# LED circuit analysis problem

I'm currently learning for an upcoming exam about diodes. I have the following exercise about LED, but I'm really having a hard time to figure it out (Sadly, the teacher isnt willing to help us). I have an AC source with Us= 230 V at 50Hz The LED is Uf= 1,7V and Iled= 20mA The Diode is a BAY61. I need to determine the resistor R1.

I don't really know where to start. The teacher never explained how to use datasheet so i googled BAY61 and found a datasheet, but what information do i need?

• split your problem in two, by replacing the AC source mentally by +V constnt voltage source first, then by -V – Marcus Müller Aug 14 '16 at 11:30

Since you're working with diodes with an AC source it is far simpler to split it into two circuits with different DC sources. First convert the RMS voltage into a PEAK voltage since you are effectively rectifying. $250V_{RMS} = 707V_{P-P}$ or $±353V_{PEAK}$.

Now make two circuits:

simulate this circuit – Schematic created using CircuitLab

The purpose of the diode is purely to protect the LED during the reverse voltage phase of the AC supply, so all we need to know is that whatever current flows through it is within acceptable levels (200mA according to the specs). So we just need to solve R1 for the LED and check that the same resistance will be OK for the diode. So a simple LED resistance calculation:

$R_1 = \frac{V_S - V_F}{I_F} = \frac{353 - 1.7}{0.02} = 17565\Omega$

Now feed that resistance into the same formula rearranged with the diode instead (the datasheet just says "<1V" for $V_F$ so I will assume 0.7V):

$I = \frac{V_S - V_F}{R_1} = \frac{353 - 0.7}{17565} = 0.02006A$

So that value is well within tolerance for the diode. Now, what if I was wrong about the diode's forward voltage? Just out of interest, what would the current be with the diode completely shorted out? Simple:

$I = \frac{V_S}{R_1} = \frac{353}{17565} = 0.0201A$

Still well within tolerance.

• Thank you very much! I'm still having a hard time to understand because I never really had an explanation how to convert AC voltage to DC. But in the second scheme, the diode comes before the resistor. Isn't the diode going to burn due to too much voltage? And why is it 250V rms? – Curious student Aug 14 '16 at 11:41
• In a series circuit there is no "before" - the current is the same through all the elements, and the voltage is divided between them. If your lecturer never told you about RMS and PEAK with AC then your lecturer is rubbish (or you were away that day). – Majenko Aug 14 '16 at 11:43
• The lecturer has been sick most of the time. Most of the students did a course about electronics before but i'm one of the few who never did so it's a real struggle to learn it – Curious student Aug 14 '16 at 11:46
• @Curious: In a simple series circuit the current simultaneously moves through the whole circuit in the same way a bicycle chain moves. It's not as though you're pouring electrons in and filling up the wire. The electrons are all there already and if you push one in one end one is pulled out at the other simultaneously. That's why it doesn't matter whether the resistor comes before or after the LED. – Transistor Aug 14 '16 at 14:50