# Homework Help: Zener diode as voltage regulator

1. May 12, 2010

### logearav

1. The problem statement, all variables and given/known data

revered members,
this circuit i have attached is for voltage regulation using zener diode.
applying kirchoff's current law i get current through resistance R is IR=IZ+ IL
And the voltage drop across the resistance R is V-VOUT
HOW this equation of voltage drop is arrived? i if s it kirchoff voltage law, it should be summation of V = summation of IR. i dont understand, please help
2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. May 12, 2010

### Mindscrape

Here, let me show you how it's supposed to go. Use KVL, and go clockwise starting at the battery

-Vs+IR+Vz=0

Say you want to take a 15V source and regulate it to 12V. Well, since the zener is going to want to take whatever current, within the Zener Action limit, is necessary to maintain the 12V it would be eating a lot more current than it would need to if it were just the zener and the source. So, you want to put a resistor to drop off a little bit of voltage, make the zener happy.

From here on out it's a design problem. First off, you're going to want to know the max current your load will draw, let's say 100mA. The zener is also going to need to get some current, which depends on the zener itself, but is usually around 5mA, and we add that in for 105mA. Because there might be slop in current drawn by both the load and the zener, account for maybe 5mA more, for 110mA total. Now we know from our KVL that R=(Vs-Vz)/I=3V/.110A=27. So pick a resistor around 27Ohms. Voila.

Oh yeah, and you probably want to get the power ratings right, but you knew that already didn't you?

3. May 13, 2010

### logearav

many thanks mr. mindscrape. you have explained beautifully. but still have one doubt. IN APPLYING KVL, regarding voltage source, when we go from positive to negative, there is fall in potential, so we give -V and when we go from negative to positive, there is a rise in potential so we give + V. so when we go clockwise it should be +Vs+IR-Vz=0. BUT you have said -Vs+IR+Vz=0.
SORRY, IF i am wrong, because that is how my physics sir told when he taught KVL. my apology if i have offended you. just i wanted to clarify

4. May 13, 2010

### Mindscrape

Ah, okay, if you want to make energy decreases negative and energy gains positive that is fine, but I'm still right :p. Here's how the KVL loop goes in your case, let's start from above the battery and before the resistor, going clockwise:

The charge is moving along just fine along the wire then, whoop, drop some energy to go through the resistor
-V=-IR
Now for our particular loop, we hit the zener next, and it wants to maintain the energy drop its rated for
-Vz(=-Iz*R, though neither Iz or R are relevant because the Zener just works)
so far
-IR-Vz
Now the charge goes back through the battery, and woohoo, gains some energy, which our sign convention has established gains as positive, so
+Vs

all together
-IR-Vz+Vs=0

multiply by -1 and you have my result :)

5. May 13, 2010

### logearav

beautiful explanation sir. i understood fully. thanks for sparing your time in educating novices like me.
thanks a lot sir.