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Old 28th June 2004, 05:59 PM   #1
dhenryp is offline dhenryp  United States
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Default ESP Ground loop break diodes

Rod Elliot's article on grounding (http://sound.westhost.com/earthing.htm) shows a ground loop breaking circuit comprised of diodes, resistor and capacitor in parallel. He mentioned in the article that the difference between mains earth and circuit zero volts often gets to 1-2V. The rectifier diodes have a forward voltage drop ~.5-.7 v. It seems to me that the diodes will drop the difference voltage by that amount but they present a very low forward resistance above that voltage. Don't these diodes effectively short out the resistor and capacitor for voltages above the .5-.7v forward drop? If this is true, the circuit would not be very effective for ground differences much above the forward voltage drop.
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Old 28th June 2004, 06:38 PM   #2
sss is offline sss  Israel
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if the zero volt line and the earth will have the same potential then u'll have no ground loop .when the difference is too high then the diodes short out the cap and resistor , and its for the better , they keep the difference to approx 0.7V .u cant connect the zero volt line to the earth for safety reasons
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Old 28th June 2004, 07:44 PM   #3
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What happens here is that the R and C break the ground loop by forcing any circulating currents in their intended path and not through the zero volts line/chassis ground connection. This is good practise. However, if the differential ground gets too high it may burn out the R, and the diodes are there to prevent that. This is overall a good solution. I would select lower C value, but that depends also a lot on the particular installation.

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Old 28th June 2004, 08:17 PM   #4
dhenryp is offline dhenryp  United States
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Thanks for the replies but I'm still confused. I thought the purpose of the diodes was to provide a high current path to earth in the case of mains voltage somehow making it on to the "zero volts" lines.

I thought the purpose of the resistor was to reduce the size of the currents circulating through the shield connecting the amp and the source caused by the normal 1-2 volts differential. The capacitor is there just to short out rf level hash.

In Rod's example, he said you might have a .5 ohm resistance between zero volts and earth. If you had a 2 volt difference and a .5 ohm interconnect resistance you would end up with 4 amps (2/.5)flowing in the source cables ground shield. If you were to just add the 10 ohm resistor you would end up with .19 amps (2/10.5). This is a big improvement (~20db?) (FYI: This works out to something like .4 watts dissipated across the resistor so it seems that Rod's 10 watt rating is quite conservative.)

If you add diodes with a .5 volt (to make the math easier) forward drop in parallel with the resistor and you have a 2 v difference between earth and 0 volts you get a much smaller improvement. Now you have .5 volts across the diodes and 1.5 volts across the .5 ohm interconnection resistance. This gives you current flows between earth and 0 volts of 3 amps (1.5 v / .5 ohm). This is a 25% improvement while the resistor on its own gave you a 2000% improvement. The resistor is effectively shorted out by the diodes above .5 volts so it is only effective in reducing ground loop currents when the differential is less than .5 v.

I think there must be a flaw in my logic (and I obviously am having no luck finding it) or differences between earth and 0 v seldom get much above .5 - .7 volts.
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Old 28th June 2004, 10:03 PM   #5
sss is offline sss  Israel
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use balanced inputs/outputs - this will solve all your problems
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Old 29th June 2004, 08:51 AM   #6
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"use balanced inputs/outputs - this will solve all your problems "

Thanks, I needed a laugh.

Very bizzare things happen when the 10 ohm resistor burns open from a ground loop.

The purpose of the diodes in the ESP circuit are to generate massive buzz if you have a real bad ground loop, hopefully you will be motivated to solve the problem at that point.

When you are ready to give up, remember:

An unbalanced line with a transformer on the receiving end is (in actual use) quieter than any electronic balanced circuit.
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Old 29th June 2004, 12:52 PM   #7
dhenryp is offline dhenryp  United States
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It's like somebody asking a question how a car's transmission works on a car forum and getting told to switch to an electric car because they don't need transmissions.

Can anybody answer my actual question? Can anyone share their experience with a ground loop break with diodes?
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Old 29th June 2004, 01:06 PM   #8
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Quote:
Originally posted by dhenryp
It's like somebody asking a question how a car's transmission works on a car forum and getting told to switch to an electric car because they don't need transmissions.

Can anybody answer my actual question? Can anyone share their experience with a ground loop break with diodes?

Hi,

Did you read my post # 3? Was there a part that was not clear?
The argumentation in post # 4 is incorrect. It is not the R and V that determine the I, just the opposite. You have a given ground current due to non-optimal construction or layout. With the given R, that results in *some* V. If the V (and the I, same thing) gets too high it may burn out the R. To prevent this, the diodes shunt current past the R and limit the dissipation in the R.

In normal use, the ground current should not be too big and the diodes are really a second line defense.

And forget about the transmission..

Jan Didden
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Old 29th June 2004, 11:18 PM   #9
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Quote:
Originally posted by dhenryp
It's like somebody asking a question how a car's transmission works on a car forum and getting told to switch to an electric car because they don't need transmissions.
u right , sorry
i think Rod explains it best :
Quote:
taken from Rods site

The loop breaker works by adding a resistance in the earth return circuit. This reduces loop currents to a very small value, and thus "breaks" the loop. The capacitor in parallel ensures that the electronics are connected to the chassis for radio frequency signals, and helps to prevent radio frequency interference. Finally, the diode bridge provides the path for fault currents. The use of a large chassis mounting (35A) type is suggested, since this will be able to handle the possibly very high fault currents that may occur without becoming open circuit. Note the way the bridge is wired, with the two AC terminals shorted, and the two DC terminals shorted. Other connection possibilities are dangerous, and must be avoided.

In the event of a major fault, one (or more) of the diodes in the bridge will possibly fail. Semiconductors (nearly) always fail as short circuit, and only become open circuited if the fault current continues and "blows" the interconnecting wires. High current bridge rectifiers have very solid conductors throughout, and open circuit diodes are very rare (I have never seen a high power bridge go open circuit - so far at least). Use of the bridge means that there are two diodes in parallel for fault current of either polarity, so the likelihood of failure (to protect) is very small indeed.
one more thing
[quote]taken from Rods site
While very effective (and safe), as mentioned above such a circuit might not be legal where you live. If this is the case and hum is causing you grief, the use of balanced interconnects will solve the problem
[/qoute]



i think it is illegal to connect the zero line to the earth everywhere

i think the earth connection is not needed for amps/ pre amps because they use transformers so i dont connect it , many commercial amps/rrecievers dont use it eather .
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