Why would you need an op amp for reference voltage when the voltage divider does the trick?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways? Can the AD8031 op amp be removed? Also, to save power, can the voltage divider be replaced with a buck converter?
power-supply op-amp filter
asked Jan 10 at 1:13
Tapatio SombreroTapatio Sombrero
15217
$endgroup$
add a comment |
$begingroup$
Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways? Can the AD8031 op amp be removed? Also, to save power, can the voltage divider be replaced with a buck converter?
power-supply op-amp filter
asked Jan 10 at 1:13
Tapatio SombreroTapatio Sombrero
15217
$endgroup$
add a comment |
$begingroup$
Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways? Can the AD8031 op amp be removed? Also, to save power, can the voltage divider be replaced with a buck converter?
power-supply op-amp filter
asked Jan 10 at 1:13
Tapatio SombreroTapatio Sombrero
15217
$endgroup$
Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways? Can the AD8031 op amp be removed? Also, to save power, can the voltage divider be replaced with a buck converter?
power-supply op-amp filter
power-supply op-amp filter
asked Jan 10 at 1:13
Tapatio SombreroTapatio Sombrero
15217
asked Jan 10 at 1:13
Tapatio SombreroTapatio Sombrero
15217
edited Jan 10 at 1:37
Tapatio Sombrero
asked Jan 10 at 1:13
Tapatio SombreroTapatio Sombrero
15217
asked Jan 10 at 1:13
Tapatio SombreroTapatio Sombrero
15217
asked Jan 10 at 1:13
Tapatio SombreroTapatio Sombrero
15217
15217
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways?
The usual reason to use an op-amp to buffer a divider like this is to ensure the reference voltage doesn't change if whatever it's connected to sinks or sources current.
Can the AD8031 op amp be removed?
In this case, since the AD8544 has only 4 pA input bias current, I'd expect the AD8031 can be removed without much change in performance.
Another issue to watch for, since this reference is connected to two different signals, is whether removing the buffer could allow the two signals to crosstalk with each other. Given the high resistor values connecting the two op-amp inputs to the reference, it's unlikely this would be a real issue, but to be sure you could simply make two different dividers and use one for each of the filter stages.
Also, to save power, can the voltage divider be replaced with a buck converter?
Any buck converter will have some output ripple. If you used it here, that ripple would be coupled directly into your filtered signal. I wouldn't do it just to save something like 150 uA. (You'd also need to find a buck converter design with less than 150 uA quiescent current to make this a positve trade)
If those 150 uA are really important to your application, you might rather find an op-amp with very low quiescent current (the AD8031 has 800 uA, you'd be looking for 10's of uA), replace the AD8031 with that, and increase the resistor values in the divider to 100 kohm or more.
Aside
The AD8031(A) is only rated to drive capacitive loads up to 15 pF and maintain stability. C2 and C4 in your schematic are probably causing the op-amp to generate noise (it may even be oscillating) rather than reducing noise. I'd remove them.
answered Jan 10 at 1:50
The PhotonThe Photon
84.4k397196
$endgroup$
1
$begingroup$
We don't know what else is hanging off of the V+ line -- if there is something drawing current, then taking out the buffer would screw things up. Agreed on the output capacitance issue. Driving capacitive loads with op-amps is well documented; the OP can do a search or ask here. Here is just one example result from searching on "op-amp capacitive load".
$endgroup$
– TimWescott
Jan 10 at 2:05
1
$begingroup$
@TimWescott, true enough. OP, my answer is written assuming you've actually told us everything about your circuit. If you're hiding some part of the circuit, the answer may not actually apply.
$endgroup$
– The Photon
Jan 10 at 2:15
add a comment |
$begingroup$
Is there a reason ...
Yes. The two 10k reisistors give the voltage reference an impedance of 5k. This means that if the current drawn from the reference changes by 0.1 mA that the voltage of the reference would change by 0.1m x 5k = 0.5 V. This would be a very unstable reference.
The op-amp buffer fixes this. The output impedance of the buffer is close to zero in comparison. This is a stable reference.
Can the AD8031 op amp be removed?
Maybe, but probably not a good idea.
Also, to save power, can the voltage divider be replaced with a buck converter?
The voltage divider consumes $ I = frac VR = frac 3.320k = 165 mu text A $.
A buck converter is designed for power supplies rather than a voltage reference. The converter would likely consume more than 165 μA so there would be no advantage.
answered Jan 10 at 1:49
TransistorTransistor
82.1k778176
$endgroup$
1
$begingroup$
The OP-amp draws 800 µA too, but it doesn't change the equation that much..
$endgroup$
– pipe
Jan 10 at 13:21
add a comment |
$begingroup$
That's a horrible circuit, I wonder where you got it from.
The AD8031 is very intolerance of capacitive loads, see Figure 46 in the datasheet, so most likely that op-amp will be oscillating at high frequency, which will, at a minimum, cause increased power consumption.
You can use a TLE2426, which will consume only 170uA typically at 5V.
Below is a way to connect a conventional op-amp in a stable manner (from a TI ADC datasheet):
That's a low-noise high speed amplifier, for yours you might try increasing the resistor values by an order of magnitude.
answered Jan 10 at 2:02
Spehro PefhanySpehro Pefhany
206k5155412
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
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votes
$begingroup$
Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways?
The usual reason to use an op-amp to buffer a divider like this is to ensure the reference voltage doesn't change if whatever it's connected to sinks or sources current.
Can the AD8031 op amp be removed?
In this case, since the AD8544 has only 4 pA input bias current, I'd expect the AD8031 can be removed without much change in performance.
Another issue to watch for, since this reference is connected to two different signals, is whether removing the buffer could allow the two signals to crosstalk with each other. Given the high resistor values connecting the two op-amp inputs to the reference, it's unlikely this would be a real issue, but to be sure you could simply make two different dividers and use one for each of the filter stages.
Also, to save power, can the voltage divider be replaced with a buck converter?
Any buck converter will have some output ripple. If you used it here, that ripple would be coupled directly into your filtered signal. I wouldn't do it just to save something like 150 uA. (You'd also need to find a buck converter design with less than 150 uA quiescent current to make this a positve trade)
If those 150 uA are really important to your application, you might rather find an op-amp with very low quiescent current (the AD8031 has 800 uA, you'd be looking for 10's of uA), replace the AD8031 with that, and increase the resistor values in the divider to 100 kohm or more.
Aside
The AD8031(A) is only rated to drive capacitive loads up to 15 pF and maintain stability. C2 and C4 in your schematic are probably causing the op-amp to generate noise (it may even be oscillating) rather than reducing noise. I'd remove them.
answered Jan 10 at 1:50
The PhotonThe Photon
84.4k397196
$endgroup$
1
$begingroup$
We don't know what else is hanging off of the V+ line -- if there is something drawing current, then taking out the buffer would screw things up. Agreed on the output capacitance issue. Driving capacitive loads with op-amps is well documented; the OP can do a search or ask here. Here is just one example result from searching on "op-amp capacitive load".
$endgroup$
– TimWescott
Jan 10 at 2:05
1
$begingroup$
@TimWescott, true enough. OP, my answer is written assuming you've actually told us everything about your circuit. If you're hiding some part of the circuit, the answer may not actually apply.
$endgroup$
– The Photon
Jan 10 at 2:15
add a comment |
$begingroup$
Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways?
The usual reason to use an op-amp to buffer a divider like this is to ensure the reference voltage doesn't change if whatever it's connected to sinks or sources current.
Can the AD8031 op amp be removed?
In this case, since the AD8544 has only 4 pA input bias current, I'd expect the AD8031 can be removed without much change in performance.
Another issue to watch for, since this reference is connected to two different signals, is whether removing the buffer could allow the two signals to crosstalk with each other. Given the high resistor values connecting the two op-amp inputs to the reference, it's unlikely this would be a real issue, but to be sure you could simply make two different dividers and use one for each of the filter stages.
Also, to save power, can the voltage divider be replaced with a buck converter?
Any buck converter will have some output ripple. If you used it here, that ripple would be coupled directly into your filtered signal. I wouldn't do it just to save something like 150 uA. (You'd also need to find a buck converter design with less than 150 uA quiescent current to make this a positve trade)
If those 150 uA are really important to your application, you might rather find an op-amp with very low quiescent current (the AD8031 has 800 uA, you'd be looking for 10's of uA), replace the AD8031 with that, and increase the resistor values in the divider to 100 kohm or more.
Aside
The AD8031(A) is only rated to drive capacitive loads up to 15 pF and maintain stability. C2 and C4 in your schematic are probably causing the op-amp to generate noise (it may even be oscillating) rather than reducing noise. I'd remove them.
answered Jan 10 at 1:50
The PhotonThe Photon
84.4k397196
$endgroup$
1
$begingroup$
We don't know what else is hanging off of the V+ line -- if there is something drawing current, then taking out the buffer would screw things up. Agreed on the output capacitance issue. Driving capacitive loads with op-amps is well documented; the OP can do a search or ask here. Here is just one example result from searching on "op-amp capacitive load".
$endgroup$
– TimWescott
Jan 10 at 2:05
1
$begingroup$
@TimWescott, true enough. OP, my answer is written assuming you've actually told us everything about your circuit. If you're hiding some part of the circuit, the answer may not actually apply.
$endgroup$
– The Photon
Jan 10 at 2:15
add a comment |
$begingroup$
Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways?
The usual reason to use an op-amp to buffer a divider like this is to ensure the reference voltage doesn't change if whatever it's connected to sinks or sources current.
Can the AD8031 op amp be removed?
In this case, since the AD8544 has only 4 pA input bias current, I'd expect the AD8031 can be removed without much change in performance.
Another issue to watch for, since this reference is connected to two different signals, is whether removing the buffer could allow the two signals to crosstalk with each other. Given the high resistor values connecting the two op-amp inputs to the reference, it's unlikely this would be a real issue, but to be sure you could simply make two different dividers and use one for each of the filter stages.
Also, to save power, can the voltage divider be replaced with a buck converter?
Any buck converter will have some output ripple. If you used it here, that ripple would be coupled directly into your filtered signal. I wouldn't do it just to save something like 150 uA. (You'd also need to find a buck converter design with less than 150 uA quiescent current to make this a positve trade)
If those 150 uA are really important to your application, you might rather find an op-amp with very low quiescent current (the AD8031 has 800 uA, you'd be looking for 10's of uA), replace the AD8031 with that, and increase the resistor values in the divider to 100 kohm or more.
Aside
The AD8031(A) is only rated to drive capacitive loads up to 15 pF and maintain stability. C2 and C4 in your schematic are probably causing the op-amp to generate noise (it may even be oscillating) rather than reducing noise. I'd remove them.
answered Jan 10 at 1:50
The PhotonThe Photon
84.4k397196
$endgroup$
Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways?
The usual reason to use an op-amp to buffer a divider like this is to ensure the reference voltage doesn't change if whatever it's connected to sinks or sources current.
Can the AD8031 op amp be removed?
In this case, since the AD8544 has only 4 pA input bias current, I'd expect the AD8031 can be removed without much change in performance.
Another issue to watch for, since this reference is connected to two different signals, is whether removing the buffer could allow the two signals to crosstalk with each other. Given the high resistor values connecting the two op-amp inputs to the reference, it's unlikely this would be a real issue, but to be sure you could simply make two different dividers and use one for each of the filter stages.
Also, to save power, can the voltage divider be replaced with a buck converter?
Any buck converter will have some output ripple. If you used it here, that ripple would be coupled directly into your filtered signal. I wouldn't do it just to save something like 150 uA. (You'd also need to find a buck converter design with less than 150 uA quiescent current to make this a positve trade)
If those 150 uA are really important to your application, you might rather find an op-amp with very low quiescent current (the AD8031 has 800 uA, you'd be looking for 10's of uA), replace the AD8031 with that, and increase the resistor values in the divider to 100 kohm or more.
Aside
The AD8031(A) is only rated to drive capacitive loads up to 15 pF and maintain stability. C2 and C4 in your schematic are probably causing the op-amp to generate noise (it may even be oscillating) rather than reducing noise. I'd remove them.
answered Jan 10 at 1:50
The PhotonThe Photon
84.4k397196
edited Jan 10 at 1:58
answered Jan 10 at 1:50
The PhotonThe Photon
84.4k397196
answered Jan 10 at 1:50
The PhotonThe Photon
84.4k397196
answered Jan 10 at 1:50
The PhotonThe Photon
84.4k397196
84.4k397196
1
$begingroup$
We don't know what else is hanging off of the V+ line -- if there is something drawing current, then taking out the buffer would screw things up. Agreed on the output capacitance issue. Driving capacitive loads with op-amps is well documented; the OP can do a search or ask here. Here is just one example result from searching on "op-amp capacitive load".
$endgroup$
– TimWescott
Jan 10 at 2:05
1
$begingroup$
@TimWescott, true enough. OP, my answer is written assuming you've actually told us everything about your circuit. If you're hiding some part of the circuit, the answer may not actually apply.
$endgroup$
– The Photon
Jan 10 at 2:15
add a comment |
1
$begingroup$
We don't know what else is hanging off of the V+ line -- if there is something drawing current, then taking out the buffer would screw things up. Agreed on the output capacitance issue. Driving capacitive loads with op-amps is well documented; the OP can do a search or ask here. Here is just one example result from searching on "op-amp capacitive load".
$endgroup$
– TimWescott
Jan 10 at 2:05
1
$begingroup$
@TimWescott, true enough. OP, my answer is written assuming you've actually told us everything about your circuit. If you're hiding some part of the circuit, the answer may not actually apply.
$endgroup$
– The Photon
Jan 10 at 2:15
1
1
$begingroup$
We don't know what else is hanging off of the V+ line -- if there is something drawing current, then taking out the buffer would screw things up. Agreed on the output capacitance issue. Driving capacitive loads with op-amps is well documented; the OP can do a search or ask here. Here is just one example result from searching on "op-amp capacitive load".
$endgroup$
– TimWescott
Jan 10 at 2:05
$begingroup$
We don't know what else is hanging off of the V+ line -- if there is something drawing current, then taking out the buffer would screw things up. Agreed on the output capacitance issue. Driving capacitive loads with op-amps is well documented; the OP can do a search or ask here. Here is just one example result from searching on "op-amp capacitive load".
$endgroup$
– TimWescott
Jan 10 at 2:05
1
1
$begingroup$
@TimWescott, true enough. OP, my answer is written assuming you've actually told us everything about your circuit. If you're hiding some part of the circuit, the answer may not actually apply.
$endgroup$
– The Photon
Jan 10 at 2:15
$begingroup$
@TimWescott, true enough. OP, my answer is written assuming you've actually told us everything about your circuit. If you're hiding some part of the circuit, the answer may not actually apply.
$endgroup$
– The Photon
Jan 10 at 2:15
add a comment |
$begingroup$
Is there a reason ...
Yes. The two 10k reisistors give the voltage reference an impedance of 5k. This means that if the current drawn from the reference changes by 0.1 mA that the voltage of the reference would change by 0.1m x 5k = 0.5 V. This would be a very unstable reference.
The op-amp buffer fixes this. The output impedance of the buffer is close to zero in comparison. This is a stable reference.
Can the AD8031 op amp be removed?
Maybe, but probably not a good idea.
Also, to save power, can the voltage divider be replaced with a buck converter?
The voltage divider consumes $ I = frac VR = frac 3.320k = 165 mu text A $.
A buck converter is designed for power supplies rather than a voltage reference. The converter would likely consume more than 165 μA so there would be no advantage.
answered Jan 10 at 1:49
TransistorTransistor
82.1k778176
$endgroup$
1
$begingroup$
The OP-amp draws 800 µA too, but it doesn't change the equation that much..
$endgroup$
– pipe
Jan 10 at 13:21
add a comment |
$begingroup$
Is there a reason ...
Yes. The two 10k reisistors give the voltage reference an impedance of 5k. This means that if the current drawn from the reference changes by 0.1 mA that the voltage of the reference would change by 0.1m x 5k = 0.5 V. This would be a very unstable reference.
The op-amp buffer fixes this. The output impedance of the buffer is close to zero in comparison. This is a stable reference.
Can the AD8031 op amp be removed?
Maybe, but probably not a good idea.
Also, to save power, can the voltage divider be replaced with a buck converter?
The voltage divider consumes $ I = frac VR = frac 3.320k = 165 mu text A $.
A buck converter is designed for power supplies rather than a voltage reference. The converter would likely consume more than 165 μA so there would be no advantage.
answered Jan 10 at 1:49
TransistorTransistor
82.1k778176
$endgroup$
1
$begingroup$
The OP-amp draws 800 µA too, but it doesn't change the equation that much..
$endgroup$
– pipe
Jan 10 at 13:21
add a comment |
$begingroup$
Is there a reason ...
Yes. The two 10k reisistors give the voltage reference an impedance of 5k. This means that if the current drawn from the reference changes by 0.1 mA that the voltage of the reference would change by 0.1m x 5k = 0.5 V. This would be a very unstable reference.
The op-amp buffer fixes this. The output impedance of the buffer is close to zero in comparison. This is a stable reference.
Can the AD8031 op amp be removed?
Maybe, but probably not a good idea.
Also, to save power, can the voltage divider be replaced with a buck converter?
The voltage divider consumes $ I = frac VR = frac 3.320k = 165 mu text A $.
A buck converter is designed for power supplies rather than a voltage reference. The converter would likely consume more than 165 μA so there would be no advantage.
answered Jan 10 at 1:49
TransistorTransistor
82.1k778176
$endgroup$
Is there a reason ...
Yes. The two 10k reisistors give the voltage reference an impedance of 5k. This means that if the current drawn from the reference changes by 0.1 mA that the voltage of the reference would change by 0.1m x 5k = 0.5 V. This would be a very unstable reference.
The op-amp buffer fixes this. The output impedance of the buffer is close to zero in comparison. This is a stable reference.
Can the AD8031 op amp be removed?
Maybe, but probably not a good idea.
Also, to save power, can the voltage divider be replaced with a buck converter?
The voltage divider consumes $ I = frac VR = frac 3.320k = 165 mu text A $.
A buck converter is designed for power supplies rather than a voltage reference. The converter would likely consume more than 165 μA so there would be no advantage.
answered Jan 10 at 1:49
TransistorTransistor
82.1k778176
edited Jan 10 at 1:53
answered Jan 10 at 1:49
TransistorTransistor
82.1k778176
answered Jan 10 at 1:49
TransistorTransistor
82.1k778176
answered Jan 10 at 1:49
TransistorTransistor
82.1k778176
82.1k778176
1
$begingroup$
The OP-amp draws 800 µA too, but it doesn't change the equation that much..
$endgroup$
– pipe
Jan 10 at 13:21
add a comment |
1
$begingroup$
The OP-amp draws 800 µA too, but it doesn't change the equation that much..
$endgroup$
– pipe
Jan 10 at 13:21
1
1
$begingroup$
The OP-amp draws 800 µA too, but it doesn't change the equation that much..
$endgroup$
– pipe
Jan 10 at 13:21
$begingroup$
The OP-amp draws 800 µA too, but it doesn't change the equation that much..
$endgroup$
– pipe
Jan 10 at 13:21
add a comment |
$begingroup$
That's a horrible circuit, I wonder where you got it from.
The AD8031 is very intolerance of capacitive loads, see Figure 46 in the datasheet, so most likely that op-amp will be oscillating at high frequency, which will, at a minimum, cause increased power consumption.
You can use a TLE2426, which will consume only 170uA typically at 5V.
Below is a way to connect a conventional op-amp in a stable manner (from a TI ADC datasheet):
That's a low-noise high speed amplifier, for yours you might try increasing the resistor values by an order of magnitude.
answered Jan 10 at 2:02
Spehro PefhanySpehro Pefhany
206k5155412
$endgroup$
add a comment |
$begingroup$
That's a horrible circuit, I wonder where you got it from.
The AD8031 is very intolerance of capacitive loads, see Figure 46 in the datasheet, so most likely that op-amp will be oscillating at high frequency, which will, at a minimum, cause increased power consumption.
You can use a TLE2426, which will consume only 170uA typically at 5V.
Below is a way to connect a conventional op-amp in a stable manner (from a TI ADC datasheet):
That's a low-noise high speed amplifier, for yours you might try increasing the resistor values by an order of magnitude.
answered Jan 10 at 2:02
Spehro PefhanySpehro Pefhany
206k5155412
$endgroup$
add a comment |
$begingroup$
That's a horrible circuit, I wonder where you got it from.
The AD8031 is very intolerance of capacitive loads, see Figure 46 in the datasheet, so most likely that op-amp will be oscillating at high frequency, which will, at a minimum, cause increased power consumption.
You can use a TLE2426, which will consume only 170uA typically at 5V.
Below is a way to connect a conventional op-amp in a stable manner (from a TI ADC datasheet):
That's a low-noise high speed amplifier, for yours you might try increasing the resistor values by an order of magnitude.
answered Jan 10 at 2:02
Spehro PefhanySpehro Pefhany
206k5155412
$endgroup$
That's a horrible circuit, I wonder where you got it from.
The AD8031 is very intolerance of capacitive loads, see Figure 46 in the datasheet, so most likely that op-amp will be oscillating at high frequency, which will, at a minimum, cause increased power consumption.
You can use a TLE2426, which will consume only 170uA typically at 5V.
Below is a way to connect a conventional op-amp in a stable manner (from a TI ADC datasheet):
That's a low-noise high speed amplifier, for yours you might try increasing the resistor values by an order of magnitude.
answered Jan 10 at 2:02
Spehro PefhanySpehro Pefhany
206k5155412
edited Jan 10 at 2:10
answered Jan 10 at 2:02
Spehro PefhanySpehro Pefhany
206k5155412
answered Jan 10 at 2:02
Spehro PefhanySpehro Pefhany
206k5155412
answered Jan 10 at 2:02
Spehro PefhanySpehro Pefhany
206k5155412
206k5155412
add a comment |
add a comment |
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