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zener diode voltage divider

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aesoton

Marine/Ocean
Sep 25, 2002
42
hello all again,
quick qs

I need to get a 1.5v supply from a 9v battery.
I want to use a zener diode to do this.
attached are two drawings 1st with a 7.5V zener in series, the second with a 1.5v zener in parallel. Am I right in thinking that you cannot get a 1.5V zener diode. Therefore does the 7.5V zener in series do the same job of creating a 1.5V supply?
 
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The problem with the first option, dropping 7.5 across the zener, is that it will be fairly sensitive to variations in the "9 volts". If the 9 is actually 9.2 to start, then your output will be 1.7. As soon as the battery drops to 8.8, then your output drops to 1.3. It is very sensitive because of the 'leverage' of the approach.

The second approach needs a series resistor. It will provide a very stable voltage until the 9-v battery has rotted away. But it draws a continuous current so the battery wouldn't last very long.

Any chance of using a 1.5 volt battery?
 
Thanks for the reply VE1BLL,
The problem this is just an off-shoot of a larger ciruit. The 9V is needed to run an amplifier so I cannot use a 1.5V battery unless I add this in a almost a seperate circuit but this will take up too much space within the device. I would prefer to use the second option for the reasons that you describe but can't find a 1.5V zener diode. I need a stable 1.5v hence the use of a zener diode.
 
Some of the standard 3-terminal regulators are actually 1.4V regulators, which drop the remaining voltage across its resistor to ground.

TTFN

FAQ731-376
 
I don't need a lot of current approx 300uA.
 
For the space, I would think a 3-terminal regulator would be perfect here... a few pennies more, but rock-solid voltage. You may even get away with using its very own leakage current to power your circuit... that would be an interesting circuit to try and make work!

Dan - Owner
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Or two resistors in a voltage divider. Depending on the variation of that 300uA, you can set the current through the resistors to swamp out the voltage variations within your limits. If the 300uA doesn't vary much, then you won't waste very much current. Might even be less than some 3-terminal regulators would use.

 
Two silicon diodes in series with the band pointing to the negative and a 120K dropping resistor should be sufficient to run the microphone. They aren't very critical. Didn't see the resistor in your drawing.
 
thanks all for you replies. I did not want to use a set of resistors for the voltage divider as I wantd a very stable voltage. Also I was trying to avoid using a voltage regulator for something cheaper. What sort of silicon diodes would I need to use for this application?
Many thanks Andrew
 
Perhaps you should quantify what the specific performance requirements are. As usual, doing this will avoid further attempts at solutions that don't meet your unstated requirements.

For example, what is your cost bogey? The example here: probably runs less than $2. You've already spent that much on this thread.

TTFN

FAQ731-376
 
Any two small silicon glass signal diodes would do up to a 1N4001, 02,...07 type. It has to be cheap to do because I have a clock radio that they mounted a standard quartz clock unit in. Instead of an AA battery, that is what they did to supply power.
 
Given your low current requirement, the LMV431 from National seems like a good candidate. It integrates a 1.2V reference zener and a linear comparator to vary the ouput to anything you like, using two resistors. It can be purchased in qty one if wished at places like Digikey. Available in both SMT and lead-thru packages.
 
I love the LM31, got a couple hundred of them. There are a lot of neat things you can do with them, but one of the things they can't do is go below 2.5 volts.
 
We all know that zener diodes are only available down to 2.4V. Below that you either use one or two FORWARD biased silicon diodes (zeners are of course reverse biassed for zener operation).

If the load is 300µA then you can't use a shunt regulator diode and a 120K resistor (as suggested by operahouse) since the maximum current from that would be 62µA. clearly you need a resistor lower than (9 – 1.5)/300µ = 25K. Typically you would allow at least 200µA for the shunt regulator, but this all depends on how much variation there is in this "300µA" load and how much voltage ripple on the 1.5V rail is acceptable.

Whilst ordinary silicon diodes are not really specified for forward operation as shunt regulators, there is a diode specially designed for this purpose. Example NXP BAS17 is a single diode "stabistor". All it is, is a single silicon diode for operation forward biassed, BUT it gives data for using it in this way which you will not find for an ordinary 1N4148, BAV99 or whatever.
 
9V batteries usually have the worst energy density of any common battery types and highest energy cost. Any reason why one 1.5V cell can't be used, or two in series, or a 3V lithium cell? Regulating 9V down to 1.5V using a linear regulator seems kinda pointless when you can easily start with a supply voltage so much closer to what you need which will give you better efficiency and lower running costs.


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OOPS...off a decimal point. Thought I read 250K on the calculator. It is probably just a little goofy electret microphone. If it is a two wire type, they can operate on higher voltage. If a 3 wire type, then I would consider changing it for a two wire unit.
 
macgyver,

I believe you are right. Where did that come from? It wasn't there when I looked last night! [tongue]

Sorry folks, I'll get my coat...


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