Because the power supplies of transistor and I C are different, we use the second formula: So we choose a transistor with h FE = 200 and I C = 100mA. The transistor must have an h FE greater than 5 X 75 / 2 => h FE > 187.5. The IC can supply a maximum current of 2mA. The supply voltage is 12V for the transistor and 5V for the IC.
The output from a 74LS series TTL IC is required to operate a relay with a 160 Ohm coil. The protective diode could be the 1N4001 or any general purpose diode. Otherwise, if the component uses another power source (like V CC) then the form is: You must select it according to it's Ic that must be greater than I L and it's current gain h FE.Then you calculate the base resistor R B, If the input is taken from a component (possible an IC) that uses the same power supply as the transistor (that is Vs), then the form is: It must be at least 5 times the load current IL divided by the maximum output current from the Input to the base of the transistor The way to calculate them is:First we calculate the load current: The values for Rb and Qs vary accordingly. The protection diode Dp is used to protect the transistor from the reverse current generated from the coil of the relay during the switch off time. The relay will be actuated when the input of the circuit goes high. It is used to switch high loads or loads that needs AC current to operate. This circuit will drive a relay coil from a low power output, usually from an IC like 555 or a TTL/CMOS.