Op-amp integrator

Discussion in 'General Electronics Chat' started by newbies_hobbyist, Jun 18, 2010.

  1. newbies_hobbyist

    Thread Starter Member

    Jun 4, 2010
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    Hi Experts,

    I have question for op-amp integrator. What will happen if I have same input at the +/- input pins with same magnitude and same phase (2.5V square wave), will it amplify the output? Well you can take a look in the circuit I attached here. I want to know also if my understanding in this particular circuit is right.

    Thanks
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For my this circuit look like simple voltage regulator with digital adjustable voltages.
     
  3. Bychon

    Member

    Mar 12, 2010
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    The definition of an op-amp is that it wants its inputs to be the same. If you are already providing that, the op-amp output will do nothing or wander about aimlessly, not amplify the output.

    As for the circuit...so many mistakes...so little time.

    Q1 starts as always on. It's "most of 12 volts output" will be divided down to about 1.5 volts on the - input of the op-amp. Before the switch closes, the op-amp will be putting most of -12V on the base of Q1, overcoming is breakdown voltage and sending several volts negative to the voltage divider. The voltage on the -input will head for below zero and the op-amp will flip to a positive output, slowed down by the integrating capacitor. This should eventually balance out with 0 on both inputs. Then the switch closes and the 2.5V will drive the op-amp positive. It will turn on Q1, sending more positive to the voltage divider, which is dividing the voltage so much that the - input will never achieve 2.5 volts and the op-amp will stick in the high condition until the switch is opened.

    If the switch switches quicker than I can type, you might achieve a steady state, but I don't know what it will be.
     
  4. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    I got a bad pain in the neck, so I gave up on the circuit
     
  5. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    It looks like a PWM regulated powersupply.
    The duty cycle of the PWM may not go over 60 %, otherwise the output voltage will clip and be unregulated.

    I have taken the drawing from the PDF so others can look at it directly.

    [​IMG]

    Bertus
     
    newbies_hobbyist likes this.
  6. newbies_hobbyist

    Thread Starter Member

    Jun 4, 2010
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    Bychon and jony130 thanks for your good input I thought the same of this circuit how it is working now I am a bit confident. Well it is true that at the beginning the Q1 is always on as it is bias by base-col resistor so it will give 1.5V drop at -in then op-amp out will rail in -Vs (steady). As soon as the clock started the switch will continously work (gnd/2.5V) which will go to the +in of op-amp and op-amp out will go from -Vs (if +in is gnd) and +Vs (while +in is 2.5Vref). Well the clock runs in 8Hz only so I think the transistor will switch 8times in a second. But with this circuit my transistor is always getting damage like the b-e is always open. Now I'm thinking that the voltage from the base is varying from -Vs and 12V so this could be the reason why it is always getting open. When I checked the data sheet of 2n2222, max absolute rating of B-E is 6V. This transistor is working for 24 hrs. Correct me if my analysis is wrong
     
  7. newbies_hobbyist

    Thread Starter Member

    Jun 4, 2010
    67
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    Bertus,
    Yes it is like a PWM that is regulating only 10V square wave and duty cycle is 50%. Thanks for a good help and posting the drawing
     
  8. newbies_hobbyist

    Thread Starter Member

    Jun 4, 2010
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    Hi Experts,

    I still need your advise regarding my question. Well another question I have is about the data sheet of 2n2222a it says that Vbe max absolute rating is 6V, what does it mean? In the circuit im sending more than 6V to the base of Q1 like the 12V does it mean I'm putting my Q1 to damage because I exceeded the absolute max rating? But when I look in the web for other circuit of transistor they are sending 10V to 20V in the base. I don't know if I'm correct on what I' thinking on how it gets damage.
    1. EOS (exceeded the abs. max rating)
    2. Over used, since this circuit works in high temperature for more than 12hrs

    Thanks,
     
  9. Darren Holdstock

    Active Member

    Feb 10, 2009
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    Agreed, it's a PWM-controlled voltage regulator. Specifically, it's a PID loop, but without the P or the D. Employing just the integral term means that the steady-state DC accuracy will be good, but the transient response will be appalling. OK for an unvarying constant load though, if that's what you've got.

    Without reading the transistor datasheet, I'm guessing that the maximum Vbe spec is *minus* 6 V. Vbe is the voltage at the base with the emitter as a reference; with a positive current flowing into an NPN transistor base Vbe is always positive, usually in the region of 0.5 to 0.7 V (maybe up to about 1 V in a power transistor if the base current is very high), but if for some reason the voltage at the base goes negative relative to the emitter then you'll get away with it up to about -5 V or so, but beyond that a parasitic zener effect kicks in and if the base current drive is high enough the transistor is destroyed. If the reverse-biased base current is limited (to a few microamps say) then Vbe can go a bit beyond the spec without total destruction, but the small-signal gain of the device is permanently lowered. If you're not relying on the small-signal gain then that's OK.

    The OP circuit could easily exceed the Vbe reverse bias spec if the output load current draw drops suddenly, as the op-amp output will clang to the -ve rail to compensate due to the slow loop response of the integrator. You can avoid this by making the -12V rail 0V, raising the 100 ohm base swamp resistor to a few kohm (and lose the c-b 10k resistor altogether or the potdown will limit the emitter voltage swing), and clamping the transistor b-e junction with a reverse-biased diode (i.e cathode to base). I'd be tempted to lose the integrator function altogether, and lower that feedback cap to a few pF just to help with stability.
     
  10. newbies_hobbyist

    Thread Starter Member

    Jun 4, 2010
    67
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    Darren,
    I tried to simulate the circuit and applied a 10ohm (enough to overload) at the ouput. The +12V is capable of giving upto 6A, the collector-emitter maximum current flow of 2n2222A as per in datasheet is 800mA so I'm expecting that the collector-emitter will get open since the current will flow from C-E is 1.2A but in fact it is the B-E gets open and the collector-emitter is still good (.6V diode drop when check) . I don't know how it happen can you explain it to me. Thanks for the idea maybe i will try that.
     
  11. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    In simulations the component are indestructible. Buy in real life if you try to get more then 0.8A for 2N2222 then you mus buy a new BJT.
     
  12. newbies_hobbyist

    Thread Starter Member

    Jun 4, 2010
    67
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    What I mean in simulation is, applied it in real circuit, I constructed the circuit. simulated it in real life. Why it is the B-E gets open and not the C-E?
     
  13. Wendy

    Moderator

    Mar 24, 2008
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    Due to the nature of BJTs the BE junction is in both paths. The base sits squarely between the collector and emitter. Be happy, a common failure mode is the CE shorted with the BE junction intact. That mode can be very confusing to find.

    The big problem I see is no feedback. All voltage regulators, either switching or not, must have some feedback.

    Here is a similar unit I designed that worked really badly (but it worked).

    [​IMG]

    Simple Switching Regulator

    My next step is to make a PWM version, which is what you are trying for.

    Side note: VR1 and VR2 were LEDs I was using as voltage droppers. I found them to be really useful for visual feedback though, since they also showed the on time for the transistor.
     
    Last edited: Jun 22, 2010
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Bill, I also initially had trouble seeing this circuit for what it is.
    The input is on the + input of the op amp. It will be 1.25V at 50% duty cycle. The feedback path is through the resistive divider consisting of the 30k and the two 4.99k resistors.
     
  15. Wendy

    Moderator

    Mar 24, 2008
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    You're right, the direction of the flow fooled me.
     
  16. newbies_hobbyist

    Thread Starter Member

    Jun 4, 2010
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    Bill, I fully understand it now when you said that the the BE junction sits in both paths. But there is one more I'm concern about aside from having a big load or short at emitter, is'nt it too much for the transistor to work for 24hrs in a high temperature and switching 8times per second? I know it is design to as switch but I'm thinking that maybe the transitor is pushed fast to it's life expectancy by utilizing it like this. Correct me if I'm wrong.
     
  17. Wendy

    Moderator

    Mar 24, 2008
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    Nope, it's not switching speed, but current and heat. The hotter a transistor is the quicker it ages, and current generates heat. Thing about switching, the reason it is so efficient is the coolest (as in temperature) states a transistor has is all the way on or all the way off. Since power is current X voltage if any one of those variables is close to zero then power is way down.
     
  18. newbies_hobbyist

    Thread Starter Member

    Jun 4, 2010
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    Bill, thanks for very helpful information. Wel if that's the case I think it is just only a short in output that makes my transistor faulty. I am thinking of a protection circuit to avoid the damage on emitter of my transistor. I will remove the 12V from collector and put one more transistor. it goes like this:
    1. 1st transistor (original), collector is connected to the 2nd transistor (emitter of it)
    2. 2nd transistor, emitter is connected to the collector of 1st transistor and collector is connected to 12V, base is connected to -in of op-amp (from divider feed back 1.5V).

    If there is a short in the output, 2nd transitor will be off since there will be no drop at didvider feed back resistor thus the conduction of 12V will be off as well.
     
  19. Wendy

    Moderator

    Mar 24, 2008
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    Feel free to post updated schematics, and we'll help you tweak them.
     
  20. newbies_hobbyist

    Thread Starter Member

    Jun 4, 2010
    67
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    Here it is, well I hope this work. What I am not sure is which one will works first is it the Q2 will go off first before the Q1 emitter gets damage or is it the Q1 emitter gets damage first? I think I need to try it but before that can you please take a look on it and give me a feedback. Thanks
     
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