FET - dc path to ground

Discussion in 'Homework Help' started by Zeeus, Jun 11, 2019.

  1. Zeeus

    Thread Starter Member

    Apr 17, 2019
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    In first image there has to be dc path to ground at input yeah?

    "in" is from function generator

    In 2nd image, from the 4th to last line: from "then tune an AM"...Please can someone explain from that point
    Don't understand

    Thanks



    // WBahn : "As this is homework, show your attempt"
     
    Last edited: Jun 11, 2019
  2. BR-549

    AAC Fanatic!

    Sep 22, 2013
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    What is the question? What's the problem?
     
  3. Zeeus

    Thread Starter Member

    Apr 17, 2019
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    lol..bedtime, no?

    Thanks

    In the first image : should there be a resistor from "in" to ground? or can connect function generator directly

    2nd image : don't understand from 4th to last line to the last line. Please explain
     
  4. crutschow

    Expert

    Mar 14, 2008
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    There is through the function generator.
    The circuit will operate as a simple AM transmitter that can be detected by an AM radio tuned to 1MHz (or whatever the Vin frequency is).
     
  5. Zeeus

    Thread Starter Member

    Apr 17, 2019
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    oh really? cause in previous chapter, somewhere it says always provide dc path to ground

    so from that simple circuit, can add mic signal as the v mod then will hear from the AM radio??
    for real???
    that's what a transmitter is for, no?

    Thanks
     
  6. BR-549

    AAC Fanatic!

    Sep 22, 2013
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    1. the current from the generator must be returned to the generator on the black generator lead. This lead is commonly grounded at the generator. If the black lead is not connected to a ground potential in test circuit, an extra current can appear. Not only will this distort the signal, but it can burn things up. There are several methods the cancel this effect.

    2. It means you can use a AM receiver to listen to this transmitter circuit. You can tune or adjust the tuning dial to find it.

    We are not suppose to directly answer questions on this particular forum.
     
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  7. crutschow

    Expert

    Mar 14, 2008
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    Really.
    Assuming, of course, that the generator ground is connected to the circuit ground.
    If not, than the generator wouldn't work as a generator for the circuit. :rolleyes:
     
  8. Zeeus

    Thread Starter Member

    Apr 17, 2019
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    why you not sleeping? what you doing ?...no wife? :rolleyes:

    btw, bs transmitter circuit...
    my AM radio, no "quiet place"...the frequency not in use, always "shhhh" sound

    Anything to do? what if to connect electret mic as the Vmod (and play sound from phone to the mic)...will hear from radio?

    edit :Trying shortly
     
    Last edited: Jun 12, 2019
  9. WBahn

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    Mar 31, 2012
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    For most simulators, every node needs to have a DC path to ground otherwise there is a good change that the simulator will not be able to get an initial operating point solution to converge.

    Consider two capacitors in series across a battery. In theory, there could be ANY voltage at the junction of the two capacitors -- imagine that the junction point was previously connected to another battery that was then removed. The simulator has no way of coming up with a single value for the voltage on that node, so it can't even start. But if you put even a very large value resistor across one of the capacitors, now the simulator can get the initial operating point to converge to a single value.

    It doesn't matter how that DC path to ground comes about, it just has to be there. If the output of your function generator is at a known output voltage at the start of the simulation, then the voltage on that node is known to the simulator just as if it had been a battery connected there.

    As the description says, the circuit is a simple AM transmitter. The carrier frequency is applied to the v_in node and the signal is applied to the v_mod node. The circuit amplitude modulates the carrier with the signal. The radio, when tuned to the same frequency as the carrier, will demodulate the combined signal and recover a replica of the original modulating signal.

    It probably says this above, but unless you have a signal going into v_mod when you try to listen to it on the AM radio you will here little to nothing.
     
  10. WBahn

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    Mar 31, 2012
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    The "shhhh" sound is background static. A "quiet place" is simply a place where there is no station can be heard. Your signal is going to be very weak, so even a weak signal from an actual broadcast station is likely to overwhelm it.
     
  11. Zeeus

    Thread Starter Member

    Apr 17, 2019
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    Hmm..Seems like theory predicts the voltage 100% unless not understand this statement..

    How can remove the background static...In book, author says the signal about 200mV (guess need increase)..so can use mic then bjt amplifier then output as Vmod?..Anyways, FET is the topic, will learn RF

    Just to test though : how to remove background static?
     
  12. crutschow

    Expert

    Mar 14, 2008
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    It was 9:44pm at my house when I wrote that.
    I usually don't go to bed until after midnight.
    And what's having a wife got to do with it? :confused:
     
  13. Zeeus

    Thread Starter Member

    Apr 17, 2019
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    ....
     
    Last edited: Jun 12, 2019
  14. ericgibbs

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    Jan 29, 2010
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    Hi Zeeus,
    Mod: Please desist from making personal comments, stay on the technical Topic.
    E
     
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  15. WBahn

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    Mar 31, 2012
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    Huh????

    Put two 1 uF capacitors in series across a 12 V battery. Are you saying that "theory" predicts that the voltage across each capacitor is 12 V?
     
  16. Zeeus

    Thread Starter Member

    Apr 17, 2019
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    was thinking about AC, middle voltage peak as 6V

    Thanks

    the midpoint between cap has to be 0v ? if not 0 then dc current is flowing through cap
     
    Last edited: Jun 12, 2019
  17. ericgibbs

    Moderator

    Jan 29, 2010
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    hi Z,
    Theoretically the junction/ midpoint would be Vsupply/2.
    In a practical application is possible that the leakage current through the two capacitors will not be the same value.

    What would you consider as a possible solution if that was the case.?

    E
     
  18. crutschow

    Expert

    Mar 14, 2008
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    Yes, current flows.
    If you connect two equal sized caps in series to 12Vdc, then current will flow through both caps until there is equal voltage across both caps, or 6V at the midpoint.
     
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  19. Zeeus

    Thread Starter Member

    Apr 17, 2019
    303
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    Add high equal value resistors in parallel with each? add leakage cap?
    dunno
     
  20. Zeeus

    Thread Starter Member

    Apr 17, 2019
    303
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    I thought so to but checked it out on spice and there was 0v...only for AC then peak is 6v
     
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