What is the rating of the "Transformer" (Its mA or amp rating )?, can you take a picture of the label and post it here ?

you call it a transformer but is it really a "Wall Wart". ie a power brick you plug into the wall ?

Provide a schematic showing all the connections you making, also is anything getting warm when you are powering the arduino ?

Peter

Hi Peter

You made some important points, I have updated the post with additional info.

Thanks!

Idan

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

Hi Peter

I am intrested checking your suggestion.

What do you think is a proper load for matching 7.5v in the wall-wart? 10khom? 1kohm? Something else?

Using Fabio's solution and assigning the right values to R1 and R2 I do get 9v when arduino+BT_module are connected as a load.

Maybe the current consumption of the circuit is not high enough?

Regards

Idan

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.

Hi Fabio

I actually don't understand your assumptions.

You assume the wall wart output voltage is linear to the load (and current).

I thought the 300mA is just the maximum average current that can be drawn.

Why you assume the voltage is linear to the current?

I accept the assumption you have min threshold for reaching7.5v as Peter mentioned, but why you say this will only happen in max current?

Idan

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.

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.