Metal bar on DMM PCBIs this a cheap Chinese fuse or a current shunt?Are these two Arduino Pro Micros jumpered correctly?What is this component with metal tabs?What are these metal plates covering parts of PCBs called?What are those small metal bars going across a PCB for?What is this connector? Round 8 pins, 5 notches on metal shellHow to control this 72-LED light bar PCBIdentification of SMD componentWhat is up with this fuse?Need help identifying replacement components on PCBAdding SMA conectors to a device whose metal casing is connected to earth ground through at a single point
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Metal bar on DMM PCB
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Metal bar on DMM PCB
Is this a cheap Chinese fuse or a current shunt?Are these two Arduino Pro Micros jumpered correctly?What is this component with metal tabs?What are these metal plates covering parts of PCBs called?What are those small metal bars going across a PCB for?What is this connector? Round 8 pins, 5 notches on metal shellHow to control this 72-LED light bar PCBIdentification of SMD componentWhat is up with this fuse?Need help identifying replacement components on PCBAdding SMA conectors to a device whose metal casing is connected to earth ground through at a single point
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$begingroup$
What is this large metal bar inside my DMMs? One of them labeled it ST. They appear to be connected to the COM port or fuse. Is this just a big jumper for the 10 A ammeter?
pcb identification
asked 8 hours ago
EricEric
161
New contributor
Eric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
What is this large metal bar inside my DMMs? One of them labeled it ST. They appear to be connected to the COM port or fuse. Is this just a big jumper for the 10 A ammeter?
pcb identification
asked 8 hours ago
EricEric
161
New contributor
Eric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
@bitsmack: Either way is fine.
$endgroup$
– JRE
8 hours ago
$begingroup$
@bitsmack: I removed my close vote on this one, and put the other one up for closing.
$endgroup$
– JRE
8 hours ago
$begingroup$
@JRE Looks like Dave Tweed marked the other question as a dupe. (Thanks, Dave!) I'm removing my comments here...
$endgroup$
– bitsmack
4 hours ago
add a comment |
$begingroup$
What is this large metal bar inside my DMMs? One of them labeled it ST. They appear to be connected to the COM port or fuse. Is this just a big jumper for the 10 A ammeter?
pcb identification
asked 8 hours ago
EricEric
161
New contributor
Eric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
What is this large metal bar inside my DMMs? One of them labeled it ST. They appear to be connected to the COM port or fuse. Is this just a big jumper for the 10 A ammeter?
pcb identification
pcb identification
asked 8 hours ago
EricEric
161
New contributor
Eric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
EricEric
161
New contributor
Eric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
EricEric
161
New contributor
Eric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
EricEric
161
asked 8 hours ago
EricEric
161
161
New contributor
Eric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Eric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
@bitsmack: Either way is fine.
$endgroup$
– JRE
8 hours ago
$begingroup$
@bitsmack: I removed my close vote on this one, and put the other one up for closing.
$endgroup$
– JRE
8 hours ago
$begingroup$
@JRE Looks like Dave Tweed marked the other question as a dupe. (Thanks, Dave!) I'm removing my comments here...
$endgroup$
– bitsmack
4 hours ago
add a comment |
$begingroup$
@bitsmack: Either way is fine.
$endgroup$
– JRE
8 hours ago
$begingroup$
@bitsmack: I removed my close vote on this one, and put the other one up for closing.
$endgroup$
– JRE
8 hours ago
$begingroup$
@JRE Looks like Dave Tweed marked the other question as a dupe. (Thanks, Dave!) I'm removing my comments here...
$endgroup$
– bitsmack
4 hours ago
$begingroup$
@bitsmack: Either way is fine.
$endgroup$
– JRE
8 hours ago
$begingroup$
@bitsmack: Either way is fine.
$endgroup$
– JRE
8 hours ago
$begingroup$
@bitsmack: I removed my close vote on this one, and put the other one up for closing.
$endgroup$
– JRE
8 hours ago
$begingroup$
@bitsmack: I removed my close vote on this one, and put the other one up for closing.
$endgroup$
– JRE
8 hours ago
$begingroup$
@JRE Looks like Dave Tweed marked the other question as a dupe. (Thanks, Dave!) I'm removing my comments here...
$endgroup$
– bitsmack
4 hours ago
$begingroup$
@JRE Looks like Dave Tweed marked the other question as a dupe. (Thanks, Dave!) I'm removing my comments here...
$endgroup$
– bitsmack
4 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
That is not a simple jumper.
That is the precision resistor used to measure the current. This is also known as a "shunt" - hence the designation ST for shunt.
You measure current by passing it through a known resistance and measuring the voltage across that resistor. Using Ohm's law, you can calculate the current from the voltage and the resistance.
If you look closely, you will see that one of them has been trimmed by making nicks in the wire. That changes the resistance slightly. You measure a known current with a new meter, then whack on the shunt to make your new meter display the known current.
The thick ones like that are usually for the 10A range. The lower current shunts are usually small, precision resistors on the board.
answered 8 hours ago
JREJRE
25.6k64585
$endgroup$
add a comment |
$begingroup$
It is the current shunt.
Your meter probably has a 200 mV full scale range and will read 10.00 A with 100 mV voltage drop across the shunt. From Ohm's Law we can calculate that the shunt resistance = V/I = 0.1/10 = 0.01 Ω.
A decent meter will have a proper fuse protecting the shunt. The fuses in your photo look too small so be very careful.
answered 8 hours ago
TransistorTransistor
92.1k788199
$endgroup$
add a comment |
$begingroup$
It is hollow copper tubing or a "cheap & dirty" 1% current shunt resistor for measuring on the 10A using mV
Here's a Murata 0.25% current shunt.
$20. See the difference?
Due to the PTC characteristic of metal conductors, heat causes the resistance to increase and yield a false rise in voltage sensed as a current. Generally, Voltage drops for I sense are limited to 50mV for this reason unless extraordinary heatsinks are used.
answered 7 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
75k229106
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That is not a simple jumper.
That is the precision resistor used to measure the current. This is also known as a "shunt" - hence the designation ST for shunt.
You measure current by passing it through a known resistance and measuring the voltage across that resistor. Using Ohm's law, you can calculate the current from the voltage and the resistance.
If you look closely, you will see that one of them has been trimmed by making nicks in the wire. That changes the resistance slightly. You measure a known current with a new meter, then whack on the shunt to make your new meter display the known current.
The thick ones like that are usually for the 10A range. The lower current shunts are usually small, precision resistors on the board.
answered 8 hours ago
JREJRE
25.6k64585
$endgroup$
add a comment |
$begingroup$
That is not a simple jumper.
That is the precision resistor used to measure the current. This is also known as a "shunt" - hence the designation ST for shunt.
You measure current by passing it through a known resistance and measuring the voltage across that resistor. Using Ohm's law, you can calculate the current from the voltage and the resistance.
If you look closely, you will see that one of them has been trimmed by making nicks in the wire. That changes the resistance slightly. You measure a known current with a new meter, then whack on the shunt to make your new meter display the known current.
The thick ones like that are usually for the 10A range. The lower current shunts are usually small, precision resistors on the board.
answered 8 hours ago
JREJRE
25.6k64585
$endgroup$
add a comment |
$begingroup$
That is not a simple jumper.
That is the precision resistor used to measure the current. This is also known as a "shunt" - hence the designation ST for shunt.
You measure current by passing it through a known resistance and measuring the voltage across that resistor. Using Ohm's law, you can calculate the current from the voltage and the resistance.
If you look closely, you will see that one of them has been trimmed by making nicks in the wire. That changes the resistance slightly. You measure a known current with a new meter, then whack on the shunt to make your new meter display the known current.
The thick ones like that are usually for the 10A range. The lower current shunts are usually small, precision resistors on the board.
answered 8 hours ago
JREJRE
25.6k64585
$endgroup$
That is not a simple jumper.
That is the precision resistor used to measure the current. This is also known as a "shunt" - hence the designation ST for shunt.
You measure current by passing it through a known resistance and measuring the voltage across that resistor. Using Ohm's law, you can calculate the current from the voltage and the resistance.
If you look closely, you will see that one of them has been trimmed by making nicks in the wire. That changes the resistance slightly. You measure a known current with a new meter, then whack on the shunt to make your new meter display the known current.
The thick ones like that are usually for the 10A range. The lower current shunts are usually small, precision resistors on the board.
answered 8 hours ago
JREJRE
25.6k64585
edited 8 hours ago
answered 8 hours ago
JREJRE
25.6k64585
answered 8 hours ago
JREJRE
25.6k64585
answered 8 hours ago
JREJRE
25.6k64585
25.6k64585
add a comment |
add a comment |
$begingroup$
It is the current shunt.
Your meter probably has a 200 mV full scale range and will read 10.00 A with 100 mV voltage drop across the shunt. From Ohm's Law we can calculate that the shunt resistance = V/I = 0.1/10 = 0.01 Ω.
A decent meter will have a proper fuse protecting the shunt. The fuses in your photo look too small so be very careful.
answered 8 hours ago
TransistorTransistor
92.1k788199
$endgroup$
add a comment |
$begingroup$
It is the current shunt.
Your meter probably has a 200 mV full scale range and will read 10.00 A with 100 mV voltage drop across the shunt. From Ohm's Law we can calculate that the shunt resistance = V/I = 0.1/10 = 0.01 Ω.
A decent meter will have a proper fuse protecting the shunt. The fuses in your photo look too small so be very careful.
answered 8 hours ago
TransistorTransistor
92.1k788199
$endgroup$
add a comment |
$begingroup$
It is the current shunt.
Your meter probably has a 200 mV full scale range and will read 10.00 A with 100 mV voltage drop across the shunt. From Ohm's Law we can calculate that the shunt resistance = V/I = 0.1/10 = 0.01 Ω.
A decent meter will have a proper fuse protecting the shunt. The fuses in your photo look too small so be very careful.
answered 8 hours ago
TransistorTransistor
92.1k788199
$endgroup$
It is the current shunt.
Your meter probably has a 200 mV full scale range and will read 10.00 A with 100 mV voltage drop across the shunt. From Ohm's Law we can calculate that the shunt resistance = V/I = 0.1/10 = 0.01 Ω.
A decent meter will have a proper fuse protecting the shunt. The fuses in your photo look too small so be very careful.
answered 8 hours ago
TransistorTransistor
92.1k788199
answered 8 hours ago
TransistorTransistor
92.1k788199
answered 8 hours ago
TransistorTransistor
92.1k788199
answered 8 hours ago
TransistorTransistor
92.1k788199
92.1k788199
add a comment |
add a comment |
$begingroup$
It is hollow copper tubing or a "cheap & dirty" 1% current shunt resistor for measuring on the 10A using mV
Here's a Murata 0.25% current shunt. $20. See the difference?
Due to the PTC characteristic of metal conductors, heat causes the resistance to increase and yield a false rise in voltage sensed as a current. Generally, Voltage drops for I sense are limited to 50mV for this reason unless extraordinary heatsinks are used.
answered 7 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
75k229106
$endgroup$
add a comment |
$begingroup$
It is hollow copper tubing or a "cheap & dirty" 1% current shunt resistor for measuring on the 10A using mV
Here's a Murata 0.25% current shunt. $20. See the difference?
Due to the PTC characteristic of metal conductors, heat causes the resistance to increase and yield a false rise in voltage sensed as a current. Generally, Voltage drops for I sense are limited to 50mV for this reason unless extraordinary heatsinks are used.
answered 7 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
75k229106
$endgroup$
add a comment |
$begingroup$
It is hollow copper tubing or a "cheap & dirty" 1% current shunt resistor for measuring on the 10A using mV
Here's a Murata 0.25% current shunt. $20. See the difference?
Due to the PTC characteristic of metal conductors, heat causes the resistance to increase and yield a false rise in voltage sensed as a current. Generally, Voltage drops for I sense are limited to 50mV for this reason unless extraordinary heatsinks are used.
answered 7 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
75k229106
$endgroup$
It is hollow copper tubing or a "cheap & dirty" 1% current shunt resistor for measuring on the 10A using mV
Here's a Murata 0.25% current shunt. $20. See the difference?
Due to the PTC characteristic of metal conductors, heat causes the resistance to increase and yield a false rise in voltage sensed as a current. Generally, Voltage drops for I sense are limited to 50mV for this reason unless extraordinary heatsinks are used.
answered 7 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
75k229106
answered 7 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
75k229106
answered 7 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
75k229106
answered 7 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
75k229106
75k229106
add a comment |
add a comment |
Eric is a new contributor. Be nice, and check out our Code of Conduct.
Eric is a new contributor. Be nice, and check out our Code of Conduct.
Eric is a new contributor. Be nice, and check out our Code of Conduct.
Eric is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
@bitsmack: Either way is fine.
$endgroup$
– JRE
8 hours ago
$begingroup$
@bitsmack: I removed my close vote on this one, and put the other one up for closing.
$endgroup$
– JRE
8 hours ago
$begingroup$
@JRE Looks like Dave Tweed marked the other question as a dupe. (Thanks, Dave!) I'm removing my comments here...
$endgroup$
– bitsmack
4 hours ago