Help needed with Resistors

[Dave Ryman]

Hi,
I'm new to this group.

I did some (very) basic electronics at school, but need to step
down a DC voltage, and am getting a little unsure if I have things right.

I'm stepping 6v (4xAAA batteries) down to 5.2v. I intend to use a
potential divider with 2 120R resistors on the +ve side (in parallel) and
a 390R resistor on the 0v side. This, I calculate, will give me 5.2v on
the output.

The device being powered may draw upto 320mA. First question: Am I
right in assuming that 2W resistors will be ok for this?

Second question: I found some helpful stuff on the net. One circuit
is for a loaded PD circuit. Am I right in assuming that my device will
put an impedance of 1.6R across the output of my PD and therefore change
the voltage? Or is this irrelevant since the load is after the supply of
the correct voltage?

--
Regards,
Dave

dave_ryman@hotmailNOSPAM.com
http://welcome.to/daves.website
http://travel.to/formula.one

 

[A E]

Dave Ryman wrote:

A E <aeisenhut@videotron.ca> wrote in
news:3F33B306.4F862EB6@videotron.ca:

Dave Ryman wrote:

Hi,
I'm new to this group.

I did some (very) basic electronics at school, but need to
step
down a DC voltage, and am getting a little unsure if I have things
right.

I'm stepping 6v (4xAAA batteries) down to 5.2v. I intend to
use a
potential divider with 2 120R resistors on the +ve side (in parallel)
and a 390R resistor on the 0v side. This, I calculate, will give me
5.2v on the output.

Only with no load, for a few minutes, as the battery voltage goes
down, because of the load of the resistors.

The device being powered may draw upto 320mA. First question:
Am I
right in assuming that 2W resistors will be ok for this?

No.

Second question: I found some helpful stuff on the net. One
circuit
is for a loaded PD circuit. Am I right in assuming that my device
will put an impedance of 1.6R across the output of my PD and
therefore change

Yes.


the voltage? Or is this irrelevant since the load is after the supply
of


No, you will need resistors ten times smaller than your load to make a
divider that gives a voltage close to what you want, in other words,
practically a short circuit.

the correct voltage?


A resistive voltage divider is the absolute worst way to achieve your
goal. The divider itself is a load to the batteries and will draw
current even without anything connected.
The usual way to get a lower voltage from a higher one is to use a
regulator. A regulator drops some voltage regardless of the load
current going through it, within limits.
But you have an dropout, or difference between your input and output
voltage of only .8 volts, barely enough to get a linear LDO (low drop
out) regulator to work, and leaves you with little to no headroom as
the voltage drops. You have several choices:
1) You will need a switcher of some sort.
2) Find out if your 5.2V device will care at all if connected to 6V.
3) Use a diode in series with your batteries. A regular silicon
rectifier drops about 0.7v at 10mA and stays fairly constant at 0.7V
even as the current increases. So you put one in series with the
batteries and you get 5.3V roughly. But the drop of the diode stays
constant as your battery voltage drops, so now you need to know if
your device will work with less than 5.2V, which is why regulators
exist.


--
Regards,
Dave


I recommend you calculate how the load will change the voltage on your
divider, just slap your 1.6 ohm in parallel with your 390, and
recalculate your voltage. Fun, huh?



Thanks. This concurs with the man in the electronics shop I went into
today. The battery holder I'm using (used to be an emergency battery
holder for a mobile phone) already has such a diode connection, so life
is suddenly much easier.

Thanks again

No problem.

 

[Dave Ryman]

A E <aeisenhut@videotron.ca> wrote in
news:3F34231B.7499E68D@videotron.ca:

(snip)
Thanks again

No problem.

Well, I've tried it....

Four 1.5v AAA Batteries give me a reading of 6.5v. With the diode added,
it actually stays exactly the same: 6.5v, sometimes drops to 6.4.

I didn't realise batteries produced such a varying voltage - each one is
producing just over 1.6v.

So is a Voltage regulator the way to go? I have no idea about these
devices, how they behave or how I wire them up. I have seen voltage
regulators on the net described as "5v" - If I wire these to a 6.0 ->
6.5v supply, do I get a steady 5v out the other end?

Some help would be greatly appreciated.

--
Regards,
Dave

dave_ryman@hotmailNOSPAM.com
http://welcome.to/daves.website
http://travel.to/formula.one

 

[Dave Ryman]

Dave Ryman <dave_rymanNOSPAM@hotmail.com> wrote in
news:Xns93D2101F92F9Ddaverymanhotmail@130.133.1.4:

A E <aeisenhut@videotron.ca> wrote in
news:3F34231B.7499E68D@videotron.ca:


(snip)
Thanks again

No problem.




Well, I've tried it....

Four 1.5v AAA Batteries give me a reading of 6.5v. With the diode
added, it actually stays exactly the same: 6.5v, sometimes drops to
6.4.

I didn't realise batteries produced such a varying voltage - each one
is producing just over 1.6v.

So is a Voltage regulator the way to go? I have no idea about these
devices, how they behave or how I wire them up. I have seen voltage
regulators on the net described as "5v" - If I wire these to a 6.0 -
6.5v supply, do I get a steady 5v out the other end?

Some help would be greatly appreciated.

I think I have it sorted. I am thinking of using the LM2940CT regulator
to provide me a voltage of 5v +/- 0.15v from a 6v +/- 0.5v source (ie:
4x AAA batteries).

The circuit I'm planning on is:
http://www.maplin.co.uk/media/largeimages/8063i0.jpg

The only thing I'm not certain about is the heat generated by the
regulator: I'll be drawing a maximum of 320mA, and am unsure about
heatsink requirements.

--
Regards,
Dave

dave_ryman@hotmailNOSPAM.com
http://welcome.to/daves.website
http://travel.to/formula.one

 

[Boris Mohar]

On 9 Aug 2003 00:35:04 GMT, Dave Ryman <dave_rymanNOSPAM@hotmail.com>
wrote:

Well, I've tried it....

Four 1.5v AAA Batteries give me a reading of 6.5v. With the diode added,
it actually stays exactly the same: 6.5v, sometimes drops to 6.4.

I didn't realise batteries produced such a varying voltage - each one is
producing just over 1.6v.

So is a Voltage regulator the way to go? I have no idea about these
devices, how they behave or how I wire them up. I have seen voltage
regulators on the net described as "5v" - If I wire these to a 6.0 -
6.5v supply, do I get a steady 5v out the other end?

Some help would be greatly appreciated.

You just stuck a diode in series with the battery and measured the other
end if the diode (cathode) without hooking up the load? Read this

http://people.deas.harvard.edu/~jones/es154/lectures/lecture_2/diode_characteristics/diode_characteristics.html

--

Regards,

Boris Mohar

Got Knock? - see:
Viatrack Printed Circuit Designs http://www3.sympatico.ca/borism/
Aurora, Ontario

 

[Watson A.Name - 'Watt Sun]

In article <Xns93D219283A32Adaverymanhotmail@130.133.1.4>,
dave_rymanNOSPAM@hotmail.com mentioned...

Dave Ryman <dave_rymanNOSPAM@hotmail.com> wrote in
news:Xns93D2101F92F9Ddaverymanhotmail@130.133.1.4:

A E <aeisenhut@videotron.ca> wrote in
news:3F34231B.7499E68D@videotron.ca:


(snip)
Thanks again

No problem.




Well, I've tried it....

Four 1.5v AAA Batteries give me a reading of 6.5v. With the diode
added, it actually stays exactly the same: 6.5v, sometimes drops to
6.4.

I didn't realise batteries produced such a varying voltage - each one
is producing just over 1.6v.

So is a Voltage regulator the way to go? I have no idea about these
devices, how they behave or how I wire them up. I have seen voltage
regulators on the net described as "5v" - If I wire these to a 6.0 -
6.5v supply, do I get a steady 5v out the other end?

Some help would be greatly appreciated.


I think I have it sorted. I am thinking of using the LM2940CT regulator
to provide me a voltage of 5v +/- 0.15v from a 6v +/- 0.5v source (ie:
4x AAA batteries).

The circuit I'm planning on is:
http://www.maplin.co.uk/media/largeimages/8063i0.jpg

Umm, if the pic gets any smaller, it's going to disappear!

The only thing I'm not certain about is the heat generated by the
regulator: I'll be drawing a maximum of 320mA, and am unsure about
heatsink requirements.

You had better think about getting some bigger batteries, too! 329 mA
is an awful lot even for AA cells. In fact, with that much current,
the voltage may drop to the point where the regulator won't work.
When you use batteries, you have to decide at what point you consider
the batteries depleted or discharged. Some equipment considers this
to be 1V per cell, so your AAAs would be putting out only 4V at that
point.


--
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[Dave Ryman]

Watson A.Name - 'Watt Sun' <alondra101@hotmail.com> wrote in
news:MPG.199e9392a2ab0a72989b8e@news.inreach.net:

In article <Xns93D219283A32Adaverymanhotmail@130.133.1.4>,
dave_rymanNOSPAM@hotmail.com mentioned...
Dave Ryman <dave_rymanNOSPAM@hotmail.com> wrote in
news:Xns93D2101F92F9Ddaverymanhotmail@130.133.1.4:

A E <aeisenhut@videotron.ca> wrote in
news:3F34231B.7499E68D@videotron.ca:


(snip)
Thanks again

No problem.




(snip)

The power supply is being used to effectively charge the lithium battery
in the equipment - in the same way as an emergency backup for a mobile
phone. I do not believe that the batteries need to "live" very long to
do this: It is a messy fix for a friend who is unable to use a mains
charger where they are.

I used to use the same process for emergency charging my Nokia mobile: I
got a few full charges out of one set of AAAs (alkaline).

--
Regards,
Dave

dave_ryman@hotmailNOSPAM.com
http://welcome.to/daves.website
http://travel.to/formula.one

 

[K Wind]

"Dave Ryman" <dave_rymanNOSPAM@hotmail.com> wrote in message
news:Xns93D2101F92F9Ddaverymanhotmail@130.133.1.4...

A E <aeisenhut@videotron.ca> wrote in
news:3F34231B.7499E68D@videotron.ca:


(snip)
Thanks again

No problem.




Well, I've tried it....

Four 1.5v AAA Batteries give me a reading of 6.5v. With the diode added,
it actually stays exactly the same: 6.5v, sometimes drops to 6.4.

Do you have the batteries connected to what you plan on powering? It doesn't
sound like it. If not, you will have the same voltage reading on both sides
of the diode.

Ken

I didn't realise batteries produced such a varying voltage - each one is
producing just over 1.6v.

So is a Voltage regulator the way to go? I have no idea about these
devices, how they behave or how I wire them up. I have seen voltage
regulators on the net described as "5v" - If I wire these to a 6.0 -
6.5v supply, do I get a steady 5v out the other end?

Some help would be greatly appreciated.

--
Regards,
Dave

dave_ryman@hotmailNOSPAM.com
http://welcome.to/daves.website
http://travel.to/formula.one

 

[Dave Ryman]

"K Wind" <kwind@news-server.neo.rr.com> wrote in
newsLlZa.69551$ib2.18008923@twister.neo.rr.com:

"Dave Ryman" <dave_rymanNOSPAM@hotmail.com> wrote in message
news:Xns93D2101F92F9Ddaverymanhotmail@130.133.1.4...
A E <aeisenhut@videotron.ca> wrote in
news:3F34231B.7499E68D@videotron.ca:


(snip)
Thanks again

No problem.




Well, I've tried it....

Four 1.5v AAA Batteries give me a reading of 6.5v. With the diode
added, it actually stays exactly the same: 6.5v, sometimes drops to
6.4.

Do you have the batteries connected to what you plan on powering? It
doesn't sound like it. If not, you will have the same voltage reading
on both sides of the diode.

Ken



I didn't realise batteries produced such a varying voltage - each one
is producing just over 1.6v.

So is a Voltage regulator the way to go? I have no idea about these
devices, how they behave or how I wire them up. I have seen voltage
regulators on the net described as "5v" - If I wire these to a 6.0 -
6.5v supply, do I get a steady 5v out the other end?

Some help would be greatly appreciated.

--
Regards,
Dave

dave_ryman@hotmailNOSPAM.com
http://welcome.to/daves.website
http://travel.to/formula.one

Thanks for your help - I realise that now!

--
Regards,
Dave

dave_ryman@hotmailNOSPAM.com
http://welcome.to/daves.website
http://travel.to/formula.one

 

[Dave Ryman]

"Walter Harley" <walterh@cafewalter.com> wrote in
news:vjfnresrjjp9e4@corp.supernews.com:

"Dave Ryman" <dave_rymanNOSPAM@hotmail.com> wrote in message
news:Xns93D1863DD394Ddaverymanhotmail@130.133.1.4...

I did some (very) basic electronics at school, but need to step
down a DC voltage, and am getting a little unsure if I have things
right.

Hey, Dave, the other responses to your question have been on target,
but it sounds like there are still a few things that could be made
more clear.

1. Batteries are not ideal components. Their voltage varies
significantly depending on how much current is being drawn and how
exhausted they are (and on many other factors as well). A nominal
1.5v battery might be 1.55v, or it might be 0.8v, or less.

2. In addition to any resistances you use for dividers, you need to
take into account that the battery has internal resistance - higher
for a small battery like a AAA, and higher as the battery approaches
end of life - and the load also has resistance. These resistances may
not be constant, and may be hard to calculate or measure, but they are
for real and they will affect your calculations.

3. Diodes have a nominal voltage drop of around 0.7v, but it is more
at higher current and less at lower current.

4. A decent voltmeter draws essentially no current from the circuit it
is measuring.

5. Ohm's Law is your friend. Voltage dropped across a resistance
equals current times resistance. However, many components (such as
diodes and batteries) have different resistances depending on how much
current is passing through them, which makes the calculation more
difficult. For DC, power dissipated equals current times voltage.

By the way, whether you use a diode or a resistor or something else to
drop the voltage, if there's voltage across it and current going
through it at the same time, it's going to dissipate power and heat
up. If you really want to drop 0.8v at 325mA, that's about a quarter
of a watt, which will be too much for a small diode. Make sure your
diode is rated for at least twice the needed power, to be safe.

Thanks very much - I am more than a little inexperienced in electronics,
knowing more-or-less what I remember from school.

Many answers on this newsgroup has helped expand my understanding, and
I've done a bit of reading up on the net.

I now realise that batteries offer a voltage during their life which is
about as consistent as a bus timetable. I also realise that there are
problems introduced by the resistance of the device connected.

I've gone for a Voltage-regulator solution, using a VR with low dropout-
voltage (since I'm only dropping about 1 volt from the supply provided by
the batteries). This, I believe, should do the job. My only concern is
how much heat will be generated by the VR, and whether current will be
drawn by the VR when the device is not connected. I am hoping that a few
holes in the case will allow heat to escape, and I guess removing the
batteries when not in use will prevent the second problem (if it is a
problem at all).

--
Regards,
Dave

dave_ryman@hotmailNOSPAM.com
http://welcome.to/daves.website
http://travel.to/formula.one

 

[Walter Harley]

"Dave Ryman" <dave_rymanNOSPAM@hotmail.com> wrote in message
news:Xns93D4ECD2FA27Cdaverymanhotmail@130.133.1.4...

I've gone for a Voltage-regulator solution, using a VR with low dropout-
voltage (since I'm only dropping about 1 volt from the supply provided by
the batteries). This, I believe, should do the job. My only concern is
how much heat will be generated by the VR, and whether current will be
drawn by the VR when the device is not connected. I am hoping that a few
holes in the case will allow heat to escape, and I guess removing the
batteries when not in use will prevent the second problem (if it is a
problem at all).

You can calculate the heating. The power dissipated is V * I, so
approximately 1v * 325mA (by your numbers) which equals .325W, that is,
about a third of a watt. The regulator device's datasheet will specify a
couple of "thermal resistance" numbers measured in degrees celsius per watt,
typically one for "junction to case" and one for "junction to ambient". If
you subtract one from the other, you'll get a measure of how much the
resistance from case to ambient in free air is. So, let's say that there's
1.5 C/W from junction to case, and 100 C/W from case to ambient (I'm making
these numbers up). Then, at .325W, the case is going to be 32.5 degrees C
hotter than ambient, and the junction (inside of the device) will be about
..5 degree hotter than that. If ambient is 25C, the case will be 57C.

Be aware that junction to ambient is usually rated for free air, not a tiny
sealed enclosure.

If you mount to a heat sink, then the heat sink will be rated for some
thermal resistance, and the connection between the device and the heat sink
will be rated, and so forth. The resistances add; it's sort of like Ohm's
Law again, with electrical power filling the role of thermal current, and
temperature filling the role of thermal voltage. That is, T = P*R.

Much more intuitively: a third of a watt in a physically small device with
no heat sink is generally going to make it hot to the touch after a few
minutes of operation, but not so hot as to melt plastic.

-walter