First project! Help please??

[nerosrevenge]

Hi there.

I will start by admitting that I have no formal education in micro-electronics (which I intend to rectify very soon) but I am facinated thusfar.

I am attempting to build a voltage regulator using Zener Diodes. I need to know how to calculate the required value of a resistor for use in the circuit.

The supply voltage is 12vdc. The required output voltage is 3vdc (although I am having difficulty finding a Zener with this at this voltage, I hope I worded that right!).

Is there a calculaiton to determine the value of the resistor I need??

Thank you all very much!!

Steve.

 

[Nigel Goodwin]

Yes, 'ohms law' the most basic formula in electronics, it's simply V=I*R, where V is in volts, I in amps, and R in ohms. For your purpose you need to rearrange it, to give R=V/I - you then just need to work out the values of V and I, and put them in the formula.

V is easy, you have 12V in , and 3V out - so you need to drop 9V across the resistor.

I is more difficult (that sounds like really bad grammer!), as you don't give any indication what the current requirements are for the 3V supply. However, the current through the zener should be substanially more than that through the load - usually about 5 to 1 is suggested. So assuming you want to draw 10mA off the 3V supply, you need 50mA through the zener diode. This gives a total of 60mA through the resistor, so I=9/0.06, which gives 150 ohms.

You also need to work out the wattage requirements for both the zener and resitor - this is another nice simple formuls W=V*I, where W is in watts. So for the resistor W=9*0.06, which is 0.54W - so you need at least a 1W resistor, and it will probably run fairly warm. For the zener diode W=3*0.06, which is 0.18W (notice I use the full 60mA, as if the load is disconnected the zener takes the full current) - so you could probably get away with a 400mW zener, although I would probably use a 1.3W one.

This should give you enough information to work out the values you require, simple zener shunt regulators like this are only really suitable for fairly low currents - for higher currents, use a transistor (or more than one) to improve the power handling.

 

[Johnson777717]

Well, when a resistor is used, it is to reduce the amount of current in a circuit. Current, can be compared to the current in a stream. As you may know, electical current is measured in Amperes. More amperes, more (or stronger) current.

So you know your input and output voltage. What we need to figure out, is:

1. What input current (amperage) that is available from the 12VDC supply.
2. What output current (amperage) that you want.

If you're not familiar with Ohm's law, then you will become familiar working with electronics. Ohm's law basicly states:

Resistance = Voltage divided by Current. Or R = E / I where:
R= resistance in ohms
E= Electromotive force (Voltage) in volts
I= Intensity (Current) in Amperes.

If you're familiar with basic algebra, then you'll be able to solve for any of the variables by algebraically adjusting the equation. Thus:
E=R x I
I=E / R
Check out the "Ohms Law" Theory article on this forum main page under "Theory article".

 

[Johnson777717]

I guess you won Mr Goodwin.

 

[Nigel Goodwin]

I guess you won Mr Goodwin.

When you're fast, you're fast! :lol:

 

[nerosrevenge]

Wow! Thank you both so much! The information you have provided me with will definately help! I will try and farmiliarize myself with Ohm's law. I will go now and see if I can find out what the current requirements are for the 3v supply.

I'll be back!

Thanks again,

Steve.

 

[nerosrevenge]

I is more difficult (that sounds like really bad grammer!), as you don't give any indication what the current requirements are for the 3V supply. However, the current through the zener should be substanially more than that through the load - usually about 5 to 1 is suggested. So assuming you want to draw 10mA off the 3V supply, you need 50mA through the zener diode. This gives a total of 60mA through the resistor, so I=9/0.06, which gives 150 ohms.

I am slightly confused. If I understand it correctly I (meaning current) is measured in amps. You give the answer to I=9/0.06 as 150 ohms. Isn't ohms a measure of resistance?

Thanks,

Steve.

 

[Nigel Goodwin]

I am slightly confused. If I understand it correctly I (meaning current) is measured in amps. You give the answer to I=9/0.06 as 150 ohms. Isn't ohms a measure of resistance?

Sorry! - little typo, that should (obviously) read R=9/0.06 :lol:

 

[nerosrevenge]

Thanks Nigel. I am still working on determining the required current for this application. I will post more when I know.

Steve.

 

[Nigel Goodwin]

Thanks Nigel. I am still working on determining the required current for this application. I will post more when I know.

Steve.

What are you connecting that needs a 3V supply?, knowing what it is 'may' give us a clue as to it's power consumption. Another clue would be what it's original PSU was, if it was a tiny watch battery it's got to be pretty low - but if it was a stack of NICAD's that you couldn't pick up, it's going to be pretty high :lol:

 

[nerosrevenge]

The voltage regulator will supply a 35mm camera. It originally ran off of 2 AA batteries in series. The problem is that the camera unit will be kept outside and in my experience AA's don't like cold weather all that much :wink:. We have a 12v gel cell that we have been running a store bought unit off and it seems to last long enough (even at -22C) to suit our needs. So the thought is to run the camera off of the same power supply.

Does that help?

 

[Nigel Goodwin]

The voltage regulator will supply a 35mm camera. It originally ran off of 2 AA batteries in series. The problem is that the camera unit will be kept outside and in my experience AA's don't like cold weather all that much :wink:. We have a 12v gel cell that we have been running a store bought unit off and it seems to last long enough (even at -22C) to suit our needs. So the thought is to run the camera off of the same power supply.

Does that help?

Yes it does, a great deal! - it's likely to have a fairly high current requirement. In particular the flash is going to probably require AMPS while it's charging - a simple resistor/zener regulator isn't going to cope with that.

Also a point to bear in mind is the actual voltage required, two new AA batteries will give a good bit more than 3 volts, which is one reason why the flash charges up faster with new batteries. Aiming for 3.5V or so would probably be a good idea, and using a more complicated transistor regulator would seem to be indicated.

If you need more details, feel free to ask.

 

[nerosrevenge]

Hey there. I received some input from B-o-b on the circuit.

He suggested adding a 2N3053 transistor to help with wattage requirements and a 1.0 MFD capacitor to help with stability. He also suggested using a Zener of 3.6v as the transistor will drop some of the voltage anyway. With the addition of the capacitor it seems like it will work to me (but then again what do I know about it :lol: ).

What do you think?

Steve.

 

[Nigel Goodwin]

Hey there. I received some input from B-o-b on the circuit.

He suggested adding a 2N3053 transistor to help with wattage requirements and a 1.0 MFD capacitor to help with stability. He also suggested using a Zener of 3.6v as the transistor will drop some of the voltage anyway. With the addition of the capacitor it seems like it will work to me (but then again what do I know about it :lol: ).

What do you think?

That sounds fine, the only suggestion I would make would be to increase the zener to 4.3V, which would give 3.6V or so out. This would help the flash to charge faster, and be similar to new batteries.

It would probably be a good idea to still measure the current, and also to check how warm the transistor gets.

 

[Phasor]

Surely a series reg, eg. LM317 would be more suitable?

For a zener shunt reg, say the input voltage is 12V (gel cell battery), the output 3.6V. You draw a maximum of say, 1A (for the flash), and wish to maintain the 5:1 ratio which Nigel suggested. So, the power dissipated by the zener is (3.6V * 5A) = 18W, and by the resistor, ((12-3.6)*6A) = 50.4W, a total of 68.4W wasted. This is obviously quite inefficient and unfeasible.

A series regulator would dissipate just ((12-3.6)*1A) = 8.4W.

EDIT:

using a more complicated transistor regulator would seem to be indicated.

Ahh, Nigel, I see you have already mentioned this, but it need not be complicated :?
How about this:
https://www.aaroncake.net/circuits/supply2.htm
(though we probably don't need ALL those capacitors :lol: )

 

[Nigel Goodwin]

Ahh, Nigel, I see you have already mentioned this, but it need not be complicated :?

Not at all, just more complicated than a simple resistor and zener. But initially we had no indication of current requirements.

How about this:
https://www.aaroncake.net/circuits/supply2.htm
(though we probably don't need ALL those capacitors :lol: )

Exactly what I was thinking of ( without all those capacitors! ). However, it would be improved by the addition of an extra one, across the zener diode.

 

[Sebi]

For best results i strongly recommend a simple switcher buck regulator from Nat.Semicond. No heat, good efficiency, and the 12V battery life four times longer.