LED diode placement-help

Thread Starter

stypee

Joined Feb 9, 2021
3
shema_za_ptp1.jpeg
Hello, I created this shematic to later design PCB. I need to put somewhere LED diode that indicates that the voltage is present on the printed circuit board. I don't really know where is the right place to put it and I would like if someone could explain to me where to put it and why should I put it there.
 

LowQCab

Joined Nov 6, 2012
4,023
It depends on exactly what you want the LED to indicate.
If you only care about "voltage is present", regardless of what that voltage may be, too-high, too-low,
Input Voltage, -vs- Regulated Voltage, etc.
Is the Regulated Voltage within an acceptable range ?
How narrow or wide must the Voltage Range be ?
You did not state what problem you are trying to solve.
 

Thread Starter

stypee

Joined Feb 9, 2021
3
It depends on exactly what you want the LED to indicate.
If you only care about "voltage is present", regardless of what that voltage may be, too-high, too-low,
Input Voltage, -vs- Regulated Voltage, etc.
Is the Regulated Voltage within an acceptable range ?
How narrow or wide must the Voltage Range be ?
You did not state what problem you are trying to solve.
I'm not so good with electronics placement in electrical circuit, so I'm sorry if I wasn't so clear with my problem. I only care that LED indicates that "voltage is present" nothing more than that. Regulated voltage is in range from 0-30V.
Input voltage can be from 2.5 to 40V and the only thing that bothers me is if the high current would damage(from heat) the LED.
I hope I said something useful with this response because I'm really a newb at this kind of stuff.
 

Audioguru again

Joined Oct 21, 2019
6,672
The datasheet for the LM317 shows that if your Rx is higher value than 120 ohms then some LM317 will have a higher output voltage when their load current becomes low or zero. Then the value of the pot should also be reduced but it will get too hot.
 

LowQCab

Joined Nov 6, 2012
4,023
What you need is a "Current Regulator" like this ..... CL520K4-GCT-ND, or this, CL525K4-GCT-ND .
These are available in 20ma, and, 25ma versions .
Download the Data Sheet to learn how they can be used .
With up to 40 Volts Input, the Current Regulator may have to dissipate up to 1-Watt of heat,
so consideration must be given to proper Heat-Sinking .
No other parts are required .
Input Voltage must exceed 1 Volt, plus, the Voltage Drop of the LED, for the LED to operate at full output .
Your LED(s) must be rated for 20ma, or more, to use these Regulators.
( 2- 10ma LEDs could be used in parallel ).

If you would like to raise the Input Voltage required to light the LED .......
Add a 1-Watt rated Zener Diode to the circuit .
The Voltage Rating of the Zener Diode,
plus, 1-Volt for the Current Regulator Voltage Drop,
plus, the Voltage Drop of your LED,
will be required before the LED will put out its full light output .
.
.
 

wayneh

Joined Sep 9, 2010
17,496
I'm not so good with electronics placement in electrical circuit, so I'm sorry if I wasn't so clear with my problem. I only care that LED indicates that "voltage is present" nothing more than that. Regulated voltage is in range from 0-30V.
Input voltage can be from 2.5 to 40V and the only thing that bothers me is if the high current would damage(from heat) the LED.
I hope I said something useful with this response because I'm really a newb at this kind of stuff.
Since you have two voltages that are both widely variable, you can't just wire in an LED and a resistor like you might if you had a predictable voltage. You need to establish a reference voltage somewhere in your circuit, or a constant-current supply for the LED.

I think I'd try to use a zener diode to establish a voltage, say 4V. Place an LED and resistor in parallel to the zener. The current in the zener must be limited by a resistor in series with it. A TL431 would be even better. Neither will solve the problem of lightning the LED at low input voltages.

There may be a simple emitter-follower transistor arrangement that would work here too but I can't seem to recall the details. Getting old sucks.
 

Audioguru again

Joined Oct 21, 2019
6,672
You never said the maximum output current you need.
The minimum output voltage of an LM317 is not 0V, it is near 1.25V.
With a 2.5V input then some LM317s will produce an unregulated output voltage of about 1V which goes up and down when the current goes up and down. The minimum input is about 4V for a fairly low (2V?) regulated output at a current of 1A.
Of course the output voltage must be about 1.5V less than the input voltage.
The LM317 will overheat and shutdown if the output voltage is set low and the current is high.
 

DickCappels

Joined Aug 21, 2008
10,153
You don't need a(n exotic) current regulator. A resistor will do.

Put the LED and current limiting resistor across the DC input.

You wrote "I need to put somewhere LED diode that indicates that the voltage is present on the printed circuit board". That is the place to put the indicator.
 

DickCappels

Joined Aug 21, 2008
10,153
The LED has a specified voltage for a given current (which should be less than the maximum allowable current).
The value of the resistor is R = (Power supply voltage - Voltage across the LED)/(current through LED.)

If you supply the power supply voltage, the voltage across the LED and the desired current I will be happy to work our a resistor value for you if you like.
 

LowQCab

Joined Nov 6, 2012
4,023
"" Input voltage can be from 2.5 to 40V ""

Resistor calculation must be based on the highest expected Voltage ........ ~40 Volts .

Average generic LED specs =
1.6V Forward Voltage Drop @ Maximum Recommended Current of ~20ma.

20ma of Current is the target.

40V minus 1.6V = 38.4V,

38.4V / 20ma = 1,920 Ohms,

Resistor Wattage calculation .......
38.4V X 20ma = 0.768 Watts,
A 1-Watt rated Power Resistor is required.

Lowest Voltage calculation at 2.5V ..........

2.5V minus 1.6V = 0.9V,

0.9V / 1,920 Ohms = 0.4ma of LED Current,

LED Illumination will be ........

0.4ma / 20ma = ~2% of maximum brilliance, Light Output from the LED will not be detectable.
------------------------------------------------------------------------------------------------------------------

Minimum LED performance utilizing a Current Regulator Chip instead of a Resistor .........

Minimum Input Voltage = 2.5V,
minus 1V Forward Voltage Drop of Current Regulator = 1.5V to LED,
Specification Sheet for the LED will have to be used to determine the percentage of
Light Output at 1.5V,
in the case of the LED chosen for this example,
the LED will produce approximately ~95% of its rated light output,
with a Minimum Specified, Voltage Regulator Supply Voltage, of 2.5V .

The LED will put out its full rated brilliance at any Input Voltage greater than ~2.6 Volts,
and all the way up to the ~90V Input Voltage Rating of the Current Regulator .

This LED Current Regulating Scheme can also be used on the
Variable-Voltage-Output of the Regulator with no change in performance .

The cost of the Current Regulator is only ~$0.70US .
The cost of a 1W, 5% Resistor is roughly ~$0.22US .

And BTW, you MUST use an LM317HV for your Specified Input Voltage Range of ~40 Volts.
Also,
replace the 240R Resistor with a 1K,
2-Watt Resistor,
and replace the 10K Pot with a 25K Pot,
and add a 100R Resistor in between the Pot and the Regulator,
this will make the maximum possible Voltage Output ~32.6V ,
and the minimum possible Voltage ~1.4V .
If you must have the capability to go all the way down to "zero" Volts,
you need to add 2- generic 5-Amp Diodes in series with the Regulator Output,
otherwise the lowest possible Voltage for this Regulator is ~1.2 Volts .


"" No protection is needed for output voltages of 25 V or less and 10-μF capacitance.
Figure 15 shows an LM317HV with protection diodes included for use with
outputs greater than 25 V and high values of output capacitance "".
.
.
.

LM-317HV Flat .png
.
.
.
 

DickCappels

Joined Aug 21, 2008
10,153
There's rub - he wants a wide range. The resistor has to be sized for the maximum and will thus be dim at the minimum. It will light, however. Maybe that'd be good enough.
Some might think it better. That is the way I have it on my bench supplies. At a glance I can tell the approximate voltage (high, medium, low, etc.)
 
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