Led voltage indicator

Ron H

Joined Apr 14, 2005
7,061
geez I didnt realize what a difference not using a 3914 would be I thought IC were more interchangeable I feel bad now that I took your time for something that could have been done easier... I guess I will go with the 3914. http://www.national.com/pf/LM/LM3914.html has a schem that looks pretty simple my only question is how to use it with something thats 400v.
thank you for all your help and patience's
Here's my best shot at using the LM3914 for your purpose. The schematic came out somewhat tinier than I had planned. :mad:
 

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bloguetronica

Joined Apr 27, 2007
1,541
geez I didnt realize what a difference not using a 3914 would be I thought IC were more interchangeable I feel bad now that I took your time for something that could have been done easier... I guess I will go with the 3914. http://www.national.com/pf/LM/LM3914.html has a schem that looks pretty simple my only question is how to use it with something thats 400v.
thank you for all your help and patience's
I think that is the best alternative, since the LM3914 can be easily adapted. Aslo, why not using a resistor instead of a lightbulb? Do you need the lightbulb for indication purposes?
 

Ron H

Joined Apr 14, 2005
7,061
I think that is the best alternative, since the LM3914 can be easily adapted. Aslo, why not using a resistor instead of a lightbulb? Do you need the lightbulb for indication purposes?
The light bulb acts approximately like a constant current limiter. At the beginning of the charge cycle, when the voltage across the bulb is high, its resistance is high because it is hot. As the capacitor charges, the resistance drops. The bulb therefore limits the peak current as a resistor would, but allows the charging to proceed more rapidly than would a constant resistance. The bulb is also a cheap way of handling the peak power dissipation. A resistor with the same capabilitiy might be considerably more expensive.
Having said all that, I have no idea if the OP uses it for these reasons. :rolleyes:
 

erice1984

Joined Jun 9, 2007
16
know where someone (myself) could find a spice model for the LM3914 ? I used a LM339 in original design of my circuit until I saw that guy, and now I want to use it for space and simplicity of it.
 

bloguetronica

Joined Apr 27, 2007
1,541
The light bulb acts approximately like a constant current limiter. At the beginning of the charge cycle, when the voltage across the bulb is high, its resistance is high because it is hot. As the capacitor charges, the resistance drops. The bulb therefore limits the peak current as a resistor would, but allows the charging to proceed more rapidly than would a constant resistance. The bulb is also a cheap way of handling the peak power dissipation. A resistor with the same capabilitiy might be considerably more expensive.
Having said all that, I have no idea if the OP uses it for these reasons. :rolleyes:
Well, the resistance of a lightbulb won't dramatically change with temperature. You are better off with a current mirror if you want a constant current.
 

Ron H

Joined Apr 14, 2005
7,061
Well, the resistance of a lightbulb won't dramatically change with temperature. You are better off with a current mirror if you want a constant current.
What think you know is not always fact.
The resistance of a light bulb changes dramatically with the filament temperature, which is dependent on the current through it. I measured the resistance of a 40 watt, 120V bulb at room temperature. It measured 27 ohms. If you calculate the resistance from power and voltage, you get 360 ohms. Thats a change of over 13:1!
 

spankey666

Joined Nov 30, 2011
91
Here's my best shot at using the LM3914 for your purpose. The schematic came out somewhat tinier than I had planned. :mad:
I,m sorry for bringing up an old post, but i have been searching the internet for ages looking for just such a circuit. the only thing is i want to measure from 150v to 300v dc for the same application of charging capacitors. would it be possible to supply me some details on how to convert this circuit to work within these parameters or is it not possible to have such a broad range with this ic ?
Thanks
Sarah
 

Audioguru

Joined Dec 20, 2007
11,248
I buy electronic parts online from Digikey. They are in a different country from me and if I order before 8:00PM then the parts are delivered to me the next morning with no customs and duty fees.
 

Audioguru

Joined Dec 20, 2007
11,248
The LM3914 is a simple voltmeter. Its datasheet explains how to use it and shows many circuit examples.

The circuit you posted has its reference voltage at pin 7 and pin 6 set to about 5V and its input attenuated to a max of about 5V. Therefore LED #10 lights when the capacitor voltage is 400V.
 

spankey666

Joined Nov 30, 2011
91
so i'm assuming that if i reduce the resistors on the input from 940k to 705k this should drop the upper voltage to 300v ??? how do i control the increments between leds coming on. the data sheet says 1.2v per increment my scaling requires 15v per increment. sorry for so many questions. i have only dabbled a little in electronics in the past so its not my strong point :( all help in trying to understand how this all works is greatly appreciated :)
 

thatoneguy

Joined Feb 19, 2009
6,359
You need to create a voltage divider that will scale 0-400V to 0-12V (or whatever your supply is).

A 510kΩ resistor in series with a 15kΩ resistor would work.

Put one side of the 510k resistor on the 400V, the other side connects to the 15k resistor and the input of the 3914, the other end of the 15k resistor goes to ground.

This is assuming you have 12V powering your 3914, it will scale 400V to 0-12V roughly, maybe add a 12V zener diode on the input as well in case you go over 400V

For a 5V divider, use 1.2M in place of the 510k resistor, and a 5V zener
 

Ron H

Joined Apr 14, 2005
7,061
You need to create a voltage divider that will scale 0-400V to 0-12V (or whatever your supply is).

A 510kΩ resistor in series with a 15kΩ resistor would work.

Put one side of the 510k resistor on the 400V, the other side connects to the 15k resistor and the input of the 3914, the other end of the 15k resistor goes to ground.

This is assuming you have 12V powering your 3914, it will scale 400V to 0-12V roughly, maybe add a 12V zener diode on the input as well in case you go over 400V

For a 5V divider, use 1.2M in place of the 510k resistor, and a 5V zener
LM3914 has a built-in programmable reference which is independent of the supply voltage. The divider needs to be scaled to match it. Sorry I don't have time to calculate the voltage divider right now.
 
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