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Resistance of ammeter caused voltage drop

Discussion in 'Electronic Basics' started by Dummy, Mar 15, 2005.

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  1. Dummy

    Dummy Guest

    I have a simple setup as below. The ammeter is used to measure the
    transmit current of the radio. Power supply has an display of total
    current as well.

    Power supply (Vin=7.5V)-------ammeter--------- (Vout) radio

    When connecting the ammeter, there's a 0.3V voltage drop across the
    ammeter before transmitting. During transmit mode, total current as
    displayed at ammeter is 1.80A, matched with the power supply current
    display. But Vout is measured to be 6.14V only, thus causing the Tx
    power to be lower.

    Using direct cable connection without ammeter, the current measured is
    almost similar, but the Tx power is much more higher. Current is
    1.85A. Vout is 7.0V during transmission - a voltage drop of 0.5V.
    I guess the ammeter is giving more resistance the the cable.

    So I was wondering whether the ammeter should be used to measure any
    high current in circuit level if it can cause some voltage drop. The
    voltage drop might affect the circuit performance at the subsequent
  2. Tim Wescott

    Tim Wescott Guest

    I'd look for a better ammeter. My old Heathkit electro-mechanical unit
    only drops 1/4 volt at full scale -- I would imagine that a Fluke or
    other serious DVM would be better (but I haven't checked, so I can't
    guarantee it).
  3. It certainly looks that way.
    That would depend on how badly you want to measure
    the current and what effect the voltage drop has. I am
    going to guess about those things for now.

    Assumption: The transmitter has a linear relationship
    between supply voltage and supply current draw.

    Assumption: The transmitter does not change its mode
    of operation at the reduced voltage you see when the
    ammeter is connected. (This is nearly but not quite
    tautological with the linearity assumption. But it is
    something you can verify independently when the
    ammeter is not present.)

    Assumption: You have a way, (such as a different cable,
    a variable bench supply, or some low Ohm resistors), to
    vary the supply with the ammeter connected.

    Assumption: Your ammeter is a better instrument than
    the meter built into the supply. (Otherwise I do not
    know why we would be having this discussion.)

    Measure the current with ammeter in at the ordinary
    supply voltage. Call this Iao. Measure the voltage
    at the transmitter with same lashup. Call this Vao.
    Measure the drop across the ammeter, Vad.

    Reduce the supply voltage by an amount similar
    to what the ammeter drops, leaving the ammeter
    in place. Measure current and voltage, to be
    called Iar and Var respectively.

    Calculate Rt = (Vao - Var) / (Iao - Iar)
    This is the slope of the voltage versus current
    characteristic for the transmitter.

    Calculate Ina = Iao + Rt * Vad
    This is an approximation of the current the
    transmitter draws when you have no ammeter
    to reduce the supply voltage it sees.
    Yes, it probably does. It would help, when deciding
    what to do about this, what you are trying to achieve
    by measuring the current. Is this a one-time affair,
    or will the ammeter become part of the setup? How
    accurately do you need to know the current? (My
    guess is that this does not matter much.)

    There are current meters that impose no DC drop.
    You could rent (or buy) one if you believe the
    above procedure is too much trouble or not
    sufficiently accurate. I do not advise this unless
    you have more need for accuracy than I can see.
    You could also put a lower shunt resistor across
    your ammeter and calibrate the combination.
  4. Robert Baer

    Robert Baer Guest

    The shunt scheme would seem to be a good idea.
    Say one gets an 0.1 ohm 1% resistor and uses kelvin connections to
    read the voltage drop across it with a DVM (lowest full scale range is
    Then at one amp, the voltage drop would be 100mV; 3 times better than
    the presumed 300mV cited (which is not that much higher than "1/4 V).
    BTW, i would put the resistor on the ground side...
  5. JeffM

    JeffM Guest

  6. Mark

    Mark Guest

  7. Jamie

    Jamie Guest

    if you want to experiment a bit, you could use an OP-Amp inputs with a
    shunt ( very low value shunt). the Op-Amp can be an Amp for a simple
    of course there is more that you need to do, this is just an idea for
    you to ponder with.
  8. JeffM

    JeffM Guest

    I was wondering whether the ammeter should be used to measure...
    Why are you responding to me?
    Go up 3 levels in the thread and respond to the OP.

    (Clueless Google posters are the bane of Usenet.)
  9. Ken Taylor

    Ken Taylor Guest

    You could buy a shunt and use a voltmeter (on a suitably low range) to
    measure the drop across it.

  10. John Fields

    John Fields Guest

    so transmit, with the ammeter in the circuit, looks like this:

    Power supply (Vin=7.5V)-------ammeter--------- (Vout=6.14V) radio

    Which makes the total (ammeter + cable) resistance look like:

    Vin - Vout 1.36
    Rt = ------------ = ------ ~ 0.756 ohms
    I 1.8

    With the ammeter out of the circuit, the cable resistance looks like:

    Vin - Vout 0.5
    Rc = ------------ = ------ ~ 0.270 ohms
    I 1.85

    which makes the ammeter resistance look like:

    Ra = Rt - Rc = 0.756 - 0.270 = 0.476 ohms

    I have a Fluke 8060A which measures 0.468 ohms on the 2 amp scale, so
    0.476 ohms for Ra doesn't seem far-fetched if your ammeter is similar,
    but 0.270 ohms seems awfully high for the cable, what with 100 feet of
    #20 being about an ohm. Just to make sure that's not a problem I'd
    check out all the wiring interfaces and make sure you don't have any
    high-resistance connections anywhere.

    Bottom line though, after you get everything cleaned up and sorted
    out, if you know that what you need to make your measurements is 7.5V
    at the radio, put your ammeter in the line if that's what you need to
    do, then monitor the voltage at the radio's power input, then key the
    transmitter and adjust the power supply until the voltage at the
    radio's power input is 7.5V. If it goes up to 8V when you stop
    transmitting, so what? You can always drop it back to 7.5 if you need
    to work on the receiver, although I doubt that you'll need to, since
    I'm sure the receiver's not running unregulated.
  11. Mark

    Mark Guest

    oh no...the thread police
  12. Fred Bloggs

    Fred Bloggs Guest

    Yeah? So the F___ what? And since when is resistance x voltage equal a
    current?- and you don't even have the sign right. Here's the deal you
    worthless, pretentious son-of-a-bitch-with-VD, you are a worthless
    p.o.s.- we are wise to your dumb ass- go away.

    Or buy/modify a p.s. with external sense compensation, damned worthless
  13. (Finally, a positive contribution.)

    That line should have read, of course, thusly:
    Calculate Ina = Iao + Vad / Rt
    And, contrary to what the esteemed Mr. Boggs
    proclaims, the sign is correct. As defined, both
    Vad and Rt are positive quantities (at least if
    the transmitter draws more power at higher
    voltages, which is already in evidence.) Since
    Ina (pronounce as: I 'n'o 'a'mmeter) represents
    the current predicted when no ammeter is present,
    and that is already known to be higher, adding the
    positive ratio Vad/Rt to the current measure with
    the ammeter present is correct for getting such a
    result. I would hope that this is now obvious
    even to the most critical observer.
    I perhaps should engage in some name-calling on
    account of the above correction, but, as a low ranking
    member of the excrement class, I am not up to it.
    I already said 'No', Fred. Do you think repetition
    is going to be effective? (It would appear so.)
    Spending money was an obvious option which
    I mentioned in several of its many forms. The
    OP's questions led me to believe he might be
    interested in using the equipment he had. We
    have seen no evidence to the contrary.
    Fred, I appreciate your opinion. Honestly.
    I tried to tell you that earlier, but I suspect
    my meaning escaped your notice.
  14. Robert Baer

    Robert Baer Guest

    Again, pay attention and use kelvin connections.
  15. Robert Baer

    Robert Baer Guest

    I think i said that, and added a cautionary note to use kelvin
  16. John Fields

    John Fields Guest

    However, you never mentioned external sense compensation (using a
    Kelvin connection at the load to supply feedback to the supply in
    order to compensate for lead resistance) and had you known such a
    thing existed you would surely have mentioned it as an "obvious

    Now that the cat's out of the bag, though, I suspect you'll soon
    become the expert you'd like us to believe you already were.

    Here, I'll save you a little time on Google:

    On power supplies supplied with external sense compensation there are
    two terminals, one usually marked "+ sense" or something like that,
    and the other one marked "- sense" or something like that. In use, a
    wire is connected from the "+ sense" terminal to the + input of the
    load at the same point the supply lead is connected to the load, and
    the "- sense" terminal is connected from the "- sense" terminal to the
    - input of the load at the same point the supply lead is connected to
    the load. That way, voltage variations _at the load_ are sensed and
    fed back to the supply where the supply voltage is automatically
    adjusted upward to compensate for the voltage dropped across the
    supply leads. If sense compensation isn't needed, the sense terminals
    are shorted to their respective supply outputs at the supply, and the
    supply regulates the voltage at that point.

    You're welcome.
  17. There are two silly assumptions you've made. As I have
    stated elsewhere, I thought the OP would like a solution
    utilizing what he mentioned he had on hand. I threw out
    a few spending options without pretending to exhaust
    them, merely to let him know he was not stuck with
    just what he had. So I was not inclined to spend much
    time trying to come up with a list that none of the smart
    alecs around here would be able to "improve" upon.
    So, assuming that my non-mention reflects ignorance
    of remote sense power supplies is fatuous. The other
    laughable assumption is that it should be an obvious
    option to anybody who knew of such supplies. Why
    should the OP go spend that kind of money when he
    can simply use a shunt or measure the resistance of his
    cable and use that and a voltmeter to measure current?
    Thanks, so much John.
    I note that your little description omits mention of the
    1k or so resistors that normally obviate the need for
    those jumpers when remote sensing is not used.

    Not knowing how old you are, I may be actually wrong
    about this, but there is a good chance that the incident
    I relate below happened before you had any inkling of
    what a circuit is or what 'electronics' means.

    Before my job as an engineering tech which preceeded
    my career as an electronics design engineer, I held a
    job as a factory test tech. One day, in return for a
    similar level of prank, I connected an RC network
    between the so far unused sense terminals and the
    output terminals on the power supply that my "pal"
    would be using after lunch to continue troubleshooting
    some equipment. (These machines were battery
    operated but run off of a DC supply during most test
    and troubleshooting.) After enjoying the spectacle of
    him trying to figure out what was going on as his bench
    supply was oscillating at a low level, I went and told a
    few other techs so they could come and "help" (see).

    As for the novelty of 4 wire connections for dealing
    with cable and connection drops, you could pull up
    one of my patents detailing a system that relies on that
    very concept in order to operate effectively. (I do
    not expect any such effort from you, attached as you
    are to the notion of my ignorance. If not for that, it
    ought to suffice to plug your "soon become" spew.)
  18. John Fields

    John Fields Guest

    In light of the fact that his power supply might have been capable of
    remote sensing and in view of your statement: "I thought the OP would
    like a solution utilizing what he mentioned he had on hand." It seems
    to me that your "familiarity" with remote sensing supplies would, at
    the very least, brought forth the question of his power supply having
    that capability and, if it did, a suggestion to use that capability.
    Key phrase here is "knew of". If the OP wasn't aware that his supply
    had that capability, or how to use it, then that option would hardly
    have been obvious. You, though, knowing that such an option might
    have been available didn't bring it up either.
    He probably shouldn't, but that's not what you advised him to do, you
    sent him on some wild goose chase to measure this and that and
    calculate this and that, when all you had to tell him was to monitor
    the voltage at the input of the radio and crank the supply up to make
    that voltage go to 7.5V when he was transmitting, LOL!

    You also wrote:

    "You could also put a lower shunt resistor across
    your ammeter and calibrate the combination."

    which is one of the stupidest things I've ever read. Do you know why
    or would you like me to explain it to you while typing through fits of
    ??? Funny... all of the stuff I've got around here _requires_
    strapping the sense terminals to the output terminals if remote
    sensing isn't used, but hey, that's only HP.
  19. Fred Bloggs

    Fred Bloggs Guest

    What a simple-minded little fairy and brainless pussy you are- the OP
    already said that he gets 1.85A at 7.0V without the ammeter- so who
    needs your crap and pretentious pseudo- engineering formula? You want to
    assume linearity for voltage drops less than 10%, then that there tells
    you the effective power supply output resistance is 0.5V/1.85A=0.27
    ohms. Then because the OP also states the load voltage is 6.14V at 1.80
    with the ammeter, you have that 7.5V=1.8*0.27+1.8*Rammeter+6.14V or
    Rammeter=0.486 ohms. The OP also states that without transmitting, the
    drop due to the ammeter is 0.3V making I= 0.3V/(0.486+0.27)ohm=0.4 amps
    at 7.2V load voltage- or a load power of 7.2*0.4=2.9 Watts. The load
    power during transmit w/o ammeter is 7*1.85=13W and the load power w/
    ammeter is 6.14*1.8=11W. The ammeter deprives the circuit of 2W of load
    power,- so that the RF output will fall something like eta*2W where eta
    is the transmit efficiency. Say for example eta is 33% more or less over
    these power consumption ranges- then he must lose about 0.7W transmit
    power- if he has a lot of Class A overhead stuff then eta may only be
    10% and the power loss is 0.2W in transmit. Your little sissy model
    conveys no usable information, and why would it- you don't what you're
    doing- you are a pretentious fraud- you latch onto to same linearity
    model for load Rt- you don't have a clue what's important and what's
    not- and once again you can't understand what the OP is asking- like
    mainly how much transmit power degradation can he expect. Get a clue and
    take a hike- you are another "unwanted" NG p.o.s- just leave.
  20. So, you sincere, undiseased, wholesome paragon of virtue,
    are you able to see and acknowledge your error? Why, if
    you are such a valuable and respected contributor, do you
    not help resolve the error you introduced? Is it that your
    tenuous sense of superiority would be threatened? Or do
    your invisible friends tell you to attempt some face saving?

    [Much evidence of unfamiliarity with the paragraph cut.]

    Nothing you said contravened the linear model for
    the transmitter current consumption over a limited
    supply voltage range. Have you managed to divine
    some requirement of the OP's for accuracy that
    would be compromised by using a linear model?
    [More crap into the bit bucket.]
    You must be daft, Fred. My answer is still No.
    So maybe you better start your threatened action
    to get me booted off without my cooperation.
    Please amuse "us" with progress reports. <g>
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