I am new to electronics. I am trying to figure out voltage divider to bias a transistor in saturation mode to drive a relay coil. Here's the circuit:
The coil resistance is 270 Ω. So the collector current has to be 44 mA. With hfe = 110, Ib = 0.4 mA. So R2 = 12V / 0.4 mA = 30 kΩ. Vb has to be lower than Ve by 0.7V. Let's say 11V. So R1 = 3.3 kΩ. Am I correct so far?
Also in some tutorial it says "The amount of biasing current flowing through resistor R2 is generally set to 10 times the value of the required base current IB so that it is sufficiently high enough to have no effect on the voltage divider current or changes in Beta." What does that mean exactly?
Best Answer
They are saying that you should have Ic/Ib ~= hFE/10 or about 11 in this case.
You want to drive the transistor deep into saturation to minimize the voltage drop across it, not provide the minimum current that (under nominal conditions) might drop a volt or two across the transistor. When fully saturated, there should be a much lower drop.
The current through R1 is subtracted from the base current, so R2 must provide even more current in order to feed both.
So, let's say we want Ib = 4mA, that means that R2 must be less than (12-0.7)/0.004 = 2.83K.
R1 is used to ensure the transistor turns off fully so, you might want to make it such that the current through R1 is 1/20 of the base current, or 200uA. So R1 = 0.7V/.2mA = 3.5K (maybe you use 3.6K as a standard value).
Now R2 has to supply 4.2mA so it will be (12-0.7)/0.0042 ~= 2.7K
Note: The value of R1 above is very conservative for most situations. To get an actual maximum value you'd have to calculate the leakage from Icbo and gain at the maximum operating temperature. Even with R1 infinite it will typically leak only a few uA through the relay coil at the maximum junction temperature. You might also want to consider leakage across the switch etc. so 100K or so would probably take care of that unless you're expecting a flood. I'd probably use 10K.