Buzzer voltage drop

P

pata

Guest
I have an electronic buzzer (ICC pn: BS2316L-06) I am going to put it into
a circuit with some LED's and other components. It draws 25ma and is rated
at 6V, how do I know what the voltage drop across the buzzer is? Is it 6V?
I need to know this so I can calculate what size resistor to put in the
circuit.

Thanks
Pat
 
pata wrote:
I have an electronic buzzer (ICC pn: BS2316L-06) I am going to put it into
a circuit with some LED's and other components. It draws 25ma and is rated
at 6V, how do I know what the voltage drop across the buzzer is? Is it 6V?
I need to know this so I can calculate what size resistor to put in the
circuit.

Thanks
Pat
Click to expand...
You can probably assume that if you provide it with 6 volts it will
pass about 25 ma.

--
John Popelish
 
I understand that but how does that tell me what the voltage drop across it
is?

"John Popelish" <jpopelish@rica.net> wrote in message
news:4171B196.EA5CCE56@rica.net...
pata wrote:

I have an electronic buzzer (ICC pn: BS2316L-06) I am going to put it
into
a circuit with some LED's and other components. It draws 25ma and is
rated
at 6V, how do I know what the voltage drop across the buzzer is? Is it
6V?
I need to know this so I can calculate what size resistor to put in the
circuit.

Thanks
Pat

You can probably assume that if you provide it with 6 volts it will
pass about 25 ma.

--
John Popelish
Click to expand...
 
Thanks, I understand that better. I am an ME and I struggled in our 1
required EE class!



"John Popelish" <jpopelish@rica.net> wrote in message
news:4171BB6B.BDB30964@rica.net...
pata wrote:

I understand that but how does that tell me what the voltage drop across
it
is?

The voltage drop across it is the voltage you supply to it. If you
have 12 volts available and want 6 volts to appear across the buzzer
while it draws 25 ma, assume the buzzer looks like 240 ohms and put
another 240 in series with it. They (the buzzer and the resistor
should split the 12 volts into two approximately 6 volt drops.

--
John Popelish
Click to expand...
 
pata wrote:
I understand that but how does that tell me what the voltage drop across it
is?
Click to expand...
The voltage drop across it is the voltage you supply to it. If you
have 12 volts available and want 6 volts to appear across the buzzer
while it draws 25 ma, assume the buzzer looks like 240 ohms and put
another 240 in series with it. They (the buzzer and the resistor
should split the 12 volts into two approximately 6 volt drops.

--
John Popelish
 
pata wrote:
Thanks, I understand that better. I am an ME and I struggled in our 1
required EE class!
Click to expand...
Of course, I left a gotcha in there to see if you would ask the next
question. The buzzer draws some pulsed waveform that averages 25 ma
when you apply 6 volts across it. There may be large instantaneous
current variations. You may need to add a capacitor in parallel with
the buzzer to soak up all those variations and hold something close to
the average drop across the buzzer to make ir work correctly.

I would guess that 10 microfarads or so might be enough.

Capacitors are sort of like fly wheels (if voltage is like rotational
speed and current is like torque) in this application. Internal
engines put out an average torque at some speed but without a
flywheel, the speed varies all over the place each turn. :)

--
John Popelish
 
OK, thanks. Let's forget about the cap for now so I don't have an overload.
Let's say my simple circuit consists of a 12V source, an LED that requires
25ma and has a Vf of 3.2v, and the buzzer. To figure out the correct
resistor the calculation would be 12-(3.2+6) /.025 = 112 ohms. Is this
correct?

Also, theoretically if the my source voltage was 9v would I not need a
resistor at all?

Don't leave me any gotchas! I need it explained to me like I was 5.

Patrick


"John Popelish" <jpopelish@rica.net> wrote in message
news:4171C31C.875ECDE6@rica.net...
pata wrote:

Thanks, I understand that better. I am an ME and I struggled in our 1
required EE class!

Of course, I left a gotcha in there to see if you would ask the next
question. The buzzer draws some pulsed waveform that averages 25 ma
when you apply 6 volts across it. There may be large instantaneous
current variations. You may need to add a capacitor in parallel with
the buzzer to soak up all those variations and hold something close to
the average drop across the buzzer to make ir work correctly.

I would guess that 10 microfarads or so might be enough.

Capacitors are sort of like fly wheels (if voltage is like rotational
speed and current is like torque) in this application. Internal
engines put out an average torque at some speed but without a
flywheel, the speed varies all over the place each turn. :)

--
John Popelish
Click to expand...
 
On Sun, 17 Oct 2004 01:01:36 +0000, pata wrote:

OK, thanks. Let's forget about the cap for now so I don't have an overload.
Let's say my simple circuit consists of a 12V source, an LED that requires
25ma and has a Vf of 3.2v, and the buzzer. To figure out the correct
resistor the calculation would be 12-(3.2+6) /.025 = 112 ohms. Is this
correct?

Also, theoretically if the my source voltage was 9v would I not need a
resistor at all?

Don't leave me any gotchas! I need it explained to me like I was 5.

John Popelish explained fairly well about your series circuit, and
Click to expand...
you're spot-on, so far. :)

You can kind of think of an electric current something like the flow
of water in pipes. Pressure is voltage, current is flow rate, and
resistance is friction.

Well, an LED, as John Popelish said, doesn't act like an ordinary
resistance. In a resistance, current, or movement, increases
proportionally to pressure, or voltage, and inversely to friction,
or resistance. An LED is like a greased chute - very little
change in pressure causes a great change in flow, unless something
else restricts that flow, in this case an "ohmic" resistor.

And a capacitor is like a split tank with a rubber baffle, and
an inductor is like a positive displacement turbine/pump coupled
to a flywheel. :)

Try googling on basic electronics tutorials and stuff.

Hope This Helps!
Rich
 
pata wrote:
OK, thanks. Let's forget about the cap for now so I don't have an overload.
Let's say my simple circuit consists of a 12V source, an LED that requires
25ma and has a Vf of 3.2v, and the buzzer. To figure out the correct
resistor the calculation would be 12-(3.2+6) /.025 = 112 ohms. Is this
correct?
Click to expand...
That is he general idea.

Also, theoretically if the my source voltage was 9v would I not need a
resistor at all?
Click to expand...
Leds have very large current swings for very small voltage changes, so
they are not modeled by ohm's law, very well.

Normally you need some resistance in series with LEDs to make sure the
current is controlled. But the buzzer may perform that resistor
function, approximately. But it would worry me.

Don't leave me any gotchas! I need it explained to me like I was 5.
Click to expand...

--
John Popelish
 
Subject: Buzzer voltage drop
From: "pata" lkj@lkj.com
Date: 10/16/2004 6:29 PM Central Daylight Time
Message-id: <g9icd.1062$5l3.302@trnddc02

I have an electronic buzzer (ICC pn: BS2316L-06) I am going to put it into
a circuit with some LED's and other components. It draws 25ma and is rated
at 6V, how do I know what the voltage drop across the buzzer is? Is it 6V?
I need to know this so I can calculate what size resistor to put in the
circuit.

Thanks
Pat
Click to expand...
The 25mA listedfor the electromechanical buzzer is an average -- sometimes a
lot more, sometimes less. It will work with a series resistor, but not too
well.

Let's assume you're running off a 13.8V supply. In order to get it to run
well, you might want to use a zener like this (view in fixed font or M$
Notepad):


13.8V
+
|
1N4737A |
Vz=7.5V/-/
^
|
|
|
| .---|
'---| |
.---| |
| '---|
|
|
___ |/
o-|___|-o-|2N3904
Control 10K | |>
Signal .-. |
10K| | |
| | |
'-' |
| |
=== ===
GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Good luck
Chris
 
On Sat, 16 Oct 2004 20:55:56 -0400, John Popelish wrote:

pata wrote:

Thanks, I understand that better. I am an ME and I struggled in our 1
required EE class!

Of course, I left a gotcha in there to see if you would ask the next
question. The buzzer draws some pulsed waveform that averages 25 ma
when you apply 6 volts across it. There may be large instantaneous
current variations. You may need to add a capacitor in parallel with
the buzzer to soak up all those variations and hold something close to
the average drop across the buzzer to make ir work correctly.

I would guess that 10 microfarads or so might be enough.

Capacitors are sort of like fly wheels (if voltage is like rotational
speed and current is like torque) in this application. Internal
engines put out an average torque at some speed but without a
flywheel, the speed varies all over the place each turn. :)
Click to expand...
No! No! No! Don't get the newbie started off all on the wrong foot.
A capacitor is a _spring._ An _inductor_ is a flywheel. "current"
means "movement", after all, right? :eek:)

Cheers!
Rich
 
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