Reply by Jasen Betts ●November 10, 20142014-11-10
On 2014-11-09, edward.ming.lee@gmail.com <edward.ming.lee@gmail.com> wrote:

> On Sunday, November 9, 2014 9:09:28 AM UTC-8, rickman wrote:
>> On 11/9/2014 4:50 AM, gm wrote:
>> > On 09.11.2014 09:28, rickman wrote:
>> >> On 11/9/2014 3:09 AM, gm wrote:
>> >>> On 09.11.2014 05:10, Jasen Betts wrote:
>> >>>> On 2014-11-06, gm <notMyMail@mail.not> wrote:
>> >>>>> On 06.11.2014 16:44, Spehro Pefhany wrote:
>> >>>>>> On Thu, 06 Nov 2014 15:24:26 +0100, gm <notMyMail@mail.not> wrote:
>> >>>>>>
>> >>>>>>> Hi.
>> >>>>>>>
>> >>>>>>> I didnt play much with RPI but now we have finish one project
>> >>>>>>> where we
>> >>>>>>> are triggering GPIO. The PCB relay board should be triggered from
>> >>>>>>> RPI
>> >>>>>>> GPIO and the door should be opened. Now, before i connect PCB to
>> >>>>>>> RPI i
>> >>>>>>> want to be sure that nothing goes wrong. This is the pcb schematics
>> >>>>>>> that
>> >>>>>>> we use for the relay trigger.
>> >>>>>>>
>> >>>>>>> http://tinypic.com/view.php?pic=10rmrsh&s=8#.VFuEUa0zdag
>> >>>>>>>
>> >>>>>>> What do you think . can i connect this directly to RPI GPIO, and
>> >>>>>>> which
>> >>>>>>> pins can i use ?
>> >>>>>>>
>> >>>>>>> Thank you in advance !
>> >>>>>>
>> >>>>>> Yes, it should work fine.
>> >>>>>>
>> >>>>>> Reduce R1 to 560R to compensate for the 3.3V output of the Rpi port
>> >>>>>> pins, and use the 5V power for the relay (and a 5V relay coil).
>> >>>>>>
>> >>>>>> Double check the connections before applying power.
>> >>>>>>
>> >>>>> -----------
>> >>>>> should i reduce the resistor value ?
>> >>>>
>> >>>> yes, the too high a resistor can cause excessive heating in the
>> >>>> transistor, and then the transistor fails.
>> >>>>
>> >>>>> Am asking because i have made the pcb and i have to test it today and
>> >>>>> i dont have 0,5K near me. I have to buy it tommorow
>> >>>>
>> >>>> Anything in the 200 Ohms to 600 Ohms range will suffice,
>> >>>> eg: you could parallell two 1K resistors to get 500 Ohms.
>> >>>> or use a 330 or a 470 if you have them on-hand.
>> >>>>
>> >>> ----------------------
>> >>> OK i can do that but i dont understand one thing:
>> >>>
>> >>> Dont i need to change R1 from 1 to 1,2 KOhm ?
>> >>> Because ih i leave it at 1K the base current will be 2,6mA instead of
>> >>> 2mA. If the raspbery PI port can handle and give this much or current
>> >>> then there is no problem. Nothing should burn out ... OR am i missing
>> >>> something ?
>> >>
>> >> When I looked up the transistor you are using it says the DC current
>> >> gain is minimum of 100.  You don't say what relay you are using or what
>> >> current it draws.  The base current limited by R1 should be no less than
>> >> 100 times smaller than the collector current which is whatever the relay
>> >> draws.  If the rPi outputs pull up to 3.3 volts, subtract 0.7 and divide
>> >> by the current to size the resistor maximum value.  You can get the I/O
>> >> voltage and the max current drive from the Broadcom data sheet for the
>> >> CPU.  If the CPU output can't drive enough current you can use two
>> >> transistors connecting the emitter of the first to the base of the
>> >> second.  Then the current gain of both transistors is multiplied.
>> >>
>> > -------------------
>> >    You don't say what relay you are using or what current it draws.
>> > *** the relay should give something like "impulse" to the trigger
>> > electronics of ramp gate. So there is no current involved...
>> > The impulse will trigger ramp to open.
>> 
>> I have no idea what you are talking about here.  In the schematic 
>> diagram you gave a link to there is a relay (K1) in the collector 
>> circuit of T1 bypassed by a diode.  There are two types of relays in 
>> use.  The first type is activated by a current being on or off.  The 
>> other type is a "latching" relay which uses one current to set it and 
>> another current to reset it.  This circuit is clearly using the former. 
>>   So the relay coil must pass some minimum amount of current in order to 
>> guarantee the contacts are in the active state (pull in current).  There 
>> is also a maximum current to guarantee the contact are in the inactive 
>> state called the drop out current.  You can safely ignore the drop out 
>> current, but you need to make sure T1 is biased to provide the minimum 
>> pull in current.  What is that value for your relay?
>
> A typical 5V DC relay (8 pins in DIP 16 size) is around 20mA to 30mA
> to activate, less to hold.  So, a 1K R(base), 3mA I(base) is plenty
> and is not too much either.  Just use 1K and be done with it.  We
> don't need to engineer this to death.     


he's opening a door, typical magnetic locks are 2A at 12v or 1A at 24V

a DIP relay isn't goind to work well in that application for long,

but he hasn't said what's closing the door or how many times he wants
to open it.






-- 
umop apisdn
Reply by rickman ●November 9, 20142014-11-09
On 11/9/2014 3:55 PM, edward.ming.lee@gmail.com wrote:

> On Sunday, November 9, 2014 9:09:28 AM UTC-8, rickman wrote:
>> On 11/9/2014 4:50 AM, gm wrote:
>>> On 09.11.2014 09:28, rickman wrote:
>>>> On 11/9/2014 3:09 AM, gm wrote:
>>>>> On 09.11.2014 05:10, Jasen Betts wrote:
>>>>>> On 2014-11-06, gm <notMyMail@mail.not> wrote:
>>>>>>> On 06.11.2014 16:44, Spehro Pefhany wrote:
>>>>>>>> On Thu, 06 Nov 2014 15:24:26 +0100, gm <notMyMail@mail.not> wrote:
>>>>>>>>
>>>>>>>>> Hi.
>>>>>>>>>
>>>>>>>>> I didnt play much with RPI but now we have finish one project
>>>>>>>>> where we
>>>>>>>>> are triggering GPIO. The PCB relay board should be triggered from
>>>>>>>>> RPI
>>>>>>>>> GPIO and the door should be opened. Now, before i connect PCB to
>>>>>>>>> RPI i
>>>>>>>>> want to be sure that nothing goes wrong. This is the pcb schematics
>>>>>>>>> that
>>>>>>>>> we use for the relay trigger.
>>>>>>>>>
>>>>>>>>> http://tinypic.com/view.php?pic=10rmrsh&s=8#.VFuEUa0zdag
>>>>>>>>>
>>>>>>>>> What do you think . can i connect this directly to RPI GPIO, and
>>>>>>>>> which
>>>>>>>>> pins can i use ?
>>>>>>>>>
>>>>>>>>> Thank you in advance !
>>>>>>>>
>>>>>>>> Yes, it should work fine.
>>>>>>>>
>>>>>>>> Reduce R1 to 560R to compensate for the 3.3V output of the Rpi port
>>>>>>>> pins, and use the 5V power for the relay (and a 5V relay coil).
>>>>>>>>
>>>>>>>> Double check the connections before applying power.
>>>>>>>>
>>>>>>> -----------
>>>>>>> should i reduce the resistor value ?
>>>>>>
>>>>>> yes, the too high a resistor can cause excessive heating in the
>>>>>> transistor, and then the transistor fails.
>>>>>>
>>>>>>> Am asking because i have made the pcb and i have to test it today and
>>>>>>> i dont have 0,5K near me. I have to buy it tommorow
>>>>>>
>>>>>> Anything in the 200 Ohms to 600 Ohms range will suffice,
>>>>>> eg: you could parallell two 1K resistors to get 500 Ohms.
>>>>>> or use a 330 or a 470 if you have them on-hand.
>>>>>>
>>>>> ----------------------
>>>>> OK i can do that but i dont understand one thing:
>>>>>
>>>>> Dont i need to change R1 from 1 to 1,2 KOhm ?
>>>>> Because ih i leave it at 1K the base current will be 2,6mA instead of
>>>>> 2mA. If the raspbery PI port can handle and give this much or current
>>>>> then there is no problem. Nothing should burn out ... OR am i missing
>>>>> something ?
>>>>
>>>> When I looked up the transistor you are using it says the DC current
>>>> gain is minimum of 100.  You don't say what relay you are using or what
>>>> current it draws.  The base current limited by R1 should be no less than
>>>> 100 times smaller than the collector current which is whatever the relay
>>>> draws.  If the rPi outputs pull up to 3.3 volts, subtract 0.7 and divide
>>>> by the current to size the resistor maximum value.  You can get the I/O
>>>> voltage and the max current drive from the Broadcom data sheet for the
>>>> CPU.  If the CPU output can't drive enough current you can use two
>>>> transistors connecting the emitter of the first to the base of the
>>>> second.  Then the current gain of both transistors is multiplied.
>>>>
>>> -------------------
>>>     You don't say what relay you are using or what current it draws.
>>> *** the relay should give something like "impulse" to the trigger
>>> electronics of ramp gate. So there is no current involved...
>>> The impulse will trigger ramp to open.
>>
>> I have no idea what you are talking about here.  In the schematic
>> diagram you gave a link to there is a relay (K1) in the collector
>> circuit of T1 bypassed by a diode.  There are two types of relays in
>> use.  The first type is activated by a current being on or off.  The
>> other type is a "latching" relay which uses one current to set it and
>> another current to reset it.  This circuit is clearly using the former.
>>    So the relay coil must pass some minimum amount of current in order to
>> guarantee the contacts are in the active state (pull in current).  There
>> is also a maximum current to guarantee the contact are in the inactive
>> state called the drop out current.  You can safely ignore the drop out
>> current, but you need to make sure T1 is biased to provide the minimum
>> pull in current.  What is that value for your relay?
>
> A typical 5V DC relay (8 pins in DIP 16 size) is around 20mA to 30mA to activate, less to hold.  So, a 1K R(base), 3mA I(base) is plenty and is not too much either.  Just use 1K and be done with it.  We don't need to engineer this to death.


Yes, it shouldn't be under-engineered either.  The OP asked a question 
about how much current the rPi output can handle.  I suppose you would 
answer that the same way you responded to the question about the 
relay... *most* MCU outputs are fine with 2.6 mA.  lol

Obviously the OP is new to this.  Why teach him sloppy habits rather 
than teaching him to find the basic facts and figure it out for himself?

BTW, if the relay really is 20 to 30 mA, then there is no need to drive 
the base with 2.6 mA.  With a minimum current gain of 100 just 0.3 mA is 
sufficient and 1 mA is more than enough.  So a 2 to 3 kohm resistor is a 
better choice if you don't know the output drive spec on the I/O pin.

-- 

Rick
Reply by ●November 9, 20142014-11-09
On Sunday, November 9, 2014 9:09:28 AM UTC-8, rickman wrote:

> On 11/9/2014 4:50 AM, gm wrote:
> > On 09.11.2014 09:28, rickman wrote:
> >> On 11/9/2014 3:09 AM, gm wrote:
> >>> On 09.11.2014 05:10, Jasen Betts wrote:
> >>>> On 2014-11-06, gm <notMyMail@mail.not> wrote:
> >>>>> On 06.11.2014 16:44, Spehro Pefhany wrote:
> >>>>>> On Thu, 06 Nov 2014 15:24:26 +0100, gm <notMyMail@mail.not> wrote:
> >>>>>>
> >>>>>>> Hi.
> >>>>>>>
> >>>>>>> I didnt play much with RPI but now we have finish one project
> >>>>>>> where we
> >>>>>>> are triggering GPIO. The PCB relay board should be triggered from
> >>>>>>> RPI
> >>>>>>> GPIO and the door should be opened. Now, before i connect PCB to
> >>>>>>> RPI i
> >>>>>>> want to be sure that nothing goes wrong. This is the pcb schematics
> >>>>>>> that
> >>>>>>> we use for the relay trigger.
> >>>>>>>
> >>>>>>> http://tinypic.com/view.php?pic=10rmrsh&s=8#.VFuEUa0zdag
> >>>>>>>
> >>>>>>> What do you think . can i connect this directly to RPI GPIO, and
> >>>>>>> which
> >>>>>>> pins can i use ?
> >>>>>>>
> >>>>>>> Thank you in advance !
> >>>>>>
> >>>>>> Yes, it should work fine.
> >>>>>>
> >>>>>> Reduce R1 to 560R to compensate for the 3.3V output of the Rpi port
> >>>>>> pins, and use the 5V power for the relay (and a 5V relay coil).
> >>>>>>
> >>>>>> Double check the connections before applying power.
> >>>>>>
> >>>>> -----------
> >>>>> should i reduce the resistor value ?
> >>>>
> >>>> yes, the too high a resistor can cause excessive heating in the
> >>>> transistor, and then the transistor fails.
> >>>>
> >>>>> Am asking because i have made the pcb and i have to test it today and
> >>>>> i dont have 0,5K near me. I have to buy it tommorow
> >>>>
> >>>> Anything in the 200 Ohms to 600 Ohms range will suffice,
> >>>> eg: you could parallell two 1K resistors to get 500 Ohms.
> >>>> or use a 330 or a 470 if you have them on-hand.
> >>>>
> >>> ----------------------
> >>> OK i can do that but i dont understand one thing:
> >>>
> >>> Dont i need to change R1 from 1 to 1,2 KOhm ?
> >>> Because ih i leave it at 1K the base current will be 2,6mA instead of
> >>> 2mA. If the raspbery PI port can handle and give this much or current
> >>> then there is no problem. Nothing should burn out ... OR am i missing
> >>> something ?
> >>
> >> When I looked up the transistor you are using it says the DC current
> >> gain is minimum of 100.  You don't say what relay you are using or what
> >> current it draws.  The base current limited by R1 should be no less than
> >> 100 times smaller than the collector current which is whatever the relay
> >> draws.  If the rPi outputs pull up to 3.3 volts, subtract 0.7 and divide
> >> by the current to size the resistor maximum value.  You can get the I/O
> >> voltage and the max current drive from the Broadcom data sheet for the
> >> CPU.  If the CPU output can't drive enough current you can use two
> >> transistors connecting the emitter of the first to the base of the
> >> second.  Then the current gain of both transistors is multiplied.
> >>
> > -------------------
> >    You don't say what relay you are using or what current it draws.
> > *** the relay should give something like "impulse" to the trigger
> > electronics of ramp gate. So there is no current involved...
> > The impulse will trigger ramp to open.
> 
> I have no idea what you are talking about here.  In the schematic 
> diagram you gave a link to there is a relay (K1) in the collector 
> circuit of T1 bypassed by a diode.  There are two types of relays in 
> use.  The first type is activated by a current being on or off.  The 
> other type is a "latching" relay which uses one current to set it and 
> another current to reset it.  This circuit is clearly using the former. 
>   So the relay coil must pass some minimum amount of current in order to 
> guarantee the contacts are in the active state (pull in current).  There 
> is also a maximum current to guarantee the contact are in the inactive 
> state called the drop out current.  You can safely ignore the drop out 
> current, but you need to make sure T1 is biased to provide the minimum 
> pull in current.  What is that value for your relay?


A typical 5V DC relay (8 pins in DIP 16 size) is around 20mA to 30mA to activate, less to hold.  So, a 1K R(base), 3mA I(base) is plenty and is not too much either.  Just use 1K and be done with it.  We don't need to engineer this to death.
Reply by rickman ●November 9, 20142014-11-09
On 11/9/2014 4:50 AM, gm wrote:

> On 09.11.2014 09:28, rickman wrote:
>> On 11/9/2014 3:09 AM, gm wrote:
>>> On 09.11.2014 05:10, Jasen Betts wrote:
>>>> On 2014-11-06, gm <notMyMail@mail.not> wrote:
>>>>> On 06.11.2014 16:44, Spehro Pefhany wrote:
>>>>>> On Thu, 06 Nov 2014 15:24:26 +0100, gm <notMyMail@mail.not> wrote:
>>>>>>
>>>>>>> Hi.
>>>>>>>
>>>>>>> I didnt play much with RPI but now we have finish one project
>>>>>>> where we
>>>>>>> are triggering GPIO. The PCB relay board should be triggered from
>>>>>>> RPI
>>>>>>> GPIO and the door should be opened. Now, before i connect PCB to
>>>>>>> RPI i
>>>>>>> want to be sure that nothing goes wrong. This is the pcb schematics
>>>>>>> that
>>>>>>> we use for the relay trigger.
>>>>>>>
>>>>>>> http://tinypic.com/view.php?pic=10rmrsh&s=8#.VFuEUa0zdag
>>>>>>>
>>>>>>> What do you think . can i connect this directly to RPI GPIO, and
>>>>>>> which
>>>>>>> pins can i use ?
>>>>>>>
>>>>>>> Thank you in advance !
>>>>>>
>>>>>> Yes, it should work fine.
>>>>>>
>>>>>> Reduce R1 to 560R to compensate for the 3.3V output of the Rpi port
>>>>>> pins, and use the 5V power for the relay (and a 5V relay coil).
>>>>>>
>>>>>> Double check the connections before applying power.
>>>>>>
>>>>> -----------
>>>>> should i reduce the resistor value ?
>>>>
>>>> yes, the too high a resistor can cause excessive heating in the
>>>> transistor, and then the transistor fails.
>>>>
>>>>> Am asking because i have made the pcb and i have to test it today and
>>>>> i dont have 0,5K near me. I have to buy it tommorow
>>>>
>>>> Anything in the 200 Ohms to 600 Ohms range will suffice,
>>>> eg: you could parallell two 1K resistors to get 500 Ohms.
>>>> or use a 330 or a 470 if you have them on-hand.
>>>>
>>> ----------------------
>>> OK i can do that but i dont understand one thing:
>>>
>>> Dont i need to change R1 from 1 to 1,2 KOhm ?
>>> Because ih i leave it at 1K the base current will be 2,6mA instead of
>>> 2mA. If the raspbery PI port can handle and give this much or current
>>> then there is no problem. Nothing should burn out ... OR am i missing
>>> something ?
>>
>> When I looked up the transistor you are using it says the DC current
>> gain is minimum of 100.  You don't say what relay you are using or what
>> current it draws.  The base current limited by R1 should be no less than
>> 100 times smaller than the collector current which is whatever the relay
>> draws.  If the rPi outputs pull up to 3.3 volts, subtract 0.7 and divide
>> by the current to size the resistor maximum value.  You can get the I/O
>> voltage and the max current drive from the Broadcom data sheet for the
>> CPU.  If the CPU output can't drive enough current you can use two
>> transistors connecting the emitter of the first to the base of the
>> second.  Then the current gain of both transistors is multiplied.
>>
> -------------------
>    You don't say what relay you are using or what current it draws.
> *** the relay should give something like "impulse" to the trigger
> electronics of ramp gate. So there is no current involved...
> The impulse will trigger ramp to open.


I have no idea what you are talking about here.  In the schematic 
diagram you gave a link to there is a relay (K1) in the collector 
circuit of T1 bypassed by a diode.  There are two types of relays in 
use.  The first type is activated by a current being on or off.  The 
other type is a "latching" relay which uses one current to set it and 
another current to reset it.  This circuit is clearly using the former. 
  So the relay coil must pass some minimum amount of current in order to 
guarantee the contacts are in the active state (pull in current).  There 
is also a maximum current to guarantee the contact are in the inactive 
state called the drop out current.  You can safely ignore the drop out 
current, but you need to make sure T1 is biased to provide the minimum 
pull in current.  What is that value for your relay?

-- 

Rick
Reply by ●November 9, 20142014-11-09
> Dont i need to change R1 from 1 to 1,2 KOhm ? Because ih i leave it at 1K=

 the base current will be 2,6mA instead of 2mA. If the raspbery PI port can=
 handle and give this much or current then there is no problem. Nothing sho=
uld burn out ... OR am i missing something ?

Don't know about the rPi, but if any controller can't handle 3mA GPIO outpu=
t, there is something wrong with it.  Most should handle 5mA.  Some handle =
20mA.


>=20
> ///////////////
> BTW. is someone of you working in Python ?
> ///////////////


Yes, Python, Perl, Php, Java, C and C++.  Why do you ask?
Reply by gm ●November 9, 20142014-11-09
On 09.11.2014 09:28, rickman wrote:

> On 11/9/2014 3:09 AM, gm wrote:
>> On 09.11.2014 05:10, Jasen Betts wrote:
>>> On 2014-11-06, gm <notMyMail@mail.not> wrote:
>>>> On 06.11.2014 16:44, Spehro Pefhany wrote:
>>>>> On Thu, 06 Nov 2014 15:24:26 +0100, gm <notMyMail@mail.not> wrote:
>>>>>
>>>>>> Hi.
>>>>>>
>>>>>> I didnt play much with RPI but now we have finish one project
>>>>>> where we
>>>>>> are triggering GPIO. The PCB relay board should be triggered from RPI
>>>>>> GPIO and the door should be opened. Now, before i connect PCB to
>>>>>> RPI i
>>>>>> want to be sure that nothing goes wrong. This is the pcb schematics
>>>>>> that
>>>>>> we use for the relay trigger.
>>>>>>
>>>>>> http://tinypic.com/view.php?pic=10rmrsh&s=8#.VFuEUa0zdag
>>>>>>
>>>>>> What do you think . can i connect this directly to RPI GPIO, and
>>>>>> which
>>>>>> pins can i use ?
>>>>>>
>>>>>> Thank you in advance !
>>>>>
>>>>> Yes, it should work fine.
>>>>>
>>>>> Reduce R1 to 560R to compensate for the 3.3V output of the Rpi port
>>>>> pins, and use the 5V power for the relay (and a 5V relay coil).
>>>>>
>>>>> Double check the connections before applying power.
>>>>>
>>>> -----------
>>>> should i reduce the resistor value ?
>>>
>>> yes, the too high a resistor can cause excessive heating in the
>>> transistor, and then the transistor fails.
>>>
>>>> Am asking because i have made the pcb and i have to test it today and
>>>> i dont have 0,5K near me. I have to buy it tommorow
>>>
>>> Anything in the 200 Ohms to 600 Ohms range will suffice,
>>> eg: you could parallell two 1K resistors to get 500 Ohms.
>>> or use a 330 or a 470 if you have them on-hand.
>>>
>> ----------------------
>> OK i can do that but i dont understand one thing:
>>
>> Dont i need to change R1 from 1 to 1,2 KOhm ?
>> Because ih i leave it at 1K the base current will be 2,6mA instead of
>> 2mA. If the raspbery PI port can handle and give this much or current
>> then there is no problem. Nothing should burn out ... OR am i missing
>> something ?
>
> When I looked up the transistor you are using it says the DC current
> gain is minimum of 100.  You don't say what relay you are using or what
> current it draws.  The base current limited by R1 should be no less than
> 100 times smaller than the collector current which is whatever the relay
> draws.  If the rPi outputs pull up to 3.3 volts, subtract 0.7 and divide
> by the current to size the resistor maximum value.  You can get the I/O
> voltage and the max current drive from the Broadcom data sheet for the
> CPU.  If the CPU output can't drive enough current you can use two
> transistors connecting the emitter of the first to the base of the
> second.  Then the current gain of both transistors is multiplied.
>

-------------------
   You don't say what relay you are using or what current it draws.
*** the relay should give something like "impulse" to the trigger 
electronics of ramp gate. So there is no current involved...
The impulse will trigger ramp to open.
Reply by rickman ●November 9, 20142014-11-09
On 11/9/2014 3:09 AM, gm wrote:

> On 09.11.2014 05:10, Jasen Betts wrote:
>> On 2014-11-06, gm <notMyMail@mail.not> wrote:
>>> On 06.11.2014 16:44, Spehro Pefhany wrote:
>>>> On Thu, 06 Nov 2014 15:24:26 +0100, gm <notMyMail@mail.not> wrote:
>>>>
>>>>> Hi.
>>>>>
>>>>> I didnt play much with RPI but now we have finish one project where we
>>>>> are triggering GPIO. The PCB relay board should be triggered from RPI
>>>>> GPIO and the door should be opened. Now, before i connect PCB to RPI i
>>>>> want to be sure that nothing goes wrong. This is the pcb schematics
>>>>> that
>>>>> we use for the relay trigger.
>>>>>
>>>>> http://tinypic.com/view.php?pic=10rmrsh&s=8#.VFuEUa0zdag
>>>>>
>>>>> What do you think . can i connect this directly to RPI GPIO, and which
>>>>> pins can i use ?
>>>>>
>>>>> Thank you in advance !
>>>>
>>>> Yes, it should work fine.
>>>>
>>>> Reduce R1 to 560R to compensate for the 3.3V output of the Rpi port
>>>> pins, and use the 5V power for the relay (and a 5V relay coil).
>>>>
>>>> Double check the connections before applying power.
>>>>
>>> -----------
>>> should i reduce the resistor value ?
>>
>> yes, the too high a resistor can cause excessive heating in the
>> transistor, and then the transistor fails.
>>
>>> Am asking because i have made the pcb and i have to test it today and
>>> i dont have 0,5K near me. I have to buy it tommorow
>>
>> Anything in the 200 Ohms to 600 Ohms range will suffice,
>> eg: you could parallell two 1K resistors to get 500 Ohms.
>> or use a 330 or a 470 if you have them on-hand.
>>
> ----------------------
> OK i can do that but i dont understand one thing:
>
> Dont i need to change R1 from 1 to 1,2 KOhm ?
> Because ih i leave it at 1K the base current will be 2,6mA instead of
> 2mA. If the raspbery PI port can handle and give this much or current
> then there is no problem. Nothing should burn out ... OR am i missing
> something ?


When I looked up the transistor you are using it says the DC current 
gain is minimum of 100.  You don't say what relay you are using or what 
current it draws.  The base current limited by R1 should be no less than 
100 times smaller than the collector current which is whatever the relay 
draws.  If the rPi outputs pull up to 3.3 volts, subtract 0.7 and divide 
by the current to size the resistor maximum value.  You can get the I/O 
voltage and the max current drive from the Broadcom data sheet for the 
CPU.  If the CPU output can't drive enough current you can use two 
transistors connecting the emitter of the first to the base of the 
second.  Then the current gain of both transistors is multiplied.

-- 

Rick
Reply by gm ●November 9, 20142014-11-09
On 09.11.2014 05:10, Jasen Betts wrote:

> On 2014-11-06, gm <notMyMail@mail.not> wrote:
>> On 06.11.2014 16:44, Spehro Pefhany wrote:
>>> On Thu, 06 Nov 2014 15:24:26 +0100, gm <notMyMail@mail.not> wrote:
>>>
>>>> Hi.
>>>>
>>>> I didnt play much with RPI but now we have finish one project where we
>>>> are triggering GPIO. The PCB relay board should be triggered from RPI
>>>> GPIO and the door should be opened. Now, before i connect PCB to RPI i
>>>> want to be sure that nothing goes wrong. This is the pcb schematics that
>>>> we use for the relay trigger.
>>>>
>>>> http://tinypic.com/view.php?pic=10rmrsh&s=8#.VFuEUa0zdag
>>>>
>>>> What do you think . can i connect this directly to RPI GPIO, and which
>>>> pins can i use ?
>>>>
>>>> Thank you in advance !
>>>
>>> Yes, it should work fine.
>>>
>>> Reduce R1 to 560R to compensate for the 3.3V output of the Rpi port
>>> pins, and use the 5V power for the relay (and a 5V relay coil).
>>>
>>> Double check the connections before applying power.
>>>
>> -----------
>> should i reduce the resistor value ?
>
> yes, the too high a resistor can cause excessive heating in the
> transistor, and then the transistor fails.
>
>> Am asking because i have made the pcb and i have to test it today and
>> i dont have 0,5K near me. I have to buy it tommorow
>
> Anything in the 200 Ohms to 600 Ohms range will suffice,
> eg: you could parallell two 1K resistors to get 500 Ohms.
> or use a 330 or a 470 if you have them on-hand.
>

----------------------
OK i can do that but i dont understand one thing:

Dont i need to change R1 from 1 to 1,2 KOhm ?
Because ih i leave it at 1K the base current will be 2,6mA instead of 
2mA. If the raspbery PI port can handle and give this much or current
then there is no problem. Nothing should burn out ... OR am i missing 
something ?

///////////////
BTW. is someone of you working in Python ?
///////////////
Reply by Jasen Betts ●November 9, 20142014-11-09
On 2014-11-06, gm <notMyMail@mail.not> wrote:

> ---------------
> btw. what pins should i use ?


If you already had it blinking a LED, use that pin (if you don't need
to blink the the LED anymore)

-- 
umop apisdn
Reply by Jasen Betts ●November 9, 20142014-11-09
On 2014-11-06, gm <notMyMail@mail.not> wrote:

> On 06.11.2014 16:44, Spehro Pefhany wrote:
>> On Thu, 06 Nov 2014 15:24:26 +0100, gm <notMyMail@mail.not> wrote:
>>
>>> Hi.
>>>
>>> I didnt play much with RPI but now we have finish one project where we
>>> are triggering GPIO. The PCB relay board should be triggered from RPI
>>> GPIO and the door should be opened. Now, before i connect PCB to RPI i
>>> want to be sure that nothing goes wrong. This is the pcb schematics that
>>> we use for the relay trigger.
>>>
>>> http://tinypic.com/view.php?pic=10rmrsh&s=8#.VFuEUa0zdag
>>>
>>> What do you think . can i connect this directly to RPI GPIO, and which
>>> pins can i use ?
>>>
>>> Thank you in advance !
>>
>> Yes, it should work fine.
>>
>> Reduce R1 to 560R to compensate for the 3.3V output of the Rpi port
>> pins, and use the 5V power for the relay (and a 5V relay coil).
>>
>> Double check the connections before applying power.
>>
> -----------
> should i reduce the resistor value ?


yes, the too high a resistor can cause excessive heating in the
transistor, and then the transistor fails.


> Am asking because i have made the pcb and i have to test it today and
> i dont have 0,5K near me. I have to buy it tommorow


Anything in the 200 Ohms to 600 Ohms range will suffice,
eg: you could parallell two 1K resistors to get 500 Ohms. 
or use a 330 or a 470 if you have them on-hand.

-- 
umop apisdn