Lowering Voltage

A

Alan Smithee

Guest
Hi,

I'm trying to use this circuit diagram
(http://ourworld.compuserve.com/homepages/Bill_Bowden/page4.htm#whistle.gif)
to control a laser. The circuit runs using +12v, but I'd like to use a
simple +9v square battery. What do I have to consider to drop the voltage?

Can I just drop the resistance? 9 volts is 3/4 of 12. Therefore reduce
resistance by 3/4 - is this right?

My main concern is that the filters will be thrown out and the signal needed
to trigger the flip flop will be too great - I can only whistle so loud.

Also, do I need a relay? The laser I plan to use operates at 3 volts -
could I just add resistance to drop the voltage? or does it not work like
this.

Many thanks
 
"Alan Smithee" <paul@eurocomposer.freesrve.co.uk> wrote in message
news:bp9430$7vh$1@news8.svr.pol.co.uk...
Hi,

I'm trying to use this circuit diagram

(http://ourworld.compuserve.com/homepages/Bill_Bowden/page4.htm#whistle.gif)
to control a laser. The circuit runs using +12v, but I'd like to use a
simple +9v square battery. What do I have to consider to drop the
voltage?

Can I just drop the resistance? 9 volts is 3/4 of 12. Therefore reduce
resistance by 3/4 - is this right?

My main concern is that the filters will be thrown out and the signal
needed
to trigger the flip flop will be too great - I can only whistle so loud.

Also, do I need a relay? The laser I plan to use operates at 3 volts -
could I just add resistance to drop the voltage? or does it not work like
this.

Many thanks
Click to expand...
I don't see anything that depends on the voltage, really. The filters depend
on the frequency, but not the voltage.

Thus, I'd say you could probably just run it off of 9V.

Regards,
Bob Monsen
 
Thanks Bob,

Apparently at 9 volts I'll get about 7.3 vots out of the emitter at 50-100
mA - the laser I plan to use runs at 3volts and 85mA - what do I need to
know to make the change?

Thanks


"Robert Monsen" <postmaster@BulkingPro.com> wrote in message
news:L_Wtb.219700$Fm2.219185@attbi_s04...
"Alan Smithee" <paul@eurocomposer.freesrve.co.uk> wrote in message
news:bp9430$7vh$1@news8.svr.pol.co.uk...
Hi,

I'm trying to use this circuit diagram


(http://ourworld.compuserve.com/homepages/Bill_Bowden/page4.htm#whistle.gif)
to control a laser. The circuit runs using +12v, but I'd like to use a
simple +9v square battery. What do I have to consider to drop the
voltage?

Can I just drop the resistance? 9 volts is 3/4 of 12. Therefore reduce
resistance by 3/4 - is this right?

My main concern is that the filters will be thrown out and the signal
needed
to trigger the flip flop will be too great - I can only whistle so loud.

Also, do I need a relay? The laser I plan to use operates at 3 volts -
could I just add resistance to drop the voltage? or does it not work
like
this.

Many thanks




I don't see anything that depends on the voltage, really. The filters
depend
on the frequency, but not the voltage.

Thus, I'd say you could probably just run it off of 9V.

Regards,
Bob Monsen
Click to expand...
 
On Mon, 17 Nov 2003 11:32:55 -0000, "Alan Smithee"
<paul@eurocomposer.freeserve.co.uk> wrote:

Thanks Bob,

Apparently at 9 volts I'll get about 7.3 vots out of the emitter at 50-100
mA - the laser I plan to use runs at 3volts and 85mA - what do I need to
know to make the change?
Click to expand...
---
Instead of an emitter follower driving a solid-state relay as the
output, I'd blow off the relay and use the output transistor as a
saturated switch. Like this:

9V
|
[7R5]
|
|A
[LASER]
|
C
Q>--[1K]---B 2N4401
E
|
GND

Q is the output from the 4013, and for a 9V supply what you need to
know to limit the current through the LASER diode to 85 mA is that you
subtract the LASER diode voltage and the transistor saturation voltage
from the supply voltage and then divide that difference by the LASER
diode current. That number will be the reqired resistance.

So, R = (Vcc - (Vlaser+Vcesat))/85mA = (9V-3.3V)/0.085A ~ 67.06 ohms.

The closest standard 5% value available which will keep the current from
going over 85mA is 68 ohms, which will limit the current to
I = E/R = 5.7V/68R = 84mA. You will be dropping 5.7V across the
resistance and allowing 84mA to flow through it, so the power it will
dissipate will be P = IE = 84mA*5.7V ~ 479mW, so I'd use a 1 Watt
resistor.

The 9V battery will be supplying 9V*84mA = 756mW, so looking at the
curves for Service Life VS Power Drain for a Duracell alkaline 9V
battery at

http://www.duracell.com/oem/Pdf/MX1604.pdf

reveals that you can expect about 4 hours of continuous operation until
the battery voltage decays to 6V.

--
John Fields
 
Thanks for that John - I'm almost there !!!! Just a couple more questions if
you don't mind

9V So, I take a feed directly from the 9Volt
battery
|
[7R5] Is this a resistor? Sorry for being dumb.
|
|A
[LASER] The laser of course
|
C
Q>--[1K]---B 2N4401 Q and Data to base? Is there an alternative to
4401?
E Maplin don't seem to stock it
|
GND

dissipate will be P = IE = 84mA*5.7V ~ 479mW, so I'd use a 1 Watt
resistor.
Curiously Maplin doesn't stock a 1w resistor - they do a 2w and a 0.6 -
Click to expand...
what's best?


The 9V battery will be supplying 9V*84mA = 756mW, so looking at the
curves for Service Life VS Power Drain for a Duracell alkaline 9V
battery at

http://www.duracell.com/oem/Pdf/MX1604.pdf

reveals that you can expect about 4 hours of continuous operation until
the battery voltage decays to 6V.

--
John Fields
Click to expand...
 
On Mon, 17 Nov 2003 18:21:49 -0000, "Alan Smithee"
<paul@eurocomposer.freeserve.co.uk> wrote:

Thanks for that John - I'm almost there !!!! Just a couple more questions if
you don't mind

9V So, I take a feed directly from the 9Volt
battery
|
[7R5] Is this a resistor? Sorry for being dumb.
|
|A
[LASER] The laser of course
|
C
Q>--[1K]---B 2N4401 Q and Data to base? Is there an alternative to
4401?
E Maplin don't seem to stock it
|
GND

dissipate will be P = IE = 84mA*5.7V ~ 479mW, so I'd use a 1 Watt
resistor.
Curiously Maplin doesn't stock a 1w resistor - they do a 2w and a 0.6 -
what's best?


The 9V battery will be supplying 9V*84mA = 756mW, so looking at the
curves for Service Life VS Power Drain for a Duracell alkaline 9V
battery at

http://www.duracell.com/oem/Pdf/MX1604.pdf

reveals that you can expect about 4 hours of continuous operation until
the battery voltage decays to 6V.

--
John Fields
Click to expand...
9V So, I take a feed directly from the 9Volt
|
[7R5] Is this a resistor? Sorry for being dumb.
|
|A
[LASER] The laser of course
|
C
Q>--[1K]---B 2N4401 Q and Data to base?
E
|
GND

---

1. Yes, take the feed directly from the battery

2. Yes, it's a resistor, but it should read 68R, (_not_ 7R5)
since it's the 68 ohm resistor referred to in the text.
Sorry about that...

3. Yes, the LASER. The 'A' refers to the anode connection.

4. Yes, through a 1000 ohm 1/4 watt resistor.

5. Any NPN capable of dissipating 50mW while carrying a collector
current of 84MA and sporting a beta of 100 or better should work.
A 2N2222A (PN2222A) should be fine.

6. you can make the equivalent of a 68 ohm 1 watt resistor by connecting
four 68 ohm 1/4 watt resistors in series-parallel:

+-----+---->
| |
[68R] [68R]
| |
+ +
| |
[68R] [68R]
| |
+-----+---->

--
John Fields
 
On Mon, 17 Nov 2003 18:21:49 -0000, "Alan Smithee"
<paul@eurocomposer.freeserve.co.uk> wrote:


Curiously Maplin doesn't stock a 1w resistor - they do a 2w and a 0.6 -
what's best?
Click to expand...
---
If you don't want to build one out of 1/4 watters, go for the 2W.

--
John Fields
 
Your an absolute star John - I can now begin to design the PCB layout - gulp


"John Fields" <jfields@austininstruments.com> wrote in message
news:1g7irv0lk2489qcelngd4kp60l6s0nqml3@4ax.com...
On Mon, 17 Nov 2003 18:21:49 -0000, "Alan Smithee"
paul@eurocomposer.freeserve.co.uk> wrote:


Curiously Maplin doesn't stock a 1w resistor - they do a 2w and a 0.6 -
what's best?

---
If you don't want to build one out of 1/4 watters, go for the 2W.

--
John Fields
Click to expand...
 
+-----+----
| |
[68R] [68R]
| |
+ +
| |
[68R] [68R]
| |
+-----+----
Click to expand...
I take it that the 1 watt resistor would be better for battery consumption?

This looks like 2 lots of 2 resistors in series - is this right? Would just
like to be sure.

Thanks again
 
"Alan Smithee" <paul@eurocomposer.freesrve.co.uk> wrote in message
news:bpboc1$kqv$1@news6.svr.pol.co.uk...
+-----+----
| |
[68R] [68R]
| |
+ +
| |
[68R] [68R]
| |
+-----+----

I take it that the 1 watt resistor would be better for battery
consumption?

This looks like 2 lots of 2 resistors in series - is this right? Would
just
like to be sure.

Thanks again
Click to expand...
FYI, the max power value of a resistor doesn't affect anything except the
heat it can expect to dissipate safely. So, the power consumption will be
identical for the 2W or 4 1/2 watt resistors. Thus, you can use the 1/2W
resistors to build a 1W resistor wihtout affecting the power consumption.

Regards,
Bob Monsen
 
In article <bp9430$7vh$1@news8.svr.pol.co.uk>,
paul@eurocomposer.freesrve.co.uk mentioned...
Hi,

I'm trying to use this circuit diagram
(http://ourworld.compuserve.com/homepages/Bill_Bowden/page4.htm#whistle.gif)
to control a laser. The circuit runs using +12v, but I'd like to use a
simple +9v square battery. What do I have to consider to drop the voltage?

Can I just drop the resistance? 9 volts is 3/4 of 12. Therefore reduce
resistance by 3/4 - is this right?
Click to expand...
The LM1458, 741, etc. opamps do poorly at low voltages, they just
can't put out much. You could replace the 1458 with a LM358, which
has identical pinout. The rest of the circuit looks like it should
work okay off a lower voltage.

My main concern is that the filters will be thrown out and the signal needed
to trigger the flip flop will be too great - I can only whistle so loud.

Also, do I need a relay? The laser I plan to use operates at 3 volts -
could I just add resistance to drop the voltage? or does it not work like
this.

Many thanks
Click to expand...

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
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@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
In article <bpb3ju$4jp$1@newsg3.svr.pol.co.uk>,
paul@eurocomposer.freeserve.co.uk mentioned...

[snip]
Q and Data to base? Is there an alternative to 4401?
Maplin don't seem to stock it

Use a BC337-25.


--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
 
Thanks to everyone!

I now have my circuit completely planned - save the layout :) or rather
:-( since I've not done this before !

Thanks anyhow


"Watson A.Name - Watt Sun, Dark Remover" <alondra101@hotmail.com> wrote in
message news:MPG.1a292cf46788acd498994f@news.dslextreme.com...
In article <bp9430$7vh$1@news8.svr.pol.co.uk>,
paul@eurocomposer.freesrve.co.uk mentioned...
Hi,

I'm trying to use this circuit diagram

(http://ourworld.compuserve.com/homepages/Bill_Bowden/page4.htm#whistle.gif)
to control a laser. The circuit runs using +12v, but I'd like to use a
simple +9v square battery. What do I have to consider to drop the
voltage?

Can I just drop the resistance? 9 volts is 3/4 of 12. Therefore reduce
resistance by 3/4 - is this right?

The LM1458, 741, etc. opamps do poorly at low voltages, they just
can't put out much. You could replace the 1458 with a LM358, which
has identical pinout. The rest of the circuit looks like it should
work okay off a lower voltage.

My main concern is that the filters will be thrown out and the signal
needed
to trigger the flip flop will be too great - I can only whistle so loud.

Also, do I need a relay? The laser I plan to use operates at 3 volts -
could I just add resistance to drop the voltage? or does it not work
like
this.

Many thanks


--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
Click to expand...
 
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