Fader/Dimmer

R

Robert

Guest
I have a light powered by 18V. I would like the light to fade up to
full brightness when the power is turned on and dim down when the
power is shut off. I want this to happen quickly, less than 1 second.
Any ideas would be appreciated.
 
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Robert wrote:
I have a light powered by 18V. I would like the light to fade up to
full brightness when the power is turned on and dim down when the
power is shut off. I want this to happen quickly, less than 1 second.
Any ideas would be appreciated.
Click to expand...
If it's 18VDC (not AC), a fairly big capacitor across the supply lines
would probably do the job. I don't know the calculations for exactly HOW
big to make the cap. Anyway: Hit the power, the capacitor charges up and
sinks most of the power leaving only a bit to the light. As it reaches
full charge, more and more current goes through the light until finally
all the power is going to the light. When you kill power, the capacitor
powers the light for a while until it runs out of charge.

If the light runs on AC, this will not work.

Chris
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"Robert" <hedrock@insightbb.com> schreef in bericht
news:004ig11aqir5bl5q1e11ju2em1pa8bu1t0@4ax.com...
I have a light powered by 18V. I would like the light to fade up to
full brightness when the power is turned on and dim down when the
power is shut off. I want this to happen quickly, less than 1 second.
Any ideas would be appreciated.
Click to expand...
What kind of light do you use?
Some lights cannot be dimmed, for other lights you need a special circuit.
And as Chris points out what kind of currunt/voltage AC or DC?
 
"Robert" <hedrock@insightbb.com> wrote in message
news:004ig11aqir5bl5q1e11ju2em1pa8bu1t0@4ax.com...
I have a light powered by 18V. I would like the light to fade up to
full brightness when the power is turned on and dim down when the
power is shut off. I want this to happen quickly, less than 1 second.
Any ideas would be
Click to expand...


Do you know the wattage of the bulb? From that you can get an idea of the
resistance in Ohms. and then you can find the value of the cap needed.

watts / 18 = current in amps

18 / current in amps = resistance in Ohms

Time = resistance X Capacitance in farads.

Example,

Say your bulb is 5 Watt

5 / 18 = .277 Amps

18 / .277 = 65 Ohms

So, 1 Sec. / 65 Ohms = .01538 Farad. Or 15,000 uF (Micro Farads)

This will be a close approximation as the filament in the bulb changes
resistance as it heats up. Caps are cheap so try a few values near the
calculated value to see what gives you the effect you are after. Use 35
Volt capacitors to give you headroom and remember to observe polarity.
 
"DBLEXPOSURE" <celstuff@hotmail.com> wrote in message
news:SKmdnQOhHPjIbJTeRVn-2g@rapidnet.com...
"Robert" <hedrock@insightbb.com> wrote in message
news:004ig11aqir5bl5q1e11ju2em1pa8bu1t0@4ax.com...
I have a light powered by 18V. I would like the light to fade up to
full brightness when the power is turned on and dim down when the
power is shut off. I want this to happen quickly, less than 1 second.
Any ideas would be



Do you know the wattage of the bulb? From that you can get an idea of the
resistance in Ohms. and then you can find the value of the cap needed.

watts / 18 = current in amps

18 / current in amps = resistance in Ohms

Time = resistance X Capacitance in farads.

Example,

Say your bulb is 5 Watt

5 / 18 = .277 Amps

18 / .277 = 65 Ohms

So, 1 Sec. / 65 Ohms = .01538 Farad. Or 15,000 uF (Micro Farads)

This will be a close approximation as the filament in the bulb changes
resistance as it heats up. Caps are cheap so try a few values near the
calculated value to see what gives you the effect you are after. Use 35
Volt capacitors to give you headroom and remember to observe polarity.
Click to expand...
*** Assuming this is a DC bulb ***!!
 
Thanks for the replies.
The 18V is DC. It powers an inverter to run some EL wire. I currently
have it running on a battery powered variable power supply. I can
control the brightness with a potentiometer. I would like do it
automatically. The current draw on the DC side is 350mA.
 
"Robert" <hedrock@insightbb.com> wrote in message
news:nd1lg1lg94ipu4neouskrr2nquo6tist8a@4ax.com...
Thanks for the replies.
The 18V is DC. It powers an inverter to run some EL wire. I currently
have it running on a battery powered variable power supply. I can
control the brightness with a potentiometer. I would like do it
automatically. The current draw on the DC side is 350mA.
Click to expand...

Okay so,

18 / .350 = 51 Ohms

T = R x C

Or C = T / R

So

1 / 51 = .0196 Farads ( 19,600 uF )

You can probably find 20,000 & 18,000 uF @ 35Vdc

Obviously the larger the uF value the longer the charge time will be...
 
So the cap should go from the positive lead to ground before the load.
Can I use several small ones in parallel?

On Mon, 22 Aug 2005 23:09:10 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> wrote:

"Robert" <hedrock@insightbb.com> wrote in message
news:nd1lg1lg94ipu4neouskrr2nquo6tist8a@4ax.com...
Thanks for the replies.
The 18V is DC. It powers an inverter to run some EL wire. I currently
have it running on a battery powered variable power supply. I can
control the brightness with a potentiometer. I would like do it
automatically. The current draw on the DC side is 350mA.



Okay so,

18 / .350 = 51 Ohms

T = R x C

Or C = T / R

So

1 / 51 = .0196 Farads ( 19,600 uF )

You can probably find 20,000 & 18,000 uF @ 35Vdc

Obviously the larger the uF value the longer the charge time will be...
Click to expand...
 
"Robert" <hedrock@insightbb.com> wrote in message
news:bdcmg1tlke2n4dip8f9n0ird2o1qicas2f@4ax.com...
So the cap should go from the positive lead to ground before the load.
Can I use several small ones in parallel?

On Mon, 22 Aug 2005 23:09:10 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:


"Robert" <hedrock@insightbb.com> wrote in message
news:nd1lg1lg94ipu4neouskrr2nquo6tist8a@4ax.com...
Thanks for the replies.
The 18V is DC. It powers an inverter to run some EL wire. I currently
have it running on a battery powered variable power supply. I can
control the brightness with a potentiometer. I would like do it
automatically. The current draw on the DC side is 350mA.



Okay so,

18 / .350 = 51 Ohms

T = R x C

Or C = T / R

So

1 / 51 = .0196 Farads ( 19,600 uF )

You can probably find 20,000 & 18,000 uF @ 35Vdc

Obviously the larger the uF value the longer the charge time will be...
Click to expand...

Nope, that won't work. I think I have mislead you. The RC circuit needs
to be in series. In your situation this would mean the bulb would light and
then slowly dim out as the cap charges and current stops. Not what you are
after.

You will need to do a series RC circuit in parallel with your lamp.
:-( The smaller the resistor the dimmer the light will start off. But,
this means you will need huge capacitance to get your 1 second... Hmmm...

18v
| |
r1 L1
| |
c1 |
| |
Gnd


If R1 was 10 Ohms C1 would be 100,000uf Hmmm.... In this case it might be
better to use several smaller caps in parallel to get to the value of C1.
But, There has to be a better way ???

I'll be back
 
"DBLEXPOSURE" <celstuff@hotmail.com> wrote in message
news:Fpadnd_NWum28pbeRVn-og@rapidnet.com...
"Robert" <hedrock@insightbb.com> wrote in message
news:bdcmg1tlke2n4dip8f9n0ird2o1qicas2f@4ax.com...

So the cap should go from the positive lead to ground before the load.
Can I use several small ones in parallel?

On Mon, 22 Aug 2005 23:09:10 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:


"Robert" <hedrock@insightbb.com> wrote in message
news:nd1lg1lg94ipu4neouskrr2nquo6tist8a@4ax.com...
Thanks for the replies.
The 18V is DC. It powers an inverter to run some EL wire. I currently
have it running on a battery powered variable power supply. I can
control the brightness with a potentiometer. I would like do it
automatically. The current draw on the DC side is 350mA.



Okay so,

18 / .350 = 51 Ohms

T = R x C

Or C = T / R

So

1 / 51 = .0196 Farads ( 19,600 uF )

You can probably find 20,000 & 18,000 uF @ 35Vdc

Obviously the larger the uF value the longer the charge time will be...










Nope, that won't work. I think I have mislead you. The RC circuit needs
to be in series. In your situation this would mean the bulb would light
and then slowly dim out as the cap charges and current stops. Not what
you are after.

You will need to do a series RC circuit in parallel with your lamp.
:-( The smaller the resistor the dimmer the light will start off.
But, this means you will need huge capacitance to get your 1 second...
Hmmm...

18v
| |
r1 L1
| |
c1 |
| |
Gnd


If R1 was 10 Ohms C1 would be 100,000uf Hmmm.... In this case it might
be better to use several smaller caps in parallel to get to the value of
C1. But, There has to be a better way ???

I'll be back
Click to expand...

After thinking about it I don't think it is to practical to use and RC
circuit. As I said the you would need a large cap. As I said above, 10
Ohms and 100,000uF would start your lamp a 1/5th normal and rise up to full
brightness. You could use ten 10,000uf in parallel but this seems
impractical.


I think the best approach would be a RC circuit driving a transistor which
in turn would drive you lamp.

I am not much on transistor circuit design but I am sure someone will chime
in with a simple circuit..
 
I tried a 4700uf across the power supply. As you said, no effect on
power up but a nice fade on power down. If there was a way to slow the
charging it might work.
 
"Robert" <hedrock@insightbb.com> wrote in message
news:2ttng1dkshnup6bubglii4gu0o4irske35@4ax.com...
I tried a 4700uf across the power supply. As you said, no effect on
power up but a nice fade on power down. If there was a way to slow the
charging it might work.
Click to expand...


As I said..


L1 is your lamp. In this circuit the light will not start completely off
but about 1/5th of full brightness.. Use ten 10,000uF is parallel.. to get
C1..

18v
| |
r1 L1
| |
c1 |
| |
Gnd


If R1 was 10 Ohms C1 would be 100,000uf Hmmm.... In this case it might
be better to use several smaller caps in parallel to get to the value of
C1. But, There has to be a better way ???
Click to expand...
I still think the more elegant approach is a series RC circuit on the base
of a transistor. This would allow you to use a larger value resistor and
thus decreasing the size of the cap needed. The transistor would control
the light bulb...

You could also use a choke in series with the lamp. Problem is it would need
to be huge... 50 Henrys... You would need a wheel barrel to carry it
around... :)
 
On Wed, 24 Aug 2005 04:32:53 GMT, Robert <hedrock@insightbb.com>
wrote:

I tried a 4700uf across the power supply. As you said, no effect on
power up but a nice fade on power down. If there was a way to slow the
charging it might work.
Click to expand...
---
This works:


+18>--> | Q1
|S1 PNP
O--+------E C---+----> >---+
|+ B | |
[470ľF] | | |
| | |+ |
+--------+ [4700ľF] [50R]
| | |
[510R] | |
| | |
GND>-------+--------------+----> >---+

I tried a 2N4403 for Q1 and it worked OK, although it got quite
warm, momentarily, during power up and power down when it was
dissipating power during the transitions. I then tried a Zetex
ZTX549 and it worked much better. The 50 ohm resistor is an
inverter-equivalent load for DC, but your inverter may act
differently on power-up and power-down.

--
John Fields
Professional Circuit Designer
 
Thanks for the post. Let me see if I got this right.

18V (+)____.____E_(Q1)_C__.__(+)LOAD
+ B l
470uf l +
l l 4700uf
._______.l l
l l
510R l
l l
(-)-------.----------------------------.------(-)LOAD

Do I still vary time with just C2(4700)?
Thanks again.


On Wed, 24 Aug 2005 10:37:20 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Wed, 24 Aug 2005 04:32:53 GMT, Robert <hedrock@insightbb.com
wrote:

I tried a 4700uf across the power supply. As you said, no effect on
power up but a nice fade on power down. If there was a way to slow the
charging it might work.

---
This works:


+18>--> | Q1
|S1 PNP
O--+------E C---+----> >---+
|+ B | |
[470ľF] | | |
| | |+ |
+--------+ [4700ľF] [50R]
| | |
[510R] | |
| | |
GND>-------+--------------+----> >---+

I tried a 2N4403 for Q1 and it worked OK, although it got quite
warm, momentarily, during power up and power down when it was
dissipating power during the transitions. I then tried a Zetex
ZTX549 and it worked much better. The 50 ohm resistor is an
inverter-equivalent load for DC, but your inverter may act
differently on power-up and power-down.
Click to expand...
 
In article <2ttng1dkshnup6bubglii4gu0o4irske35@4ax.com>, Robert wrote:
I tried a 4700uf across the power supply. As you said, no effect on
power up but a nice fade on power down. If there was a way to slow the
charging it might work.
Click to expand...
you'd need to limit the curren that's flowing out of the power supply
one way to do that is to use a resistor but that'll also reduce the maximum
voltage the inverter sees, I think a better solution can be done a transistor
and a resistor two, but I forget the details.

Bye.
Jasen
 
On Thu, 25 Aug 2005 01:59:58 GMT, Robert <hedrock@insightbb.com>
wrote:


On Wed, 24 Aug 2005 10:37:20 -0500, John Fields
jfields@austininstruments.com> wrote:

On Wed, 24 Aug 2005 04:32:53 GMT, Robert <hedrock@insightbb.com
wrote:

I tried a 4700uf across the power supply. As you said, no effect on
power up but a nice fade on power down. If there was a way to slow the
charging it might work.

---
This works:


+18>--> | Q1
|S1 PNP
O--+------E C---+----> >---+
|+ B | |
[470ľF] | | |
| | |+ |
+--------+ [4700ľF] [50R]
| | |
[510R] | |
| | |
GND>-------+--------------+----> >---+

I tried a 2N4403 for Q1 and it worked OK, although it got quite
warm, momentarily, during power up and power down when it was
dissipating power during the transitions. I then tried a Zetex
ZTX549 and it worked much better. The 50 ohm resistor is an
inverter-equivalent load for DC, but your inverter may act
differently on power-up and power-down.
Click to expand...


Thanks for the post. Let me see if I got this right.

18V (+)____.____E_(Q1)_C__.__(+)LOAD
+ B l
470uf l +
l l 4700uf
._______.l l
l l
510R l
l l
(-)-------.----------------------------.------(-)LOAD

Do I still vary time with just C2(4700)?
Thanks again.
Click to expand...
---
I can't read your schematic.

If you can't read mine you need to set your newsreader to display
text in a non-proportional font like Courier or Courier new.

But, to answer your question, it turns out the 470ľF cap isn't
needed at all so, yes, changing the 4700ľF cap will change the
timing.

Q1
PNP
18V (+)------E C---+---(+) LOAD
B |
| |
| |+
| [4700ľF]
| |
[510R] |
| |
GND>-----------+-----+---(-) LOAD


--
John Fields
Professional Circuit Designer
 
Thanks again for the help. Finally got the time and parts to try it.
The fade out is great, but it doesn't fade on. If anything, there
might be a very slight delay before it comes on at full brightness. I
can't tell for sure, but not the fade I get when I power down. Any
more thoughts?

On Thu, 25 Aug 2005 05:53:09 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Thu, 25 Aug 2005 01:59:58 GMT, Robert <hedrock@insightbb.com
wrote:


On Wed, 24 Aug 2005 10:37:20 -0500, John Fields
jfields@austininstruments.com> wrote:

On Wed, 24 Aug 2005 04:32:53 GMT, Robert <hedrock@insightbb.com
wrote:

I tried a 4700uf across the power supply. As you said, no effect on
power up but a nice fade on power down. If there was a way to slow the
charging it might work.

---
This works:


+18>--> | Q1
|S1 PNP
O--+------E C---+----> >---+
|+ B | |
[470ľF] | | |
| | |+ |
+--------+ [4700ľF] [50R]
| | |
[510R] | |
| | |
GND>-------+--------------+----> >---+

I tried a 2N4403 for Q1 and it worked OK, although it got quite
warm, momentarily, during power up and power down when it was
dissipating power during the transitions. I then tried a Zetex
ZTX549 and it worked much better. The 50 ohm resistor is an
inverter-equivalent load for DC, but your inverter may act
differently on power-up and power-down.



Thanks for the post. Let me see if I got this right.

18V (+)____.____E_(Q1)_C__.__(+)LOAD
+ B l
470uf l +
l l 4700uf
._______.l l
l l
510R l
l l
(-)-------.----------------------------.------(-)LOAD

Do I still vary time with just C2(4700)?
Thanks again.

---
I can't read your schematic.

If you can't read mine you need to set your newsreader to display
text in a non-proportional font like Courier or Courier new.

But, to answer your question, it turns out the 470ľF cap isn't
needed at all so, yes, changing the 4700ľF cap will change the
timing.

Q1
PNP
18V (+)------E C---+---(+) LOAD
B |
| |
| |+
| [4700ľF]
| |
[510R] |
| |
GND>-----------+-----+---(-) LOAD
Click to expand...
 
"Robert" <hedrock@insightbb.com> wrote in message
news:eek:uo9h1lkfclg97nken75k601orplporho8@4ax.com...
Thanks again for the help. Finally got the time and parts to try it.
The fade out is great, but it doesn't fade on. If anything, there
might be a very slight delay before it comes on at full brightness. I
can't tell for sure, but not the fade I get when I power down. Any
more thoughts?

On Thu, 25 Aug 2005 05:53:09 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 25 Aug 2005 01:59:58 GMT, Robert <hedrock@insightbb.com
wrote:


On Wed, 24 Aug 2005 10:37:20 -0500, John Fields
jfields@austininstruments.com> wrote:

On Wed, 24 Aug 2005 04:32:53 GMT, Robert <hedrock@insightbb.com
wrote:

I tried a 4700uf across the power supply. As you said, no effect on
power up but a nice fade on power down. If there was a way to slow the
charging it might work.

---
This works:


+18>--> | Q1
|S1 PNP
O--+------E C---+----> >---+
|+ B | |
[470ľF] | | |
| | |+ |
+--------+ [4700ľF] [50R]
| | |
[510R] | |
| | |
GND>-------+--------------+----> >---+

I tried a 2N4403 for Q1 and it worked OK, although it got quite
warm, momentarily, during power up and power down when it was
dissipating power during the transitions. I then tried a Zetex
ZTX549 and it worked much better. The 50 ohm resistor is an
inverter-equivalent load for DC, but your inverter may act
differently on power-up and power-down.



Thanks for the post. Let me see if I got this right.

18V (+)____.____E_(Q1)_C__.__(+)LOAD
+ B l
470uf l +
l l 4700uf
._______.l l
l l
510R l
l l
(-)-------.----------------------------.------(-)LOAD

Do I still vary time with just C2(4700)?
Thanks again.

---
I can't read your schematic.

If you can't read mine you need to set your newsreader to display
text in a non-proportional font like Courier or Courier new.

But, to answer your question, it turns out the 470ľF cap isn't
needed at all so, yes, changing the 4700ľF cap will change the
timing.

Q1
PNP
18V (+)------E C---+---(+) LOAD
B |
| |
| |+
| [4700ľF]
| |
[510R] |
| |
GND>-----------+-----+---(-) LOAD
Click to expand...



Try John's original circuit with the rc circuit on the input??


+18>--> | Q1
|S1 PNP
O--+------E C---+----> >---+
|+ B | |
[470ľF] | | |
| | |+ |
+--------+ [4700ľF] [50R]
| | |
[510R] | |
| | |
GND>-------+--------------+----> >---+
Click to expand...
 
On Tue, 30 Aug 2005 18:34:40 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> wrote:

He said the 470uf cap wasn't needed.

Try John's original circuit with the rc circuit on the input??


+18>--> | Q1
|S1 PNP
O--+------E C---+----> >---+
|+ B | |
[470ľF] | | |
| | |+ |
+--------+ [4700ľF] [50R]
| | |
[510R] | |
| | |
GND>-------+--------------+----> >---+
Click to expand...
 
"Robert" <hedrock@insightbb.com> wrote in message
news:ipbah1pinoehli670beeh6j1t8brpu22fp@4ax.com...
On Tue, 30 Aug 2005 18:34:40 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:

He said the 470uf cap wasn't needed.

Try John's original circuit with the rc circuit on the input??


+18>--> | Q1
|S1 PNP
O--+------E C---+----> >---+
|+ B | |
[470ľF] | | |
| | |+ |
+--------+ [4700ľF] [50R]
| | |
[510R] | |
| | |
GND>-------+--------------+----> >---+
Click to expand...

I know he did, But it isn't working for you. Can't hurt to try. It may
slow down the transistor coming on.
 
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