Resistance of ammeter caused voltage drop

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).

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

Yup. There are even ready-made units.

make sure your connections are all tight and low resistance and you use heavy wire. It may be the connections to the meter causing the problem. Or it may be the meter. What kind of meter is it?

Mark

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

200mV). 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...

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 meter. of course there is more that you need to do, this is just an idea for you to ponder with.

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.)

oh no...the thread police

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

1.8A--->

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.

-- John Fields

current?- and you don't even have the sign right.

(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.

worthless p.o.s.- we are wise to your dumb ass-

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.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

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 idiot.

--- 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 option".

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.

-- John Fields

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.)

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

[Some text and a load of crap cut.]

worthless p.o.s.- we are wise to your dumb ass-

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.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

"Fred Bloggs" wrote in message news: snipped-for-privacy@nospam.com... [More invective and refusal to admit clear error cut.]

The answer is still No. Do you imagine it will change at some magical repetition number? If so, why? If not, why not stop asking?

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

--
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.

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

Ken

Again, pay attention and use kelvin connections.

I think i said that, and added a cautionary note to use kelvin connections.