Chip with simple program for Toy

[Tom Biasi]

<upgrdman@mindspring.com> wrote in message
news:1116232488.194543.169740@g14g2000cwa.googlegroups.com...

Thanks guys, so now I have a few more questions:

To clarify, is this what Don is reccomending:

[+]LED,resistor[-]
[+]LED,resistor[-]
[+]LED,resistor[-]
...

and I'm still a bit confused about what resistor(s) to choose. I was
reading this page on how to power LEDs:

http://wolfstone.halloweenhost.com/TechBase/litlpo_PoweringLEDs.html

and it leads me to believe that resistors are specified for both
resistance (ohms) and wattage (watts). Since now I will be using
resistors, I was thinking of going with 4 AA NiMH batteries to increase
runtime. So this leads me to the question of which resistors to choose.
Is the 39ohms, or the 22ohms resistor better suited to this task? and
what wattage? and just so I can better understand this, is it better to
pick a resistor rated to handle more ohms and more wattage? or less?

I noticed a section on that website I linked to above about combining
both the series and parallel methods,
http://wolfstone.halloweenhost.com/TechBase/litlpo_PoweringLEDs.html#WiringLargeNumbersOfLEDsInSeriesParallel
and was thinking of maybe going that route. Is that a good idea, and if
so, how would that affect my resitor selections.

If I should go with the hybrid serial-parallel route, won't I need to
have a battery that supplies more volage? 4 NiMH AA's would be 4.8V...
not even enough for 2 LED's. (I plan to use all 20 LED's) I was
thinking of perhaps buying a pair of rechargable 9V batteries, but 18V
is only enough for about 5 LED's. I do not want to use too many
batteries as it would weight-down my RC car... and since I have a large
supply of rechargable AA's I would prefer to use them. Would I need to
get another device to increase the voltage... what is it called, a
transformer?

Sorry, I'll bet I'm really confusing stuff up right about now. So back
to basics... Is there a nice webpage or book that describes the basic
electronics vocabulary, and basic devices (diodes, resistors,
transformers, capacitors, transistors, etc...) I think this is my main
problem, because the webpage I linked to above did a great job of
defining how the equations worked, it's just that I don't quite know
what voltage, wattage, current, power, etc. are, and I'm going off of
guesses and past expierences.

Thanks again,
--Farrell F.

Hi,
Try 39 Ohms and if the LED is not bright enough try a lower value.
1/4 watt resistors are readily available and are more than sufficient for
your case.

Let's educate you a little so that you may better understand the suggestions
that are given to you.

The LEDs are diodes not light bulbs.
The junction of the diode will emit light when current passes through it.
The voltage across the junction will rise as current increases up to a
point; at that point the voltage will remain fairly constant. This is the
forward voltage the spec talks about.
If you supply more voltage than that the junction will overheat and die.
The resister in series will allow the extra voltage to "have somewhere to
be".
So the resister that you need is determined by the extra voltage that you
have and the current that you need, R=V/I.
You mentioned a 3.8 volts supply and LEDs that could max at 3.0 volts.Your
resister would be .8V/.020A= 40 Ohms.
39 Ohms is the nearest standard value. The power dissipated would be P=VI.
..8V*.020 A. = .016 Watts.
For physical convenience I would use a 1/4 watt resister. If you use more
voltage in your supply you would need to replace the .8v with the excess
that you have.
Forget about a transformer right now. The theory doesn't apply to what you
are doing.
Good Luck,
Tom



>

 

[John Larkin]

On 16 May 2005 06:09:36 -0700, "BobG" <bobgardner@aol.com> wrote:

Anyone know how fast you can turn these on and off? Can I make a motor
controller out of em and run em at 1KHz?

Well, the datasheets have the specs. But if you're thinking PWM at 1
KHz, probably not; the optocoupled SSRs are pretty slow.

John

 

[Yoda]

Hard to say but you'll probably be okay if it's just a basic electronics
course. You'll probably get into ohm's law, basic power equations and maybe
loop and nodal analysis which requires solving systems of equations. I
wouldn't think it would go much farther than that. But if you're seriously
going into electronics or electrical engineering the math gets intense later
so more math is always better as a general rule.

Good luck


"Top gear winner 2" <esrabb@cox.net> wrote in message
news:1116286041.804869.205390@g49g2000cwa.googlegroups.com...

I am going to take a basic electronics course at my home from Thompson
Education Direct. This will be a great first start for me to learn
basic electronics and after I get a diploma in basic electronics, I am
going to do my own electronics projects and moving up in the
electronics field. Also I am going to learn about 555 timers and
digital electronics in the future. Also my mom is teaching me college
intermediate algebra math at this time. I have one Question? Is the
math in this basic electronics course harder then college intermediate
algebra?

Thanks,

Chris

 

[Tom Biasi]

<upgrdman@mindspring.com> wrote in message
news:1116315610.968710.97290@g47g2000cwa.googlegroups.com...

Thanks Tom!

So since you reccomend that I should not go with a transformer, I guess
my assumption of how to wire my LED's in my previous post is correct?
e.g.

[+]LED,resistor[-]
[+]LED,resistor[-]
[+]LED,resistor[-]
...

No? and does it matter which "side" of the LED the resistor will be
attached to?

And now I think I will go with a 4 pack of 1.2V NiMH AA's as my power
source, so will you please check my math and tell me if I am messing
things up:

* 3.0V is the min my LED's are spec'd for, so would it be bad if I aim
to provide them with 3.6V, just a little below their 3.8V max?
* 3.6V per LED, 4.8V power source leaves me with 1.2V to resist.

1.2V/.02A=60 Ohms
1.2V*.02A=.024 Watts

And since I am not familiar with standard resistor specifications, I'm
guess I should still go with a 1/4th Watt resistor, for approx. 60
Ohms... and should I round up or down?

I looked online at Radioshack.com and found 100 Ohm 1/4 Watt
resistors... do they sound about right?
http://www.radioshack.com/category.asp?catalog%5Fname=CTLG&category%5Fname=CTLG%5F011%5F002%5F014%5F001&Page=1

Thanks,
--Farrell F.

Hi,
Your math is correct.
You can get a standard resistor in 10% of 56 Ohms and in 5% of 62 Ohms. 1/4
watt is fine.
It doesn't matter which side the resister goes on but it does matter which
side gets the plus and minus.
Just some additional information: You indicate that you may still think the
3.0 to 3.8 is a min/max supply rating for your devices.
It is a range of expected forward drop voltage. The forward drop can be
anywhere in that range for any given device. Measure them in operation. You
may need to calculate a different resistor for each LED if you want uniform
brightness.
Regards,
Tom

>

 

[Jim Douglas]

I am about ready to take a course from these guy's also, although different
has anyone had any experience with this company Thompson Education Direct? I
don't want a degree just want to learn more about the subject so that I can
kick someone's ass at the office!


"Top gear winner 2" <esrabb@cox.net> wrote in message
news:1116286041.804869.205390@g49g2000cwa.googlegroups.com...

I am going to take a basic electronics course at my home from Thompson
Education Direct. This will be a great first start for me to learn
basic electronics and after I get a diploma in basic electronics, I am
going to do my own electronics projects and moving up in the
electronics field. Also I am going to learn about 555 timers and
digital electronics in the future. Also my mom is teaching me college
intermediate algebra math at this time. I have one Question? Is the
math in this basic electronics course harder then college intermediate
algebra?

Thanks,

Chris

 

[Rich Grise]

On Tue, 17 May 2005 00:40:11 -0700, upgrdman wrote:

So since you reccomend that I should not go with a transformer, I guess my
assumption of how to wire my LED's in my previous post is correct? e.g.

[+]LED,resistor[-]
[+]LED,resistor[-]
[+]LED,resistor[-]
...

No?

No, Yes. ;-) I've always put the resistor in the + lead, just 'cause
it feels better to have the LED cathode closer to ground.

and does it matter which "side" of the LED the resistor will be
attached to?

No, they're in series, and every point in a series circuit passes the
same current.

And now I think I will go with a 4 pack of 1.2V NiMH AA's as my power
source, so will you please check my math and tell me if I am messing
things up:

* 3.0V is the min my LED's are spec'd for, so would it be bad if I aim to
provide them with 3.6V, just a little below their 3.8V max? * 3.6V per
LED, 4.8V power source leaves me with 1.2V to resist.

NO!

LEDs are rated by _current_ - the voltage is just what happens to
appear across a given LED when the rated current is passed through it.
In this case, you'd calculate the resistor value at the low end of
that range, for a current of, say 20 mA. So 4.8V - 3.0V is 1.8V, divided
by .02A is 90 ohms. 91 is the closest standard value. If your LED
happens to be one of the ones that drops 3.8V at the rated 20 mA current,
then there will be 1.0V across your 90 ohm resistor, which will give
you 11 mA, but (A) The only LEDs I've done the experiment were so
close to as bright at 11 mA as they were at 20 mA that it was practically
the same, and (B) you're on the safe side for any LED within that range.

As Tom Biasi has suggested, it would probably be a good idea to
set up a circuit that provides a constant 20 mA, and measure the
voltage drop across a batch of LEDs. It could be very educational.

This is all assuming that your LEDs are rated for 20 mA, of course.
If yours have a different spec, just use that value in the formulas.

If you use a 12V supply, then for 20 mA with an LED that drops 3V,
you'd use a (12 - 3)/.02 = 450 ohm resistor, which if your LED
drops 3.8V, would still give you 18.2 mA, which is much closer
to a constant current supply for your tests.

Hope This Helps!
Rich

 

[steamer]

--All parts available at Radio Shack; I'm doing this to run
various Stamp applications on a small art car. Getcher self one of
those cigarette lighter 3-way splitters, then plug in the various
transformers that RS also has, to change voltages as needed.

--
"Steamboat Ed" Haas : The other night I
Hacking the Trailing Edge! : dreamed about wasabi...
http://www.nmpproducts.com/intro.htm
---Decks a-wash in a sea of words---

 

[dan]

Saran <saranyan13@gmail.com> wrote:

So, let me restate what I understood. The current mirror's output
current is used to charge the gate of the transistor. So, depending on
the magnitude of the current, the gate voltage increaser faster or
slower (sorry about my layman language, I just want to be very very
clear).

If I understood this correctly, then:
1) If a current mirror is charging the gate of a transistor, or in
other words, changing its Vgs, the maximum Vgs value is determined by
the saturation voltage for the mirror.
2) We are controlling the rate at which gate is charged in a circuit.

Overall, I'd say you didn't understand it, and it wasn't explained real
well to you, but maybe I just don't understand what you and others have
said.

First off, MOSFETs are generally considered to be voltage controlled
current sources, at least when operated in the saturation regime (Vds >
Vgs - Vt). BJTs on the other hand are current controlled current sources.
Each of these transistors have respective advantages.

In analog circuitry you often deal with both voltage and current, and
since MOSFETs are voltage controlled current sources its important to
consider all aspects. MOSFETs are complicated devices, entire books have
been written simply discussing aspects of how MOSFETs work. To try to
boil them down into one or two sentences will surely leave you with an
incorrect understanding. Even the most cursory treatment of them in an
engineering text book would be at least several pages long.

The purpose of a current mirror is to duplicate (or mirror) a current with
some sort of multiplicative factor (M/N). If you are following good
layout practices, M and N should both be integers. Let's say we have a
simple N-channel mirror, where M and N are both 1. So we have two
identical transistors (M1 and M2). M1 and M2's gates are both tied
together and tied to the drain of M1. M1 and M2's source is tied to
ground. A current, I1 is pushed into the drain of M1. M1 is said to be
"diode connected" because its drain and gate are shorted together.

Now, let's ignore M2 for the moment. If we take M1 to be an ideal
transistor, no current will enter the gate of M1, so all of I1 must enter
the drain. The gate however is capacitive, as such its voltage can be
increased by pumping current in, or decreased by pumping current out.
The gate has access to a current source (I1) to increase or decrease the
gate voltage so as to allow all the current to flow through the drain.
It should be obvious that as its adjusting the gate voltage the current
through the drain will not perfectly match I1, but it should get closer
and closer over time. Eventually the gate voltage will be forced to
settle to whatever voltage will produce drain current equal to I1. Assume
for the moment that M1 is biased into saturation:

Id = K * (W/L) * (Vgs - Vth)^2 * (1 + Lambda*Vds)

Lambda is the channel length modulation and is often quite small. It can
intentionally be made small by increasing L (gate length). K and Vth are
constants that are fixed by the processing. W and L are the gate width
and length respectively and are chosen by the designer.

Now, let's add M2 back into the circuit. M2's gate is shorted to M1's.
M2's source is shorted to M1's. If we look at the Id equation, assuming
identical W/L, the current through the drain of M2 should closely match
that of M1. In real designs the designer may use different W's for M1
and M2 in order to achieve the multiplicative effect I previously
mentioned.

I don't think all that much thought is given to the amount of current
required to charge the gate of the MOSFET, unless speed is a concern in
which case there are much more advanced and accurate topologies that can
be used. Typically current mirrors are used for providing a static bias
current to multiple points within a circuit based on a single reference.
In this case, the reference rarely moves, and speed simply isn't a
concern, and hence no one thinks about the charge rate. There are other
instances when an AC signal is being mirrored in which case it may be more
of a concern however.

So the current mirror's *INPUT* current is used to charge the gates of the
mirror transistors. The output current could be used for a variety of
things depending on why a mirror was even employed to begin with.
Regarding your assertion about the maximum voltage... The only criteria
on a current mirror and voltage is that the transistors (both of them)
must remain in saturation (i.e. Vds > Vgs - Vth). If the transistors
slip into triode region, then the transistors cease to act as very good
current sources, and start acting more like resistors, as such the mirror
becomes far less effective if the loads on the mirror branches differ
substantially.

dan

 

[John Popelish]

Andrew Holme wrote:

I found this surrey.ac.uk past paper whilst surfing:

http://www.ee.surrey.ac.uk/Teaching/Exams/pastpapers/01-02/Spring/Level4/EE4.rfs02.pdf

I'm a bit confused by the 0.5 ohm loss resistance shown in figure A1 on
page 2.

I don't see where the .5 ohm resistor is labeled as being a loss
resistance. I think it is probably the internal series resistance of
the varactor.

The capacitive reactances add up to -j160. If the inductor
was +j160 to resonate, the unloaded Q would be 160/0.5 = 320, which
sounds a bit high even for unloaded Q. It says Q=80 on the diagram,
which is more like it. Did they calculate 80/160=0.5 by mistake? Is
80 supposed to be the loaded Q? What am I missing??


BTW if you go up a few directories on that URL, you can browse all
their past papers.

 

[John Popelish]

pdrunen@aol.com wrote:

Hi All,

I was looking at a '70 era circuit book and found that devices like the
7400 nand gate could be used as a linear amplifier of small signals.

You could use one of the inputs as a control signal to turn on/off the
output and the other would be bias with a resistor-divider and the ac
signal injected via a cap. Gain was like 15db.

Can other device families such as the HCT or HCTLS or LS, S, be used
and can I get a higher frequency than standard TTL?

How about ECL, can ECL be used in the linear small-signal mode?

I look forward to your input.

pdrunen.

Almost any logic function that inverts, can be used as a linear

amplifier, with the addition of a bias (DC feedback) arrangement, and
possibly a phase compensation network, (to make sure the total loop
gain falls below 1 before the loop phase shift increases more than
about 145 degrees). Even multistage CMOS inverters, like the buffered
HC family can be stabilized as linear amplifiers. You may have to be
concerned about power consumption for some types, though.

 

[John Larkin]

On 18 May 2005 07:05:37 -0700, pdrunen@aol.com wrote:

Hi All,

I was looking at a '70 era circuit book and found that devices like the
7400 nand gate could be used as a linear amplifier of small signals.

You could use one of the inputs as a control signal to turn on/off the
output and the other would be bias with a resistor-divider and the ac
signal injected via a cap. Gain was like 15db.

Can other device families such as the HCT or HCTLS or LS, S, be used
and can I get a higher frequency than standard TTL?

Multistage logic, like the HC/HCT/4000B stuff, tends to oscillate if
used like this; too many poles, too much gain. The unbuffered "UB"
CMOS types generally work. Most real TTL (bipolar) parts will work.

How about ECL, can ECL be used in the linear small-signal mode?

ECL usually works fine as a fast linear amp, especially the simple
gates. The nuclear folks like to use ecl in things like wire chambers
that need many channels of fast gain cheap.


John

 

[steamer]

--Um, yeah, I think so... ;-)

--
"Steamboat Ed" Haas : The other night I
Hacking the Trailing Edge! : dreamed about wasabi...
http://www.nmpproducts.com/intro.htm
---Decks a-wash in a sea of words---

 

[Bob Masta]

On 18 May 2005 07:05:37 -0700, pdrunen@aol.com wrote:

Hi All,

I was looking at a '70 era circuit book and found that devices like the
7400 nand gate could be used as a linear amplifier of small signals.

You could use one of the inputs as a control signal to turn on/off the
output and the other would be bias with a resistor-divider and the ac
signal injected via a cap. Gain was like 15db.

Can other device families such as the HCT or HCTLS or LS, S, be used
and can I get a higher frequency than standard TTL?

How about ECL, can ECL be used in the linear small-signal mode?

I look forward to your input.

pdrunen.

Don Lancaster has some amp circuits in his classic
"CMOS Cookbook".

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Home of DaqGen, the FREEWARE signal generator

 

[John Bokma]

wrote:

hi
i wish to wire 6 (or x number) of the same led, using a psu - 12v line.
Most probaly it will be an orange led, thought i should say that as
i've read that the actual colour will effect the resistors ..is that so
?
So simply whats the best way .... ?

find out the Vforward and Iforward for the LEDs you are going to use


(12 - x * Vf)
R = -------------
If

--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html

 

[Tsu Do Nimh]

tim wrote:

I have the pickit 1 programmer. I want to go through the
tutorial examples, starting with Debounce. Tha samples
come in an asm and C versions. My question, what do I need
to compile the C files?
Hi,

Try this:

http://www.micahcarrick.com/v2/content/view/14/8/

Tsu
--
Tsu|
Do|
Nimh|
|

 

[Bob Monsen]

colin2000@talk21.com wrote:

find out the Vforward and Iforward for the LEDs you are going to use


(12 - x * Vf)
R = -------------
If

--
John


thanks.....one thing though
could you elaberate on Vforward and Iforward ((If Vf)?!

Put them all end-to-end (in series) and put a 1k resistor in series too.
That should light them up with 3.6mA. If you want them brighter, use a
lower value resistor, but too low will cause them to fail. The lowest
you probably want to go is 220 ohms.

---
Regards,
Bob Monsen

 

[Joe McElvenney]

Hi,

I'm a bit confused by the 0.5 ohm loss resistance shown in figure A1 on
page 2. The capacitive reactances add up to -j160. If the inductor
was +j160 to resonate, the unloaded Q would be 160/0.5 = 320, which
sounds a bit high even for unloaded Q. It says Q=80 on the diagram,
which is more like it. Did they calculate 80/160=0.5 by mistake? Is
80 supposed to be the loaded Q? What am I missing??

By giving you the inductor's Q (= 80), they are indirectly telling you
its total loss resistance (RF and DC) which I make to be 2 ohms. Add-
in the 0.5 ohm parasitic and you finish up with Q(tot) = 64. This value
doesn't include any loading by the bias and emitter resistors but then
that is the beauty of the Clapp. This oscillator dates back to the
30's but J. K. Clapp published an article on it in QST as late as
October 1948.

BTW if you go up a few directories on that URL, you can browse all
their past papers.

What I love about these old exam papers is that they remind one just
how much has been forgotten over the years and forces you to dig out
the books once again - good old Terman, et al...


Cheers - Joe

 

[colin]

<u035m4i02@sneakemail.com> wrote in message
news:1116516241.689332.57000@o13g2000cwo.googlegroups.com...

It seems like a simple question, but I've gotten drastically different
responses from different people. How do you calculate the differential
input impedance for an amplifier like this:

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opampvar6.html

What if you keep your source between the two inputs, but ground one?
Does the impedance now change?

If you connected two separate sources from ground to each input (or a
single common-mode source?), what input impedances do they see?

And so on. I have been very confused with people telling me I'm not
taking "negative impedance" into account and such.

Is it possible to measure the input impedance with a standard ohmmeter
or RCL meter? (As long as the amp doesn't clip, of course.)

the non inverting input just sees 2 100k resistors in sreries asuming the op
amp impedance can be neglected so is just purly 200k.

the inverting input just sees the first 100k as the op amp tries to keep the
end of that resistor at half the potential of the non inverting input,
however a change in voltage on just the non inverting input cuases a change
in curent on the inverting input.

Colin =^.^=

 

[Ban]

u035m4i02@sneakemail.com wrote:

It seems like a simple question, but I've gotten drastically different
responses from different people. How do you calculate the
differential input impedance for an amplifier like this:

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opampvar6.html

What if you keep your source between the two inputs, but ground one?
Does the impedance now change?

If you connected two separate sources from ground to each input (or a
single common-mode source?), what input impedances do they see?

And so on. I have been very confused with people telling me I'm not
taking "negative impedance" into account and such.

Is it possible to measure the input impedance with a standard ohmmeter
or RCL meter? (As long as the amp doesn't clip, of course.)

Differential input impedance is 200k on the +in and 66.66k on the -in
common mode input impedance is equal and 200k on both inputs.
Let us have +Vin= V1 and -Vin= V2
You can find that out by first looking at the +in, where the level is
1/2Vin. The opamp steers its output in such a way that there is the same
voltage on the -in. This means the voltage across R1 is 1.5Vin. since the
resistor is 100k this corresponds to R= 100k/1.5 =66.66666k
Now do the same with V1 and V2 both +Vin to obtain the common mode input
impedance.
This is taking the opamp as ideal, which is pretty much what you can also
measure.
I think you should measure the current from a fixed voltage source. The
ohms range it is pretty meaningless and might damage the meter depending on
the supply voltage of ther opamp. The additional current can lead to
unexpected output voltages.

--
ciao Ban
Bordighera, Italy

 

[John Bokma]

wrote:

find out the Vforward and Iforward for the LEDs you are going to use


(12 - x * Vf)
R = -------------
If

--
John

thanks.....one thing though
could you elaberate on Vforward and Iforward ((If Vf)?!

Those values are given in a datasheet. Vf is different for LEDs with
different colours. If is often around 20 mA, but look it up.

E.g. http://www.lsdiodes.com/5mm/5mmorange.htm

Vf = 1.7
If = 20 mA

( 12 - 6 * 1.7 ) / 0.02 A = 90 Ohm.

So use a 100 Ohm resistor in this case.

--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html