Hi
Have a setup where i need a transistor to "not work" when B is GROUND.

If "floating" the transistor should open up and draw the relay.

Is this setup working ? or do i need another size of R1 ?

 

Have a setup where i need a transistor to "not work" when B is GROUND.

Where is "B"? Is it the transistor base?

If "floating" the transistor should open up and draw the relay.

What does "draw the relay" mean?

Is this setup working ? or do i need another size of R1 ?

What is the coil resistance?

 

B = Base of Transistor,
Draw relay, sorry for bad English, When the transistor opens up and the relay get working,
Coil resistance is 360 Ohm,
All is 12 V DC.

Need to have as little mA to my "switch" a possible, The one to the left, where Ground is one end and Base is the other.
So when switch opens, the transistor gets active,...…

 

The circuit is ok, but you should consider the draw current of relay, because the Ib only could provides about 1.13mA and the Ic can get 11.3mA, is that enough to drive the realy? (hFE = 10)

I_Relay = 12V/360Ω = 33.33 mA
So you need to reduce the values of R1.

 

B = Base of Transistor,

To avoid confusion, you should have labeled that point on the schematic.

Draw relay, sorry for bad English, When the transistor opens up and the relay get working,

It would be more descriptive if you used ON and OFF for the transistor state because you're using it like a switch.

Coil resistance is 360 Ohm,
All is 12 V DC.

Coil current will be
\( \small I = \frac{V}{R} = \frac{12V}{360 ohms} = 33.3mA \)

When operating in saturation mode, we typically assume a beta of 10, so you need to provide 3.3mA of base current to the transistor. You're providing 1.2mA.

 

When the transistor opens up and the relay get working

The normal syntax for that is that the transistor is "on" and the relay is energized.
"Opens up" means the transistor is open or "off".

R1 should be 10 to 20 times the relay coil resistance or a minimum of 7kΩ.
If that's to much, could could use an N-MOSFET such as the 2N7000, which requires no gate current, only a voltage to turn on, so R1 could be 100kΩ or higher.

 

5 k for R1 ? or how to calc this one ?

 

Little confused. One Say a lower value transistor and Another says a higher value

What to do?

 

1. I_Relay(Ic) = 12V/360Ω = 33.33 mA, counting it as 34mA.
2. Ib = Ic/10 = 34mA/10 = 3.4 mA
3. R1 = (12V-0.7V)/3.4 mA = 11.3V/3.4mA = 3323Ω, choose 3.3K.

 

Instead of guessing, look at the datasheet for the European BC547. It shows that the weakest one saturates fairly well when its base current is 1/20th the collector current. Then when the collector current is 32mA, the base current must be at least 32/20= 1.6mA and the base resistor must be no more than 7288 ohms but 6.8k is the closest standard value.
American 2Nxxxx transistors are usually spec'd with more base current so that the saturation voltage drop is less than the BCxxx European transistors so they use a base current of 1/10th the collector current.

 

The factor of Ib = Ic /10 is called "forced beta" in the lingo of
circuit design. Basically if one wants to turn on a transistor
"hard" causing it to have min Vce and lots of Ic, one forces
~ 1/10 the Ic into the base.

You can see that in this datasheet curve, for example, for the 2N3904 -




Regards, Dana.

 

Data for BC547:


Now I'll have to remember that the rule of thumb is different on the other side of the pond... Or not and just use 10.

 

Watch out for data on graphs. A graph shows a "typical" transistor that you cannot buy. A transistor you buy might have minimum spec's that are much worse than typical.

 

To make the circuit as published in post #1 function as desired, use an MPSA13 transistor, which is a darlington type, 500Ma max Ic and an effective gain of over 1000.