Power arduino using lm338k

Question

Hi

I have a transformer 13.8v DC.

I also have arduino nano. According to the spec, arduino can be powered from 7-12v.

I thought using http://www.ti.com/lit/ds/symlink/lm338.pdf in-order to step down the voltage to around 12v.

I have used the default scheme as mentioned in the datasheet

R1 is 1Kohm, R2 is 1Mohm.

When Vout is un-connected to arduino - If I sample Vout using a multi-meter I get 12.19v which is great!
When I connect Vout to arduino's Vin port - the voltage drops down and continuously changes between 8.5v and 9.2v.
I want to mention the circuit is not getting hot, not the arduino and not the regulator

If I understand lm338k functionality, it supposed to provide a stable voltage even with load connected.

Even though ~9v is valid for powering arduino, I don't understand the phenomena.

Someone have an idea what I am doing wrong?

How can I continue the debug?

Wall-Wart photo:

I don't know why it says 7.5V, it actually outputs 13.75v when I test it with multimeter.

Measurement

I have just received The specified item was not found.

It has an osciliscope of up to 12v (Good enough for this task)

I put the measurment on Vout of the circuit above.

I have catched a trigger (8v fall edge) when I put higher load (I provde a pulse for enabling a DC motor driver) in parallel to the arduino.

You can see the voltage it pushed from 9.5v to 6.5v during the DC activation (Higher load is attached)

When aduino is reaching towards 6.9v it then power down and the motor stops.

The signal comming back to 9.5v is because both arduino and DC motor are offline since the arduino cannot contorl over the DC anymore...

Vout on the circuit is connected to Vin of arduino.

I then put big capacitor between the 5v voltage reference of the arduino (Regulated power supply) and GND.

This keeps ardunio on 5v during this hickup.

Thanks!

Idan

gecoz · Accepted Answer

Hi Idan,

The behaviour of Vout with your original set up, I'm not sure if the choice of the values for R1 and R2 @ Vout=12V were correct. From the formula, setting R1 to 100K you get I1=12uA, as you correctly stated. But when calculating R2, you need to remember that the current flowing through R2 is I1 plus Iadj, and the typical value for Iadj is 45uA (again, the datasheet is the source). Now, the voltage V2 across R2 is the difference between the choosen Vout and the voltage across R1, which is Vref:

V2=Vout−Vreƒ =12V−1.25V=10.75V

which gives for R2 the value of:

R2=V2I2 =10.75V57uA ≈189Kohm

Choosing 1Mohm for R2 might have caused the instability of the Vref you have seen (not able to settle on about 1.25V).

I'm pleased to see that with your new values, calculated @ Vout=9V (giving the LM338 the needed minimum headroom of 3V), the regulation works as expected.

Fabio.

jw0752 · Answer

Hi Idan,

You might consider using a regular 7809 regulator to bring the voltage down to 9 volts. You will not have to bother with resistors or adjustments. The Arduino has its own regulator to take the voltage down to the voltage needed by the processor and by bringing the voltage down externally you take some stress off the internal regulator.

John

Robert Peter Oakes · Answer

While this is correct regarding the remediation of the external regulator. Thus is not the primary reason for the posters symptoms. It is bease the transformer is actually only a 7.5v one and the higher volts is simply because it is unregulated and to all intents open circuit when reading the 13v value. As soon as the load is connected it drops to its correct voltage

gecoz · Answer

Hi Idan,

I think the problem you are experiencing could be due to the little headroom you are giving to the LM338. For the regulation to be effective, the datasheet advises (pag.10, par 7.4.2) for a minimum 3-V headroom Vin-Vout, otherwise the device may drop out and Vout will be Vin minus drop out voltage. In your case, the headroom is only 1.8V, so you'd get Vout = 12.19 - 3 = 9.19V, which is consistent with your reading.

You might want to recalculate R1 and R2, using the formula on pag.11 of the datasheet, this time for a Vout of 9V, and verify the voltage will remain within the specifications given on the datasheet after you plug the Arduino in.

Fabio.

idanre1 · Answer

Hi

I have made some expirements acording to gecoz suggestions.

First I want to mention no matter what values I choose, In case no load is connected I always see ~12V on the output.

Probably you need to load the circuit in-order to see its real behavior.

Acording to the datasheet, as Fabio suggested:

I need to see about 1.25V in Vref.

I see about 1.2-1.9V depends on other values of R1.

If I choose R1 as 100Kohm I get i1=1.2/100000=0.012mA

If I choose R2 as 1Mohm I get V(R2)=12v

I have chosen other values below:

R1 = 1K

R2 = 10K

And I see about 9V feeding the arduino.

This still falls into 3V of Vout-Vin but I don't see flactuations anymore. Just steady 9V.

I guess the lm338k having trouble keeping 1.2v over 1Mohm instead of 1Kohm. Maybe that is why I see less steady voltage using original values.

Idan

Robert Peter Oakes · Answer

A thing to mention here is that the "transformer " is 7.5v when under load. When you measure with a multimeter it only has about a 10Mohm load and so the output goes quite high fooling you into thinking it is marked wrong. What you have is an UN-REGULATED supply.

Therefor you don't need any regulator to use it with the ardiino. As soon as you connect it he volts will drop to the nominal 7.5 and all will be good.

Hope this clarifies things for you

Robert Peter Oakes · Answer

See my other response. You don't even need a regulator external to the arduino

gecoz · Answer

Hi Idan, According to your wall-wart specs, you get 7.5V @ 300mA (2.25W), therefore theoretically a load of 25ohm would do (mind, you need a 3W resistor if you want to test it without smoke!). Considering that the LM338 is working fine even with this unregulated supply (I never noticed you were using it before), my guess would be that the load offered by the LM338+Arduino is high, forcing the wall-wart to reduce the output current, thus increasing its output voltage to something > 12V. The only way to verify this is to actually measure the current draw from the wall-wart. Anyway, because of the behaviour of the unregulated power supply, I would be careful using it without any regulation circuitry: if for any reason the current draw from the wall-wart goes above the maximum 300mA, the voltage provided by the supply starts dropping below the nominal 7.5V, which could cause troubles to the Arduino Nano (Arduino needs 7V to 12V source for its internal regulator to work properly). Fabio.

gecoz · Answer

Hi Peter, Indeed you are right, the unregulated wall-wart is definitely causing the drop from 12V to 9V once load is connected. I was focussed more on the schematic, which even with a regulated voltage source of a nominal value of 13.8V used instead of the wall-wart, would still not work properly, because of the way the LM338 was employed. (Moreover I completely missed the update to the post, showing the wall-wart - my bad!).Thanks for pointing that out. Fabio.

Robert Peter Oakes · Answer

The solution right now is working but the minute you increase the load further it will fail again as the transformer is nominally rated for 7.5v at upto 300mA but as it is unregulated it's voltage will vary with load, more load less volts..... The load needed to see if the volts is correct is simple i=v/r or r = v/I The arduino will never see the 13v ish you measured as it will always be placing some kind of load on the transformer. It may provide more than 7.5 but not more than 12 to 13 as this is the no load voltage

gecoz · Answer

Hi Idan, My assumption is perhaps overly simplified. Unregulated power supply typically is designed for "matching" loads, where you know what voltage and current the load will need (within known tolerance). This "matching" is needed, as unregulated power supply output voltage is affected both by the input voltage variation and varying load conditions. Therefore, to simplify, I used the nominal values for voltage and current to calculate a nominal power of the wall-wart. With unregulated supply, the current rating you find on the label is indeed the max output current (independently from the output voltage), but the nominal output voltage 7.5V does give you the voltage rating for your load, thus I wrote the nominal power is 2.25W. This does not mean that 7.5V are only guaranteed at @ 300mA, but the values written on the label do give you an indication on what to expect. What I was suggesting is that, using a purely resistive load, you can test this condition by "forcing" the wall-wart to give the load its max current, and the 25ohm resistor would make sure the voltage across the load is matching the nominal output voltage. I'm not assuming a linear dependency between output voltage and current to the load, the test I was suggesting above is only a "single point" test of the supply on the V-I curve for the power supply, and in the special case of purely resistive load. In general, the dependency V-I is non linear, but can be approximated as linear for output currents from 0 up to max output current, but becomes strongly non-linear as the current grows greater than the max output current. If you look at the plots below, perhaps my suggestion will become clearer (this is just an example): I hope this explains my thinking Fabio.

Power arduino using lm338k

Measurement

Top Replies