Help in Identifying loading issue with 555 timer chip

Discussion in 'General Electronics Chat' started by gomavs123, Jun 20, 2011.

  1. gomavs123

    Thread Starter New Member

    Jun 10, 2011
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    Hey guys,


    I am right now designing a self powered circuit for a project of mine, using generated energy from an ac source to power and operate different circuits. I am using a full wave bridge rectifier circuit to charge a capacitor (schematic is attached). In implementation, I am using a low power ICM7566 timer chip from maxim.

    However, when I hooked this circuit up and measured the voltage at the output, I get a very small output reading (see attached image). However, my AC pulse generator is generating an amplitude of 10V. I was at least expecting the 555 timer to give a 5V output pulse (or bigger).

    At first I thought it might be high power dissipation from the chip, but I found out from measurement and fellow helpers that power dissipation from CMOS chips are very small by convention (in the μW range- my measurement was even smaller).

    Therefore, I am lost as to why this is happening- I imagine there is loading of some sort, and therefore measured the input impedance. However, I found out that its input impedance was 2Mohms, which is relatively high, high enough to not load up the circuit.

    by the way, the ground is not at earth ground as seen in the schematic- that 0 gnd was put in so pspice could run the simulation. It is a unified "location", connected to the negative output terminal of the rectifier.

    I am using a standard oscilloscope from tektronix for measurement, with probes at 10Mohms impedance applied to it.

    Any help and/or suggestions will be greatly appreciated. Thanks!
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    I could not find specs on a Maxim ICM7566. However there is an Intersil ICM7555/7556 CMOS version of the ubiquitous 555 Timer. This CMOS version can only source 1mA and sink 3mA. So even a 1K-ohm load is a bit heavy on this chip.
     
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  3. gomavs123

    Thread Starter New Member

    Jun 10, 2011
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    Thanks for your reply-
    here is the link to the datasheet of the chip that I am using:
    http://www.alldatasheet.com/datasheet-pdf/pdf/72716/MAXIM/ICM7556IPD.html

    I assumed what you meant by the load being "a bit heavy on this chip" meant that the maximum sourcing current of the chip is not enough to supply the 1kohm resistor for about 5V.

    Looking at the datasheet, it seems that for an output voltage of 5V, I can expect about 10mA of source current. wouldn't that mean that the chip that I am using should be able to display up to 10V? (10mA * 1Kohm)

    Thanks for the quick continued help!
     
    Last edited: Jun 20, 2011
  4. iONic

    AAC Fanatic!

    Nov 16, 2007
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    If you are measuring the voltage as in the schematic you provided, then you will not have any voltage. You need to measure pin 3 to ground. As it stands now you are measuring pin with respect to V+ (something less than 10V).
     
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  5. Wendy

    Moderator

    Mar 24, 2008
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    The lower the supply voltage to these chips the less current they can provide. They require very little power, but they have little drive too. If you move up in the supply voltage their drive gets a bit better, 10ma and 100ma respectively for 12V.

    There are lots of way to improve this, simply by using transistors to increase their drive.
     
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  6. MrChips

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    Oct 2, 2009
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    Sorry, I misread the spec sheets on the ICM7555.

    I did not have any of these but I put on a breadboard a TLC555CP from Texas Instruments and a LMC555CN from National Semiconductor. The specs say CMOS 555 should source 10mA and sink 100mA.

    With 5V supply and 1K-ohm load I am getting 4V high pulse, i.e. a source of 4mA.

    With 500-ohm load, the pulse is 3.5V, i.e. 7mA.

    On sink, with 100-ohm, the pulse amplitude is 3V, i.e. 30mA sink.

    Thus you must be doing something wrong if you cannot get a 4V pulse with a 1K-ohm load.
     
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  7. gomavs123

    Thread Starter New Member

    Jun 10, 2011
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    Thanks for all the help and feedback guys.


    Ionic-
    It is my mistake for not making the image clear- The two “v” probe figures indicate two locations of my oscilloscope probe- the ground probe were connected respectively to its lowest common potential.

    Bill marsden & mrchips-

    I think the reason my circuit is causing such different behavior than when it is connected to a normal dc power supply is because it is charged from an ac voltage source.

    In fact, looking at my circuit, I think maybe the reason for my charging capacitor to not go above 1V is because the 555 timer IC, though low power, exhibits its own intrinsic capacitance. Therefore, in terms of the ac source, it will be as if it is charging two capacitors in parallel. (Though the capacitance from the 555 timer IC should be negligibly small? This part I am not sure of)

    I think what I have to do is a transistor design that will completely block out the IC stage from the charging stage until it reaches a certain voltage- like a button based switch that is automatically pressed when it reaches a certain voltage, as seen in the circuit diagram in figure 12 of pg 7 of this paper:
    http://www.media.mit.edu/resenv/pubs/papers/98_08_PP_wearcon_final.pdf
    I am lost on how I can incorporate a similar design for a simpler circuit like mine.

    A lot of stuff to digest- I appreciate if it you read through it all, thanks again.
     
    Last edited: Jun 21, 2011
  8. Wendy

    Moderator

    Mar 24, 2008
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    There are a couple of things you can do to increase the efficiency of such rigs, though not as much as you might like. The power has to be there to begin with though.

    Look into Shottky diodes. A typical silicon diode drops between 0.6 and 0.7VDC. This is a lot of power right there. A shottky diode tends to be slightly larger than it's silicon equivalent, but it only drops 0.2 to 0.3VDC. That is a major improvement right there.

    Another thing is storage. Almost all solid state electronics (555 included) needs pure DC. You could charge up a super cap and use it. If the power generation is greater than power out it will work well. A super cap is relatively small physically, but you can have several farads at 5V, a really deep storage system.
     
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  9. gomavs123

    Thread Starter New Member

    Jun 10, 2011
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    some good suggestions here. Thanks! I will definitely try it out.
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    The output high current of an LMC555, TLC555 or ICM7555 is 10mA only when it has a 12V supply and the loaded output voltage is as low as 10.5V.

    Intersil's datasheet shows detailed curves of the output current. With a 5V supply the output high current is typically 7mA into a dead short or 2.2mA when the loaded output voltage has dropped to 4V.

    Some might produce currents that are half.
     
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  11. MrChips

    Moderator

    Oct 2, 2009
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    Try rewiring your 555 timer as follows:

    [​IMG]

    Make R1 = 1K, R2 = 1K, C1 = 10uF as a starting point.
     
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  12. Wendy

    Moderator

    Mar 24, 2008
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    [​IMG]

    Duty cycle of your design will be 67%, which isn't the best. Better is to make R2 10X larger. This will put the duty cycle at 52%. Also, don't forget this needs to be very low power, so R1 = 10KΩ (it is shorted to ground on half the duty cycle), R2 = 100KΩ, and C1 =1µF.

    Source: My Cookbook
     
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