[R-390] Question about Carrier Level Meter

[Bob Camp]

Hi

Here's how to at least begin the process of figuring out what's going 
on:

1) Grab a D cell and a 20 K ten turn pot ( or something similar)
2) Set the pot to max resistance and hook it up in series with the new 
meter and no shunt
3) Adjust the pot so you have a full scale reading on the meter
4) attach the shunt and see if the meter goes to around 1/10th scale

You won't get exactly 1/10th scale because of the meter resistance but 
you should get close. If you get something lots lower then the new 
meter is well above 100 ohms resistance. If the reading is 
significantly higher then the meter is well below 100 ohms resistance 
(my bet is on low).

If you now rig the new meter and shunt in series with a 1 ma reading 
VOM you can adjust the shunt to give you 1 ma full scale. If that works 
out to a 4 ohm shunt, then that 's what you use. There is only one 
value of shunt that will do the trick. I keep a couple spools of 
resistance wire around for this kind of thing.

The next part is magic. The shunt will always be 1/9th of the meter's 
resistance. If you know the shunt you know the resistance of the meter 
plus shunt.

The meter will do odd things in parallel with an existing meter. 
Putting a second meter in drops the impedances and messes up the 
circuit. Simply measuring the voltage across the meter does not mess up 
the circuit. When you put the new meter with no shunt across the 
existing meter you are using the new meter as a high impedance volt 
meter. Strange to be talking about a 100 ohm current meter as a high 
impedance voltmeter, but that's what it is compared to a 10 ohm system.

	Take Care!

		Bob Camp
		KB8TQ



On Mar 24, 2005, at 11:13 PM, Barry wrote:

> I have some 100uA meters that I'd like to use as replacements for the 
> meters
> in my R390A.  They are supposed to have an internal resistance of 100 
> ohms.
> Supposedly, by shunting with an 11 ohm resistor and then placing an 8 
> ohm
> resistor in series, this will make them effectively 1mA meters with
> approximately the same internal resistance as the original meters.
>
> I experimented a bit tonight with this and I didn't get the results I
> thought I would.  I shunted one of them with 11 ohms and placed it in
> parallel with the existing meter.  I expected both meters to read
> approximately 1/2 of what they normally would in the circuit.  
> Instead, the
> shunted meter barely moved and the original meter hardly noticed the 
> new
> meter in the circuit.  I removed the shunt and the new meter read a bit
> more, but not nearly as high as the original meter that it was in 
> parallel
> with.
>
> I then placed an original meter in parallel with the existing meter 
> and I
> got the action I expected.  For example, a 60dB signal would show as 
> about
> 30dB when they were connected in parallel.
>
> Is this happening due to the fact that the internal resistance of the 
> new
> meter is so much higher than the original meter and the current is 
> simply
> taking the path of least resistance?  If I remove the original meter 
> from
> the circuit, can I expect the new meter to perform better or will it 
> be low
> reading no matter what?
>
> If I have to, I'll build Jan's circuit and tune them up that way, but 
> I was
> hoping to simply use a resistance network and use them as is.  Maybe 
> not?
>
> Thanks,
>
> Barry(III) - N4BUQ
>
> _____________________________________________________________
> R-390 mailing list
> Home: http://mailman.qth.net/mailman/listinfo/r-390
> Help: http://mailman.qth.net/faq.htm
> Post: mailto:R-390 at mailman.qth.net
> Unsubscribe: http://mailman.qth.net/mailman/options/r-390
>