Simple transistor Q...my head isnt working

[Coyoteboy]

BC183 NPN taking base input from a 5v IC output (through a resistor,
assuming 10K for now) but need it to switch on a 12v 500 ohm relay.

According to the spec sheet all voltage/pin differences are within tolerance
and the collector-emitter current will be around 20mA (BC183 can handle
100mA) - i am ok to just throw the relay (with emf protection diode) in
between a 12v rail and the collector and tie the emitter to the 12v ground
am i not? I'm not going to back feed my IC with 12v or anything stupid am i?

Thanks guys, my ideas are great, my knowledge small

J

 

[Willco]

Don't worry, there isn't enough current through the 10K resistor
to damage the the IC and most IC outputs are voltage protected.
Unless the transistor is damaged the most voltage on the base is
..65 volts.


Jack
Willco Electronics

 

[Willco]

Don't worry, there isn't enough current through the 10K resistor
to damage the the IC and most IC outputs are voltage protected.
Unless the transistor is damaged the most voltage on the base is
..65 volts.


Jack
Willco Electronics

 

[Coyoteboy]

Smashing, thanks Jack.

James

Don't worry, there isn't enough current through the 10K resistor
to damage the the IC and most IC outputs are voltage protected.
Unless the transistor is damaged the most voltage on the base is
.65 volts.


Jack
Willco Electronics

 

[Coyoteboy]

OK, so it didnt work - for some odd reason it just wouldnt switch on with
the output from the picaxe. So i substitued in a big IRF620 mosfet, no gate
resisor, relay and diode on teh drain side and the source hooked straight to
0v, as described EXACTLY in the picaxe manual. The IC's 5V output now wont
get above 3v and the transistor wont turn on. Any ideas?! I assumed the
mosfet is drawing too much current but it has a high internal resistance so
shouldnt?? The IC can source 20mA@5v.
Anyone with a clue please help me!

James

 

[Winston]

OK, so it didnt work - for some odd reason it just wouldnt switch on with
the output from the picaxe. So i substitued in a big IRF620 mosfet, no gate
resisor, relay and diode on teh drain side and the source hooked straight to
0v, as described EXACTLY in the picaxe manual. The IC's 5V output now wont
get above 3v and the transistor wont turn on. Any ideas?! I assumed the
mosfet is drawing too much current but it has a high internal resistance so
shouldnt?? The IC can source 20mA@5v.
Anyone with a clue please help me!

James

Try a pullup resistor between your MOSFET gate and +12V. Use Ohm's law to
size the resistor so that it sources less current into the output of your
PIC processor than the output can tolerate. Put less clumsily, if your
processor can sink say 10 mA, choose a 1.2K ohm 1/4 W resistor to limit
logic '0' current at or below 10 mA. That way, your PIC need not 'source'
all the current needed to forward - bias the gate when it outputs a logic '1'.

Have a look at:
http://www.irf.com/product-info/datasheets/data/irf620.pdf

Third page down, labeled page '181'.
Lower left corner. The chart labeled 'Fig 3. Typical Transfer Characteristics'.
Shows that with a gate bias of 12 V, the MOSFET will be very well saturated.
The IRF620 can tolerate up to 20 V of Vgs so don't worry about the 12 V level.

If the gate voltage does not increase with the resistor in place, consider
that the MOSFET might have a gate - to - source short or your PIC output might
be shorted to ground.

--Winston

 

[CoyoteBoy]

Try a pullup resistor between your MOSFET gate and +12V. Use Ohm's law to

size the resistor so that it sources less current into the output of your
PIC processor than the output can tolerate. Put less clumsily, if your
processor can sink say 10 mA, choose a 1.2K ohm 1/4 W resistor to limit

Ta for the pointer, but is that safe - the max input at the IC is 5v,
surely if I put in a 12v input through a resistor I might blow the
damn thing, unless I diode protect it ? I'm just being over-protective
as I only have on IC left until my next order comes through in a week
lol!

James

 

[Winston]

Ta for the pointer, but is that safe - the max input at the IC is 5v,
surely if I put in a 12v input through a resistor I might blow the
damn thing, unless I diode protect it ? I'm just being over-protective
as I only have on IC left until my next order comes through in a week
lol!

Yeah, if the processor is not capable of handling 12 V,
we need to accomodate that. Lets scratch the MOSFET.

Have a look at the datasheet for your processor.
The PICaxe hardware looks to be manufactured by Microchip

http://www.microchip.com

If your output pin can source a few milliamps, consider
connecting the base of an NPN Darlington transistor through
a resistor of say, 680 ohms. Parallel the resistor with
a diode so that the transistor is solidly turned off when
the pin goes to logic '0'.


With the emitter grounded, you should be able to drive your
12 V load connected between the 12 V rail and the collector
of the Darlington.
A diode connected across the emitter and collector will
protect the transistor from flyback current from the relay.

That would be way better.

--Winston

 

[crzndog]

OK, for a NMOSFET to work yo uwill need to make sure that the Vth of the MOSFET is around below 3V! Now the Gate Voltage (wrt to the Source [which should be grounded] ) should be positive. You'll prbably see about 3-5V from te PIC. Pretty normal. I would also add a parallel resistor (1K shoudl be OK) between the Gate and Source terminals of the transistor since it is such a high impedance/resistance that it could pick up noise.

The drain of the transistor should connect to the LOAD and the other end of the load shoudl connect to 12V. This is very important!! Do not try and drive the load by making the Source and GND the "load" connections.


ie - the correct way is:

+12V
|
LOAD
|
|- d
-----g------|
|- s
|
GND



THe reason it would not work the other way round is that you are effectively moving the Vgs voltage to a higher level if the load was connected in the Source leg. The load would develop a voltage across it and hence the voltage on G would have to be higher to achieve Vgs>= Vth.

 

[Coyoteboy]

Cheers for that, ive copied all my replies into a FAQ for myself lol, i always make the same mistakes!

Cheers again
James

OK, for a NMOSFET to work yo uwill need to make sure that the Vth of the MOSFET is around below 3V! Now the Gate Voltage (wrt to the Source [which should be grounded] ) should be positive. You'll prbably see about 3-5V from te PIC. Pretty normal. I would also add a parallel resistor (1K shoudl be OK) between the Gate and Source terminals of the transistor since it is such a high impedance/resistance that it could pick up noise.

The drain of the transistor should connect to the LOAD and the other end of the load shoudl connect to 12V. This is very important!! Do not try and drive the load by making the Source and GND the "load" connections.


ie - the correct way is:

+12V
|
LOAD
|
|- d
-----g------|
|- s
|
GND



THe reason it would not work the other way round is that you are effectively moving the Vgs voltage to a higher level if the load was connected in the Source leg. The load would develop a voltage across it and hence the voltage on G would have to be higher to achieve Vgs>= Vth.