resistor ohm ?

[Jak24]

hi !

i have this transistor :
http://www.fairchildsemi.com/ds/BC%2FBC547.pdf (546)
and i need to use it in here :
http://interactive.usc.edu/members/phoberman/relayCircuit.gif
how much ohm should the resistor be, from base to the digital pin?
thanks

Regards

Jak24

[waltr]

Ok, a bipolar transistor is a current controlled device. You set the current into the base (to the emitter) to control the current flow into the collector (reverse from a pnp).

The current gain is in the data sheet as hfe (DC current gain) which is between 110 and 800. When using the transistor as a switch use the lowest hfe value in the calculations to ensure transistor saturation. So the base current times hfe equals collector current (Ibe * hfe = Ic), where Ic is the max current needed to power the relay, or relay coil voltage divided by the relay coil resistance, Icoil = Vcoil/Rcoil.
The base resistor then sets (or limits) the base current. So base resistor value is the IO high voltage minus the base-emitter voltage (about 0.6V) divided by the required base current, R = ((ioV - Vb)/Ib).

Since I don't know the relay's coil resistance I can't calculate the base resistor value but instead gave you the information required to calculate the base resistor value.

[Pratheek]

Read these two excellent tutorials -
http://www.kpsec.freeuk.com/components/tran.htm
http://www.kpsec.freeuk.com/trancirc.htm

You should then be able to find the answer to your question.

[Jak24]

HI

thanks for replies, this is the relay I'm using :
http://sea.com.ua/img/info/ningbo%20forward/JZC-21F%28NT70%29.pdf
(jzc-20f) its resistance is 400 ohm

Ok, a bipolar transistor is a current controlled device. You set the current into the base (to the emitter) to control the current flow into the collector (reverse from a pnp).

The current gain is in the data sheet as hfe (DC current gain) which is between 110 and 800. When using the transistor as a switch use the lowest hfe value in the calculations to ensure transistor saturation. So the base current times hfe equals collector current (Ibe * hfe = Ic), where Ic is the max current needed to power the relay, or relay coil voltage divided by the relay coil resistance, Icoil = Vcoil/Rcoil.
The base resistor then sets (or limits) the base current. So base resistor value is the IO high voltage minus the base-emitter voltage (about 0.6V) divided by the required base current, R = ((ioV - Vb)/Ib).

Since I don't know the relay's coil resistance I can't calculate the base resistor value but instead gave you the information required to calculate the base resistor value.

uhm so my high is 4.8 - 0.6 =4.2 divided by base current?, I'm not sure how to calculate that...
this is the battery I'm using :http://www.all-battery.com/12v2600mahnimhbatterypack11620.aspx
thanks

Regards

Jak24

[waltr]

The base current needed is the Collect current divided by the Hfe (DC gain). The collector current will equal the relay coil current which can be calculated from Ohms law, I = E/R. So 5V across a 400 Ohm coil is 12.5mA. From the Data sheet (and my post above) base current = 12.5 /110 = 114uA. Base resistor = 4.2V/114uA = 36.96k Ohm max. This could be smaller since the wish is the have the transistor in saturation. So any resistor from about 10k to 36k Ohm will work.

[Jak24]

HI

The base current needed is the Collect current divided by the Hfe (DC gain). The collector current will equal the relay coil current which can be calculated from Ohms law, I = E/R. So 5V across a 400 Ohm coil is 12.5mA. From the Data sheet (and my post above) base current = 12.5 /110 = 114uA. Base resistor = 4.2V/114uA = 36.96k Ohm max. This could be smaller since the wish is the have the transistor in saturation. So any resistor from about 10k to 36k Ohm will work.

OK, but my relay is 12v so could you tell me if my calculations are correct?
12v / 400 ohm = 30mA     
30 / 110 = 270uA
(i re-check the high volt it was 5.14) so 5.14/ 270 = 19 k Ohm max so 19 - 8 k should work
right?
thanks

Regards
Jak24

[waltr]

Ah....the schematic you posted showed +5V on the relay and you didn't say it was otherwise.
But you now did go through the calculations yourself.

19 to 8 k should work

yep.