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$begingroup$
In a schematic I've been reviewing I see in only one spot that there is a 1/2v going somewhere? What does that mean?
simulate this circuit – Schematic created using CircuitLab
ground schematics
asked 1 hour ago
greyBowgreyBow
1085
New contributor
greyBow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
In a schematic I've been reviewing I see in only one spot that there is a 1/2v going somewhere? What does that mean?
simulate this circuit – Schematic created using CircuitLab
ground schematics
asked 1 hour ago
greyBowgreyBow
1085
New contributor
greyBow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
The schematic you've laid out here doesn't make sense, because a ground symbol is what we consider to be 0 V so marking it with a voltage is contradictory. Could you provide a photo/screenshot of the original schematic rather than your redrawing? There might be some subtlety missing. Other context such as what the schematic is supposed to be a part of, or demonstrate, might also be useful. (But thanks for taking the time to do the embedded schematic — it's usually better than alternatives!)
$endgroup$
– Kevin Reid
58 mins ago
$begingroup$
@KevinReid I've added the full schematic and highlighted the spots that the 1/2v shows up (it in two places actually)
$endgroup$
– greyBow
54 mins ago
2
$begingroup$
@greyBow It just means, in this case, $4.5:textV$.
$endgroup$
– jonk
50 mins ago
$begingroup$
@KevinReid Oh okay, I see, so the 1M resistor brings it down to 4.5v? But then where would it go?
$endgroup$
– greyBow
49 mins ago
1
$begingroup$
:( that symbol choice was pretty bad (especially in addition with this ground symbol instead of 3 lines)
$endgroup$
– Wesley Lee
31 mins ago
|
show 1 more comment
$begingroup$
In a schematic I've been reviewing I see in only one spot that there is a 1/2v going somewhere? What does that mean?
simulate this circuit – Schematic created using CircuitLab
ground schematics
asked 1 hour ago
greyBowgreyBow
1085
New contributor
greyBow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In a schematic I've been reviewing I see in only one spot that there is a 1/2v going somewhere? What does that mean?
simulate this circuit – Schematic created using CircuitLab
ground schematics
ground schematics
asked 1 hour ago
greyBowgreyBow
1085
New contributor
greyBow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
greyBowgreyBow
1085
New contributor
greyBow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 54 mins ago
greyBow
asked 1 hour ago
greyBowgreyBow
1085
New contributor
greyBow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
greyBowgreyBow
1085
asked 1 hour ago
greyBowgreyBow
1085
1085
New contributor
greyBow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
greyBow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
The schematic you've laid out here doesn't make sense, because a ground symbol is what we consider to be 0 V so marking it with a voltage is contradictory. Could you provide a photo/screenshot of the original schematic rather than your redrawing? There might be some subtlety missing. Other context such as what the schematic is supposed to be a part of, or demonstrate, might also be useful. (But thanks for taking the time to do the embedded schematic — it's usually better than alternatives!)
$endgroup$
– Kevin Reid
58 mins ago
$begingroup$
@KevinReid I've added the full schematic and highlighted the spots that the 1/2v shows up (it in two places actually)
$endgroup$
– greyBow
54 mins ago
2
$begingroup$
@greyBow It just means, in this case, $4.5:textV$.
$endgroup$
– jonk
50 mins ago
$begingroup$
@KevinReid Oh okay, I see, so the 1M resistor brings it down to 4.5v? But then where would it go?
$endgroup$
– greyBow
49 mins ago
1
$begingroup$
:( that symbol choice was pretty bad (especially in addition with this ground symbol instead of 3 lines)
$endgroup$
– Wesley Lee
31 mins ago
|
show 1 more comment
1
$begingroup$
The schematic you've laid out here doesn't make sense, because a ground symbol is what we consider to be 0 V so marking it with a voltage is contradictory. Could you provide a photo/screenshot of the original schematic rather than your redrawing? There might be some subtlety missing. Other context such as what the schematic is supposed to be a part of, or demonstrate, might also be useful. (But thanks for taking the time to do the embedded schematic — it's usually better than alternatives!)
$endgroup$
– Kevin Reid
58 mins ago
$begingroup$
@KevinReid I've added the full schematic and highlighted the spots that the 1/2v shows up (it in two places actually)
$endgroup$
– greyBow
54 mins ago
2
$begingroup$
@greyBow It just means, in this case, $4.5:textV$.
$endgroup$
– jonk
50 mins ago
$begingroup$
@KevinReid Oh okay, I see, so the 1M resistor brings it down to 4.5v? But then where would it go?
$endgroup$
– greyBow
49 mins ago
1
$begingroup$
:( that symbol choice was pretty bad (especially in addition with this ground symbol instead of 3 lines)
$endgroup$
– Wesley Lee
31 mins ago
1
1
$begingroup$
The schematic you've laid out here doesn't make sense, because a ground symbol is what we consider to be 0 V so marking it with a voltage is contradictory. Could you provide a photo/screenshot of the original schematic rather than your redrawing? There might be some subtlety missing. Other context such as what the schematic is supposed to be a part of, or demonstrate, might also be useful. (But thanks for taking the time to do the embedded schematic — it's usually better than alternatives!)
$endgroup$
– Kevin Reid
58 mins ago
$begingroup$
The schematic you've laid out here doesn't make sense, because a ground symbol is what we consider to be 0 V so marking it with a voltage is contradictory. Could you provide a photo/screenshot of the original schematic rather than your redrawing? There might be some subtlety missing. Other context such as what the schematic is supposed to be a part of, or demonstrate, might also be useful. (But thanks for taking the time to do the embedded schematic — it's usually better than alternatives!)
$endgroup$
– Kevin Reid
58 mins ago
$begingroup$
@KevinReid I've added the full schematic and highlighted the spots that the 1/2v shows up (it in two places actually)
$endgroup$
– greyBow
54 mins ago
$begingroup$
@KevinReid I've added the full schematic and highlighted the spots that the 1/2v shows up (it in two places actually)
$endgroup$
– greyBow
54 mins ago
2
2
$begingroup$
@greyBow It just means, in this case, $4.5:textV$.
$endgroup$
– jonk
50 mins ago
$begingroup$
@greyBow It just means, in this case, $4.5:textV$.
$endgroup$
– jonk
50 mins ago
$begingroup$
@KevinReid Oh okay, I see, so the 1M resistor brings it down to 4.5v? But then where would it go?
$endgroup$
– greyBow
49 mins ago
$begingroup$
@KevinReid Oh okay, I see, so the 1M resistor brings it down to 4.5v? But then where would it go?
$endgroup$
– greyBow
49 mins ago
1
1
$begingroup$
:( that symbol choice was pretty bad (especially in addition with this ground symbol instead of 3 lines)
$endgroup$
– Wesley Lee
31 mins ago
$begingroup$
:( that symbol choice was pretty bad (especially in addition with this ground symbol instead of 3 lines)
$endgroup$
– Wesley Lee
31 mins ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Look at the top of R20 - that is labeled V and is the supply rail. (V is also connected to 9VDC which is the power input - see the DC connector and battery, towards the top left of the schematic.)
Therefore, as commented by jonk, the node at the junction of equal resistors R20 and R21 must be half of V hence 1/2V means exactly that.
Also, looking carefully, the arrow symbols labeled 1/2V are slightly smaller than the arrows which are the ground symbol. On the full schematic you can compare their size and see the difference - but otherwise, that choice of arrow by the designer could easily be confusing! As kindly pointed out by Kevin in the comments, that smaller arrow is being used here as the symbol for a named node.
answered 36 mins ago
SamGibsonSamGibson
11.8k41739
$endgroup$
1
$begingroup$
Might be worth pointing out that the same arrow symbol is used to connect "9VDC". So it's a general named node symbol.
$endgroup$
– Kevin Reid
34 mins ago
$begingroup$
Thank you for the explanation, so then does that mean that 1/2v coming off the junction between r20 and r21 would then connect to the 1/2v coming off of R2?
$endgroup$
– greyBow
29 mins ago
$begingroup$
@greyBow Yes. Those nets are tied together.
$endgroup$
– jonk
27 mins ago
$begingroup$
@greyBow - Yes, exactly that. That "half the supply voltage" is being used to bias the gate ofQ1viaR2.
$endgroup$
– SamGibson
26 mins ago
1
$begingroup$
I see. Thank you SO much for the help, this has been extremely helpful. Many, many thanks!
$endgroup$
– greyBow
25 mins ago
add a comment |
$begingroup$
The down arrow below R2 in the original schematic is not a Ground symbol - it indicates "this point is connected to something that-a-way" - it connects to an upward-pointing arrow in the power supply section below. That point will be at half the 9 V power supply voltage due to R20 and R21. It provides an appropriate bias voltage for Q1.
Drawing a line to show the connection would make much more sense.
Edit: as @kevin pointed out, that narrow arrow symbol is used as a general named signal marker - all such arrows with the same name will be connected together.
answered 32 mins ago
Peter BennettPeter Bennett
38.2k13068
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Look at the top of R20 - that is labeled V and is the supply rail. (V is also connected to 9VDC which is the power input - see the DC connector and battery, towards the top left of the schematic.)
Therefore, as commented by jonk, the node at the junction of equal resistors R20 and R21 must be half of V hence 1/2V means exactly that.
Also, looking carefully, the arrow symbols labeled 1/2V are slightly smaller than the arrows which are the ground symbol. On the full schematic you can compare their size and see the difference - but otherwise, that choice of arrow by the designer could easily be confusing! As kindly pointed out by Kevin in the comments, that smaller arrow is being used here as the symbol for a named node.
answered 36 mins ago
SamGibsonSamGibson
11.8k41739
$endgroup$
1
$begingroup$
Might be worth pointing out that the same arrow symbol is used to connect "9VDC". So it's a general named node symbol.
$endgroup$
– Kevin Reid
34 mins ago
$begingroup$
Thank you for the explanation, so then does that mean that 1/2v coming off the junction between r20 and r21 would then connect to the 1/2v coming off of R2?
$endgroup$
– greyBow
29 mins ago
$begingroup$
@greyBow Yes. Those nets are tied together.
$endgroup$
– jonk
27 mins ago
$begingroup$
@greyBow - Yes, exactly that. That "half the supply voltage" is being used to bias the gate ofQ1viaR2.
$endgroup$
– SamGibson
26 mins ago
1
$begingroup$
I see. Thank you SO much for the help, this has been extremely helpful. Many, many thanks!
$endgroup$
– greyBow
25 mins ago
add a comment |
$begingroup$
Look at the top of R20 - that is labeled V and is the supply rail. (V is also connected to 9VDC which is the power input - see the DC connector and battery, towards the top left of the schematic.)
Therefore, as commented by jonk, the node at the junction of equal resistors R20 and R21 must be half of V hence 1/2V means exactly that.
Also, looking carefully, the arrow symbols labeled 1/2V are slightly smaller than the arrows which are the ground symbol. On the full schematic you can compare their size and see the difference - but otherwise, that choice of arrow by the designer could easily be confusing! As kindly pointed out by Kevin in the comments, that smaller arrow is being used here as the symbol for a named node.
answered 36 mins ago
SamGibsonSamGibson
11.8k41739
$endgroup$
1
$begingroup$
Might be worth pointing out that the same arrow symbol is used to connect "9VDC". So it's a general named node symbol.
$endgroup$
– Kevin Reid
34 mins ago
$begingroup$
Thank you for the explanation, so then does that mean that 1/2v coming off the junction between r20 and r21 would then connect to the 1/2v coming off of R2?
$endgroup$
– greyBow
29 mins ago
$begingroup$
@greyBow Yes. Those nets are tied together.
$endgroup$
– jonk
27 mins ago
$begingroup$
@greyBow - Yes, exactly that. That "half the supply voltage" is being used to bias the gate ofQ1viaR2.
$endgroup$
– SamGibson
26 mins ago
1
$begingroup$
I see. Thank you SO much for the help, this has been extremely helpful. Many, many thanks!
$endgroup$
– greyBow
25 mins ago
add a comment |
$begingroup$
Look at the top of R20 - that is labeled V and is the supply rail. (V is also connected to 9VDC which is the power input - see the DC connector and battery, towards the top left of the schematic.)
Therefore, as commented by jonk, the node at the junction of equal resistors R20 and R21 must be half of V hence 1/2V means exactly that.
Also, looking carefully, the arrow symbols labeled 1/2V are slightly smaller than the arrows which are the ground symbol. On the full schematic you can compare their size and see the difference - but otherwise, that choice of arrow by the designer could easily be confusing! As kindly pointed out by Kevin in the comments, that smaller arrow is being used here as the symbol for a named node.
answered 36 mins ago
SamGibsonSamGibson
11.8k41739
$endgroup$
Look at the top of R20 - that is labeled V and is the supply rail. (V is also connected to 9VDC which is the power input - see the DC connector and battery, towards the top left of the schematic.)
Therefore, as commented by jonk, the node at the junction of equal resistors R20 and R21 must be half of V hence 1/2V means exactly that.
Also, looking carefully, the arrow symbols labeled 1/2V are slightly smaller than the arrows which are the ground symbol. On the full schematic you can compare their size and see the difference - but otherwise, that choice of arrow by the designer could easily be confusing! As kindly pointed out by Kevin in the comments, that smaller arrow is being used here as the symbol for a named node.
answered 36 mins ago
SamGibsonSamGibson
11.8k41739
edited 22 mins ago
answered 36 mins ago
SamGibsonSamGibson
11.8k41739
answered 36 mins ago
SamGibsonSamGibson
11.8k41739
answered 36 mins ago
SamGibsonSamGibson
11.8k41739
11.8k41739
1
$begingroup$
Might be worth pointing out that the same arrow symbol is used to connect "9VDC". So it's a general named node symbol.
$endgroup$
– Kevin Reid
34 mins ago
$begingroup$
Thank you for the explanation, so then does that mean that 1/2v coming off the junction between r20 and r21 would then connect to the 1/2v coming off of R2?
$endgroup$
– greyBow
29 mins ago
$begingroup$
@greyBow Yes. Those nets are tied together.
$endgroup$
– jonk
27 mins ago
$begingroup$
@greyBow - Yes, exactly that. That "half the supply voltage" is being used to bias the gate ofQ1viaR2.
$endgroup$
– SamGibson
26 mins ago
1
$begingroup$
I see. Thank you SO much for the help, this has been extremely helpful. Many, many thanks!
$endgroup$
– greyBow
25 mins ago
add a comment |
1
$begingroup$
Might be worth pointing out that the same arrow symbol is used to connect "9VDC". So it's a general named node symbol.
$endgroup$
– Kevin Reid
34 mins ago
$begingroup$
Thank you for the explanation, so then does that mean that 1/2v coming off the junction between r20 and r21 would then connect to the 1/2v coming off of R2?
$endgroup$
– greyBow
29 mins ago
$begingroup$
@greyBow Yes. Those nets are tied together.
$endgroup$
– jonk
27 mins ago
$begingroup$
@greyBow - Yes, exactly that. That "half the supply voltage" is being used to bias the gate ofQ1viaR2.
$endgroup$
– SamGibson
26 mins ago
1
$begingroup$
I see. Thank you SO much for the help, this has been extremely helpful. Many, many thanks!
$endgroup$
– greyBow
25 mins ago
1
1
$begingroup$
Might be worth pointing out that the same arrow symbol is used to connect "9VDC". So it's a general named node symbol.
$endgroup$
– Kevin Reid
34 mins ago
$begingroup$
Might be worth pointing out that the same arrow symbol is used to connect "9VDC". So it's a general named node symbol.
$endgroup$
– Kevin Reid
34 mins ago
$begingroup$
Thank you for the explanation, so then does that mean that 1/2v coming off the junction between r20 and r21 would then connect to the 1/2v coming off of R2?
$endgroup$
– greyBow
29 mins ago
$begingroup$
Thank you for the explanation, so then does that mean that 1/2v coming off the junction between r20 and r21 would then connect to the 1/2v coming off of R2?
$endgroup$
– greyBow
29 mins ago
$begingroup$
@greyBow Yes. Those nets are tied together.
$endgroup$
– jonk
27 mins ago
$begingroup$
@greyBow Yes. Those nets are tied together.
$endgroup$
– jonk
27 mins ago
$begingroup$
@greyBow - Yes, exactly that. That "half the supply voltage" is being used to bias the gate of
Q1 via R2.$endgroup$
– SamGibson
26 mins ago
$begingroup$
@greyBow - Yes, exactly that. That "half the supply voltage" is being used to bias the gate of
Q1 via R2.$endgroup$
– SamGibson
26 mins ago
1
1
$begingroup$
I see. Thank you SO much for the help, this has been extremely helpful. Many, many thanks!
$endgroup$
– greyBow
25 mins ago
$begingroup$
I see. Thank you SO much for the help, this has been extremely helpful. Many, many thanks!
$endgroup$
– greyBow
25 mins ago
add a comment |
$begingroup$
The down arrow below R2 in the original schematic is not a Ground symbol - it indicates "this point is connected to something that-a-way" - it connects to an upward-pointing arrow in the power supply section below. That point will be at half the 9 V power supply voltage due to R20 and R21. It provides an appropriate bias voltage for Q1.
Drawing a line to show the connection would make much more sense.
Edit: as @kevin pointed out, that narrow arrow symbol is used as a general named signal marker - all such arrows with the same name will be connected together.
answered 32 mins ago
Peter BennettPeter Bennett
38.2k13068
$endgroup$
add a comment |
$begingroup$
The down arrow below R2 in the original schematic is not a Ground symbol - it indicates "this point is connected to something that-a-way" - it connects to an upward-pointing arrow in the power supply section below. That point will be at half the 9 V power supply voltage due to R20 and R21. It provides an appropriate bias voltage for Q1.
Drawing a line to show the connection would make much more sense.
Edit: as @kevin pointed out, that narrow arrow symbol is used as a general named signal marker - all such arrows with the same name will be connected together.
answered 32 mins ago
Peter BennettPeter Bennett
38.2k13068
$endgroup$
add a comment |
$begingroup$
The down arrow below R2 in the original schematic is not a Ground symbol - it indicates "this point is connected to something that-a-way" - it connects to an upward-pointing arrow in the power supply section below. That point will be at half the 9 V power supply voltage due to R20 and R21. It provides an appropriate bias voltage for Q1.
Drawing a line to show the connection would make much more sense.
Edit: as @kevin pointed out, that narrow arrow symbol is used as a general named signal marker - all such arrows with the same name will be connected together.
answered 32 mins ago
Peter BennettPeter Bennett
38.2k13068
$endgroup$
The down arrow below R2 in the original schematic is not a Ground symbol - it indicates "this point is connected to something that-a-way" - it connects to an upward-pointing arrow in the power supply section below. That point will be at half the 9 V power supply voltage due to R20 and R21. It provides an appropriate bias voltage for Q1.
Drawing a line to show the connection would make much more sense.
Edit: as @kevin pointed out, that narrow arrow symbol is used as a general named signal marker - all such arrows with the same name will be connected together.
answered 32 mins ago
Peter BennettPeter Bennett
38.2k13068
edited 19 mins ago
answered 32 mins ago
Peter BennettPeter Bennett
38.2k13068
answered 32 mins ago
Peter BennettPeter Bennett
38.2k13068
answered 32 mins ago
Peter BennettPeter Bennett
38.2k13068
38.2k13068
add a comment |
add a comment |
greyBow is a new contributor. Be nice, and check out our Code of Conduct.
greyBow is a new contributor. Be nice, and check out our Code of Conduct.
greyBow is a new contributor. Be nice, and check out our Code of Conduct.
greyBow is a new contributor. Be nice, and check out our Code of Conduct.
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The schematic you've laid out here doesn't make sense, because a ground symbol is what we consider to be 0 V so marking it with a voltage is contradictory. Could you provide a photo/screenshot of the original schematic rather than your redrawing? There might be some subtlety missing. Other context such as what the schematic is supposed to be a part of, or demonstrate, might also be useful. (But thanks for taking the time to do the embedded schematic — it's usually better than alternatives!)
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– Kevin Reid
58 mins ago
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@KevinReid I've added the full schematic and highlighted the spots that the 1/2v shows up (it in two places actually)
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– greyBow
54 mins ago
2
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@greyBow It just means, in this case, $4.5:textV$.
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– jonk
50 mins ago
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@KevinReid Oh okay, I see, so the 1M resistor brings it down to 4.5v? But then where would it go?
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– greyBow
49 mins ago
1
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:( that symbol choice was pretty bad (especially in addition with this ground symbol instead of 3 lines)
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– Wesley Lee
31 mins ago