Blown meter shunt

[kb3ouk]

Yesterday I finally attempted to fire up the Millen 90800 exciter for my homebrew rig. Couldn't get it to do anything though because I believe the B+ was too low (only 300 volts the way I had my power supply configured). Today I redone the supply to do about 650 volts (which is still less than what the manual says to run, it suggests 750 on the HV). As soon as I hit the switch to turn the voltage on, there's a flash and smoke pouring out of it. It appears that the 807's plate current shunt exploded. Now I do know that the meter works, because I did get movement yesterday in the 6L6 plate current position, but could never get the thing to oscillate (I think the voltage was way too low). I believe these shunts are 100 ohms. The 6L6 shunt measures 95 ohms, so I would say it should be ok. What I'm trying to figure out is what caused it to blow and was it fairly common for these rigs to blow the 807's shunt? I asked because when I goggled images of the 90800 I found at least one other one that had the same resistor blown in two. Also, I'm not sure what the purpose of this is or not, but my 90800 has it and so do the rest of them I've seen, but there is a small coil of wire that runs between the two meter terminals, what is it for?

[N8ETQ]

  Yo'

     Step one would be to strap a pair of back to back
diodes across that thang, before your looking for his
replacement. 100 ohms sounds pretty large to me, shunts
typically run in the 1/10 to 1/100 ohm range.

     Best bet would be to hook your 2 dollar Harbor
Fright DMM in the circuit and read what's going on.

GL

/Dan

[W7TFO]

That little coil of wire IS your shunt, not a 100-Ohm resistor.

73DG

[kb3ouk]

Then what are the resistors for? The meter is switched so that it can be used on both the 6L6 and 807, and it is measuring the current across both of those resistors. I'm guessing it just died of old age though, I got a temporary fix together for it and tried it again and was able to get it to oscillate and put out 25-30 watts.

[N8ETQ]

  Yup,  your right, those 100 ohm "R"'s are the shunts for the 150ma meter.
Good job, I would still put diodes across the meter to protect it.  At 750
Volts it would probably make rated output and has a better chance of blowing
the meter.

Good job

/Dan

[kb3ouk]

I'm going to add the diodes when I do the permanent repair on the shunt. I'm not going to run it at full plate voltage, 25 watts is plenty of power for what I need it for.

[w4bfs]

whoa  ... not so fast

if you have a 150 mA current flow thru a 100 ohm resistor, what will be the voltage drop ?

the diodes will conduct in normal operation and will only get a small deflecton

something is not right here !!

time to backtrack and use a small power supply and a decade resistor box to characterize the meter and find out what is going on ... it is likely that the meter may have an internal shunt and/or multiplier resistance ....you will need to know to be able to make good desision (s)

n.b. I just looked at schizmatic .... yes it appears that the meter has an internal shunt and if the 100 ohm resistors values are not critical as compared to the value of that internal resistance of the meter (say 2 or 3 orders of magnitude) then the back-to-back diodes are fb ...

*********

here is an example:

a 0 - 1 mA meter with an internal movement resistance of 50 ohms is to be internally shunted to form a 0 - 150 mA meter ... what shunt resistance is needed ?     use Ohms law twice to find the answer

(1) for the meter movement -  E = IxR  = .001 x 50  = 50 mV for full scale

(2) for the shunt - R = E / I = .05 / (150 -1) mA = .05 / .149  = 0.33 ohm (approx)

so as you can see the 100 ohm resistor in parallel with the meter network is almost 3 orders of magnitude higher and is not critical

******



an unneutralized 807 with no parasitic suppression at 750V is very iffy for stability ... big parasite may have eaten up yer metering system ... that circuit used special shielding, is it in place ?

I wonder if that 100 ohm resistor is a 100 milliohm resistor as Dan said ... would make a difference

[KA2DZT]

Sometimes a gold band can age and look brown.  If it's under 10 ohms most all those resistors are wire wound and the first color band will be twice as wide as the rest.


I just took a look at the schematic that is posted here,  those two 100ohm resistors are there to complete the circuit path when the meter switch is in either of the two positions.  The 100 ohms has very little to do with the meter shunts and the value is not critical.

The resistance of a 150ma meter is probably less than an ohm.


Another look at the schematic,  notice there is no grid current meter.  Also look at the 24K and 51K resistors.  They form a voltage divider to ground off the 807 screen resistors.  Wonder what's the wattage rating of those two resistors?  They would take a high dissipation hit if either or both of the tubes failed to light up.  Even under normal operating conditions they need to be at least 2 watt resistors.   

Fred

[KC4VWU]

Now might be just as good a time as any to dig out the 90800 here and have a look see.

[kb3ouk]

Everything appears stock and looks like every other 90800 underneath that I've seen. Other than a little bit of drift (probably from the power supply voltage not being very stable, it bounces around a little), it seems to run ok at just over 600 volts.

[kb3ouk]

Ok so I replaced the blown 100 ohm resistor and added back to back diodes across the meter and now it will not read current when I have the switch thrown to read the 807's plate current, but it will read the plate current of the 6L6.when I had it working the other day, I just ran the switch over on the 807 side and let the current pass through the meter without any resistor in the circuit and it worked fine. Could the 100 ohms be too high? I have a 1943 ARRL Handbook here that has an article with the original transmitter that the 90800 was based on, and in that one they used 10 ohm resistors. The 6L6 side works fine and reads current with the original resistor for it (that one didn't blow and reads 95 ohms on my meter) but the one I just replaced doesn't let the meter read plate current on the 807.

[KC4VWU]

Bad switch? Iv'e seen it many times where one set of poles work fine and the other set will not.

Also, I had a look in the one here, which is original, and both resistors are intact and 100 ohms.

[KA2DZT]

Get rid of the back to back diodes.  The current is passing through the diodes instead of the meter.  150ma meter should read below 1 ohm,  MEASURE IT.  I told you before the 100 ohm resistors are not critical.  They are there to complete the circuit path when the meter switch is in the opposite position.  When the meter switch is in either of two positions you have a less than a 1 ohm meter across the 100 ohms.  What part of this are you not understanding??  Look at the schematic, you should be able to see this.

If one of the 100 resistors was burned out, it had nothing to do with the meter circuit even though you may think it did.  If one of the 100 ohm resistor was burnt open the xmtr would not work unless you moved the meter switch to that resistor.  When you do that the 1 ohm meter completes the circuit and the xmtr would begin working again.  The only reason the two resistors are there is because you have one meter measuring two circuit currents.  If the xmtr had two meters, one for each circuit, the two 100 ohm resistors and the switch would not even be there.

I'm going to test you on all of this next Friday (I give all my tests on Fridays), if you fail, we'll be forced to come there and tear up your license.

Fred

PS,  it's a little too cold to travel that far.  So, you'll have to mail your license here, we'll tear it up and send the pieces back to you.

[KA2DZT]

Bad switch? Iv'e seen it many times where one set of poles work fine and the other set will not.

Also, I had a look in the one here, which is original, and both resistors are intact and 100 ohms.

Check the switch,  if it's a slide switch they can be troublesome, even toggles can act the same way if they haven't been switched back and forth for years.  Use some WD=40 on the switch contacts and it should clean the contacts.

The diodes are really not needed on this type of meter.  Diode forward voltage drop is about .7 volts which is greater than the voltage drop across a less-than-1 ohm meter with only a few hundred milliamps through it.

I still would remove them.

Fred

The diodes can be a safety feature only if the meter shunt (inside the meter) was to open.  Since the schematic shows a 150ma meter, the shunt should be inside the meter.  The reason there would be a shunt is because most common meter movements are 1ma and the resistance of most 1ma meters is around 50-60 ohms.  Hence, the shunt to make it read 150 ma full scale.  The shunt would be less than an ohm and most likely a short piece of calibrated resistance wire.  Rarely to they open.

[kb3ouk]

Bad switch? Iv'e seen it many times where one set of poles work fine and the other set will not.

Also, I had a look in the one here, which is original, and both resistors are intact and 100 ohms.

Turns out thats what it was, the side that the 807 is on must be dirty or something inside the toggle switch.

Get rid of the back to back diodes.  The current is passing through the diodes instead of the meter.  150ma meter should read below 1 ohm,  MEASURE IT.  I told you before the 100 ohm resistors are not critical.  They are there to complete the circuit path when the meter switch is in the opposite position.  When the meter switch is in either of two positions you have a less than a 1 ohm meter across the 100 ohms.  What part of this are you not understanding??  Look at the schematic, you should be able to see this.

So I have one person telling me use the diodes and you telling me not to, which one do I listen to? I do know how to read a schematic, so I do know exactly what those resistors are doing. I was just asking a question if those might've been more critical than what you have brought up more than once now, since it seems like about half a dozen people have chimed in here and every one of them has had something different to add that isn't related to what the last person said.

If one of the 100 resistors was burned out, it had nothing to do with the meter circuit even though you may think it did.  If one of the 100 ohm resistor was burnt open the xmtr would not work unless you moved the meter switch to that resistor.  When you do that the 1 ohm meter completes the circuit and the xmtr would begin working again.  The only reason the two resistors are there is because you have one meter measuring two circuit currents.  If the xmtr had two meters, one for each circuit, the two 100 ohm resistors and the switch would not even be there.

How do you think I got the transmitter to work in the first place? set the switch on the 807 side to allow the current to pass through the meter. For my homebrew transmitter I was going to use a single meter to read current in two different circuits and was going to use a setup just like that and decided just to go with dedicated meters.

[KA2DZT]

I was just busting your chops a little.  I've never put diodes across meters, but some builders use them.  They protect the meter movement if the internal shunt was to open but you don't  really need them.  The 100 ohm resistors are not critical,  because once the very low resistance meter is across them (via the switch) the 100 ohms is too great a value to alter the resistance of the meter.  OTOH if they were say only 10 ohm resistors, then they would begin to alter the meter readings.  So, 100 ohms or 120 ohms it's not going to change any meter reading.  OTOH lets say they were 500 or 1000 ohm resistors, those certainly are even higher causing even much less of any affect on meter readings.  So why not use 500 ohm resistors?  500, 1000 ohm or higher values will cause too much of a voltage drop to the plate of the tube.

So you can see something in the range of a 100 ohms is going to have no affect on meter readings and cause no real voltage drop to the tube.  Say the tube has .050 amps (50ma) of plate current.  .050amps x 100 ohms = 5 volt drop in voltage to the plate. Use 1000 ohms and the voltage drop becomes 50 volts.

Hope some of this helps. BTW I was just kidding about tearing up your license (maybe)

Fred

[kb3ouk]

I'm not sure if that meter has an internal shunt or not, there is a coil of insulated wire, looks like maybe 20 gauge, connected between the two terminals of the meter. I figured that was the shunt, because that was the only logical explanation I could think of for why that piece of wire was there, and I've seen it on other 90800s. I guess Millen probably took whatever meters they could get ahold of and just used their own custom scale on them, since there are no marks on the meters that say who made them.

[KA2DZT]

I'm not sure if that meter has an internal shunt or not, there is a coil of insulated wire, looks like maybe 20 gauge, connected between the two terminals of the meter. I figured that was the shunt, because that was the only logical explanation I could think of for why that piece of wire was there, and I've seen it on other 90800s. I guess Millen probably took whatever meters they could get ahold of and just used their own custom scale on them, since there are no marks on the meters that say who made them.

What you say is very possible, hard to know what some companies did in the early days.  You can tell with an ohm meter.  You have to be somewhat careful trying to measure the resistance of a meter movement.  An ohm meter will put a few volts on the meter (without the shunt wire) and that will damage the meter.  If your meter doesn't have an internal shunt then you will need the external shunt that was on the meter terminals.  Hopefully you still have it.

To measure the meter movement resistance you must put a high value resistor in series with the meter.  Measure the sum of the movement resistance and the series resistor.  Subtract the series resistor leaves the movement resistance.  I use 10-turn pots to do these type measurements.  Start with about 20K for the series resistance.  Reduce the series resistance until the meter reads full scale current.  You need a digital ohm meter to get accurate results. You have to be careful with the digital ohm meter on different ranges, the amount of voltage coming from the ohm meter could be different on different ranges.  So, don't use the auto-range feature.  If you need to change meter ranges, first run the 20K pot back up, change ranges, and then start reducing it again.

20K is a good series resistor to start with for most meters like 1ma, 1/2ma meter movements.  Meter movements that are less, like 50uamp, you must start with much higher series resistors.

OK, why do you need to know the movement resistance??  You may need to replace that shunt wire.  If you do, you must first know the internal resistance of the meter movement itself.  You must also know the real full scale current of the meter movement.

Another simple test would be to put a 1.5 volt battery in series with the 20K pot along with your digital milliamp meter in series.  Adjust the pot to get full scale current reading on the meter.  Read the measured current on the digital meter.  Check to see how close the two meters track.

You may need to do this current test to check the accuracy of the shunt with the meter (150ma full scale)

Fred

[w4bfs]

I'm not sure if that meter has an internal shunt or not, there is a coil of insulated wire, looks like maybe 20 gauge, connected between the two terminals of the meter. I figured that was the shunt, because that was the only logical explanation I could think of for why that piece of wire was there, and I've seen it on other 90800s. I guess Millen probably took whatever meters they could get ahold of and just used their own custom scale on them, since there are no marks on the meters that say who made them.

What you say is very possible, hard to know what some companies did in the early days.  You can tell with an ohm meter.  You have to be somewhat careful trying to measure the resistance of a meter movement.  An ohm meter will put a few volts on the meter (without the shunt wire) and that will damage the meter.  If your meter doesn't have an internal shunt then you will need the external shunt that was on the meter terminals.  Hopefully you still have it.

To measure the meter movement resistance you must put a high value resistor in series with the meter.  Measure the sum of the movement resistance and the series resistor.  Subtract the series resistor leaves the movement resistance.  I use 10-turn pots to do these type measurements.  Start with about 20K for the series resistance.  Reduce the series resistance until the meter reads full scale current.  You need a digital ohm meter to get accurate results. You have to careful with the digital ohm meter on different ranges, the amount of voltage coming from the ohm meter could be different on different ranges.  So, don't use the auto-range feature.  If you need to change meter ranges, first run the 20K pot back up, change ranges, and then start reducing it again.

20K is a good series resistor to start with for most meters like 1ma, 1/2ma meter movements.  Meter movements that are less, like 50uamp, you must start with much higher series resistors.

OK, why do you need to know the movement resistance??  You may need to replace that shunt wire.  If you do, you must first know the internal resistance of the meter movement itself.  You must also know the real full scale current of the meter movement.

Another simple test would be to put a 1.5 volt battery in series with the 20K pot along with your digital milliamp meter in series.  Adjust the pot to get full scale current reading on the meter.  Read the measured current on the digital meter.  Check to see how close the two meters track.

You may need to do this current test to check the accuracy of the shunt with the meter (150ma full scale)

Fred

what Fred said