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I'm designing a PCB to filter and amplify a differential body signal with amplitude 5 mV up to 5 V. The first part of the circuit consists of a first order highpass filter to remove any DC componen...
#5: Post edited
by
Carl
·
2025-02-24T10:46:51Z (1 day ago)
- I'm designing a PCB to filter and amplify a differential body signal with amplitude 5 mV up to 5 V. The first part of the circuit consists of a first order highpass filter to remove any DC components from the signal. Then it moves into an instrumentation amplifier for amplification of the differential signal and attenuation of the common mode voltage. The instrumentation amplifier has a common-mode rejection ratio of 120 dB. The first stage of my circuit is seen below.
- 
- The signal-to-noise ratio before the instrumentation amplifier is $ \frac{5 \: \text{mV}}{5 \: \text{V}} = -60 \: \text{dB}$. After the amplifier, the SNR is 60 dB.
- I want to maintain the signal integrity of my differential signal. To do this, there should be as little voltage drop of my signal across any impedances on the path to the amplifier as possible. A $0.5 \: \text{mV}$ of "lost signal" is not acceptable. For that reason, I chose to route this first stage with a trace width of 1 mm. According to KiCad's calculator tools, a trace of length 5 mm with this thickness has $2.5 \: \text{m}\Omega$ resistance, which I suppose is good enough.
- 
My question is: After the amplification stage, the signal now has a 5 V amplitude, and the SNR is much greater. To what trace width am I allowed to go down to? Routing with 1 mm trace width on the entire board is not doable. Would trace width of 0.2 mm be appropriate? Is there anything I should be aware of when changing the trace width?
- I'm designing a PCB to filter and amplify a differential body signal with amplitude 5 mV up to 5 V. The first part of the circuit consists of a first order highpass filter to remove any DC components from the signal. Then it moves into an instrumentation amplifier for amplification of the differential signal and attenuation of the common mode voltage. The instrumentation amplifier has a common-mode rejection ratio of 120 dB. The first stage of my circuit is seen below.
- 
- The signal-to-noise ratio before the instrumentation amplifier is $ \frac{5 \: \text{mV}}{5 \: \text{V}} = -60 \: \text{dB}$. After the amplifier, the SNR is 60 dB.
- I want to maintain the signal integrity of my differential signal. To do this, there should be as little voltage drop of my signal across any impedances on the path to the amplifier as possible. A $0.5 \: \text{mV}$ of "lost signal" is not acceptable. For that reason, I chose to route this first stage with a trace width of 1 mm. According to KiCad's calculator tools, a trace of length 5 mm with this thickness has $2.5 \: \text{m}\Omega$ resistance, which I suppose is good enough.
- 
- My question is: After the amplification stage, the signal now has a 5 V amplitude, and the SNR is much greater. To what trace width am I allowed to go down to? Routing with 1 mm trace width on the entire board is not doable. Would a trace width of 0.2 mm be appropriate? Is there anything I should be aware of when changing the trace width? The board is entirely analog. Not digital circuitry.
#4: Post edited
by
Carl
·
2025-02-24T10:46:08Z (1 day ago)
- I'm designing a PCB to filter and amplify a differential body signal with amplitude 5 mV up to 5 V. The first part of the circuit consists of a first order highpass filter to remove any DC components from the signal. Then it moves into an instrumentation amplifier for amplification of the differential signal and attenuation of the common mode voltage. The instrumentation amplifier has a common-mode rejection ratio of 120 dB. The first stage of my circuit is seen below.
- 
- The signal-to-noise ratio before the instrumentation amplifier is $ \frac{5 \: \text{mV}}{5 \: \text{V}} = -60 \: \text{dB}$. After the amplifier, the SNR is 60 dB.
- I want to maintain the signal integrity of my differential signal. To do this, there should be as little voltage drop of my signal across any impedances on the path to the amplifier as possible. A $0.5 \: \text{mV}$ of "lost signal" is not acceptable. For that reason, I chose to route this first stage with a trace width of 1 mm. According to KiCad's calculator tools, a trace of length 5 mm with this thickness has $2.5 \: \text{m}\Omega$ resistance, which I suppose is good enough.
- 
My question is: After the amplification stage, the signal now has a 2.5 V amplitude, and the SNR is much greater. To what trace width am I allowed to go down to? Routing with 1 mm trace width on the entire board is not doable. Would trace width of 0.2 mm be appropriate? Is there anything I should be aware of when changing the trace width?
- I'm designing a PCB to filter and amplify a differential body signal with amplitude 5 mV up to 5 V. The first part of the circuit consists of a first order highpass filter to remove any DC components from the signal. Then it moves into an instrumentation amplifier for amplification of the differential signal and attenuation of the common mode voltage. The instrumentation amplifier has a common-mode rejection ratio of 120 dB. The first stage of my circuit is seen below.
- 
- The signal-to-noise ratio before the instrumentation amplifier is $ \frac{5 \: \text{mV}}{5 \: \text{V}} = -60 \: \text{dB}$. After the amplifier, the SNR is 60 dB.
- I want to maintain the signal integrity of my differential signal. To do this, there should be as little voltage drop of my signal across any impedances on the path to the amplifier as possible. A $0.5 \: \text{mV}$ of "lost signal" is not acceptable. For that reason, I chose to route this first stage with a trace width of 1 mm. According to KiCad's calculator tools, a trace of length 5 mm with this thickness has $2.5 \: \text{m}\Omega$ resistance, which I suppose is good enough.
- 
- My question is: After the amplification stage, the signal now has a 5 V amplitude, and the SNR is much greater. To what trace width am I allowed to go down to? Routing with 1 mm trace width on the entire board is not doable. Would trace width of 0.2 mm be appropriate? Is there anything I should be aware of when changing the trace width?
#3: Post edited
by
Carl
·
2025-02-24T10:45:41Z (1 day ago)
- I'm designing a PCB to filter and amplify a differential body signal with amplitude 5 mV up to 5 V. The first part of the circuit consists of a first order highpass filter to remove any DC components from the signal. Then it moves into an instrumentation amplifier for amplification of the differential signal and attenuation of the common mode voltage. The instrumentation amplifier has a common-mode rejection ratio of 120 dB. The first stage of my circuit is seen below.
- 
- The signal-to-noise ratio before the instrumentation amplifier is $ \frac{5 \: \text{mV}}{5 \: \text{V}} = -60 \: \text{dB}$. After the amplifier, the SNR is 60 dB.
I want to maintain the signal integrity of my differential signal. To do this, there should be as little voltage drop of my signal across any impedances on the path to the amplifier as possible. A $0.5 \: \text{mV}$ of "lost signal" is not acceptable. For that reason, I chose to route this first stage with a trace width of 1 mm. According to KiCad's calculator tools, a trace of length 5 mm with this thickness has $2.5 \: \text{m}\Omega$ resistance.- 
- My question is: After the amplification stage, the signal now has a 2.5 V amplitude, and the SNR is much greater. To what trace width am I allowed to go down to? Routing with 1 mm trace width on the entire board is not doable. Would trace width of 0.2 mm be appropriate? Is there anything I should be aware of when changing the trace width?
- I'm designing a PCB to filter and amplify a differential body signal with amplitude 5 mV up to 5 V. The first part of the circuit consists of a first order highpass filter to remove any DC components from the signal. Then it moves into an instrumentation amplifier for amplification of the differential signal and attenuation of the common mode voltage. The instrumentation amplifier has a common-mode rejection ratio of 120 dB. The first stage of my circuit is seen below.
- 
- The signal-to-noise ratio before the instrumentation amplifier is $ \frac{5 \: \text{mV}}{5 \: \text{V}} = -60 \: \text{dB}$. After the amplifier, the SNR is 60 dB.
- I want to maintain the signal integrity of my differential signal. To do this, there should be as little voltage drop of my signal across any impedances on the path to the amplifier as possible. A $0.5 \: \text{mV}$ of "lost signal" is not acceptable. For that reason, I chose to route this first stage with a trace width of 1 mm. According to KiCad's calculator tools, a trace of length 5 mm with this thickness has $2.5 \: \text{m}\Omega$ resistance, which I suppose is good enough.
- 
- My question is: After the amplification stage, the signal now has a 2.5 V amplitude, and the SNR is much greater. To what trace width am I allowed to go down to? Routing with 1 mm trace width on the entire board is not doable. Would trace width of 0.2 mm be appropriate? Is there anything I should be aware of when changing the trace width?
#2: Post edited
by
Carl
·
2025-02-24T10:45:13Z (1 day ago)
- I'm designing a PCB to filter and amplify a differential body signal with amplitude 5 mV up to 5 V. The first part of the circuit consists of a first order highpass filter to remove any DC components from the signal. Then it moves into an instrumentation amplifier for amplification of the differential signal and attenuation of the common mode voltage. The instrumentation amplifier has a common-mode rejection ratio of 120 dB. The first stage of my circuit is seen below.
- 
- The signal-to-noise ratio before the instrumentation amplifier is $ \frac{5 \: \text{mV}}{5 \: \text{V}} = -60 \: \text{dB}$. After the amplifier, the SNR is 60 dB.
I want to maintain the signal integrity of my differential signal. To do this, there should be as little voltage drop of my signal across any impedances on the path to the amplifier as possible. A $0.5 \: \text{mV}$ of "lost signal" is not acceptable. For that reason, I chose to route this first stage with a trace width of 1 mm. According to KiCad's calculator tools, a trace of 5 mm with this thickness has $2.5 \: \text{m}\Omega$ resistance.- 
- My question is: After the amplification stage, the signal now has a 2.5 V amplitude, and the SNR is much greater. To what trace width am I allowed to go down to? Routing with 1 mm trace width on the entire board is not doable. Would trace width of 0.2 mm be appropriate? Is there anything I should be aware of when changing the trace width?
- I'm designing a PCB to filter and amplify a differential body signal with amplitude 5 mV up to 5 V. The first part of the circuit consists of a first order highpass filter to remove any DC components from the signal. Then it moves into an instrumentation amplifier for amplification of the differential signal and attenuation of the common mode voltage. The instrumentation amplifier has a common-mode rejection ratio of 120 dB. The first stage of my circuit is seen below.
- 
- The signal-to-noise ratio before the instrumentation amplifier is $ \frac{5 \: \text{mV}}{5 \: \text{V}} = -60 \: \text{dB}$. After the amplifier, the SNR is 60 dB.
- I want to maintain the signal integrity of my differential signal. To do this, there should be as little voltage drop of my signal across any impedances on the path to the amplifier as possible. A $0.5 \: \text{mV}$ of "lost signal" is not acceptable. For that reason, I chose to route this first stage with a trace width of 1 mm. According to KiCad's calculator tools, a trace of length 5 mm with this thickness has $2.5 \: \text{m}\Omega$ resistance.
- 
- My question is: After the amplification stage, the signal now has a 2.5 V amplitude, and the SNR is much greater. To what trace width am I allowed to go down to? Routing with 1 mm trace width on the entire board is not doable. Would trace width of 0.2 mm be appropriate? Is there anything I should be aware of when changing the trace width?
#1: Initial revision
by
Carl
·
2025-02-24T10:44:02Z (1 day ago)
Changing PCB trace width once signal-to-noise ratio is high
I'm designing a PCB to filter and amplify a differential body signal with amplitude 5 mV up to 5 V. The first part of the circuit consists of a first order highpass filter to remove any DC components from the signal. Then it moves into an instrumentation amplifier for amplification of the differential signal and attenuation of the common mode voltage. The instrumentation amplifier has a common-mode rejection ratio of 120 dB. The first stage of my circuit is seen below.
The signal-to-noise ratio before the instrumentation amplifier is $ \frac{5 \: \text{mV}}{5 \: \text{V}} = -60 \: \text{dB}$. After the amplifier, the SNR is 60 dB.
I want to maintain the signal integrity of my differential signal. To do this, there should be as little voltage drop of my signal across any impedances on the path to the amplifier as possible. A $0.5 \: \text{mV}$ of "lost signal" is not acceptable. For that reason, I chose to route this first stage with a trace width of 1 mm. According to KiCad's calculator tools, a trace of 5 mm with this thickness has $2.5 \: \text{m}\Omega$ resistance.
My question is: After the amplification stage, the signal now has a 2.5 V amplitude, and the SNR is much greater. To what trace width am I allowed to go down to? Routing with 1 mm trace width on the entire board is not doable. Would trace width of 0.2 mm be appropriate? Is there anything I should be aware of when changing the trace width?