Beginner circuit help.

D

Dan Schuman

Guest
I'm working on a circuit to help learn all this jazz, and wonder how
this could be accomplished:

Two devices to be used on the circuit are a speaker and an LED.

Source: 12V battery (Two 1.5V + 9V batteries)

LED: 3V / 20mA
Speaker: 12V / 15mA

(I am aware these devices operate at varied amperes and such, but I
would ideally like these to work at these specific values for
educational
purposes.)

How could I design a circuit that houses both these devices at their
rated amps? Right now I'm putting a parallel circuit near the
beginning to cut up the amperes, but then I seem to get stuck when
trying to use the lower voltage device at 20mA.

Any help?
 
In article <fe83eaaa-b162-4247-b3cc-
bac617af40c2@a18g2000pra.googlegroups.com>, danschuman@gmail.com says...

[snip]

How could I design a circuit that houses both these devices at their
rated amps? Right now I'm putting a parallel circuit near the
beginning to cut up the amperes, but then I seem to get stuck when
trying to use the lower voltage device at 20mA.
Click to expand...
Assuming you mean to use the LED across the 12v
supply, you use Ohm's Law like this:

Supply voltage - Led voltage 12 - 3
R = ---------------------------- = ------ = 450
LED current .02

Then use a 470 ohm standard value resistor in
series with the LED, like so:

+12v -----[R]----|>|---- Gnd
| LED
Spkr
|
Gnd


HTH
 
Dan Schuman wrote:

I'm working on a circuit to help learn all this jazz, and wonder how
this could be accomplished:

Two devices to be used on the circuit are a speaker and an LED.

Source: 12V battery (Two 1.5V + 9V batteries)

LED: 3V / 20mA
Speaker: 12V / 15mA

(I am aware these devices operate at varied amperes and such, but I
would ideally like these to work at these specific values for
educational
purposes.)

How could I design a circuit that houses both these devices at their
rated amps? Right now I'm putting a parallel circuit near the
beginning to cut up the amperes, but then I seem to get stuck when
trying to use the lower voltage device at 20mA.
Click to expand...
You haven't even said what you want to do with the speaker and led !

Graham
 
Randy Day wrote:

+12v -----[R]----|>|---- Gnd
| LED
Spkr
|
Gnd
Click to expand...
And putting DC through the speaker will achieve what exactly ? Other than
heating the voice coil and draining the battery.

Graham
 
On Fri, 26 Sep 2008 16:36:07 -0700 (PDT), Dan Schuman
<danschuman@gmail.com> wrote:

I'm working on a circuit to help learn all this jazz, and wonder how
this could be accomplished:

Two devices to be used on the circuit are a speaker and an LED.

Source: 12V battery (Two 1.5V + 9V batteries)

LED: 3V / 20mA
Speaker: 12V / 15mA

(I am aware these devices operate at varied amperes and such, but I
would ideally like these to work at these specific values for
educational
purposes.)

How could I design a circuit that houses both these devices at their
rated amps? Right now I'm putting a parallel circuit near the
beginning to cut up the amperes, but then I seem to get stuck when
trying to use the lower voltage device at 20mA.

Any help?
Click to expand...
---
Sure! :)

First off, to get the batteries to give you 12 volts you'll need to
connect them in series, like this: (View in Courier)



+------->+12V
|+
[1.5V]
|-
|
|+
[1.5V]
|-
|
|+
[9V]
|-
+------->0V


Next, since the LED is rated for 3V at 20 mA, we'll need to get rid of
9V while limiting the current through the LED to 20mA by using a
resistor in series with the LED:


+--------+---->+12V E1
|+ |
[1.5V] |
|- [R]
| |
|+ |
[1.5V] +----> E2
|- |
| |A
|+ [LED]
[9V] |
|- |
+--------+---->0V


In order to do that we can rearrange Ohm's law:


E = IR


to:

E
R = ---
I

where R is the resistance of the resistor,
E is the voltage to be dropped across the resistor, and
I is the current through the LED and the resistor.


Since the voltage dropped across the resistor will be the supply voltage
minus the voltage dropped across the LED, we can say:


E1 - E2 12V - 3V
R = --------- = ---------- = 300 ohms
I 0.03A


And we'll have:


+--------+
|+ |
[1.5V] |
|- [300R]
| |
|+ |
[1.5V] |
|- |
| |A
|+ [LED]
[9V] |
|- |
+--------+


Since you say your speaker is rated for 12V an it'll draw 15mA when
connected to a 12V source, then all that's necessary is for you to
connect it across the battery.

your circuit then, with everything connected, will look like this:


+--------+-------+
|+ | |
[1.5V] | |
|- [300R] |
| | |
|+ | |+
[1.5V] | [SPKR]
|- | |
| |A |
|+ [LED] |
[9V] | |
|- | |
+--------+-------+


JF
 
In article <48DDDA9C.C1C6A560@hotmail.com>,
rabbitsfriendsandrelations@hotmail.com says...
Randy Day wrote:


+12v -----[R]----|>|---- Gnd
| LED
Spkr
|
Gnd

And putting DC through the speaker will achieve what exactly ? Other than
heating the voice coil and draining the battery.
Click to expand...
He referred to a '12v' speaker; I'm guessing he
means a powered speaker/amplifier. If I'm wrong,
it's still a learning experience.
 
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