How to make two independent voltage source from single 12V 2Amps power supply

Question

Well, I am cranking my head around this one. I have a 5V microcontroller (AtTiny2313a-pu) which drives a 12V(max) DC motor(ampere rating unknown).

I have programmed my uC correctly and caused the motor to "blink" for one second then turn off for one second and so on(not resorting to PWM). I tested using this using this schematic and it works. (I am using a TIP120 npn transistor)

Instead of using two batteries/ supplies, I tried to use the same battery for powering my microcontroller and Load and I am getting very strange behavior. I set the blink time to 2sec but here with this schematic(below), I observe random switching.

I have read about a phenomenon called ground bounce from the noise generated by motor that is temporarily resetting my microcontroller.

I followed the instruction from a similar post that appeared in the forum and I tried to connect the battery ground to collector first then to my microcontroller and its regulator as mentioned in the answer but I observe the same strange behavior of resetting. How would I get rid of this behavior?

Also, I have a 12V 2amps adapter and I want to power my project through this. Is there a way to split 12V into 5V(for microcontroller) and 9V for Motor as an independent power source as described in schematic 1?

I read that linear voltage regulators are very bad for powering motors,Will buck converter serve my purpose?

PS: I am very new to electronics please don't vote me down. In case I miss something I will modify in the post and abide by the standards of this forum.

Thanks you

Answer 1

"But is it possible to make two supplies from the 12V 2amps adapter so that schematic 1 like circuit may fix this issue?"

Without knowing the current draw of the motor, the answer is, "Maybe, maybe not." It all depends on how much current the motor draws, especially on turn-on.

Presumably you are interested in trying something like

^{simulate this circuit – Schematic created using CircuitLab}

Here's the problem: when motors turn on, they draw much more current than normal. As the motor starts turning and gets up to speed, the current required gradually drops to normal. When this happens, it will cause the output of the power supply to drop, and this will cause the uC voltage to drop - and you get weird behavior as the uC briefly stops working correctly.

You might think that the solution would be to put a big capacitor on the output of the 7805, to provide voltage for a brief period when there is a drop in the supply voltage. Unfortunately, that's not a good idea - 7805s don't like big capacitors on their output and may start oscillating.

Instead, you should try something like

^{simulate this circuit}

There are a few caveats.

1 - I have no idea how much current your uC circuit draws, so the value of C1 is entirely speculative. 1000 uF may be higher than you need, or it may be lower. It depends on how much current the uC draws, and how long a dropout you need to operate through.

2 - When power is turned on, D1 has to provide a high-current spike to charge up C1. Depending on what value of C1 you have, and how gradually the power supply turns on, it's possible that turn-on will blow up D1. In that case you would need a beefier diode.

3 - (optional) It's a good idea to start working with less obsolete chips. Learn to use something like an LM317 if you're willing to stick with linear regulators. At the expense of a couple more resistors, you can use a single part number for different voltages - and the performance of newer regulators is better than ancient, first-generation devices like the 7800 series.

Answer 2

Your diagrams show no capacitors on the input or output of the 7805. So it may not be working correctly. Follow the recommendation in the manufacturer's datasheet. Adding them may be enough to solve your problem.

Motors take a large current surge on start-up. How large that surge is depends on the design of the motor and what load it is driving. This can make the supply voltage drop when you try to start it. A normal 9V battery may not cope with the extra current. If using a 12V power supply, again you need to know if the current output of the power supply is enough to handle the motor surge. If you have a datasheet for the motor, see if it lists a stall current.

PS. Your diagram would be a lot clearer if you drew the motor above the transistor. And drew the transistor with the terminals labelled correctly!

Answer 3

You can't power this circuit from two independent (isolated) power supplies. You must have a common ground for this circuit to function, and once you have the grounds of two different power supplies connected, it is no different and no more independent than what you already have.

The TIP120 only turns on if there is current flowing through the base and out the emitter. For current to flow, the base voltage must share a common ground with the motor drive voltage.

To use two isolated power supplies would require something like an opto-isolator between your ATTiny and the transistor base.

Different voltage potentials with a common ground are, well, different voltage potentials with common ground. There is nothing you will gain (and indeed, it will be functionally equivalent) by any circuit you might use to derive two different voltages from a single power supply.

So yes, you can split your 12V power supply into a 9V and a 5V power supply. You would do so by using an LM7805 and LM7809.

The only way would require creating an isolated voltage source using something like a flyback converter and a transformer, but this is far more advanced than is even reasonable to build for something like this.

I promise you that your problem has nothing to do with how you're getting the 5V rail from a single power source, and has everything to do with your complete lack of any capacitance anywhere. You don't even have a ceramic capacitor for your microcontroller.

When the motor turns on, it causes a voltage sag that drives the 7805 out of regulation (the voltage drops briefly) and this triggers the brown-out detection on the ATTiny, causing it to reset itself. Additionally, the 7805 isn't even stable without input and output capacitance in the first place, so I'm amazed the circuit worked at all.

Either way, whether it is with a 9V battery or a 12V power brick, you need some bulk capacitance there to handle the large current spikes switching on the motor will demand.

That will be far far easier than any possible way to generate an isolated output from an existing power supply.

But if you really want two power supplies... just power the motor from the 12V power brick and power the microcontroller and the 7805 from a 9V battery, with their grounds joined. Batteries work as isolated power supplies in a pinch, albeit ones with limited capacity.

Another note: There is no point in powering the motor from a 12V power supply, at least with the circuit as you have it now. The TIP120 is an NPN darlington transistor, yet you're using it as a high-side switch. For base current to flow out of the emitter, the voltage at the emitter must be less than the base voltage minus 0.7V for the base voltage drop. This means you won't ever be able to turn on the transistor more than a few volts (or less) at the emitter, and that is the most the motor will see. The rest of the 12V will be dropped across the transistor, making it get very hot very fast.

You need to use NPN transistors as low-side switches. Connect the collector the motor's negative lead, and connect the emitter to ground. This will let you properly power the motor from 12V.

Final note: motors are noisy. The pull-up on the reset pin on AVRs is usually pretty weak and it is possible that noise could couple in and trigger a spurious reset. You should consider using a stronger external pull-up resistor, like 10kΩ to VCC on the reset pin (pin 1).