Resistors: what will give me a constant voltage

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SUMMARY

The discussion centers on achieving a constant voltage output, specifically 1.5V or 3V, from varying input voltages using voltage regulation techniques. Participants recommend using an Active Voltage Regulator, such as the LM317, or a DC-DC Buck Converter for efficient voltage regulation. The conversation highlights the inefficiencies of linear regulators and the importance of considering transient current demands in applications like powering cameras. Additionally, the use of zener diodes is mentioned but deemed impractical for significant power regulation.

PREREQUISITES
  • Understanding of Active Voltage Regulators, specifically LM317
  • Knowledge of DC-DC Buck Converter circuits
  • Familiarity with transient current demands in electronic devices
  • Basic concepts of power dissipation in voltage regulation
NEXT STEPS
  • Research the specifications and applications of the LM317 voltage regulator
  • Learn about designing and implementing DC-DC Buck Converter circuits
  • Explore the effects of transient current on voltage regulation in electronic devices
  • Investigate power dissipation calculations for linear and switching regulators
USEFUL FOR

Electronics engineers, hobbyists working on battery-powered devices, and anyone involved in designing voltage regulation systems for mobile applications.

Steven Ellet
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I would like to know what will give me a constant voltage no matter what the input is (within reason of course) Example: input 9 volts, output 1.5 volts; input 6 volts, output 1.5 volts; input 1 volt, output <1.5 volts.
 
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You will need an active circuit like a voltage regulator. This has nothing to do with resistors other than the fact that a voltage regulator will have some resistors in it.
 
Steven Ellet said:
I would like to know what will give me a constant voltage no matter what the input is (within reason of course) Example: input 9 volts, output 1.5 volts; input 6 volts, output 1.5 volts; input 1 volt, output <1.5 volts.

As phinds says, you would use an Active Voltage Regulator: http://en.wikipedia.org/wiki/Voltage_regulator

(skip down the wikipedia page until you get to the "Active Regulators" part...) :-)
 
Steven Ellet said:
On a related note, would dimmer switch technology (https://en.wikipedia.org/wiki/Dimmer) do a simalar (not identical) job if hard wired?

P.S. would a dimmer work on dc power?

Welcome back!

No, a dimmer will not do what you have asked for. Did you read the voltage regulator links? Why won't that work for you? :smile:
 
berkeman said:
Welcome back!

No, a dimmer will not do what you have asked for. Did you read the voltage regulator links? Why won't that work for you? :smile:
Sorry for the long delay, I have kind of been all over the place. As far as I could tell that requires a "check voltage" and that is a little inconvenient. For example, if I have a car battery and want to get only 1.5v from it, I would effectively need a 1.5 voltage battery to make it work.
 
Steven Ellet said:
Sorry for the long delay, I have kind of been all over the place. As far as I could tell that requires a "check voltage" and that is a little inconvenient. For example, if I have a car battery and want to get only 1.5v from it, I would effectively need a 1.5 voltage battery to make it work.
as others have said, using a 1.5V voltage regulator will be the best way to get 1.5V from a higher voltage battery. Why do you think a regulator will not work?
 
Steven Ellet said:
Sorry for the long delay, I have kind of been all over the place. As far as I could tell that requires a "check voltage" and that is a little inconvenient. For example, if I have a car battery and want to get only 1.5v from it, I would effectively need a 1.5 voltage battery to make it work.

Not sure why you think that.

A linear voltage regulator can give you 1.5V out from an input of 15V down to about 4V. See the LM317 voltage regulator, for example. The disadvantage of a linear regulator is that it burns power doing its regulation job. If you have 12V in and 1.5V out, the regulator is dropping the difference voltage 10.5V across it. Multiply that voltage by the current through the regulator to get how much power it has to dissipate.

A better solution is to use a DC-DC converter circuit, like a Buck Regulator.

What is your application? How much application current do you need at 1.5V out? Will you input voltage ever be below about 4V?
 
berkeman said:
What is your application?
I have a camera that requires 3v (2 AA/1.5). unfortunately once the battery strength drop below 2.5v, the camera dies, I found a way to input ac and output a constant 3v, I want to do this with a battery(s) to make it more mobile.
 
  • #10
Steven Ellet said:
I have a camera that requires 3v (2 AA/1.5). unfortunately once the battery strength drop below 2.5v, the camera dies, I found a way to input ac and output a constant 3v, I want to do this with a battery(s) to make it more mobile.
No problem. Get a 12V battery, a voltage regulator that runs off of 12V and outputs 3v. Done.
 
  • #11
phinds said:
No problem. Get a 12V battery, a voltage regulator that runs off of 12V and outputs 3v. Done.
I'll try that and get back to you.
 
  • #13
phinds said:
No problem. Get a 12V battery, a voltage regulator that runs off of 12V and outputs 3v. Done.

I consider that to be irresponsible advice. It does not take the realities of the system into consideration and makes the task seem much simpler than it really is.

There are two basic regulator architectures to get 3V from 12V.

1. Linear regulator. This just drops the output voltage from 12V to 3V, dissipating the power dropped across the regulator. For example, if your camera draws 200ma (I have no idea if that is realistic) then the camera uses 3V* 0.2A = 600mW. The regulator dissipates 9V*0.2A = 1.8W, or three times what the camera uses. You need to understand what the camera draws and whether your regulator will safely dissipate the power it needs to. The higher your battery voltage above 3V, the more power you waste regulating it down to 3V. The most efficient is a large 3V battery pack and no regulator.

2. Switching (buck) regulator. This uses inductors and active switching technology to efficiently (90%, maybe) convert 12V to 3V. So, in the case above you would only draw 660mW from the source. But, this is a complicated design. I just want you to understand it exists.

In both of the cases above, your circuit needs to deal with transient current demands. For example when zooming, focusing, etc the camera draws a lot more current. If your circuit is not designed to consider those occasional demands then the camera may malfunction as the voltage dips due to excessive load.

Again, I recommend a large 3V battery pack.
 
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  • #14
buck regulator modules are very cheap and easily obtainable, with no circuit construction to do, other than wire the module into your system

eg ...
https://www.amazon.com/dp/B00C64ZJYU/?tag=pfamazon01-20

and this one ...

https://www.amazon.com/dp/B00CEUN0VU/?tag=pfamazon01-20

you couldn't build one for the cost of just buying one

I would put fuses in the input and output positive leads rated at around 500mA more than what the camera is supposed to drawDave
 
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  • #15
I guess the decision on whether to go with a 3V battery pack or a 12V battery with a buck switcher comes down to how much battery you want to lug around. A 12V car battery provides lots of power, but is heavy. A bunch of D cells is lighter and easy to replace. There are lots of other options, but those are the big two. It depends on how long you want the batteries to last.
 
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  • #16
Jeff Rosenbury said:
I guess the decision on whether to go with a 3V battery pack or a 12V battery with a buck switcher comes down to how much battery you want to lug around. A 12V car battery provides lots of power, but is heavy. A bunch of D cells is lighter and easy to replace. There are lots of other options, but those are the big two. It depends on how long you want the batteries to last.

yup, exactly, Jeff,

the OP hasn't really been forthcoming about the overall project requirements
we know it's a camera and he wants it more mobile. But he hasn't defined what "more mobile" means to him :wink:
As is a common case, we all get many posts into a thread, bouncing all sorts of ideas around, not knowing what is really needed

So @Steven Ellet ...for us to help you better, you need to be clearer on your project requirements :smile:

cheers
Dave
 
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  • #17
use a zener diode if you want the simplest method.
 
  • #18
Don't_ban_me! said:
use a zener diode if you want the simplest method.

Uhhhh ---- that's not a very practical suggestion. I suggest you try to design a zener diode solution that allows for 50ma steady state currents and 1A transients and maintains 3 or 3.3V +- 10%. Don't forget to deal with power dissipation in the zener and series resistor, and the change in zener voltage with varying current.

The simplest is a voltage regulator, which is designed to deal with a wide range of output currents.
 
  • #19
meBigGuy said:
Uhhhh ---- that's not a very practical suggestion. I suggest you try to design a zener diode solution that allows for 50ma steady state currents and 1A transients and maintains 3 or 3.3V +- 10%. Don't forget to deal with power dissipation in the zener and series resistor, and the change in zener voltage with varying current.

The simplest is a voltage regulator, which is designed to deal with a wide range of output currents.
A three terminal voltage regulator is simple, but inefficient. I would not recommend it for battery applications. A buck mode switcher or simply a couple of D cells seems better.

A zener(ish) diode is usually bad as a power regulator, but might be used a a reference voltage with some other components. But wrap those components in one package and it's called a linear voltage regulator. It's simple to use and usually has lots of added features like 50/60 Hz noise rejection.
 

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