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Discussion in 'General Electronics Discussion' started by Roelof, Mar 7, 2014.

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  1. Roelof

    Roelof

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    Mar 7, 2014
    I want to remotely activate a pepper spray sprayer ( I have got it working, just have a few questions on the temperature of the transistors and "ringing").

    I have removed the motor and gear set of a broken "sprayer" (the kind that sprays a burst of insecticide or fragrance every say 15 min) and built a custom setup to accommodate the pepper spray canister as the dimensions differ from the OEM canisters.

    What I also have is an IP camera that I will connect to through the internet which has a relay switch built in which I can activate via an app or the web browser. So the idea is to activate the sprayer with this relay, and "protect" the camera should the thief decide to steal or try and damage the camera when he/she notices it. I then want to spray the pepper spray for a period of 3 seconds and have it off for say 5 seconds for as long as I keep the camera relay switch closed.

    To do this I implemented a simple 555 timer circuit to give me the pulses. So for as long as I keep the relay in the camera closed it will spray for 3 seconds and be off for 5 seconds. I have used variable resistors (pots) to adjust the timing of the sprays. Up to this part I am fine.

    The original circuit board would give a much shorter burst of spray, so the motor never really becomes "stationary", and as the entire system will run of battery power (12V) and the original system used 3V, I want to adjust the ampere to the motor via a transistor, and therefore to "fine tune" the amount of power the motor has. Directly via the 12V it is to powerful and will damage the gears, or damage the nozzle to the canister, and I also suspect the timing of the original circuit board was such that it would cut power just before the hammer gets "stationary" (nozzle fully depressed) so even with 3V with a duration of more than a second it will be to powerful.

    So I have fed the output from pin 3 through a transistor with a variable resistor between pin 3 and the base of the transistor, and the emitter of this transistor to the base of a larger transistor (again with a variable resistor in between). With these pots I can then adjust the amount of current going through the motor. I adjusted these pots to get about 150mA - 200mA, 12V through the motor and this works perfectly.

    However, from my understanding, this is only 1.8W - 2.4W, and the transistors gets, in my opinion way to hot, to the point that it burns when I touch it. And then another thing I don't understand is that if there is no "load" on the motor the pulses become erratic, the frequency increases a lot (to show the actual frequency I connected a LED in parallel with the motor just to confirm that it is not the motor just spinning down while the power is actually off). Why is this happening? Is this the "ringing" I have read about? This happen even with a diode connected in parallel with the motor, as I have read this combats this?

    So my actual questions are, why is the transistors getting that hot? and what causes the erratic pulses from the 555? I thought the circuits will be separated through the transistors? Is the transistors getting hot since it is not "fully turned on"?

    I know I might have gone overboard with the transistors and that I could just probably feed the motor directly from pin 3 with a current limiting pot (but the ones I have a of low wattage and I think they will probably fry with that amount of power and probably the 555 to? (200mA at 12V directly from the 555).

    I have attached a schematic of my circuit and below is the components I use:

    R1 - 47k pot
    R2 - 47k pot
    R3 - 1M pot
    R4 - 47k pot
    C1 - 220uF 25V

    tr1 - 2N3904
    tr2 - BD241C


    I have several other transistors bipolar and mosfets available for instance -
    TIP31C
    IRF740
    IRF224N
    IRF840
    BD139 - 10
    IRF630
    2N3053

    If you think I should rather use another transistor, but I think the BD241C is already an overkill considering it can handle 3A and I am only measuring 200mA for Ice?

    Does it get to hot since I am giving the base 12V instead of 5V?

    I tried this circuit to learn more about transistors so if you could explain where I am going wrong with this circuit then hopefully I will have a better understanding... :)
     

    Attached Files:

  2. BobK

    BobK

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    Have you checked with a lawyer about the legality of this? I suspect that it would be considered a booby trap, which might be illegal.

    Bob
     
  3. Roelof

    Roelof

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    Mar 7, 2014
    I live in South Africa, and have seen devices that spray pepper spray if a sensor is activated for sale. Pepper Spray is non lethal.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Your circuit is poor for a number of reasons, but I won't address those parts which seem to work.

    I imagine you have no heatsink on TR2 and that is why it is getting hot.

    To fix the mis-triggering of the 555, place a 0.1uF capacitor across the motor, another directly across the power pins of the 555 and another between pin 5 of the 555 and ground.
     
  5. Roelof

    Roelof

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    Mar 7, 2014
    No I don't have a heatsink on TR2, but for the low power passing through it, should it really need a heatsink?

    I am trying to learn the use of transistors with this circuit apart from the function I wish for it to perform.

    If you don't mind, would you please enlighten me with regards to the poor circuit design you are referring to. Keep in mind I am trying to learn.

    I will try the capacitors where you said I should, but why is this happening? I have read on a few forums about "ringing" and I suppose this causes the miss-triggering? From my understanding this only happens on high frequencies, or is this feedback from the motor (being "on and "off" at a high frequency while spinning)? Are the circuits not "separated" through the transistors? I understand there is a diode between b and e, and that should not cause current to flow "back" to the 555 and cause interference? Or is it flowing back to TR2? I am open to any constructive criticism, that way I learn
     
    Last edited: Mar 7, 2014
  6. Roelof

    Roelof

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    Mar 7, 2014
    This is how you want me to add the capacitors?

    Thank you for your assistance thus far Steve, I appreciate it, from what I have read on the forums I know you know your stuff and appreciate that I can get some of your assistance
     

    Attached Files:

  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The power rating of a transistor assumes good (in some cases *perfect*) heatsinking.

    In summary:

    1) you use a darlington where you don't need one
    2) you limit base current as a means of controlling collector current in a way that is not stable
    3) the load is in the emitter rather than the collector (probably increases the stability in this case, but generally not used)
    4) you have failed to bypass the power supply at all.
    5) you have failed to place a capacitor on pin 5 of the 555 (this stabilises the control voltage and helps prevent false triggering)

    And construction-wise, the ground wire to the motor and the power connection from the collector of Q2 should be wired directly back to the battery.

    There may be several reasons

    1) noise from the motor is upsetting the 555
    2) a voltage drop caused by the motor starting (or the rise when it stops) may drop (increase) the voltage to the 555 and cause the trigger points to change and thus the unusual timing errors.

    "ringing" bay not be the correct term and there are multiple different issues in your reasoning above. The diode across the motor acts against one if them. Yes, the motor turns on and off at least twice on each revolution and the commutators change the polarity of the voltage to the rotor. Transistors do not really "separate" things from the perspective of noise.

    You probably want to control the voltage to the motor rather than the current. (Or even better use PWM, but it sounds like that would be overkill here)
     
  8. Roelof

    Roelof

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    Mar 7, 2014
    Okay I accept that I should use a heatsink, I just thought a heatsink is "overkill" in this case as TR2 is rated as a total of 40W on the datasheet I read, and I am only using only about 2W which is 5% of the total? I thought I would have been able to get away without using a heatsink? (as I have seen on many circuits, but maybe I have mistaken those components for transistors, I have not checked their model number back then).

    With regards to the darlington pair, I agree that is not necessary, but at the time I though this might reduce the "noise" or "ringing" as I called it, and the circuit kind of stayed like that. I initially only used one transistor.

    I also thought about reducing the voltage instead of the ampere, but I was not sure how to do this other than with a voltage divider, but I did not have high enough transistors to handle the 2W, unless I coupled 8 or 10 transistors in parallel for each single resistor one would normally put in series? So it thought I have transistors that can handle the 2W, and my understanding the amount of ampere to the base determines the amount of ampere flowing from C to E (or E to C depending on how you look at it). Afterall I have to reduce the power of the motor, voltage remaining the same if I reduce the ampere I reduce the power. Is this wrong as you say I am limiting base current as a means of controlling collector current?

    Would you be so kind to show me a way you would have reduced the voltage instead of the ampere, considering the motor uses about 2W when at the desired "power" level?

    Then, I made a mistake in my circuit, the motor is currently actually connected between 12V and c of tr2, not e to 0V as on my circuit diagram... What is the difference between the two ways? Has it got to do with the Vbe and Vcb differences if the V at c is "lower" after the load (with reference to 0V) that it would be if it is before the load? So after the load the voltage "may" be 3V compared to 0V (due to voltage drop over the load) , and feeding it 12V and if it is compared to say 9V at e if the and feeding it 12V if the transistor is before the load?
     
    Last edited: Mar 8, 2014
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yep, that looks right.

    Don't equate number of posts with knowledge, but thanks anyway :)

    The thermal properties of the package mean that it gets about 35C hotter for every watt it dissipates. Let's assule it's dissipating 2W, that means a temperature rise of 70C. If the ambient temperature is 30C then the device will be at 100C -- hot enough to boil water, and easily enough to burn you.

    The issue is the power dissipated by the transistor, not the power switched by it. Yes, you can probably get away without a heatsink, as long as a transistor at 100C isn't going to melt anything.

    It won't do that, 1 transistor is plenty.

    The way you have it wired up now, you could use a voltage divider to the base of the transistor. This might actually be a reason to use a darlington pair. If I have time I'll draw it up for you.

    Give me a few minutes...

    Lots actually. connected to the emitter, the transistor has negative feedback and operates in a linear mode. Connected to the collector the transistor acts as a switch. In your case, you have worked very hard to get it to work in a linear manner even when connected as a switch.
     
  10. Roelof

    Roelof

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    Mar 7, 2014
    I edited this after you commented, and I am not sure whether you saw the change. Does the above sound correct to you? I will play around with putting the load before and after the transistor and see what happens? can I just put a lamp (I have a normal car break lamp I also use in testing to draw more current) for this purposes? and a pot to the base and adjust the resistance?
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Here is an alternate way of driving the motor using parts you already (mostly) have.

    It's not perfect, but it should be better.

    [​IMG]

    The main difference is that the base current is derived from a potential divider (R2) which you can vary to get the voltage you desire.

    The motor voltage will be about 1.2V lower than the voltage at the base of Q1.

    Because I have used your darlington pair, the current required by the base of Q1 will be small and as the current drawn by the motor varies, the voltage to it will remain fairly constant.

    The voltage will drop as the battery goes flat, but I have not considered this as I assume the battery will be connected to a charger.

    Note that the +12V power and Power ground should be separate connections back to your battery.
     

    Attached Files:

    Last edited: Mar 8, 2014
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Not sure if I noticed, but I think my answer covered it.

    It's all to do with the difference between common emitter and common collector amplifiers.

    Normally I would recommend common emitter (which is not what you drew, but what you are apparently using) unless you want to control the voltage to the load.
     
  13. Roelof

    Roelof

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    Mar 7, 2014
    Okay, I see what you have done with R2. Specifically connecting the entire resistor from pin 3 to ground and the middle pin to the base with R1. Is R1 there just to protect the transistor if R2 is turned such that it has no resistance to the base from pin 3, or is there another specific reason?

    Just to properly understand, adjusting R2, will cause a voltage difference to the base in relation to ground from say 12V to 0V, but also since the resistance is increasing, vary the current going to base? which is then adjusting the current that can flow from C to E? So it will go something like 11V at 100mA to 1V at 10mA (just to show the correlation of increase of V together with increase in A).

    If you don't mind, this brings me to another question... Vbe as defined in datasheets. For the BD241C I see that it is 5V? what does this really mean, I find conflicting explanations for this. If I supply just more than 0.7V to base and vary the current this will vary the current through CE depending on the hFE? (0.7V because of the internal diode from B to E?). Then will there be a difference in the current from C to E if I give base a 2V or 5V or 12V supply at say 1mA? in other words, since my understanding is that the bipolar transistor is a current sensing or current driven component, the actual voltage at B would not have an effect (as long as it is above 0.7V and below Vc, in this case 12V?
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yep, that's the reason.

    No, this is an emitter follower. The voltage on the emitter of Q2 will be 1.4 volts lower than the voltage on the base of Q1 (the 1.4 volts is made up of the sum of the Vbe values).

    If we assume no current flows into the base -- it will be pretty small because the combined gain of Q1 and Q2 is large -- then the base voltage is the same as the voltage on the wiper of R1.

    That means, as you adjust R1, the voltage across the motor changes.

    That is the maximum reverse voltage you can place across this junction without it breaking down. In your circuit the value will always be zero or greater.

    Yes, that's pretty much correct. It's not a magic 0.7V, the effect will start with less than 0.7V and if you force a lot of base current it can rise slightly, but for most uses it's between 0.6V and 0.7V.

    Note that this is forward biased. This is Vbe, and it is positive for an NPN transistor. The breakdown voltage specified above if positive should be Veb, or if specified as Vbe as a negative voltage. (And indeed in this datasheet it is shown as Veb)

    No, if you give 1mA of base current, the voltage from which it is derived makes no difference. You can think of the transistor as a current amplifier.

    The voltage will have a *HUGE* effect, but transistors are not used that way. The voltage at the base while the transistor is switched on varies very little from 0.7V.

    Essentially you're looking at a diode here. If you look at the voltage vs current of a diode you'll see that the current can vary by a huge amount and the voltage hardly changes.
     
  15. Roelof

    Roelof

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    Mar 7, 2014
    Sorry I meant, there will be little difference in the current from C to E whether the voltage supplied to the base is 2V, 5V, or 12V, as you have corrected me in my statement. But then I don't understand why you say the voltage will have a huge effect, or is it only if you use the transistor in some other way (not the way I am using it or how one normally use a transistor).

    Then to just clarify on the Vbe. I did not notice that it said Veb (devil is in the detail...), but my understanding is that Veb +5V would also mean Vbe -5V (as you have stated), and that would only happen if you also have a negative voltage rail (for instance 2 batteries connected in series with the "open positive" connected to C, E connected to the middle (where the pos and neg of the batteries connect) and B to the "open negative"?


    Thank you so much Steve, I have really learned a lot with this session combined with the prior reading I have done, and I will go and read these stuff again and I am sure this will help me understand them much better. I will go and test your circuit to see and measure the voltage and ampere and different spots to see what is happening and then perhaps return with some more questions.

    Just one last one for now... I was starving the motor of ampere as the TR2 simply did not allow more current to flow while the V of the motor was 12V? (in effect increased the resistance of the motor?) and the way you proposed, which I like, is reducing the voltage over the motor causing the motor to just draw less current from an "infinite" source of current?
     
  16. Roelof

    Roelof

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    Mar 7, 2014
    I have quickly tested your circuit as a emitter follower, and it seems like this is a very efficient voltage divider?

    Since from what I can see the only real "loss" is the current flowing through the entire resistor of R2, so if you make this a 100k pot (even 1M) then you only get current of about 100uA? and considering that you can have several amps through the last transistor in the darlington pair, it is very efficient? Since the current flowing through Ibe of the first transistor and Ibe of the second transistor in the darlington pair would end up flowing through your load? (And perhaps some power lost through heat of the transistors, on which I will put a heat sink... ;) ) I hope I am not wrong... and if I am not I don't know why I haven't realised that before...
     
  17. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Normally you apply a voltage to a resistor connected to the base. The resistor drops all but about 0.7V and this determined the current.

    So, no matter what voltage you use, if you pick a suitable value resistor to set the current to the same value, the results are the same.

    With the same value resistor. the current will vary almost linearly with voltage, so twice the voltage is close to twice the current.

    OK, the preceeding also assumes the emitter is grounded.

    If, however, you were to place a voltage source on the base directly, the current would be determined by the V-I characteristics of the emitter junction. That might be almost zero up to 0.4V, then rise through a couple of uA to 0.5V, then to a couple of mA at 0.65V. I have seen simulations where the magnitude of the current at (say) 5V is GA (that's 1,000,000,000's of amps). Clearly not linear, and clearly not going to happen (the transistor will emit a puff of smoke first) .

    There are other ways it could happen, but you normally take steps to ensure it won't.

    If you've seen the two transistor multivibrator circuit, with a 9V power supply the base emitter junction can easily be reverse biased by 8V or so. That's due to the capacitors, but inductors (and your motor contains one of these) can also do the same thing.

    No problems, happy to help.

    Good to hear you're going to keep investigating.

    The voltage and the current are related. limiting the current will typically happen by reducing the voltage and limiting the voltage will typically be done by reducing the current.

    The difference is that a circuit with a limited current will see a fall in voltage (and hence power) as more current is demanded (and not supplied), whereas with a limited voltage as more current is demanded it is allowed to be supplied in order to maintain the voltage.

    If you think that the speed of the motor is determined by the voltage and the torque by the current, it is better to slow the motor by controlling the voltage than by limiting the current.

    It's actually more complex than that, so don't take the analogy too far. The main thing is that with a limited voltage, as the motor demands more power, the power to it increases rather than decreases.

    This is not a perfect voltage source, and as the motor demands more current the voltage will fall slightly.
     
  18. Roelof

    Roelof

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    Mar 7, 2014
    I think I understand that now, the resistor drops all the voltage but the 0.7V (which in an ideal world with perfect components) is the voltage drop over a diode (which in this case is the diode connected internally between B and E) .... the same as one resistor will have a voltage drop of say 9V of connected directly over the poles of a 9V battery... (2 of the same resistance will drop 4.5V each) which comes from kirch..someone's law. but in this instance the diode always drops 0.7V (irrespective of the total supply voltage and the rating of the resistor).

    With regards to directly applying voltage to base:
    Then if you apply a voltage directly to base, like you said at 0.5V the diode will "leak" tiny tiny amounts of current at 0.5V increasing as the voltage goes up to "0.7V" (because it is not the ideal world) where it will "fully" allow current to flow, and this tiny leak would be the same as if we limited the supply voltage with the correct resistor calculated through ohm's law for instance.


    And with regards to your last explanation, it is still the voltage that drop due to the limiting current, the resistance remained the constant value with regards to ohm's law. I cant remember actually testing the voltage over the motor but I will do that to see it for myself and better understand this before altering my circuit to test the voltage divider circuit you proposed...

    And will you please just quickly comment on my post about the efficiency of your circuit as a voltage divider as opposed to 2 resistors in series and then taking the output from between the 2 resistors to your lower voltage load.

    You have opened my eyes, feels like I had a few light bulb going on moments (the little pun in there was intended...).
     
  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yeah, pretty much. Over the normal range of currents you can often assume it's constant.

    Yep. There is an equation that gives you the current at any voltage. See here. But typically you'd just look at the datasheet.

    A motor doesn't have a fixed resistance (or appears not to). While the motor is turning it generates a voltage in opposition to the voltage applied to it. This means that it seems to have a higher resistance. As you slow the motor (increase load) the induces voltage drops and the current rises. So you can't treat it like a simple resistance.

    Yes, there is some power lost in the voltage divider in my circuit. However the vast bulk of the inefficiency comes from the use of what is essentially a linear regulator to reduce the power going to the motor. A highly efficient way to control the motor speed would be to use PWM, but that would require a lot more complex circuit. It doesn't seem like this circuit warrants that.
     
  20. Roelof

    Roelof

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    Mar 7, 2014
    Yes I know the motor does not have a fixed resistance per se, for the sake of the argument, if we replaced the motor with say a 10 ohm resistor, we would see a voltage drop, or if I were to hold the motor stationary the resistance would "settle" and cause a voltage drop if we were limiting the current available (200mA).

    I am thinking of using the circuit you showed me at a different scenario, or even in the current project. The IP camera requires 5V, at idle it draws 0.5A, but with the PTZ actions going on the max is 1.7A, I could have used this circuit as a fairly efficient way to supply the camera with power from a 12V battery?. Just to give some background, this will be used on a farm without mains electricity. The electricity we use is from solar panels into some battery banks. Power therefore is "costly" as it is not freely available. (I have wired the lights to the 12V battery and we use 12V energy saving bulbs to light the house (not permanently occupied) and we have an inverter which I have connected a plug to the existing plug wires that runs through the house, and this I plug into the inverter which is only really on when it is required, which at this stage is almost never as we charge cell phones through car chargers connected to the battery as well as a 12V laptop charger which I have purchased). So the whole system will run off these batteries. For the 5V required I purchased a 2A car battery charger, so that is pretty much sorted already, my question is just this would also been a very effective way to get that 5V? compared to 2 resistors of a suitable wattage considering they would have had to be round about 6ohm give or take to still be able to supply 2A at the 5V (or 6 if they were of similar resistance) as you would have had a constant waste of 2A @ 12V through the last resistor?

    I have tested your circuit replacing pin 3's input with the normal 12V line just to see what happens, and without the lightbulb even getting really bright (its a 12V brake light bulb of a car) the transistor got to hot to touch, so I am happy given your circuit that the transistor was supposed to get that hot without a heat sink, I was worried that somehow I shorted something that caused the transistor to heat up really quickly, but you have answered that question now.

    (((Just on a side note, when I rewired (or basically just connected the light wiring to the batteries) I found an old incandescent light bulb in one of the fixtures rated at 36V... We purchased the farm about 7 years ago and before that we estimated the last time anyone really used the house (physically lived on the farm) was about 30 years ago, so that was the first time in 30 years a light went on in that house and electricity was flowing....)))
     
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