voltage regulator

hi! im faizan my problem is related to design of voltage regulator. i need to design a voltage regulator which gives an output of 15 volts and 350mA current. i need to know that what would be the specification of the zener diode and transformer used in the circuit(i mean the type of zener diode usedi.e its value etc). can anybody plz help me.

On a sunny day (20 Dec 2006 09:00:28 -0800) it happened shiningstar snipped-for-privacy@hotmail.com wrote in :

Difficult to say unles you specify voltage tolerances of the input and voltage of the input, size of the capacitor (so we know the ripple), and why such a huge zener. Zeners work by sharing some of the currrent, so as input voltage increases zener current will increase. We also need to know output tolerance and temperature range (not all zeners are made equal). In fact I would not use a zener but a switcher or a 7815 chip in this case, with some more componenents.

hi!plz reply to me on my e-mail adress:- shininstar snipped-for-privacy@hotmail.com all those who want to help(regarding voltage regulator) are requested to reply as soon as possible.i'll be greatful.

lookin forward

faizan

hi guyz!how r u doin?i've a little problem with an assignment i've got regarding voltage regulator. ASSIGNMENT:- my assignment is to design a voltage regulator which gives an output voltage of

15 volts & a current of 350mA(mili-ampere).the components used should be a step-down transformer, a fullwave rectifier bridge,2 resistors(Rin,Rload ),a capacitor & a zener diode. PROBLEM:- the problem is that i am not able 2 adjust the zener diode accordingly.when i use an ideal one the current & voltage are not exact."can any one of you tell me the specifications of the zener diode and the transformer.'ll be very thankful.plz help. .

------------------------ Fun thought: Could you cheat and shop for a regulated linear DC wallwart? I'll guess with the above ratings maybe....$15.00 canadian... Take the labels off and say you also learned plastic molding too :) Crack it open and study the design..

Design Tip: Take a look at a zener data sheet. The IV transfer is not that of an ideal regulator. I don't know why but I fk'n hate zeners... Everytime I try to apply, it gets shot down on tolerance spec, drift, heat, impedance...

D

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On a sunny day (20 Dec 2006 09:19:38 -0800) it happened shiningstar snipped-for-privacy@hotmail.com wrote in :

May I suggest you change 'shininstar' to 'blackhole'. Are you sending this from your cellphone during the exams?

Are you restricted to *only* the components you listed? Are you restricted to real zeners, or are you allowed to make up the specs for a theoretical zener? Are you *sure* it is 350 mA and not 35 mA? Are there any other facts given in the assignment?

Ed

else to do my homework for me for free.

Sorry, we don't do lazy kids' homework.

Good Luck! Rich

hi! im not a lazy kid at work.i was just lookin to get some verification about the circuits i made myself.thnk u verh for ur help,.

--- I'll help you, but I'd appreciate it if you didn't use chat room slang and if you bottom posted.

First, here's a circuit, which you can view in Courier, which shows how your system should be wired up

15V / +---+---[R1]--+-------+ +------+ | | | | MAINS>---P||S--|~ +|--+ |+ |+ | R||E | | [BFC] [ZENER] [R2] MAINS>---I||C--|~ -|--+ |C1 | | +------+ | | | | +---+---------+-------+

With this kind of problem you have to work backwards, so your load resistor has to allow 350mA through it with a voltage drop of 15 volts across it. That means the value of the resistor, R2, has to be:

E 12V R = --- = ------- ~ 34.3 ohms I 0.35A

and the power it must dissipate will be:

P = IE = 12V * 0.35A = 4.2 watts

Now, to select the Zener you have to determine what the load's dynamic characteristics are going to be like. That is, what will the change of current through the load look like as a function of time?

In this case, there's no problem since the load is just a resistor and its resistance shouldn't change much when it gets hot.

Now, the deal with Zeners is that when you reverse bias them so that what's called the "test current", Izt, passes through them, the voltage across them is guaranteed to be within a certain range, usually 5 or 10% of the nominal voltage, so your job is to make sure that that test current will be going through the Zener while the load current is going through the load.

Now you have to choose a Zener. Since your load isn't going to vary much, Izt is only going to vary because of changes in the mains voltage and because of the ripple in C1. For a first cut, let's try a 1N5352B, which is a 15V 5W Zener with an IZT of 75mA. The data sheet is here:

Since we have 350mA in the load and 75mA through the Zener, that current is all having to go through R1, but now, before we can figure out what value of R1 to choose, we have to consider the "front end" of the supply. That is, the transformer, rectifier bridge, and filter.

To start, let's say that we never want the voltage on the C1 end of R1 to go below 16V. That'll give us a volt of headroom with the load current and Izt through R1, so for a first cut,

Ein - Vz 16V - 15V R1 = ---------- = ---------------- ~ 38 ohms IL + Iz 0.35A + 0.075A

Now, looking at the capacitor, we know that since:

IdT C = ----- dV

Where C is the capacitance in farads, I is the steady-state load current in amperes, dT is the reciprocal of the rectified mains frequency, and dV is the desired ripple voltage,

if we don't want dV to go below 16V when the capacitor is discharging into the load and the mains voltage cycle is below the point where it's high enough to charge the cap and feed the load, then we must choose the capacitance of the capacitor so that its output voltage never falls below 16V during that mains "dead time."

Assuming we allow one volt of ripple, then the capacitor will charge to a peak of 17V and discharge to 16V before the next mains cycle comes along and starts charging it up again for the dead time after the present peak. if we plug some numbers into the equation, we'll wind up with:

IdT 0.425A * 8.33E-3s C = ----- = ------------------ = 3.54E-3F = 3540µF. Not bad... dV 1V

But now we've got to get that 17V from somewhere. The bridge. If you want to use a full-wave bridge, then there will always be two rectifiers in series in front of the reservoir cap, and assuming a forward voltage drop of 0.7V per rectifier, that means that if you want 17V out of the bridge you've got to put:

Vin = Vout + 2Vf = 17V + 1.4V = 18.4V into it.

But now we've got to get that 18.4V from somewhere. That'll be the transformer, and we've got to make sure that we can get it when the mains are 10% under their nominal 120V rating, so that's at 108V.

So, we need a transformer than can put out 18.4 volts, peak, with

108 volts, RMS, on the primary. Since transformers are rated for RMS in and out, we'll have to convert that 18.4VPK into RMS, and we do that by dividing it by the square root of 2:

VP 18.4 VRMS = --------- = ------- ~ 13VRMS sqrt(2) 1.414

Transformers are also rated with 120V on the primary, so since that

13V is with 108V on the primary, what we need to do is up the secondary voltage by the ratio of nominal mains to low mains:

Vmains nom 120V Vsnom = Vslow ------------ = 13V ------ = 14.4 volts, RMS Vmains low 108V

So we need a transformer with a 120V primary and a 14.4V secondary, and it'll need to be able to supply about 1.8 times the current into the load and the Zener because it also has to charge up the cap so, since the load needs 350mA and the Zener needs 75mA, the transformer secondary needs to be rated for:

Is = 1.8* (Il + Izt) = 0.765 ampere

What we need then, is a transformer with a 120V primary and a secondary rated to supply 0.765 amps at 14.4 volts.

Unfortunately, a rare beast to find.

Fortunately, there are lots of standard transformers out there which are pretty close, among them a Triad FP30-400:

which can be found at Digi-Key for $14.00. That's a little high and you can do better if you shop around for different vendors and different manufacturers.

In any case, the transformer is rated to put out 15VRMS at 800mA with a 120V input, so now we can work out the value of R1.

With low mains, the transformer will put out 13.5VRMS which, after the 1.4V lost in the bridge, will charge the cap up to about 17.7V so,with the Zener end of R1 at 15V and the cap end at 16.7V (the low voltage point of the ripple waveform) the resistor will need to drop

1.7V at 475mA, so its value needs to be:

E 16.7V -15V R = --- = ------------ ~ 3.6 ohms I 0.475A

and it'll be dissipating roughly:

P = IE = 0.475A * 1.7V ~ 0.8 watt.

So, finally, we have:

15V R1 / +---+--[3R6]--+-------+ +------+ | | | | MAINS>---P||S--|~ +|--+ |+ |+ | R||E | | [4700] [1N5352] [34R3] MAINS>---I||C--|~ -|--+ |C1 | |R2 +------+ | | | | +---+---------+-------+

Now we have to check that nothing's going to fry at high line.

With high mains, the voltage into the transformer will be 132V, so the secondary will rise to 16.5VRMS, which is 23.3 peak, so after the diode loss of 1.4V through the bridge the cap will charge up to about 22 volts, so C1 should be 4700µF at 35V. It's 4700µF instead of the 3540 we originally figured because of the 20% tolerance, and

4700µF is the closest standard value that'll assure us at least 3540µF if the cap is 20% low. Now, with 22V on the cap end of R1 and 15V on the Zener end, the current through the resistor will be:

E 22V - 15V I = --- = ----------- ~ 1.94A R 3.6R

of which 350mA will be going through the load and the remainder through the Zener. That's about 1.6 amps through the Zener, which means it'll be dissipating about 24 watts, so the scheme is flawed if it needs to run from low mains to high mains.

However, if your assignment only deals with a single mains voltage, use 120VRMS and the procedure outlined above to find the proper value for R1.

-- JF