyist, I'd go with
1/10th (beta=10 for switching) for an estimate, though, and risk it to
see how it works out.
>So, is the 5 mA for I_ebo the minimum current needed to "activate" the
>base (fully open the "gate" from collector to emitter)?
No, not at all. I frankly don't recall ever thinking about that
parameter, though. Perhaps someone here can explain it more fully. I
could guess, but I'd rather here from someone about this. But keep in
mind that this is a maximum specification (not a minimum, as you seem
to suggest) and it isn't the parameter to look at for turning on the
BJT. Basically, you need to go from about 0mA to about 500mA into the
base. During part of this time, the BJT will be in its linear region
and the load (motor) will experience a rising current that is kind of
proportional to your ramping-up base drive current (consistent with
what the effective motor inductance and voltage rail permits.) While
this is taking place, your Vce will be much larger and the BJT will
dissipate more for part of this short time.
>So, would applying 1 mA to 5 mA activate the base linearly (for
>example, 1 mA at the base = 20% collector-emitter, 2 mA base = 40%
>collector-emitter?)
No. In saturation, the Ic isn't a linear function of Ib. Only out of
saturation might that be roughly true. Look closely at that Figure 6.
Imagine you have a supply voltage of 10V and we can treat (I don't
want to get into unloaded and loaded motor behaviors, which I can
probably not do well at explaining) your motor as resistive and
exhibits 2.5 ohms. This gets is to about the 4A curve in Figure 6. So
let's say you are driving 400mA into the base. You will see Vce=0.5V.
This means 9.5V across your motor. That works out to a little less
than 4A (9.5V/2.5ohms). Now, if you take the curve literally and you
look at reducing the base drive by half, to 200mA, then your
Vce=0.55V. This means that your resistance (motor) experiences 9.45V.
The current is still about the same, now 9.45A/2.5ohms. Not much
change in current. So the 4A curve probably still applies, so far.
Cutting down to 150mA leads to a Vce=0.6V. Again, perhaps not so much
change. But enough that you can imagine that the Ic=4A curve may be
beggining to not apply so well, anymore. As you drop to Ib=100mA, you
can see that the 4A curve goes almost straight up. This tells you
that the 4A curve isn't applying anymore, at all. You are getting
into the more linear region where the beta=30 will start to dominate
(30*100mA is 3A, which you might use as a rough guess here), but keep
in mind that as Ic continues to decline the beta rises, too. So it's
not perfectly linear here, either. For example, in the case of 100mA
for your Ib, you might guess that beta=30 because you've gone into the
linear (non-saturated) region, because beta=27 at 25C at Ic=5A (Figure
4.) This would suggest Ic=3A. But if you look at Figure 4 closely,
at Ic=3A, you see that beta=40. So that would suggest, at Ib=100mA,
that Ic=4A. But at Ic=4A on Figure 4, beta=30, not 40, which again
suggests Ic=3A. The actual Ic will be something between 3A and 4A,
trusting these curves, with beta that is between 30 and 40. Where
exactly, you won't really know without testing. At Ib=50mA, you might
guess a beta=40 and compute Ic=2A, but at Ic=2A the beta is listed at
more than 50. So this suggests a little higher than Ic=2A at Ib=50mA.
So you can see it is 'kind of' linear with Ib, but not strictly so as
we get into lower currents.
What you can consider as your minimum base current is roughly what
value it is, at which your motor won't run at all even with you
turning it slightly with your fingers, divided by about 100 or maybe
even 150. If you get below that, the motor won't have much drive
current.
Of course, keep in mind I'm not experienced in these things.
>I guess I need help figuring out how to read these spec sheets.
I hope that helps some.
Jon
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Author: Jonathan KirwanDate: 19:33 21-12-06
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On Fri, 22 Dec 2006 00:29:11 GMT, Jonathan Kirwan
<jkirwan@easystreet.com> wrote:
>but I'd rather here from someone about this
... hear ...
Oh, well.
Jon
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Jonathan Kirwan wrote:
...
> >So here's a spec sheet of a 2N3055 (what a beauty):
> >http://www.st.com/stonline/books/pdf/docs/4079.pdf
>
> No curves to speak of. I think you'll find this better:
>
> http://www.onsemi.com/pub/Collateral/2N3055-D.PDF
>
> Take a look at Figure 6. To read it, just look at the curve for "4.0
> A". This is your collector current curve. Now, note the y-axis is
> your Vce you can expect to get. Lower is better, here. Then look at
> the x-axis. This is your base current. You can see it knee-over
> around 200mA (1/20th of collector current) at around Vce=0.55V and
> that if you push this to 400mA (1/10th of the collector current) you
> get about 0.50V. As you can see, cramming in more base current pushes
> the Vce down a little bit (50mV), but that you have to double the base
> current to do it. I think that curve will help you think about this.
Actually, by "what a beauty" I'd meant the pic of the TO-3. ;-)
Ok. I found Figure 6, and I was a little puzzled by it. Let's say I
have a light bulb (or, for argument's sake, simply a fuse!) in series
between +12V and the collector, and the emitter is grounded.
Won't V_ce be 12V? Or is this a different V_ce?
Ok, so the name of the game is to minimize V_ce (as much as is
practical; the knee-point is probably good enough.)
(snip lots of great stuff that will require more time for me to study.)
Thank you,
Michael
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Author: Phil AllisonDate: 21:38 21-12-06
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<mrdarrett@gmail.com
> When would one use a transistor? mosfet? hexfet?
>
> Basically looking into PWM switching of medium-sized DC motors (12V,
> ~5A or so).
>
> Which should I use...? A beefy transistor like a TIP36C? Or a hexfet
> like IRLZ34NPBF (98c at Mouser)?
** Forget bi-polar devices completely for this app - they have high heat
loses and take high power to drive.
Go with MOSFETS, the lower the on resistance the better.
You can simply parallel two or more if needed to reduce heat or increase
current.
Test and find the "stall current" of the motor - it is WAAAAAY higher than
the normal running current. This IS the current that will flow if power
is suddenly applied or the motor is jammed and cannot turn.
Your drive circuit must be able to withstand the "stall current" of the
motor or it will soon fail.
Switching frequency can be from 200HZ to 5kHz - what is best it depends
on the motor and what you are doing with it.
Don't forget to put a fast diode across the motor.
....... Phil
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Author: Jonathan KirwanDate: 17:38 22-12-06
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On 21 Dec 2006 16:45:57 -0800, mrdarrett@gmail.com wrote:
>Jonathan Kirwan wrote:
>
>...
>
>> >So here's a spec sheet of a 2N3055 (what a beauty):
>> >http://www.st.com/stonline/books/pdf/docs/4079.pdf
>>
>> No curves to speak of. I think you'll find this better:
>>
>> http://www.onsemi.com/pub/Collateral/2N3055-D.PDF
>>
>> Take a look at Figure 6. To read it, just look at the curve for "4.0
>> A". This is your collector current curve. Now, note the y-axis is
>> your Vce you can expect to get. Lower is better, here. Then look at
>> the x-axis. This is your base current. You can see it knee-over
>> around 200mA (1/20th of collector current) at around Vce=0.55V and
>> that if you push this to 400mA (1/10th of the collector current) you
>> get about 0.50V. As you can see, cramming in more base current pushes
>> the Vce down a little bit (50mV), but that you have to double the base
>> current to do it. I think that curve will help you think about this.
>
>Actually, by "what a beauty" I'd meant the pic of the TO-3. ;-)
I was kind of thinking that.
>Ok. I found Figure 6, and I was a little puzzled by it. Let's say I
>have a light bulb (or, for argument's sake, simply a fuse!) in series
>between +12V and the collector, and the emitter is grounded.
>
>Won't V_ce be 12V?
That depends. If you are using a dead short to +12, then probably the
dead short will win out and then, yes, the Vce will be near 12V, for a
short time, until it blows up. However, that isn't the usual case.
Your BJT has to be big enough to handle the current needed to allow
its Vce to be low when turned on in the way I was discussing earlier.
One way to explain what I'm saying is this: You have some kind of
load attached between the collector and some supply rail. Mentally
assume for a moment that this load has some fixed resistance to it
that we can use for talking purposes. That figure can be 100 ohms or
1000 ohms or 1 ohm or .1 ohm. Doesn't matter. Just pick a number and
stick with it for a moment. Then select a voltage for your supply
rail. 12V is fine. Or pick something else. We'll assume that the
emitter is connected to the ground reference, for talking purposes.
Whatever you pick for the supply voltage and the load resistance, you
can now compute a current from them, I=Vsupply/Rload, that tells you
the maximum current possible if the BJT were a perfect switch and the
Vce across it was exactly 0V. If your transistor isn't able to stand
this current, then you need a bigger one. If your transistor _can_
stand that level of current, then when you drive enough base current
into it to equal about 1/10th or 1/5th of that level, you can be
fairly sure that the transistor has a Vce that is small -- say .4V, or
.6V, or .8V, or even .03V. But with that much base current flooding
through, it will be able to pull that collector down pretty far.
All this assumes that the BJT can stand it.
So now, think about a fuse. Call it 0.01 ohms. If you had a 12V
rail, then the maximum current would be 12/.01=1200A. Now, assuming
you had a fuse that was rated for 1500A, let's say, and a BJT that was
rated for a collector current of 2000A, and assuming you supplied some
1/5th of that, or 400A of base current, and assuming you had a power
supply that could meet these requirements and more and that everything
could easily dissipate 12*1200A = 14,400 watts without going up in
smoke (this, I need to see), then your BJT might just be able to pull
its collector down, less than 1V.
Of course, none of that is very likely except that the fuse might be
.01 ohms and your voltage rail could be 12V. If you supplied a MJ2955
with a base current of 1A (a not unreasonable possibility), then you
could expect that the MJ2955's collector would be pulled down below 1V
or so if the load resistance were such that if 10A flowed through it
then the voltage drop across it would similarly allow the collector to
go below 1V. But if you were using a fuse, the fuse would probably
just blow, since it's load resistance probably isn't an ohm or more.
So suppose you had a 10 ohm resistor as the collector load, here? Then
if the collector current was 1.15A, which it could easily be if you
were sinking in 1A of base current, then the load resistance would be
dropping 11.5V, which means the collector would be .5V. The actual
case here, looking over Figure 6 on the Ic=1A curve (close to 1.15A)
and trying to follow it all the way over to where Ib=1A, would
probably be more like a Vce of .1V. So actually, you'd get a
collector current of (12-.1)/10 = 1.19A with a base current of 1A. And
yes, the Vce would be very low. But if you knew your load was 10 ohms
and you could estimate that the current should be about 1.1A to 1.2A,
then you can also see that a base current of 15mA-20mA would suffice
to get the Vce low enough for your needs. Just to be sure, you could
push it to 50mA. As you can see, this is about 1/50th to 1/20th of
the collector current. Which is why 1/10th is an extremely safe bet
in most circumstances.
>Or is this a different V_ce?
Yes and no. I hope the above explanation clears this up.
>Ok, so the name of the game is to minimize V_ce (as much as is
>practical; the knee-point is probably good enough.)
When you are thinking of the transistor as a switch, then you want the
voltage dropped by the switch itself to be as low as possible so that
it mimics a switch as well as it can. Another reason for getting the
Vce low is that then the transistor itself reduces its dissipation (so
long as you aren't going nuts in over-driving the base.) The total
dissipation in the BJT switch is roughly Vce*Ic+Vbe*Ib. That is what
you often want to minimize in your switch. In saturation (BJT used as
a switch), Vce <= Vb. Ic is usually a lot larger than Ib (by a factor
of 10 or more), and Vce usually is NOT 10 times smaller than Vbe, so
the first part of the equation more often dominates. You could focus
on it. On the other hand, as you crank up Ib to larger values, you
will drop Vce to smaller values -- which helps... up to a point. At
some further point, though, increasing Ib doesn't materially change
Vce anymore and it just makes dissipation worse. You can see this in
Figure 6, as you see the Ic curve flattening out towards the right.
Driving up Ib more than a certain point ceases to have much impact on
Vce.
Now, these curves are 'typical' which means that many times these
curves are close. But if you want to tolerate worst cases, you will
need to push Ib further up to the right on Figure 6 than just
eyeballing it may tell you. Just so that if you get a part that is on
the wrong side of the distribution it will still be adequately driven.
So although the curves may suggest to you that Ib=Ic/20 is more than
good enough, it may not be. You need to look over the minimums and
maximums and compare them with the typicals to get an idea about it.
Jon
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