Power supply problems (too many amps)

[yuri1993]

I'm new to electrical so everyday I'm learning something new, I recently switched from using motor drivers to using transistors to controlling my motors and through this I think I've found a problem with my power supply.

My power supply is 13.5v dc and 1.0 amp but when i put it through my transistor set up I only get about 180ma (and the drill doesn't run). When I connect the power supply directly to my motor I get a reading of about 3 amps from my power supply from the voltimeter.


I'm at a lost as to why my power supply is giving any less or anymore than 1amp of current, is it a bad power supply?

here is my schematic everything is the same except it is a pi connected to gpio 18 instead of an arduino, the diode is a 1N4001

[redhawk]

What happens when you short C and E of your transistor, as well as B and C ??

Richard S.

[karlkiste]

The power supply tries it's best to keep the output at 13.5V. It gives the current it is asked for. If the motor asks for 3A, 3A will be running until the power supply switches off or dies. The power supply could reduce the current by reducing the voltage, but that's not its job. The supply is fully OK, based on what you're writing.

[FLYFISH TECHNOLOGIES]

Hi,

I'm at a lost as to why my power supply is giving any less or anymore than 1amp of current, is it a bad power supply?

Information about Current for your Power supply is the maximum current the device is capable to supply (in the long term and within specification). The number means that your device was designed to provide up to 1A.

When the current is below 1A, you're in the "safe zone", when it is above this value, then it is overloaded. The result of overloead is one (or all) of the following: the voltage is not stable anymore, it typically drops and the device temperature increases, which can burn your power supply.

Current parameter is something what could be compared with car's maximum carriage weight specification - you can always drive car loaded below this maximum capacity value, but it is not a good idea to put more load on it then specified by a manufacturer.

(What we're discussing here are power supplies, which are voltage generators - they try to keep constant voltage.
Current generators also exist, which try to maintain constant current, but they are out of scope here.)


Best wishes, Ivan Zilic.

[yuri1993]

Hi,

Information about Current for your Power supply is the maximum current the device is capable to supply (in the long term and within specification). The number means that your device was designed to provide up to 1A.

When the current is below 1A, you're in the "safe zone", when it is above this value, then it is overloaded. The result of overloead is one (or all) of the following: the voltage is not stable anymore, it typically drops and the device temperature increases, which can burn your power supply.

Current parameter is something what could be compared with car's maximum carriage weight specification - you can always drive car loaded below this maximum capacity value, but it is not a good idea to put more load on it then specified by a manufacturer.

(What we're discussing here are power supplies, which are voltage generators - they try to keep constant voltage.
Current generators also exist, which try to maintain constant current, but they are out of scope here.)


Best wishes, Ivan Zilic.

Ok that makes sense but then why does it choose to provide only 180ma when using transistor set up. Also if i wanted to decrease the amount of amps being drawn by the motor should I add some resistors?

[FLYFISH TECHNOLOGIES]

Hi,

but then why does it choose to provide only 180ma when using transistor set up.

I'd need to analyze the circuit and measure signals... I also don't know anything about software commands driving the transistor.
Therefore, I can write just a very generic answer (and risking, that it does not apply):
Transistor is here in the configuration acting as a "current valve". It obviously is controlled in a way that this valve is not fully open, so you measure only 180mA... Possible reason is that there it is controlled with PWM signal, but let's forget about this now.

Also if i wanted to decrease the amount of amps being drawn by the motor should I add some resistors?

Yes,... but not where you're intending to.
Adding resistors in series to motor is not a good idea. The reason is that you will waste a lot of power on this resistor, it will also get hot, and probably got quickly danaged due to improper power rating (each resistor have its maximum power dissipation capability).
The proper location to put resistor is (in your case) in the control gate of the transistor. By increasing/decreasing the resistor you control how much this "current valve" is opened. This control is not linear (so decreasing the resistor value by 2 will almost newer double the current).

Nevertheless, my warm advice: Start learning these electronics fundamentals with basic circuits, forget about arduino and other boards with processors - take just "simple" elements - resistors, transistors, motors, power supply, LED diode, capacitors. Start learning what is the purpose of each, what their characteristics (eg. resistance) represent, then verify understanding by experiments this on this basic set.


Best wishes, Ivan Zilic.

[Tage]

I'm new to electrical so everyday I'm learning something new, I recently switched from using motor drivers to using transistors to controlling my motors and through this I think I've found a problem with my power supply.

My power supply is 13.5v dc and 1.0 amp but when i put it through my transistor set up I only get about 180ma (and the drill doesn't run). When I connect the power supply directly to my motor I get a reading of about 3 amps from my power supply from the voltimeter.


I'm at a lost as to why my power supply is giving any less or anymore than 1amp of current, is it a bad power supply?

here is my schematic everything is the same except it is a pi connected to gpio 18 instead of an arduino, the diode is a 1N4001

there are several explanations.
if you have a 1A power supply and you manage to get it to supply 3A when you connect your motor to it, you should check the output voltage.
perhaps the motor is not intended to run from 13V. make sure you do not apply more voltage than it is designed for.

the TIP120 or any Darlington is not suitable to use with the Pi that has only 3.3V and very weak GPIO outputs.
even if you supply the gate with 12mA you will typically have 2V across the Darlington at 3A, whcih means it will overheat quickly. and you can only get a very low current from you GPIO. you probably have much less than 3.3V at the GPIO if you try to drive this Darlington, which also has a high voltage drop on the base-emitter.
it is unlikely you can drive enough base current into the TIP120 to run this motor, and if you could, the transistor will overheat.

you need to use a n-channel MOSFET that has very low threshold voltage. the best thing is to get one that has Rdson specified at 2.5V. then you can drive the motor without overheating the transistor.

[emgi]

The output signal from the Pi might not be enough to directly steer the TIP120 transistor fully into saturation.
Explanation: GPIO out ~3V (at least less than 3.3) The treshold voltage for the transistor is ~0.6-0.7V.
According to Ohm's Law, the 2k2 resistor will limit the current to I=U/R = (3 - 0.6)/2.2 equals approximately 1mA.
Power transistors in this configuration do not have a very high amplification factor, 1mA input is simply not enough to switch a 1 A output current.

You need to add an extra driver stage to steer enough current into the base of the transistor.

Edit: I quickly looked at the specs for GPIO and this confirms my suspicion:
http://www.mosaic-industries.com/embedd ... ifications

The limit per port is 16mA but preferrably you should draw less to prevent damaging the CPU.
Even 16mA will still not suffice for the TIP120 so a driver stage is definitely recommended.
Typically this could be an general purpose NPN transistor (BC107, 2N3904, 2N2222, ...) steering a PNP transistor (BC556, 2N2907, ...) which has its emitter connected to the 12V line. The collector of that PNP transistor can then be used to steer a decent amount of current through the TIP120.
That should do the trick!

/emgi

[grahamed]

Hi

The TIP120 is a Darlington - essentially two transistors in series.

Referring to http://www.fairchildsemi.com/ds/TI/TIP120.pdf throughout.

Typically a transistor needs VBE something between 0V7 and 1V to turn it on. Each transistor in the Darlington needs the same so its base must rise to something around 2V for saturation (saturation means it is turned fully on which is what you want) [see VBE(ON) line in Electrical Characteristics, gives VBE=2V5 but that's for IC=3A].

The Pi GPIO high output under load is a bit less than 3V3 - call it 3V to allow for the output resistance of the pin - and the required base voltage is 2V (or a bit more), and the required base current is about 4mA [see test conditions in VCE(SAT) line in Electrical Characteristics]. So Ohm's Law says you need a resistor of (3-2)/4 k or 250R. A Darlington is not a happy match to any 3V3 device - 3V is rather too close to 2V..

Use a meter to check these voltages - bear in mind that the datasheet gives typical values - YMMV.

Does the Arduino example work? Well the Arduino puts out 5V so the base resistor has 3V across it. For 4mA that gives 750R. The circuit shows 2k2 so I guess someone thinks the gain is rather closer to the quoted hFE of 1000. But then hFE is the small- signal forward gain and a transistor in saturation is hardly in small-signal operation.

The TIP120 can disipate 2W max with no heatsink at an ambient 25C [see PC, Ta=25C line in Absolute Maximum Ratings] so 1A and 2V is pushing it a bit - fit a smallish heat-sink perhaps? The dissipation is high due to the VCE of 2V - typical for a Darlington - a single transistor might pull its collector down to 100mV and dissipation at 1A would be 0W1.

Use a logic level FET? Sure why not, just make sure you choose one specified for 3V on the gate.

Bear in mind I usually manage to get something wrong - no doubt I'll be told.

OK I just noticed that the required motor current is 3A! So it's all a good bit worse that I said and the base resistor needs to be 40R!