I've tried taking out resistors from below the switch, adding higher resistor values, using the schematic ScottWang did, but nothing seems to work for me.

I'm sure what you mixed up two ways and what is the real circuit?

There is a problem there and that is the pin 3 Vout will be like as Vout = Vcc - 1.4V = 9V-1.4V=7.6V.
Using cmos type ICM7555 can be reach up to 9V, but the output current will be more less as 0.8 mA.

 

This is the LED I'm trying to use.

Emitted Color: Red
Output Power: 10W
Intensity Typical:900~1000 Lumen
Forward Voltage:9V-12V
Forward Current:900mA
Color Temperature:6000~6500K

 

Concerned about the usage life and brightness of LED, you should using the current less than 80% as 720 mA.
I'm not sure what kind of power are you using (?mA)?
I think you should buy a AC/DC adaptor as 90~245Vac to 12Vdc/1.5A or 2A.

I will redraw the circuit.

 

Concerned about the usage life and brightness of LED, you should using the current less than 80% as 720 mA.
I'm not sure what kind of power are you using (?mA)?
I think you should buy a AC/DC adaptor as 90~245Vac to 12Vdc/1.5A or 2A.

I will redraw the circuit.

I used a 120v ac/dc power supply 12vdc 2.1amp to do my testing. I plan to use 12.5vAC that I have available and can convert to dc hopefully. The LED will be used in a flash situation so the on time will not be very long. The LED seemed to be bright enough at 8.5 volts. I'd like to run it at around 8.5 v. Anything over 9.5v might cause it to heat up and anything below might not light it enough. Thanks for your help.

 

I was attached two circuits, the one is use BJT and another one is use N ch mosfet IRF540, if you want to use n mosfet and you could get the Rds lower than IRF540, it's better.

I was assumed that the Vcc is 12V and the current draw of led is 720 mA, Vce=0.2V, Vbe=0.7V, Vds of mosfet as 0V(when it turn on), so if you want to change anything then you can according to the condition to recalculate the values of resistor, the Zener Zd1 was designed to protecting the Vgs if it has any high voltage input.

And the current from pin 3 of Ne555 is 7.2 mA, and Ic2 of Q2 is 72 mA, the IC3 of Q3 is 720 mA, and the LED count it as 8.5V/720 mA, the pratical wattage of resistor is counts at least 4 times of calculation values.

 

Awesome. I'll try them out. Thanks for doing that. Maybe I can get it right. I'll let you know what happens.

 

I was forgot those two diodes 1N4007 are used for another application of lower current, here is the new circuit.

 

Just a couple of questions. Are the Rc2 just a 150 ohm 5watt resistor? Rb2 says 1.2k which I assume is a 1.2k resistor but I don't understand +180. Rc3 looks like a 5ohm 15watt resistor. Am I assuming correctly ? Thanks for your help.

 

RC2 = 150Ω/5W
RC3 = 5Ω/15W
Rb2 = 1.2 K+180Ω=1.38K

Because I can't find a properly values for the Rb2 resistor, so it used two resistor in series.

 

RC2 = 150Ω/5W
RC3 = 5Ω/15W
Rb2 = 1.2 K+180Ω=1.38K

Because I can't find a properly values for the Rb2 resistor, so it used two resistor in series.

O.K. Thanks. That's what I was thinking, but wanted to make sure. I'll put it together tomorrow hopefully. I'll let you know.

 

Hey ScottWang. Awesome that it worked and was wired just like I need it. (Switch and Bulb both to ground.) I put together the circuit in your Post #28 figure #1. I was excited to see it work. I've tried a lot of different resistors on the left side attached to pin #2. 1k, 47k and 100k. I have 2- 10 Watt 5 ohm resistors in series (rather than the 1- 15watt 5ohm, they didn't have it locally) on your schematic. I also tried using just 1 10watt 5 ohm. Here's what happened. The LED did flash, but the light was very dim. I am only flashing it for a short time, so I think it needs more voltage. Is there a way to get more power to the bulb to make it brighter ? The switch only has .74v when it's left open. It does get 1.3v when I flash the bulb. It worked correctly otherwise, the voltage remaining off when the switch was closed. The reason I need the 7.5 - 8 v running to the switch is because I'm splitting the wire off on the + side to run to a gauge that requires the 7.5 - 8v. Do you have PayPal? Thanks.

 

The reason I need the 7.5 - 8 v running to the switch is because I'm splitting the wire off on the + side to run to a gauge that requires the 7.5 - 8v.

I still don't get it because you didn't talk much about it and this - The switch only has .74v when it's left open. It does get 1.3v
You need to draw the circuit, otherwise I can't help for it.

Here is how I calculated the static state current and resistor for LED:

Rc3 = (12V - Q3_Vce - V_Led1)/0.72 A
= (12V - 0.2V - 8.5V)/0.72 A
= 3.3V/0.72 A
= 4.583 Ω

I chose the Rc3 for 5Ω.
And calculated the current as:
I_Rc3 = 3.3V/5Ω
= 0.66A
(Here, I was calculated 4.58 Ω not 5Ω, so it reduced about 60mA, because it needs in series with some resistor, so I didn't calculated too much)

I didn't realize that you have mentioned that you will use it with flash led in posted #24.

How are the frequency and duty cycle of led flashing? (High Duty cycle)

The average current of LED as :
I = 0.72 A(High Duty cycle/100%)
= 0.72A (50%/100%)
= 0.72A (0.5)
= 0.36A

If the duty cycle is 50% then it could be only 330mA when it is flashing,
I = 660mA*0.5 = 330mA

High duty cycle means that the percentage of led in the turn on status.

So you can measure the current and adjust the values of Rc3 little by little.

 

If calculate the led current is 0.72A and duty cycle is 0.5% then the instantaneous current will be up to 144A.

If do not calculate the current that high, maybe only use 10A, and the Q2_Ic2 as 1A, and the Q2_Ib2 is 100mA,

Using Tip41 to replace the 2sc1384
Q2_Ib2 = (12V - 1.4V - Q2_Vbe2)/Rb2.
Rb2 = (12V - 1.4V - Q2_Vbe2)/Q2_Ib2.
= (10.6V-0.7V)/100mA
= 9.9V/100mA
= 99 Ω, it can be choose 100Ω

W_Rb2 = 9.9V * 100mA
= 0.99 W
If using 5% for duty cycle not 0.5% then the W = 0.99W*5% = 0.05W, so use 0.25W is ok.

IC2_Q2 = (12V - Q3_Ibe3 - Vce2_Q2)/Rc2
Rc2 = (12V - Q3_Vbe3 - Vce2_Q2)/IC2_Q2
= (12V - 0.7V - 0.2V)/1A
= 11.1V/1A
= 11.1Ω, it can be choose 11Ω

W_Rc2 = 11.1V * 1A
= 11.1 W
If using 5% for duty cycle not 0.5% then the W = 11.1W*5% = 0.555W, so use 2.5W is ok. (if counts the duty cycle for 0.5% then use 1/4W is ok)

Using 2N2955 to replace the tip42.
Ic3_Q3 = I_Led1
IC3_Q3 = (12V - Q3_Vce3 - V_led1)/Rc3
Rc3 = (12V - Q3_Vce3 - V_led1)/IC3_Q3
= (12V - 0.2V - 8.5V)/10A
= 3.3V/10A
= 0.33Ω.

W_Rc3 = 3.3V * 10A
= 33 W
If using 5% not 0.5% then the W = 33W*5% = 1.65W, so use 2W is ok.
(if counts the duty cycle for 0.5% then use 1/4W is ok)

 

I'm using the diagram in Post #4. I have tried NPN Mosfet irf 510, 540, PNP Mosfet, 2n2222, 2n3904 and others. Can't get the circuit to light the bulb when the switch opens. In fact only once did I get the switch to light the bulb when I closed it (I need it to do just the opposite, lights when open). I really need 7.5 vdc running to the switch.

The circuit in Post #1 does what I need when R1 is swapped with S1, but I need S1 connected to ground as in Post #4.

Can anybody suggest a transistor that might work in this situation ? Or a modification to the above schematics. Thanks

Almost any small-signal NPN will work as long as you push enough current into its base to saturate the collector-to-emitter junction, and if you're using a bipolar 555 with C1 equal to 100nF, R1 and R2 should be about 10k in order to get the voltage into pin 2 low enough to reliably trigger the timer when the trigger source goes low.

 

I still don't get it because you didn't talk much about it and this - The switch only has .74v when it's left open. It does get 1.3v
You need to draw the circuit, otherwise I can't help for it.

Here is how I calculated the static state current and resistor for LED:

The LED doesn't appear to need a ballast since its output power is rated with a Vf of between 9 and 12 volts, those vagaries being taken care of internally.

 

View attachment 120001 View attachment 120001
I have this circuit and all I need is to somehow change S1 so that it is a normally closed switch and as soon it opens it triggers the 555. Anybody have any ideas? I have the timing correct in terms of lighting my LED. I have 12vdc running through S1. Thanks.

It's a 555, so there are zillions of ways to get it to do what you want.

Can you elaborate a little about what you want to put into your widget versus what you want to get out of it?

Also, can / will you post the LED's data sheet, or a link to it, please?

 

The LED doesn't appear to need a ballast since its output power is rated with a Vf of between 9 and 12 volts, those vagaries being taken care of internally.

Did you see what the TS said in #24 as these:
The LED seemed to be bright enough at 8.5 volts. I'd like to run it at around 8.5 v. Anything over 9.5v might cause it to heat up and anything below might not light it enough.

Do you know what is light fades in led?

 

Did you see what the TS said in #24 as these:
The LED seemed to be bright enough at 8.5 volts. I'd like to run it at around 8.5 v. Anything over 9.5v might cause it to heat up and anything below might not light it enough.

Do you know what is light fades in led?

Oops...
I missed #24, and I don't understand what you mean by: "Do you know what is light fades in led?" but since English doesn't seem to be your first language, perhaps I'm misconstruing your attitude as being confrontational.

In any case, instead of using a power-hungry resistive dimmer, I'd suggest that the OP PWM the LED using a second 555 in astable mode, with its RESET connected to the output of the first 555, like this:

 

Oops...
I missed #24, and I don't understand what you mean by: "Do you know what is light fades in led?" but since English doesn't seem to be your first language, perhaps I'm misconstruing your attitude as being confrontational.

In any case, instead of using a power-hungry resistive dimmer, I'd suggest that the OP PWM the LED using a second 555 in astable mode, with its RESET connected to the output of the first 555, like this:

View attachment 120458

I'm considering the brightness attenuation of led, and you probably missed this --
The LED will be used in a flash situation so the on time will not be very long .

The flashing time could be a few mS. so that's why I recalculated the values of resistors in #33, do you think the pwm mode is a useful way for this?

 

I'm considering the brightness attenuation of led, and you probably missed this --
The LED will be used in a flash situation so the on time will not be very long .

Actually, I picked up on it but decided to let it slide for a while since neither the ON time nor the duty cycle was defined.

The flashing time could be a few mS. so that's why I recalculated the values of resistors in #33, do you think the pwm mode is a useful way for this?

Flash time isn't usually expressed in terms of millisiemens, but other than that it could be extremely useful in that it'll eliminate the ballast resistor and its attendant waste of energy.

Unfortunately, the OP has addressed neither the application, the flash duration, nor the flash period, so it's difficult to determine whether PWM will work for him in, say, a dynamic photographic application.