# Need help with calculations for my project

#### Spiderjay2

Joined Jun 24, 2017
9
Hi I need to wire together two motors and a solenoid actuator together but have no idea what batteries would be best or if I should wire in parallel or in a series/parallel hybrid. The motors are rated at 6V, 300ma. The solenoid is rated at 4.5 V and the current draw is unknown. They need to all run at the same time and the motors at full speed. Thank you for reading, for any answers, and for your time.

#### wayneh

Joined Sep 9, 2010
16,398
Hi I need to wire together two motors and a solenoid actuator together but have no idea what batteries would be best or if I should wire in parallel or in a series/parallel hybrid. The motors are rated at 6V, 300ma. The solenoid is rated at 4.5 V and the current draw is unknown. They need to all run at the same time and the motors at full speed. Thank you for reading, for any answers, and for your time.
You should wire in parallel unless you have a good reason not to. Four AAs would likely be fine. I suspect the motors would draw the voltage down to a safe level for the solenoid. Otherwise add one or two diodes in series with only the solenoid, to drop the voltage.

#### Spiderjay2

Joined Jun 24, 2017
9
You should wire in parallel unless you have a good reason not to. Four AAs would likely be fine. I suspect the motors would draw the voltage down to a safe level for the solenoid. Otherwise add one or two diodes in series with only the solenoid, to drop the voltage.
I was reading something about wiring leds in parallel being a bad idea because of the lower voltage leds drawing all of the current and eventually burning out, then causing the next ones in line to burn out, then the next, etc. This doesn't apply for other types of circuits wired in parallel?

#### wayneh

Joined Sep 9, 2010
16,398
I was reading something about wiring leds in parallel being a bad idea because of the lower voltage leds drawing all of the current and eventually burning out, then causing the next ones in line to burn out, then the next, etc. This doesn't apply for other types of circuits wired in parallel?
Nope. Consider the lamps in your home. They're all in parallel and so is everything else you plug in. Many or most items self-limit the current they draw as long as they are supplied the proper voltage. LEDs are an exception and sadly a very, very common one. They have no built-in regulation of current. But your motors and solenoid do.

#### Spiderjay2

Joined Jun 24, 2017
9
Nope. Consider the lamps in your home. They're all in parallel and so is everything else you plug in. Many or most items self-limit the current they draw as long as they are supplied the proper voltage. LEDs are an exception and sadly a very, very common one. They have no built-in regulation of current. But your motors and solenoid do.
Makes perfect sense! Thank you for your effort and fast replies! One more quick question, would there be anything to worry about if I wire an led in series with the solenoid to consume the remaining 1.5V after the solenoid uses 4.5V, like you suggested?

#### dl324

Joined Mar 30, 2015
10,932
would there be anything to worry about if I wire an led in series with the solenoid to consume the remaining 1.5V after the solenoid uses 4.5V
That would depend on the LED and solenoid current.

#### AnalogKid

Joined Aug 1, 2013
8,487
would there be anything to worry about if I wire an led in series with the solenoid to consume the remaining 1.5V after the solenoid uses 4.5V, like you suggested?
Yes.

An LED is not a power device. A typical small LED is rated for 20 mA typical, 50 mA max current, which probably is less than the solenoid current. Better to use two 1 A rectifiers in series, like a 1N4002 through 1N4008.

ak

#### Spiderjay2

Joined Jun 24, 2017
9
Is there any reason not to just use a resistor? Also, why 2 1A diodes?

#### LesJones

Joined Jan 8, 2017
2,663
The probable reason that "weyneh" has suggested diodes instead of a resistor to drop the voltage to the solenoid is that you have given no information about the solenoid. (Current or it's resistance.) Without this information the value of the required series resistor cannot be calculated. Silicon diodes give a faily constant voltage drop of about 0.7 volts over a faily wide current range so two diodes in series would drop the voltage by about 1.4 volts.

Les.

#### Spiderjay2

Joined Jun 24, 2017
9
The probable reason that "weyneh" has suggested diodes instead of a resistor to drop the voltage to the solenoid is that you have given no information about the solenoid. (Current or it's resistance.) Without this information the value of the required series resistor cannot be calculated. Silicon diodes give a faily constant voltage drop of about 0.7 volts over a faily wide current range so two diodes in series would drop the voltage by about 1.4 volts.

So is the idea that components will only take the current that they need false? If not then why would the current rating of the led matter?

#### Picbuster

Joined Dec 2, 2013
1,008
Hi I need to wire together two motors and a solenoid actuator together but have no idea what batteries would be best or if I should wire in parallel or in a series/parallel hybrid. The motors are rated at 6V, 300ma. The solenoid is rated at 4.5 V and the current draw is unknown. They need to all run at the same time and the motors at full speed. Thank you for reading, for any answers, and for your time.
The information is not complete.
When a motor is not rotating and suddenly switched on a high start-up current will occur. It's duration is depending on the load.( this include its own rotor mass).
About 4 x the given current In practice in your case 1.2 Amp till rotation speed is reached.
We need the mass/torque needed.
Picbuster

#### Spiderjay2

Joined Jun 24, 2017
9
The information is not complete.
When a motor is not rotating and suddenly switched on a high start-up current will occur. It's duration is depending on the load.( this include its own rotor mass).
About 4 x the given current In practice in your case 1.2 Amp till rotation speed is reached.
We need the mass/torque needed.
Picbuster
The no load speed is 16,500 and the stall torque in g/cm is 28

#### wayneh

Joined Sep 9, 2010
16,398
Is there any reason not to just use a resistor? Also, why 2 1A diodes?
Each diode will provide a reliable voltage drop of ~0.7V under just about any current. The drop across a resistor depends on current, and will provide the lowest voltage to the solenoid at the instant it is calling for the most current, which is probably the instant you ask it to move. A resistor might work if you knew which value to use, but the diode will work for sure under just about any circumstance, as long as it is rated for at least the solenoid current.

#### AnalogKid

Joined Aug 1, 2013
8,487
So is the idea that components will only take the current that they need false?
No. But that is a poor way to think about what is going on, and can lead to some incorrect conclusions. For example, a 100 ohm rated for 1 W resistor can operate continuously with 10 volts across it because that is a power dissipation of 1 W. If you put 100 V across it it will burst like a small firecracker, because it is trying to dissipate 100 W

All components have impedance, and all impedance "impedes" or restricts electron flow. However, the characteristics of the impedance vary greatly with different types of components. For example, the impedance of an inductor increases as the frequency of the signal increases, while the impedance of a capacitor decreases and the impedance of a resistor is unchanged. Then come diodes, transistors, and other devices whose impedance (or effective impedance) is not nearly as simple.
If not then why would the current rating of the led matter?
The current rating of the LED comes from the power rating of the LED chip and its package, similar to the power rating of a resistor. Because an LEDs forward voltage is nearly constant across a wide current range, a more convenient way to state the part's limitations is by translating the max power rating to a max current rating (Watt's Law). As with the resistor example, if you force too much current through the LED, it will fail.

ak

#### Spiderjay2

Joined Jun 24, 2017
9
Each diode will provide a reliable voltage drop of ~0.7V under just about any current. The drop across a resistor depends on current, and will provide the lowest voltage to the solenoid at the instant it is calling for the most current, which is probably the instant you ask it to move. A resistor might work if you knew which value to use, but the diode will work for sure under just about any circumstance, as long as it is rated for at least the solenoid current.
Wait then if that is the case... Why not always use a diode instead of a resistor?

#### Spiderjay2

Joined Jun 24, 2017
9
The current rating of the LED comes from the power rating of the LED chip and its package, similar to the power rating of a resistor. Because an LEDs forward voltage is nearly constant across a wide current range, a more convenient way to state the part's limitations is by translating the max power rating to a max current rating (Watt's Law). As with the resistor example, if you force too much current through the LED, it will fail.

ak
That's the part I not getting. How does one "force" current through something if the part only uses the current the it needs?

#### wayneh

Joined Sep 9, 2010
16,398
Wait then if that is the case... Why not always use a diode instead of a resistor?
Resistors are only useful voltage regulators when the current is predictable, making the voltage also predictable. Some devices such as LEDs (while conducting), filament lightbulbs and basically resistive loads fall in this category. A resistor works as a way to drop voltage for these devices. The current flowing in many other circuits or devices is not as predictable, and using a diode provides a rock solid reliable and simple way to shave off some voltage.

#### wayneh

Joined Sep 9, 2010
16,398
That's the part I not getting. How does one "force" current through something if the part only uses the current the it needs?
Applying a voltage above the something's rated voltage may force more current through it than it can handle - like too high a pressure on a garden hose. That's the problem with LEDs: Their entire range of working voltage might be as narrow as 0.5V. Too low and it won't light at all. Less than one volt more, and you've destroyed it with excess current. A filament lightbulb will light to some degree over a much wider range.

#### AnalogKid

Joined Aug 1, 2013
8,487
That's the part I not getting. How does one "force" current through something if the part only uses the current the it needs?
You are taking what is true in a defined situation and generalizing it to other, undefined situations. All standard US light bulbs run on 110 Vac, from 25 W to 200 W. In this narrow case, each bulb "uses the current it needs" only because the voltage is defined and regulated, and the bulb's thermal characteristics are designed for that voltage (only!). But if you connect a 110 V bulb to a 220 V line in England, it will draw much more than it needs and burn out almost instantly.

Sticking to DC to keep the math simple, an Ohm's Law permutation: I = E / R. The current through a resistor is directly proportional to the voltage across it. If you design a 1 W resistor into a 5 V circuit in such a way that the power dissipated in the resistor is 1/2 W, it will run a bit warm to the touch but be very reliable for a long time. In this case, the resistor is drawing what it needs - actually, what you think it needs. But raise the circuit voltage to 12 V, and the resistor is in trouble.

Devices (light bulbs, subway cars, whatever) do not draw what they need. They draw what they are designed to draw in a specific operating environment. Those two things are not the same.

ak