I've an arduino uno and an easy driver v4 stepper motor driver. now the driver needs 12 V power, and I don't want to have two power cables one for the arduino and one for the driver.
Is it possible to connect both by using only one power plug as on my attached circuit? or would there be an other simpler solution? Is it possible to power the arduino directly with 12 V and then use the GND and Vin on the arduino and connect it to the stepper driver itself?
Ah and I don't want to use the usb, because my plan is to control the stepper motor over bluetooth.
Is it possible to connect both by using only one power plug as on my attached circuit? or would there be an other simpler solution?
Yes it is possible, but probably not the best way to go. It will result in the onboard 5V regulator running hot on the arduino dissipating the extra 4.5Vdc (the optimum input voltage for the external power barrel jack is 7.5Vdc)
Also, the motor may result in spikes on the 12V line . The 5V regulator will filter those out but you should add 0.1 uF and 1 uF decoupling caps on the 5V power bus to compensate.
Is it possible to power the arduino directly with 12 V and then use the GND and Vin on the arduino and connect it to the stepper driver itself?
Electrically that is the same as using a "Y" cable to split the 12V to both the driver motor power input and the arduino external power barrel jack.
a lot. that's what i thought that one element would run hot. so i'll go for the solution you proposed.
First i had another idea, using a arduino nano, according to a sketch i found online. so i draw it accordingly. on the "rx" cable the resistors are still missing to get there 3.3v.
So my question is would this be possible too? or is it better to use like an Y cable to power the easy driver separate from the arduino nano?
And is it the best way to power the arduino over VIN?
I saw there is a kickstarter project where they produce an arduino nano with bluetooth on board but i'm not sure if its available soon.
I know you are trying to avoid using two power supplies but the truth is that is the best solution. If you must use only a 12V supply then use a power resistor to drop the voltage to 7.5V . The only way to calculate the value correctly is to measure the current of the 12V supplying the arduino external barrel jack. This is because the normal method of measuring current is to put the meter in series with the load. Since the load is everything that runs on 5V, there is no way to disconnect those circuits on the arduino. By putting the meter in series with the 12V , we can measure the same load using power (P= I*V)
For example, my arduino UNO draws 50mA @ 5Vdc, if I bypass the onboard regulator and connect to the 5V pin directly.
P= I x V = 0.05A x 5.00V =0.250W = 250mW = 1/4 W
That same board would draw how much current at 12V ?
P = I (A) x V (V) ==>I (A) = P (W)/V (V)
I = P/V = 0.250W/12V =0.0208333 A = 20.8 mA.
To convert the current at 12V to the current on the 5V load side, simply use the same formula
ie:
Given ILOAD = 20.8mA @ 12V
What is the current for the 5V load (the arduino) ?
P = I x V = > I = P/V
I= P/V = 0.250W/5V
I = 0.05 A = 50 mA. @ 5V
This is how you can calculate the 5V load current (which you need to know to calculate the dropping resistor value)
To calculate the value and wattage of a dropping resistor to drop the 12V down to 7.5V,
Rdropping= (VIN-VLOAD)/ ILOAD
=( 12V-7.5V)/(0.05+0.015)=4.5V/0.065 A = 69.2 ohms (adding 15 mA for onboard regulator overhead)
PRdropping= Iload*Vload= 0.065 A * 5V = 0.325 W (use 1/2 W resistor)
So we have a minimum and maximum value for the resistor.
69.2 is close to 68 so you can use 68 ohms , 1/2 W if you add 50mA of load to the arduino.
If you add nothing then R = 4.5V/0.05A = 90 ohms, 1/2 W.
Rdropping = 68 to 90 ohms , 1/2 W.
To allow addition of extra circuitry , you can pick a value near the middle of this range, like 82 ohms , 1/2W and use that in series
with the 12V to drop the voltage down to 7.5V at the input of the arduino external barrel jack. (The voltage can ONLY be measured with the plug connected to the arduino, to obtain a valid voltage. Measuring the voltage with no load is invalid. After you add the resistor and plug the connector into the barrel jack, measure the voltage on the side of the dropping resistor CLOSEST to the arduino. If it is ABOVE]
7.5V, DECREASE the resistor value to the next lower standard value. If the voltage is BELOW 7.5 V, INCREASE the resistor value to the next higher standard value. Since the dropping resistor dissipates 325mW of heat this is the price you are paying to protect the onboard 5V regulator, (which I assume you probably could not replace yourself if it failed) which if it failed , could subject the entire arduino UNO to 12V, damaging everything. I don't know what the likelyhood of that is but is it worth taking the chance ?
