# noob question - resistance through a device + LED

A

#### anonymous

Jan 1, 1970
0
OK - so now Im wondering:

seems ok.. yes?

what if I want to have an led light when the plug is inserted?

+9v ----> + lead ---> device ----> - lead -----> R1 ----> led ----> ground

but if the LED needs 2V then I would be running 11 V through my device..
right? I can figure R1 with V= IR.. but what voltage should I put there?
will it be 9?

Since the 2 voltages are in series they are additive. But I don't want more
than 9 V at my device.

I think this 2nd diagram isn't correct.

J

#### John Popelish

Jan 1, 1970
0
anonymous said:
OK - so now Im wondering:

seems ok.. yes?

what if I want to have an led light when the plug is inserted?

+9v ----> + lead ---> device ----> - lead -----> R1 ----> led ----> ground

but if the LED needs 2V then I would be running 11 V through my device..
right? I can figure R1 with V= IR.. but what voltage should I put there?
will it be 9?

Since the 2 voltages are in series they are additive. But I don't want more
than 9 V at my device.

I think this 2nd diagram isn't correct.

The sum of all voltage around any loop always add to zero. If you
loop a 9 volt battery with a series combination of an LED and a
resistor, the battery voltage must equal the sum of the LED voltage
and the resistor voltage. So you assume that there is about 9-2 volts
across the resistor, and calculate its value based on the current you
want to have passing through the LED and resistor while the resistor
drops 7 volts. For instance, if you want about 10 ma through the led
and resistor, the value would be about 7v/.01A=700 ohms. A 680 ohm
would be the next lower common value.

A

#### anonymous

Jan 1, 1970
0
So indeed with my diagram #2 I would be pushing excess voltage through my
device. I would imagine that it would be damaged by having 11 volts so thats
not an option.

I need 2 volts at the LED - lets say 20 mA. This gives me R1 of 2 / .02 =
100 ohms.

Whats a better layout for this? The end result should be 9V at the device
and an LED that turns on when the device is plugged in.

A

#### andy

Jan 1, 1970
0
OK - so now Im wondering:

seems ok.. yes?

what if I want to have an led light when the plug is inserted?

+9v ----> + lead ---> device ----> - lead -----> R1 ----> led ----> ground

but if the LED needs 2V then I would be running 11 V through my device..
right?

no - if anything it would reduce the voltage through the device by 2V or
more, depending on what the device is.
I can figure R1 with V= IR.. but what voltage should I put there?
will it be 9?

In this circuit, the way I'd work it out is to say the voltage across the
led will be around 2V for most currents, so take that off 9V to leave 7V
for the rest of the circuit. Then find the resistance of the device, Rd
(if it has a stable resistance at all), add it to R1, and divide 7V by
R1+Rd. This gives the current through the whole thing. Then multiply by Rd
again to get the voltage across the device.
Since the 2 voltages are in series they are additive. But I don't want more
than 9 V at my device.

I think this 2nd diagram isn't correct.

If you just want the led as an indicator, you'd be better doing it like
this:

| |
+--> led --> R1 --+

Then you know that you have 9V across the device, and across the led and
R1 together, so R1 should be easy to work out.

A

#### anonymous

Jan 1, 1970
0
Ah v fine.

So by putting the resistor + LED at the + terminal I then have 2 resistances
in parallel? They would both see 9V and I could adjust the LED resistor to
get the appropriate voltage?? maybe??

andy said:
OK - so now Im wondering:

seems ok.. yes?

what if I want to have an led light when the plug is inserted?

+9v ----> + lead ---> device ----> - lead -----> R1 ----> led ---->
ground

but if the LED needs 2V then I would be running 11 V through my device..
right?

no - if anything it would reduce the voltage through the device by 2V or
more, depending on what the device is.
I can figure R1 with V= IR.. but what voltage should I put there?
will it be 9?

In this circuit, the way I'd work it out is to say the voltage across the
led will be around 2V for most currents, so take that off 9V to leave 7V
for the rest of the circuit. Then find the resistance of the device, Rd
(if it has a stable resistance at all), add it to R1, and divide 7V by
R1+Rd. This gives the current through the whole thing. Then multiply by Rd
again to get the voltage across the device.
Since the 2 voltages are in series they are additive. But I don't want
more
than 9 V at my device.

I think this 2nd diagram isn't correct.

If you just want the led as an indicator, you'd be better doing it like
this:

| |
+--> led --> R1 --+

Then you know that you have 9V across the device, and across the led and
R1 together, so R1 should be easy to work out.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
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[/QUOTE]

A

#### anonymous

Jan 1, 1970
0
Unf I don't know the resistance across the device. Its a 9v powered
'accessory' that is turned on/off on the unit. When on it draws 9V at
anywhere from 20-200mA.

andy said:
OK - so now Im wondering:

seems ok.. yes?

what if I want to have an led light when the plug is inserted?

+9v ----> + lead ---> device ----> - lead -----> R1 ----> led ---->
ground

but if the LED needs 2V then I would be running 11 V through my device..
right?

no - if anything it would reduce the voltage through the device by 2V or
more, depending on what the device is.
I can figure R1 with V= IR.. but what voltage should I put there?
will it be 9?

In this circuit, the way I'd work it out is to say the voltage across the
led will be around 2V for most currents, so take that off 9V to leave 7V
for the rest of the circuit. Then find the resistance of the device, Rd
(if it has a stable resistance at all), add it to R1, and divide 7V by
R1+Rd. This gives the current through the whole thing. Then multiply by Rd
again to get the voltage across the device.
Since the 2 voltages are in series they are additive. But I don't want
more
than 9 V at my device.

I think this 2nd diagram isn't correct.

If you just want the led as an indicator, you'd be better doing it like
this:

| |
+--> led --> R1 --+

Then you know that you have 9V across the device, and across the led and
R1 together, so R1 should be easy to work out.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML:
 or [attachment] in the subject line.
[/QUOTE]

J

#### John Popelish

Jan 1, 1970
0
anonymous said:
So indeed with my diagram #2 I would be pushing excess voltage through my
device. I would imagine that it would be damaged by having 11 volts so thats
not an option.

There is no 11 volts? You have 9 volts available, and you get ot
consume this in parts with all the components connected across that 9
volts.
I need 2 volts at the LED - lets say 20 mA. This gives me R1 of 2 / .02 =
100 ohms.

That calculation is meaningless, since an LED is not a resistance, but
a very nonlinear (current not proportional to voltage) device. The
manufacturer warns you that its typical voltage drop is about 2 volts
when 20 milliamps passes through it. It is also very nearly 2 volts
when half that current passes through it. Not a resistor.
Whats a better layout for this? The end result should be 9V at the device
and an LED that turns on when the device is plugged in.

You layout is fine.

Assume that if you get the current about right (near 20 ma) the LED
will, somehow, drop about 2 volts (that is the given in the circuit,
based on the LED datasheet). That leaves about 7 extra volts from the
battery to be wasted in the series resistor, while it also passes the
same 20 ma of current. Ohm's law applies to resistors (they do drop a
voltage proportional to their current) so you use ohm's law to
calculate what resistance will drop (waste) 7 volts when 20 ma is
passing through. 7/.02=350 ohms. 360 and 390 are standard 5% values,
so one of these should be easy to find and will pass almost the
desired 20 ma when the extra 7 volts is impressed across it.

P

#### Peter Bennett

Jan 1, 1970
0
Ah v fine.

So by putting the resistor + LED at the + terminal I then have 2 resistances
in parallel? They would both see 9V and I could adjust the LED resistor to
get the appropriate voltage?? maybe??

You want to connect the LED+resistor _in parallel_ with your device,
so that your device and the LED+resistor combination will each see the
full battery voltage.

You select the resistor in series with the LED to set the _current_
through that branch of the circuit. I find that most LEDs are bright
enough at 10 mA, so I would use a 680 or 750 ohm resistor.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter

A

#### andy

Jan 1, 1970
0
Ah v fine.

So by putting the resistor + LED at the + terminal I then have 2 resistances
in parallel? They would both see 9V and I could adjust the LED resistor to
get the appropriate voltage?? maybe??
yes.

andy said:
OK - so now Im wondering:

seems ok.. yes?

what if I want to have an led light when the plug is inserted?

+9v ----> + lead ---> device ----> - lead -----> R1 ----> led ---->
ground

but if the LED needs 2V then I would be running 11 V through my device..
right?

no - if anything it would reduce the voltage through the device by 2V or
more, depending on what the device is.
I can figure R1 with V= IR.. but what voltage should I put there?
will it be 9?

In this circuit, the way I'd work it out is to say the voltage across the
led will be around 2V for most currents, so take that off 9V to leave 7V
for the rest of the circuit. Then find the resistance of the device, Rd
(if it has a stable resistance at all), add it to R1, and divide 7V by
R1+Rd. This gives the current through the whole thing. Then multiply by Rd
again to get the voltage across the device.
Since the 2 voltages are in series they are additive. But I don't want
more
than 9 V at my device.

I think this 2nd diagram isn't correct.

If you just want the led as an indicator, you'd be better doing it like
this:

| |
+--> led --> R1 --+

Then you know that you have 9V across the device, and across the led and
R1 together, so R1 should be easy to work out.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML:
 or [attachment] in the subject line.
[/QUOTE][/QUOTE]

R

#### Robert C Monsen

Jan 1, 1970
0
anonymous said:
OK - so now Im wondering:

seems ok.. yes?

what if I want to have an led light when the plug is inserted?

+9v ----> + lead ---> device ----> - lead -----> R1 ----> led ---->
ground

but if the LED needs 2V then I would be running 11 V through my
device..
right? I can figure R1 with V= IR.. but what voltage should I put
there?
will it be 9?

Since the 2 voltages are in series they are additive. But I don't
want more
than 9 V at my device.

I think this 2nd diagram isn't correct.

I think the problem is that you want some kind of indicator when the
device is active, meaning its passing some current. You don't want to
put the LED in series; that'll make the LED brightness depend on how
much current the device draws, and if the device wants more than 20mA,
the LED will get fried quickly.

If the switch is external to your device, then you can just use the
fact that voltage is applied:

T
---
VCC-o o---o----------,
| |
| V =>
.-------. -
| device| |
| | .-.
| | | | 620R
'-------' | |
| '-'
| |
GND------o----------'

created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de

When the switch is closed, your thing turns on, and so does the LED.

However, I think the real problem is that the switch is inside the box
labeled "your device" above. If this is the case, the only thing you
can really do is measure whether current is flowing through the
device. To do that, you put a resistor in series with it, and measure
the voltage across that resistor. If its non-zero, then the current
through it is non-zero.

Generally, sense resistors are quite small so they don't interfere
with the circuit they are measuring (ie, 1 ohm or less). This means
that the voltage across them is also quite small. If your device draws
10mA, then the voltage across a 1 ohm resistor is 10mV. Thats a pretty
small difference to sense using your typical transistor circuit, due
to variations in transistors and resistors.

You can use a comparator to measure differences, and also power your
led from the output:

VCC
------o-------------o------o--------------.
| | | |
| | | |
.----o-----. | | V => (Your LED)
| your | .-. | -
| device | | |R1 | |
| | | | | .-.
| | '-' | | |R4
'----o-----' | | | |
| | | '-'
| | |\| |
o-------------|----|-\ |
| | | >-------------o
| o--o-|+/ |
| | | |/| |
.-. .-. | | ___ |
| |Rs R2| | '--------|___|-----'
| | | | | R3
'-' '-' |
| | |
GND | | |
------o-------------o------'

R1 = 18k
R2 = 10 ohms
R3 = 100k
R4 = 620
Rs = 1 ohm

comparator is an LM339 or something like it

created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de

This circuit measures the current across Rs, and compares it to the
5mV reference at the V+ input. Thus, if the current through your
device is more than 5mA, the LED will turn on. When it drops below
5mA, the LED will turn off. R3 gives you a little bit of hysteresis,
in case your circuit draw wants to hover around 5mA.

R1 and R2 are chosen based on the current you want to turn the LED on
at. Its set up for 5mA, but you can change that if you want. Using a
22k trimmer in place of R1 might be a good substitution, because then
you can adjust the point where the current turns on the LED.

Regards,
Bob Monsen

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