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LM317 Driver - No output voltage
- Thread starter Lgzaqq
- Start date
panic mode
- Joined Oct 10, 2011
- 4,867
not everything... your resistor seem to be 330Ohm instead of 10Ohm
if the value is 330 Ohm that means LM317 output current is 1.25V/330 Ohm = 3.78mA
the circuit in video is made to output 125mA
if the value is 330 Ohm that means LM317 output current is 1.25V/330 Ohm = 3.78mA
the circuit in video is made to output 125mA
Jon Chandler
- Joined Jun 12, 2008
- 1,560
The black wire is V+ input, and the red wire output?
Welcome to AAC!
Is black connected to your power supply and red connected to the load?
Did you measure the voltages on the 3 terminals to see what the regulator was doing? LM317 aren't guaranteed to operate with less than a 10mA load current.
Is black connected to your power supply and red connected to the load?
Jon Chandler
- Joined Jun 12, 2008
- 1,560
In the constant current configuration the constant-current-output is from the ADJUSTMENT pin, not the OUTPUT pin.
Jon Chandler
- Joined Jun 12, 2008
- 1,560
From the TI datasheet.
No.
The ADJUSTMENT pin may be connected directly to the output but the current is still from the OUTPUT through R1.
The ADJUSTMENT pin is just connected for reference, and has less than a mA of current coming out.
It maintains a constant 1.25V across the resistor to generate the constant-current.
Jon Chandler
- Joined Jun 12, 2008
- 1,560
The datasheets all show the constant current connection point is the adjustment pin. Semantics aside, I will defer to their recommendation.
What size is that resistor? I see orange orange red (3300 ohms) - correct? This means your current will be 1.25v/3300 = some microamps.
you should be using a 10 ohm resistor for 125mA. Or 5 ohm for a 250mA.
So you're only getting 0.00038 Amps (0.378 milliamperes).
same answer was given in post #2.
i have one more question
I see 0.25 watt and 1 watt resistors
Does it matter in this build?
Connecting two 2.2om 0.25w resistors give Me 4.4om 0.5w?
DickCappels
- Joined Aug 21, 2008
- 10,661
Power = E^2/R
Power = 1.25^2 / 10
Power = 160 milliwatts.
You can use a 1/4 watt(0.25 watt) resistor.
Power = 1.25^2 / 10
Power = 160 milliwatts.
You can use a 1/4 watt(0.25 watt) resistor.
If you are referring to my answer, there are no "Semantics" involved.
I agree the connection point is the Adjustment pin.
But the current comes from the Output pin, not the Adjustment pin.
Your problem was already pointed out by @panic mode. You're using the wrong resistor value.
\( \large R = \frac{V}{I} = \frac{1.25V}{200mA} = 6.25\Omega\)
It does.
\( \large P = I^2R = 200mA*200mA*6.25\Omega = 0.25W\)
You should use at least a 1/2W resistor.
Jon Chandler
- Joined Jun 12, 2008
- 1,560
The problem is that post #4 incorrectly said the output to the LED should come from the output pin – this is incorrect for a constant current regulator.
In post #6 and with the figure in post #7, I pointed out that the output (to the LED) needed to be connected to the Adjustment pin, not the output pin. I was not trying to explain the inner workings of the LM317, just clarifying where the output connection had to be made. This is the central point in this thread.
Your reply in post #8 is technically correct. But it only adds to the confusion about the application under discussion here.
Okay.
I missed what post #4 was saying and see know what you meant.
