About transistors
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If I want to control high current by using transistors,let's say
5a,how much curren do I need to give from the PIN,any formular for
this calcualtion?
Thank you.
5a,how much curren do I need to give from the PIN,any formular for
this calcualtion?
Thank you.
Comments
I hope someone will make sure I am putting this correctly, but you
need to find the Hfe for that unit.
take the total current thru the transistor in your case 5Amps.
assuming 12 volts as 5A at 5V is a lot of transformer.
figure your total amps as the supplied amps less the load. for
instance a relay coil has resistance and will use slighty less. just
calculate for ohms law.
but, for a simple start, assume 5 amps.
divide the total current thru the transistor by the Hfe of the
transistor. that is the current needed to turn on the transistor.
assume a Darlingotn with an Hfe of 1,000.
5 Amps divided by 1,000hfe = 5 mA
5/1000 = 5mA
If you had a high power transistor on hand, but it has an hfe of 50
then 5/50=100mA. you could use that transistor AND drive IT with a
pn2222. that 100mA needed to drive the primary transistor from the
2222 would be 100/50 = 2mA (50 being the hfe for the 2222) The high
power transistor would use the power supply current and not draw it
from the Stamp.
This is called a Darlington pair, one transistor driving another.
Maybe someone will post how to select a MOSFET ?
Dave
--- In basicstamps@yahoogroups.com, "baosheng_zhu <ted616@v...>"
<ted616@v...> wrote:
> If I want to control high current by using transistors,let's say
> 5a,how much curren do I need to give from the PIN,any formular for
> this calcualtion?
> Thank you.
davemucha@j... writes:
> Hi Bao,
>
> I hope someone will make sure I am putting this correctly, but you
> need to find the Hfe for that unit.
>
> take the total current thru the transistor in your case 5Amps.
> assuming 12 volts as 5A at 5V is a lot of transformer.
>
> figure your total amps as the supplied amps less the load. for
> instance a relay coil has resistance and will use slighty less. just
> calculate for ohms law.
>
> but, for a simple start, assume 5 amps.
> divide the total current thru the transistor by the Hfe of the
> transistor. that is the current needed to turn on the transistor.
>
> assume a Darlingotn with an Hfe of 1,000.
> 5 Amps divided by 1,000hfe = 5 mA
> 5/1000 = 5mA
>
> If you had a high power transistor on hand, but it has an hfe of 50
> then 5/50=100mA. you could use that transistor AND drive IT with a
> pn2222. that 100mA needed to drive the primary transistor from the
> 2222 would be 100/50 = 2mA (50 being the hfe for the 2222) The high
> power transistor would use the power supply current and not draw it
> from the Stamp.
>
> This is called a Darlington pair, one transistor driving another.
>
> Maybe someone will post how to select a MOSFET ?
>
> Dave
>
Your analysis of how to calc required current using HFE is correct.
There are many logic level mosfets that will also do the job, such as a
IRL540N(36A). Here is one supplier of them;
http://www.glitchbuster.com
[noparse][[/noparse]Non-text portions of this message have been removed]
--- In basicstamps@yahoogroups.com, "Dave Mucha <davemucha@j...>"
<davemucha@j...> wrote:
> Hi Bao,
>
> I hope someone will make sure I am putting this correctly, but you
> need to find the Hfe for that unit.
>
> take the total current thru the transistor in your case 5Amps.
> assuming 12 volts as 5A at 5V is a lot of transformer.
>
> figure your total amps as the supplied amps less the load. for
> instance a relay coil has resistance and will use slighty less.
just
> calculate for ohms law.
>
> but, for a simple start, assume 5 amps.
> divide the total current thru the transistor by the Hfe of the
> transistor. that is the current needed to turn on the transistor.
>
> assume a Darlingotn with an Hfe of 1,000.
> 5 Amps divided by 1,000hfe = 5 mA
> 5/1000 = 5mA
>
> If you had a high power transistor on hand, but it has an hfe of 50
> then 5/50=100mA. you could use that transistor AND drive IT with
a
> pn2222. that 100mA needed to drive the primary transistor from
the
> 2222 would be 100/50 = 2mA (50 being the hfe for the 2222) The
high
> power transistor would use the power supply current and not draw
it
> from the Stamp.
>
> This is called a Darlington pair, one transistor driving another.
>
> Maybe someone will post how to select a MOSFET ?
>
> Dave
>
>
>
>
>
>
>
>
>
>
>
>
> --- In basicstamps@yahoogroups.com, "baosheng_zhu <ted616@v...>"
> <ted616@v...> wrote:
> > If I want to control high current by using transistors,let's say
> > 5a,how much curren do I need to give from the PIN,any formular
for
> > this calcualtion?
> > Thank you.
is the selection of a MOSFET done exactly the same way ?
I was not sure if the MOSFET uses hfe for the gain ?
Dave
--- In basicstamps@yahoogroups.com, smartdim@a... wrote:
> In a message dated 12/27/2002 7:58:01 AM Pacific Standard Time,
> davemucha@j... writes:
>
> > Hi Bao,
> >
> > I hope someone will make sure I am putting this correctly, but
you
> > need to find the Hfe for that unit.
> >
> > take the total current thru the transistor in your case 5Amps.
> > assuming 12 volts as 5A at 5V is a lot of transformer.
> >
> > figure your total amps as the supplied amps less the load. for
> > instance a relay coil has resistance and will use slighty less.
just
> > calculate for ohms law.
> >
> > but, for a simple start, assume 5 amps.
> > divide the total current thru the transistor by the Hfe of the
> > transistor. that is the current needed to turn on the transistor.
> >
> > assume a Darlingotn with an Hfe of 1,000.
> > 5 Amps divided by 1,000hfe = 5 mA
> > 5/1000 = 5mA
> >
> > If you had a high power transistor on hand, but it has an hfe of
50
> > then 5/50=100mA. you could use that transistor AND drive IT with
a
> > pn2222. that 100mA needed to drive the primary transistor from
the
> > 2222 would be 100/50 = 2mA (50 being the hfe for the 2222) The
high
> > power transistor would use the power supply current and not draw
it
> > from the Stamp.
> >
> > This is called a Darlington pair, one transistor driving another.
> >
> > Maybe someone will post how to select a MOSFET ?
> >
> > Dave
> >
>
> Your analysis of how to calc required current using HFE is correct.
>
> There are many logic level mosfets that will also do the job, such
as a
> IRL540N(36A). Here is one supplier of them;
> http://www.glitchbuster.com
>
>
>
> [noparse][[/noparse]Non-text portions of this message have been removed]
let you do the math.
Dave
--- In basicstamps@yahoogroups.com, "baosheng_zhu <ted616@v...>"
<ted616@v...> wrote:
> Thank you,just a small question,what is hfe?
>
>
> --- In basicstamps@yahoogroups.com, "Dave Mucha <davemucha@j...>"
> <davemucha@j...> wrote:
> > Hi Bao,
> >
> > I hope someone will make sure I am putting this correctly, but
you
> > need to find the Hfe for that unit.
> >
> > take the total current thru the transistor in your case 5Amps.
> > assuming 12 volts as 5A at 5V is a lot of transformer.
> >
> > figure your total amps as the supplied amps less the load. for
> > instance a relay coil has resistance and will use slighty less.
> just
> > calculate for ohms law.
> >
> > but, for a simple start, assume 5 amps.
> > divide the total current thru the transistor by the Hfe of the
> > transistor. that is the current needed to turn on the transistor.
> >
> > assume a Darlingotn with an Hfe of 1,000.
> > 5 Amps divided by 1,000hfe = 5 mA
> > 5/1000 = 5mA
> >
> > If you had a high power transistor on hand, but it has an hfe of
50
> > then 5/50=100mA. you could use that transistor AND drive IT with
> a
> > pn2222. that 100mA needed to drive the primary transistor from
> the
> > 2222 would be 100/50 = 2mA (50 being the hfe for the 2222) The
> high
> > power transistor would use the power supply current and not draw
> it
> > from the Stamp.
> >
> > This is called a Darlington pair, one transistor driving another.
> >
> > Maybe someone will post how to select a MOSFET ?
> >
> > Dave
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > --- In basicstamps@yahoogroups.com, "baosheng_zhu <ted616@v...>"
> > <ted616@v...> wrote:
> > > If I want to control high current by using transistors,let's
say
> > > 5a,how much curren do I need to give from the PIN,any formular
> for
> > > this calcualtion?
> > > Thank you.
ted616@v... writes:
> Thank you,just a small question,what is hfe?
>
>
> --- In basicstamps@yahoogroups.com, "Dave Mucha <davemucha@j...>"
> <davemucha@j...> wrote:
> >Hi Bao,
> >
> >I hope someone will make sure I am putting this correctly, but you
> >need to find the Hfe for that unit.
> >
> >take the total current thru the transistor in your case 5Amps.
> >assuming 12 volts as 5A at 5V is a lot of transformer.
> >
> >figure your total amps as the supplied amps less the load. for
> >instance a relay coil has resistance and will use slighty less.
> just
> >calculate for ohms law.
> >
> >but, for a simple start, assume 5 amps.
> >divide the total current thru the transistor by the Hfe of the
> >transistor. that is the current needed to turn on the transistor.
> >
> >assume a Darlingotn with an Hfe of 1,000.
> >5 Amps divided by 1,000hfe = 5 mA
> >5/1000 = 5mA
> >
> >If you had a high power transistor on hand, but it has an hfe of 50
> >then 5/50=100mA. you could use that transistor AND drive IT with
> a
> >pn2222. that 100mA needed to drive the primary transistor from
> the
> >2222 would be 100/50 = 2mA (50 being the hfe for the 2222) The
> high
> >power transistor would use the power supply current and not draw
> it
> >from the Stamp.
> >
> >This is called a Darlington pair, one transistor driving another.
> >
> >Maybe someone will post how to select a MOSFET ?
> >
> >Dave
> >
> >
HFE is a transistor paramater that describes the current gain of the
transistor. For example if a transistor has an HFE of 100 then the current
into the base lead will be amplified by a factor of 100 for the collector
current.
Typically a user knows what the collector current will be (the load current).
Divide the load current by hfe to get the bas current. Once you know the
required base current, you select the correct size resistor that will yield
the desired base current.
Example: Load current is one amp. Transistor HFE is 100. 5 volts will drive
the base.
1/100 = 10 ma
5 v / 10 ma = 500 ohms which is your value for a base resistor.
This works most of the time, but to be more precise you need to subtract the
forward biased base emitter diode drop (~0.7v) for the 5v drive....
4.3 / .01 430 ohms......
[noparse][[/noparse]Non-text portions of this message have been removed]
The MOSFETS are selected by a different concept, and are generally easier.
Traditional "BJT" transistors are generally thought of as current
amplifiers, so "hfe" comes into play.... How much input current do I need to
control the load current and have the transistor "saturated" so it is fully
on?
With the MOSFETS, you don't have any appreciable steady input current once
the gate is charged. You can think of the gate as a small capacitor. That
is simplified, but good enough for what we're doing here. We need to apply
enough voltage to the gate to turn the transistor fully on, which achieves
minimum resistance between the "drain" and "source" for the load current.
As with any switch, the lower the resistance, the better.
The IRF510 (or IRF511), which is sometimes suggested for Stamp circuits,
does not fully turn on with a 5V signal at the gate. It can work if you
aren't pushing the device to the limit, but you always have more resistance
to your load current than if it was fully turned on (which would require >5V
gate voltage. That is why I really like (and sell) the logic voltage
MOSFETS which do fully turn on with a 5V signal. With it fully turned on,
you have lower resistance which translates to more power delivered to your
load, and therefore less power turned into heat in your transistor. That
may mean not needing a heat sink, or needing a smaller one than you would
otherwise.
Darlingtons, for example, by their nature have a "one diode" voltage drop
built in. So even when a Darlington is fully turned on you have that
voltage drop, which causes power loss to your load (and heating of the
transistor). Now if you are controlling LEDs and would have a current
limiting resistor anyway, maybe it isn't a big deal. But if you are
controlling motors, or other devices that you want to receive full power
from your supply, the extra loss in a Darlington doesn't make me too happy.
You can instead select a MOSFET with a current rating and "on resistance"
value that will result in low power loss in the transistor.
The gate capacitance isn't a big deal unless you need to duty-cycle the
MOSFET at a high frequency. If that is the case, you may need to add some
additional circuitry between the Stamp pin and the MOSFET gate. For slower
switching, that is generally unnecessary. For example, in the
Microcontroller Application Cookbook, Matt Gilliland shows the MOSFET gates
connected directly to the Stamp pins. I usually prefer to add a 50 to 100
ohm resistor in series near the gate, which is suggested in some data sheets
to reduce the chance of oscillation.
When you consider that the IRL520N costs only 68 cents, is easily turned on
fully by a Stamp pin, can switch 10A continuous and up to 35A pulsed, and is
a "no brainer" connection to the Stamp I/O pin, it's pretty nice. If you
need to handle more current, or want an even lower on resistance, the
IRL530N and IRL540N will do the trick. Just pull one out and plug in the
"bigger" one -- no "hfe" to worry about.
Randy
www.glitchbuster.com
> Hi Randy,
>
> is the selection of a MOSFET done exactly the same way ?
>
> I was not sure if the MOSFET uses hfe for the gain ?
>
> Dave
>