(b) The diagram of FIGURE 2 shows a bi-directional opto coupler input interface circuit. When a supply voltage of 20 V is applied the LED carries a current and 2 V is dropped across it. Calculate the value of the LED current and the value of current through the 3 k resistance.
Separate supply used for inputs.
Note: Either polarity.
(c) The circuit shown in FIGURE 3 is part of the interface of a relay output module. Ib is 1 mA and VCC is 9 V. The relay requires a
minimum of 50 mA to energise.
Complete the values of the assumptions listed below in order to calculate:
• voltage across R1
• value of R1
• voltage across the relay coil
• voltage across R2
• value of R2
• collector of current Ic.
Assumptions: Logic '1' = V
Logic '0' = V
Transistor forward current gain hfe = LED current = 10 mA LED voltage drop at 10 mA = V Base/emitter voltage = V
Collector emitter voltage when transistor is on = 1 V
FIG. 3© BrainMass Inc. brainmass.com October 10, 2019, 8:26 am ad1c9bdddf
The diagram of FIGURE 2 shows a bi-directional opto-coupler input interface circuit. When a supply voltage of 20 V is applied the LED carries a current and 2 V is dropped across it. Calculate the value of the LED current and the value of current through the 3 k resistance.
Adding in some voltage values at strategic points in the circuit for the LED conducting 'ON'
State we obtain the following
The potential difference across the 3kΩ resistor is thus 20V-2V=18V and the resultant
current through the 3kΩ resistor is simply given by Ohm's Law as
The current flowing through the 470Ω resistor is simply the potential across it (2V) divided
2 LED (Opto-Couplers) circuits are analysed and circuit parameters determined, such as currents an, voltages and component values