Not getting full power from transistor when small motor is connected to circuit

Discussion in 'The Projects Forum' started by summersab, Apr 27, 2016.

  1. summersab

    Thread Starter Active Member

    Apr 8, 2010
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    So, I created a PWM-based ramp-up circuit after getting help on a previous forum:
    http://forum.allaboutcircuits.com/t...tor-ramp-circuit-still-have-questions.122348/

    This is my final circuit:
    http://goo.gl/jjHcfN

    I'm wanting to have a small motor connected to the emitter of a TIP3055 NPN transistor and to ground so that it will turn on when the circuit is active. The motor is very small, and based on my little multimeter, it is ~1.6ohm (not kilo, just ohm). When the motor isn't connected, the ramp-up circuit behaves as expected, and the voltage goes from zero to full when measured between the emitter and ground. However, when the motor is connected, the emitter voltage drops, and the motor barely spins. What am I doing wrong? Am I using the wrong type of transistor, or have I wired it improperly? Admittedly, I'm not very good with transistors...

    Thanks!
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    I'm guessing that the circuit cannot supply the required current for such a low impedance motor. BTW if the supply is 3.7V the motor will require 2.31 Amperes.
     
  3. summersab

    Thread Starter Active Member

    Apr 8, 2010
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    Hmm... I'm using some pretty stout Sanyo 14500 lithium ion batteries, and they most certainly power the motor just fine if I connect it directly to a single cell. I know the transistor I picked is overkill as far as the power rating (90W, 15A), but I wasn't quite sure what parts I'd be using, so I figured I'd go for high-power and then scale down once everything works. Since everything should be rated high enough to power this motor, what could be going on? Do I need to add resistors or maybe a diode to the emitter? Let me know if I need to take some readings, and I'll gladly post back.

    Thanks for the help!
     
  4. dannyf

    Well-Known Member

    Sep 13, 2015
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    Unless you have a particular reason, that kind of connection makes little sense for a switch.
     
  5. Papabravo

    Expert

    Feb 24, 2006
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    In order to get 2.31 Amperes of collector current, how much base current do you need?
     
  6. Dyslexicbloke

    Active Member

    Sep 4, 2010
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    I am no expert but given that a BJT is a current based device, IE the EC current is proportional to the BE current X gain, you are going to have problems with this arrangement because the voltage drop across the motor will directly affect VBE and thus IB.

    You would be better with an open collector arrangement... That way VBE is constant so IB is fixed by its limiting resistor, assuming a consistent output from the amp.
    IC will be IB X gain and IE will be IB +IC, amusing the voltage is sufficient to drive it/them.
    Obviously if the motor is the limiting factor RE current then IC is Unknown, which I think is saturation, but I could well be wrong about the terminology.
    That said open collector outputs are common as muck and used everywhere, or at least they were until FET's took over for most low voltage switching applications.
    Essentially the driver circuit just sets IB, completely independent of the load, and IC is what it is, dependent on the load, because V load is always VCC - VEC(Min)

    Either ground E and connect the motor from VCC to C or invert the amp and use an NPN. with the motor where it is.
    You would also be better with a packaged darlington power transistor which will have a huge gain and thus turn on faster. (I think)
    There is a load of maths here ...
    http://www.ittc.ku.edu/~jstiles/312/handouts/Example A BJT Circuit in Saturation.pdf
    My head hurts,

    If it were me, because I never learned the maths or even proper transistor theory, I would just test it...
    Check the data sheet for IB max and then dont exceed it whilst doing the following.
    Have a reasonable guess at RB - Run your motor whilst observing VEC - Adjust RB until it is just slightly smaller than the point at which making it smaller stops reducing VEC.

    You now have the minimum current you can apply to the base that achieves the minimum forward voltage drop achievable.

    (If this methodology is flawed I would welcome being corrected, because I am doing it wrong)

    Hope this helps.
     
  7. summersab

    Thread Starter Active Member

    Apr 8, 2010
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    Danny, I'm using a transistor because I'm not using it as a switch. The base is receiving a linear increasing output from a ramp circuit. I'm wanting the motor to turn on using a ramp instead of coming on suddenly.

    Papabravo, this sounds like a pop quiz... should I assume 2.31A is the answer? If so... then what's the point of using a transistor? Is there a different component that I should be using? I can't power the motor straight from the PWM/linear ramp because it doesn't have enough current, so I thought I could deliver this current by using a transistor connected to a full-power source.

    Holy crap, Dyslexic, that's a lot. My understanding of transistors is pretty crap, and I usually do the plug and test method, as well. I'm going to wait for a few more responses and mull over what you wrote...
     
  8. Papabravo

    Expert

    Feb 24, 2006
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    You have identified your problem, which is that you don't understand transistors very well. A pop quiz is a pedagogical exercise for students who have studied a body of material. Since you have not done that, the reference is not appropriate. It was a question meant to guide your inquiry into how a transistor works as a current amplifier. The whole purpose of using a transistor is to use a small current to control a larger current. So no, 2.31 Amperes in not the required base current. The required current should be smaller, and the question for you is how much smaller. In the responses so far you have received other hints about what to do. You can buckle down and solve the problem with reading and research or you can continue to whine and maybe someone will take pity on you. I have no idea how this will turn out.

    Here is an online article that you might derive some benefit from:
    https://en.wikipedia.org/wiki/Bipolar_junction_transistor
    http://www.electronics-tutorials.ws/transistor/tran_1.html
    Did I mention that this site has an electronic textbook associated with it?
    http://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/bipolar-junction-transistors-bjt/

    Don't forget to avail yourself of the references if you have access to a suitable library.
     
  9. Dyslexicbloke

    Active Member

    Sep 4, 2010
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    If you are trying to control speed you would be MUCH better using the PWM directly, via a suitable switch, a FET being the best choice with respect to losses but you don't have enough voltage to drive a FET gate.

    The motor will behave, almost, as it it was being supplied the average of the voltage pulses it gets so 50% duty cycle would give a similar result to supplying 1/2 the peek voltage of the pulse. In practice the speed is roughly what you would expect from reduced volyage but the torque curve is better so applying shaft load will have less effect on a pulsed motor than one with a simple reduced voltage.

    Interestingly if you measure the average voltage, with a low pass filter, you can get a pretty good estimate of the speed when you compare the expected voltage, based on duty cycle and motor impedance, to what you are actually measuring.
    Again in practice I wouldn't calculate this I would measure at no load or normal load, whichever is applicable, and then use the gathered data to construct a polynomial curve or even a lookup table that will tell you if you are running at the speed you expect or not.
    Its not very precise at at very high loads and isn't linear but it is consistent and therefore can be calibrated with a little effort.

    Anyway I digress...
    You could use a transistor to switch PWM pulses and at 3.7 V it is probably the only option. (read on)
    However a transistor is going to drop 0.8V or more which is about 20% of your supply and of course will dissipate some, possibly lots, of power/heat. Since I dont know the current I cant tell you the Watts dissipated but if you measure the current, just the motor and battery, the wasted power will be IX0.8
    By the way you cant measure a motors impedance like that to calculate current. (More later)

    A FET has a tiny on resistance, miliOhms and is relatively easy to drive. The low on resistance makes the voltage drop and thus power dissipated negligible, at least it is in this case.
    The issue here is voltage because you would need to apply somewhere between 10 and12 V to the gate of most FETs to turn them fully on.
    Of course you could just put a little 9V cell above your 3.7V rail and the problem would go away and if you dont want two batteries then an isolated voltage converter chip/module would also work. There are FET driver chips called Bootstrap drivers that have voltage shifting and gate isolation built in but I have never used one so I cant comment further.

    When a motor is spinning it generates a voltage which opposes the supply limiting the current.
    If you lock the rotor of you DC motor it will behave like a resistor amusing you only supply it with a constant CD voltage so the current would be very high, usually high enough to damage the motor very quickly if the voltage is its normal operating voltage.
    As it starts to spin it generates a voltage, back EMF, which limits the current thus limiting the torque and slowing its acceleration.
    at some point it balance's out and the speed stabilizes.
    When you apply a load, negative shaft torque, the motor will slow a little, which allows the current to increase again until a new balance is reached with a higher torque, higher current but a lower speed.
    This is why a loaded motor takes more current than an unloaded one and why DC resistance is not much use at all as a predictor of performance. This is true of any rotating magnetic machine, All be it in different guises.

    Al
     
  10. MaxHeadRoom

    Expert

    Jul 18, 2013
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    This will be the current at a stall, when running, the generated BEMF will reduce this current to a very low value, the current will only increase according to load. But never see 2.3 amps unless stalled at full voltage.
    I recently did an empirical test on a 2hp DC 90vdc T.M. motor and with no load the current was never higher than 2 amps across the rpm range from 0 to 90v.
    Motor rated current 18 amps
    When loaded the BEMF is lower in relation to the applied voltage, hence the current increases proportionately.
    Max.
     
    Last edited: Apr 27, 2016
  11. summersab

    Thread Starter Active Member

    Apr 8, 2010
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    Wow, um... that was incredibly rude. I don't recall ever whining, thank you. If I may, I do have a degree in mechanical engineering. I took classes in EE, but I was clearly not very good at them. I have looked at those links at one time or another, but beyond simplifying resistors and logic gates, the nomenclature is above my comprehension. I was under the impression that forums were where you go for help when you don't understand something. If I may take a second guess at answering your question, it would be "saturation current." The transistor appears to saturate just fine when the motor isn't attached, but then it drops to under a volt with the motor attached.

    Al,

    FETs, eh? I've never really known what they do. I know that they're related to transistors, but I've never messed with them. I'll go dive into reading up on them.

    As for some of your other points, the circuit that I'm using comes from here:
    http://www.edn.com/design/analog/4314544/Integrator-ramps-up-down-holds-output-level (article)
    http://m.eet.com/media/1126960/12071-figure.pdf (actual circuit image)

    It includes a PWM, but I'm only using the ramp portion of the circuit. I originally tried some other circuits online, but they were either designed to power a motor with a PWM and manually vary the speed with a potentiometer (I don't want fixed speed - I want it to ramp up and latch) or they would ramp up, drop back to 0V, and repeat (sawtooth pulses). I found this, and for the most part, it works.

    Lastly, I connected the motor to an old bench-top power supply that I have, and provided that it is working right (fingers crossed), the motor pulled 2.2A at 5V. So, 2.27ohm? Still seems really low...
     
  12. MaxHeadRoom

    Expert

    Jul 18, 2013
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    You cannot reliably measure the armature resistance like that, the recognized way is to apply a small DC voltage with a locked rotor and measure the current, test at more than one spot on the armature and take the lowest reading.
    This resistance only decides the current at stall, not when rotating.
    Max.
     
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  13. summersab

    Thread Starter Active Member

    Apr 8, 2010
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    Are you a motor expert...?
     
  14. MaxHeadRoom

    Expert

    Jul 18, 2013
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    Apart from initial electrical/electronic instruction.
    Just been working with all types for over 50yrs so some has rubbed off.
    AC, Universal, DC servo, BLDC, ECM, AC synchronous to name a few.
    And their drives.;)
    Max.
     
    Last edited: Apr 28, 2016
  15. Papabravo

    Expert

    Feb 24, 2006
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    From post #1:
    What am I doing wrong? Am I using the wrong type of transistor, or have I wired it improperly? Admittedly, I'm not very good with transistors...
    From post #3
    ...what could be going on? Do I need to add resistors or maybe a diode to the emitter?
    OK so you can't figure out what is going on with transistors by traditional means. Here's the executive summary in a nutshell.
    The configuration you are using where the load is connected between the emitter terminal and ground is called a "common-collector" or "emitter-follower" connection. This circuit has no voltage gain, in other words the emitter "follows" what ever signal appears at the base. It may be that this configuration will never operate the way you want it to for the reasons detailed in post #6.

    http://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/common-collector-amplifier/

    You can learn what you need to know or you can keep throwing components at the wall to see what sticks. It is completely up to you and we can't shove this stuff at you if you won't make even a minimal effort. You may call it rude, but it would be worse to string you along thinking you understood something that you don't.

     
  16. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Here's a simplified simulation which shows motor current for Motor1 in the common-collector arrangement which you are using, and for Motor2 in a common-emitter configuration which has been suggested above.
    RampUp.PNG
     
  17. summersab

    Thread Starter Active Member

    Apr 8, 2010
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    Ah, that makes sense! I kinda glazed over Dyslexic's post, but I get it, now. This diagram is a better approximation of what my circuit looks like:
    http://goo.gl/3Lq83z

    I have a 12V boost converter that I'm using to drive a small timer, and I decided to connect that to the ramp circuit in hopes that it would excite the base enough to saturate the transistor and give me full power. I understand that I'm going to see a voltage loss from using a transistor, but with the motor connected, the circuit still doesn't seem to be getting anywhere near 80% - I'm getting 50% at best. Would using a Darlington get me closer to full power?

    Also, FETs were mentioned, and I wondered if I should pursue that instead since I have 12V available. With a FET, would I wire it in the same manner as the NPN (common-emiter configuration)?

    So, my ramp circuit gives a linear voltage output when the motor isn't connected and ramps up to 3.7V (measured at the transistor base). With the motor connected, the base never reaches full voltage. My reasoning for "throwing components at the wall" was that perhaps current was flowing the wrong direction somehow, so I tried adding diodes - perhaps not the most electronically sound solution, but that was my logic. As for adding resistors, I'm never quite sure what resistors to add to a transistor. I've looked at some of the online calculators, but either a) they each ask for different values than the last calculator and seemingly have no consistency, or b) they ask for values using different nomenclature than what I can find in spec sheets, and they don't define the variables they're asking for. So, give me a little credit, here - I'm trying.
     
  18. Papabravo

    Expert

    Feb 24, 2006
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    I believe you, and the light usually does not go on overnight. You keep chipping away at it bit by bit and eventually you come around to an intuitive understanding. The exercise that helped me the most was setting up a test jig to characterizes an individual vacuum tube (transistor) and then using the data to plot a set of characteristic curves. Taking the data was tedious and plotting graphs by hand was even more so, but in the end I've never been confused about how a device operates.
     
  19. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Schematic from first post.
    What op amp are you using?
    What signal is on the + of the lower op amp??
    Here is what I suspect:
    In the emitter follower arrangement you loose 0.7 volts. So if your battery is 3.7 volts the highest the voltage can be across the motor is 3 volts.
    But...... Many op amps don't turn on all the way to the positive supply so it may be much less.
     
  20. summersab

    Thread Starter Active Member

    Apr 8, 2010
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    Yeah, as for me, the project that taught me the most was building a 4-bit calculator using an Alteration Quartus PLD. It could add, subtract, multiply, divide, AND, OR, XOR, etc. I could Karnaugh map the s#!% out of things... but that's digital electronics - this is analog, and I only got through Circuits II.

    I guess I didn't post the article where I found this circuit. It lists out a lot of info:

    http://www.edn.com/design/analog/4314544/Integrator-ramps-up-down-holds-output-level (article)
    http://m.eet.com/media/1126960/12071-figure.pdf (actual circuit image)

    I'm using the recommended LMC6484 which is rail-to-rail. The thing that gets me is that the voltage WILL ramp up linearly to max when the motor is removed, but when the motor is added, it only goes to 1.8V or so (measured from ground to base/output of the ramp circuit). It's as if I've somehow added a resistor to create a voltage divider. Shouldn't the collector and emitter be isolated from the base? Why is this impacting the circuit like this? That's why I tried randomly throwing diodes at the problem, but no joy.
     
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