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transistor theory problem

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philipsi

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how to make a transistor a switch in a circuit?for example if i want to turn on a motor connected to its collector,what should i do?
 

philipsi

New Member
how to connect the power mosfet to function as a switch?is it same like bipolar transistor?do i need resistor in series with its gate?
 

Russlk

New Member
The bipolar transistor is a current operated device. To drive a motor, you should supply 1/10 of the motor current to the base, or if the transistor is a darlington, 1/100 of the motor current. The mosfet is a voltage operated device. You should get the data sheet to see how much voltage is required to turn it on, but typically 10 volts will do it. Most mosfets are enhancement mode, that is, it takes a positive voltage to turn on an N type. There are a few depletion mode mosfets available, these are normally on and require a negative voltage to turn an N type off.
 

philipsi

New Member
if i want to use a output of a PIC microcontroller to turn the power mosfet such as IRFZ44, what should i do?is it possible?
 

fat-tony

Member
philipsi said:
if i want to use a output of a PIC microcontroller to turn the power mosfet such as IRFZ44, what should i do?is it possible?
most likely. give me a few minutes to look up data sheets.
 

fat-tony

Member
Using the data sheet available at http://www.remcomplekt.ru/datasheet/trz/IRFZ44.PDF

It looks like this will work just fine. The PICmicro I/O pin should be putting out a GND or a +5V signal. When grounded, the IRF shouldn't conduct, and since +5V > Vgs(thresh), it will conduct just fine when turned on. The only thing you have to be careful about is supply voltage. I'm assuming that you're just going to be using a 12V supply or something like that, so it should be OK... but if you're going higher than around 60V, you're going to have problems.
 

philipsi

New Member
i just connected a simple testing circuit on the breadboard.it is shown below.at first,i thought that the circuit work properly since the motor could run.but when i removed the the 5V power supply from its gate terminal, the motor still running!!!!???? what happen to this circuit?did i connect it wrongly?plz help
 

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Roff

Well-Known Member
The intrinsic gate capacitance will hold a charge for quite a while, holding the MOSFET on. Add a 1k resistor from gate to GND which is always connected. This will discharge the capacitance when you remove the 5 volt supply. You should also connect a 1N4001 (or equivalent) diode across the motor, cathode to +12v. This will protect the MOSFET from the flyback voltage spike that will otherwise occur when the current through the motor's inductance is interrupted by turning off the MOSFET.
 

philipsi

New Member
if i want to use the output of the PIC microcontroller to drive the power mosfet,what should i do to make the power mosfet fully ON so that power disspation is minimised???
the output of microcontroller seems not enough to fully turn on the power mosfet

is it necessary to place a current limiting resistor in series with the gate of the powe mosfet?
 

fat-tony

Member
philipsi said:
if i want to use the output of the PIC microcontroller to drive the power mosfet,what should i do to make the power mosfet fully ON so that power disspation is minimised???
the output of microcontroller seems not enough to fully turn on the power mosfet

is it necessary to place a current limiting resistor in series with the gate of the powe mosfet?
http://www.electro-tech-online.com/custompdfs/2003/12/irfz44n.pdf

Reading from the datasheet, Vgs(thresh) looks like between 2.0V and 4.0V. Presumably, the pic puts out a 5V high signal, so that should be good. The rub, however, has to do with Rds (drain-source resistance), and the Id.

What kind of resistance does the motor have? Rds(on) for the IRFZ44 is very tiny, so it shouldn't matter a whole lot, but it's still a necessary part of the analysis. As long as the motor has a resistance > .25ohm, it should be fine (ohm, not kohm).

The math behind the discharge makes me smile :). As it turns out, a resistor from the gate to ground would have to be... 13.6Mohm for a .1 second discharge. Any smaller value, and the MOSFET should just switch faster. Basically, grab a resistor out of your box that is larger than 10k, and put it there. It'll work fine.

What do you mean that you don't think the mosfet is turning on all the way? Is the motor spinning slower than it would if you hook it up straight to 12V?
 

fat-tony

Member
This is still bugging me :)

I looked into it a bit more, and it looks like (not quite the same here..) at 25V, and 5Vgs, the mosfet should conduct around 20A. That should be plenty to power your motor, no?

I'm not sure how the curve changes when Vds=12V instead of 25V, but it shouldn't be too different.
 

Roff

Well-Known Member
Fat Tony said,
The math behind the discharge makes me smile . As it turns out, a resistor from the gate to ground would have to be... 13.6Mohm for a .1 second discharge. Any smaller value, and the MOSFET should just switch faster. Basically, grab a resistor out of your box that is larger than 10k, and put it there. It'll work fine.
True, but turning off (or on) a power MOSFET (or power anything, for that matter) slowly is not a good idea. Peak power is dissipated during the transition time. For the protection of the transistor, this time needs to be short. In power switching, 100 milliseconds is not short.
 

philipsi

New Member
what i mean is that,if i want to use power mosfet as a switch,i have to drive the power mosfet into ohmic region and bypass the active region.to drive into ohmic region,according to datasheet (and wat i read from a textbook) is that the supply voltage to the gate terminal should be about 10V and this is not enough from microcontroller pin. only when the power mosfet is driven by that voltage,it dissipate less power (since the resistance is near zero and all the voltage is dropped on the load coonected to the drain)then how to let the gate receive about 10V?

i consider to use a transistor to drive it but dont know how?or any other way?

once again, do i need to connected a current limiting resistor in series to the gate terminal?
 

Roff

Well-Known Member
There are lots of power MOSFETs that will turn on with 5v Vgs. Do a Google search for "logic level mosfet".
 
The bipolar transistor is a current operated device. To drive a motor, you should supply 1/10 of the motor current to the base, or if the transistor is a darlington, 1/100 of the motor current. The mosfet is a voltage operated device. You should get the data sheet to see how much voltage is required to turn it on, but typically 10 volts will do it. Most mosfets are enhancement mode, that is, it takes a positive voltage to turn on an N type. There are a few depletion mode mosfets available, these are normally on and require a negative voltage to turn an N type off.
Depending on the bjt device & current value, 1/10 may be excessive base drive. I've found that in some instances 1/20. or even 1/50, is adequate base drive. I generally look for the worst case minimum beta value, derate (reduce) for lowest temperature, then take 1/2 to 2/3 of that value for the base drive. This results in lower base drive than the rule of thumb "1/10" value.

Of course if the beta has a minimum value of 20 or 25, derating to low temp may result in 15 to 20, so that 1/10 base drive is optimum. But many bjt parts may have a worst case min beta of 80 or more. At low temp it could drop to 50, so that a base drive of 1/25 to 1/30 of collector current is more than enough to assure saturation under all conditions.

The reason I mention this is that if the source providing the drive signal is a micro, or other low current device, a base drive value of 1/10 might exceed its capability. I advise the user to inspect the data sheet & compute the needed value. If only 1/25 base drive is needed, why use 1/10? The source delivering the unneeded high current value will get hotter and/or droop in voltage. To summarize, 1/10 may be necessary, but if not, I recommend using a value per the above.
 

Jaguarjoe

Member
1/10 is used because that takes the transistor out of the active region and into the saturated region where Vces<Vbe, where the B-C junction is forward biased.
 
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