CALCULATION OF VOLTAGE ACROSS THREE DIODES IN SERIES
Welcome back with a new quiz by Online Technologies For Beginners (NPROT).
A circuit consisting of a 5V voltage generator indicated with V1 is assigned to which a 100 ohm resistor and three 1N4001 type diodes are connected in series.
It is are asked to calculate the voltage Vout measured across the three diodes in series and the current I which circulates in the circuit.
To obtain the voltage Vout, it is sufficient to add the voltage drops across each diode, indicated in the figure with the symbols VD1, VD2 and VD3.
First of all, note that instead of the circuit symbol, their real appearance was used for the diodes. In particular, it should be noted that the white band indicates the cathode of the diodes.
As a result, the three diodes are forward biased. This allows us to assume that the voltage drop across each diode is equal to the threshold voltage that we assume around 0.7 V.
Adding the three voltage drops, we have that:
Vout = VD1 + VD2 + VD3 = 3· 0.7 V = 2.1 V
We therefore obtained the first result required by the exercise.
To calculate the current I circulating in the circuit, we can observe that this current also flows through the resistor R1 of which we know the value of its resistance.
It is also easy to determine the value of the voltage across the resistor. In particular, we observe that the voltage V1 is applied to the left terminal of the resistor, while the voltage Vout is applied on the right terminal of the resistor.
We therefore have that the voltage across R1 is equal to:
V1 – Vout = 5 V -2.1 V = 2.9 V.
Once we know the voltage across the resistor and its resistance value, we can apply Ohm’s law to calculate the current flowing through it.
It follows that:
As already observed, this current represents the current I which circulates in the circuit.
We therefore have that I = 29 mA.
And with this last observation, we therefore obtained the result required by the exercise.
Well, exercise solved!
Good electronics everyone!
