The circuit has to operate in both differential input and single ended input modes. When in single ended input mode, there's no way to ground one input, so it will float. In this circuit, the two inputs are tied through C9. So I can't see how this circuit would allow single ended operation. If, however, C9 is split in two, as a pair of 1 uF capacitors to ground, that would seem to solve that particular issue. But I'm not certain it would still operate the same way. It seems to me the split capacitor will operate both in common mode and in differential mode, while the single capacitor is only effective in differential mode. Any comments? -- Rick C. - Get 1,000 miles of free Supercharging - Tesla referral code - https://ts.la/richard11209
Two Pole, Low Pass Active Filter with Differential Inputs
Started by ●March 28, 2023
Reply by ●March 28, 20232023-03-28
On Tuesday, March 28, 2023 at 2:33:35 PM UTC-4, Ricky wrote:> The circuit has to operate in both differential input and single ended input modes. When in single ended input mode, there's no way to ground one input, so it will float. > > In this circuit, the two inputs are tied through C9. So I can't see how this circuit would allow single ended operation. If, however, C9 is split in two, as a pair of 1 uF capacitors to ground, that would seem to solve that particular issue. But I'm not certain it would still operate the same way. > > It seems to me the split capacitor will operate both in common mode and in differential mode, while the single capacitor is only effective in differential mode. > > Any comments?Care to link the circuit diagram image?> > -- > > Rick C. > > - Get 1,000 miles of free Supercharging > - Tesla referral code - https://ts.la/richard11209
Reply by ●March 28, 20232023-03-28
On Tuesday, March 28, 2023 at 3:39:12 PM UTC-4, Fred Bloggs wrote:> On Tuesday, March 28, 2023 at 2:33:35 PM UTC-4, Ricky wrote: > > The circuit has to operate in both differential input and single ended input modes. When in single ended input mode, there's no way to ground one input, so it will float. > > > > In this circuit, the two inputs are tied through C9. So I can't see how this circuit would allow single ended operation. If, however, C9 is split in two, as a pair of 1 uF capacitors to ground, that would seem to solve that particular issue. But I'm not certain it would still operate the same way. > > > > It seems to me the split capacitor will operate both in common mode and in differential mode, while the single capacitor is only effective in differential mode. > > > > Any comments? > Care to link the circuit diagram image?Sorry about that. https://www.eevblog.com/forum/beginners/high-pass-filtering-for-differential-opamp/?action=dlattach;attach=492248;image -- Rick C. + Get 1,000 miles of free Supercharging + Tesla referral code - https://ts.la/richard11209
Reply by ●March 28, 20232023-03-28
On Tuesday, March 28, 2023 at 3:42:11 PM UTC-4, Ricky wrote:> On Tuesday, March 28, 2023 at 3:39:12 PM UTC-4, Fred Bloggs wrote: > > On Tuesday, March 28, 2023 at 2:33:35 PM UTC-4, Ricky wrote: > > > The circuit has to operate in both differential input and single ended input modes. When in single ended input mode, there's no way to ground one input, so it will float. > > > > > > In this circuit, the two inputs are tied through C9. So I can't see how this circuit would allow single ended operation. If, however, C9 is split in two, as a pair of 1 uF capacitors to ground, that would seem to solve that particular issue. But I'm not certain it would still operate the same way. > > > > > > It seems to me the split capacitor will operate both in common mode and in differential mode, while the single capacitor is only effective in differential mode. > > > > > > Any comments? > > Care to link the circuit diagram image? > Sorry about that. > > https://www.eevblog.com/forum/beginners/high-pass-filtering-for-differential-opamp/?action=dlattach;attach=492248;imageThat circuit is intended for a bipolar supply on the OA, then you can GND an input. Common mode gain is 0.> > -- > > Rick C. > > + Get 1,000 miles of free Supercharging > + Tesla referral code - https://ts.la/richard11209
Reply by ●March 28, 20232023-03-28
On Tuesday, March 28, 2023 at 4:13:02 PM UTC-4, Fred Bloggs wrote:> On Tuesday, March 28, 2023 at 3:42:11 PM UTC-4, Ricky wrote: > > On Tuesday, March 28, 2023 at 3:39:12 PM UTC-4, Fred Bloggs wrote: > > > On Tuesday, March 28, 2023 at 2:33:35 PM UTC-4, Ricky wrote: > > > > The circuit has to operate in both differential input and single ended input modes. When in single ended input mode, there's no way to ground one input, so it will float. > > > > > > > > In this circuit, the two inputs are tied through C9. So I can't see how this circuit would allow single ended operation. If, however, C9 is split in two, as a pair of 1 uF capacitors to ground, that would seem to solve that particular issue. But I'm not certain it would still operate the same way. > > > > > > > > It seems to me the split capacitor will operate both in common mode and in differential mode, while the single capacitor is only effective in differential mode. > > > > > > > > Any comments? > > > Care to link the circuit diagram image? > > Sorry about that. > > > > https://www.eevblog.com/forum/beginners/high-pass-filtering-for-differential-opamp/?action=dlattach;attach=492248;image > That circuit is intended for a bipolar supply on the OA, then you can GND an input. Common mode gain is 0.You don't seem to understand the question. 1) I can't ground an input, as there is no circuitry to do that. There is no room on the board to add a switch. One input will have to float when used single ended. 2) This will run on a single power supply voltage, but you can consider the terminology of "ground" to refer to a reference voltage that is half the power supply. 3) The inputs will be coupled to the signal source through capacitors, so no DC path. 4) The concern is that when using the circuit with a single ended input, and one input floating, that C9 will cause signal to drive the floating input, impacting both the gain, and the filter response. Actually, the filter response might not be so important. The single ended mode does not use the full 20Hz to 20kHz bandwidth. It's a 1 kHz sine wave with 100 Hz amplitude modulation. Does this make more sense? -- Rick C. -- Get 1,000 miles of free Supercharging -- Tesla referral code - https://ts.la/richard11209
Reply by ●March 28, 20232023-03-28
On Tue, 28 Mar 2023 13:12:57 -0700 (PDT), Fred Bloggs <bloggs.fredbloggs.fred@gmail.com> wrote:>On Tuesday, March 28, 2023 at 3:42:11?PM UTC-4, Ricky wrote: >> On Tuesday, March 28, 2023 at 3:39:12?PM UTC-4, Fred Bloggs wrote: >> > On Tuesday, March 28, 2023 at 2:33:35?PM UTC-4, Ricky wrote: >> > > The circuit has to operate in both differential input and single ended input modes. When in single ended input mode, there's no way to ground one input, so it will float. >> > > >> > > In this circuit, the two inputs are tied through C9. So I can't see how this circuit would allow single ended operation. If, however, C9 is split in two, as a pair of 1 uF capacitors to ground, that would seem to solve that particular issue. But I'm not certain it would still operate the same way. >> > > >> > > It seems to me the split capacitor will operate both in common mode and in differential mode, while the single capacitor is only effective in differential mode. >> > > >> > > Any comments? >> > Care to link the circuit diagram image? >> Sorry about that. >> >> https://www.eevblog.com/forum/beginners/high-pass-filtering-for-differential-opamp/?action=dlattach;attach=492248;image > >That circuit is intended for a bipolar supply on the OA, then you can GND an input. Common mode gain is 0. > >It's too bizarre to think about.
Reply by ●March 28, 20232023-03-28
On Tuesday, March 28, 2023 at 4:43:21 PM UTC-4, Ricky wrote:> On Tuesday, March 28, 2023 at 4:13:02 PM UTC-4, Fred Bloggs wrote: > > On Tuesday, March 28, 2023 at 3:42:11 PM UTC-4, Ricky wrote: > > > On Tuesday, March 28, 2023 at 3:39:12 PM UTC-4, Fred Bloggs wrote: > > > > On Tuesday, March 28, 2023 at 2:33:35 PM UTC-4, Ricky wrote: > > > > > The circuit has to operate in both differential input and single ended input modes. When in single ended input mode, there's no way to ground one input, so it will float. > > > > > > > > > > In this circuit, the two inputs are tied through C9. So I can't see how this circuit would allow single ended operation. If, however, C9 is split in two, as a pair of 1 uF capacitors to ground, that would seem to solve that particular issue. But I'm not certain it would still operate the same way. > > > > > > > > > > It seems to me the split capacitor will operate both in common mode and in differential mode, while the single capacitor is only effective in differential mode. > > > > > > > > > > Any comments? > > > > Care to link the circuit diagram image? > > > Sorry about that. > > > > > > https://www.eevblog.com/forum/beginners/high-pass-filtering-for-differential-opamp/?action=dlattach;attach=492248;image > > That circuit is intended for a bipolar supply on the OA, then you can GND an input. Common mode gain is 0. > You don't seem to understand the question. > > 1) I can't ground an input, as there is no circuitry to do that. There is no room on the board to add a switch. One input will have to float when used single ended. > > 2) This will run on a single power supply voltage, but you can consider the terminology of "ground" to refer to a reference voltage that is half the power supply. > > 3) The inputs will be coupled to the signal source through capacitors, so no DC path. > > 4) The concern is that when using the circuit with a single ended input, and one input floating, that C9 will cause signal to drive the floating input, impacting both the gain, and the filter response. Actually, the filter response might not be so important. The single ended mode does not use the full 20Hz to 20kHz bandwidth. It's a 1 kHz sine wave with 100 Hz amplitude modulation. > > Does this make more sense?I suspected that's what you were saying but wasn't sure. All that amplifier is a standard differential amplifier with some RCs hung off the intermediate nodes to get the 2-pole roll off. If you leave one input open you lose the DC common mode rejection. Whatever side, (+) or (-), you choose to drive, leaving the other open, the signal average will be multiplied by a gain.> > -- > > Rick C. > > -- Get 1,000 miles of free Supercharging > -- Tesla referral code - https://ts.la/richard11209
Reply by ●March 28, 20232023-03-28
On Tuesday, March 28, 2023 at 6:28:01 PM UTC-4, Fred Bloggs wrote:> On Tuesday, March 28, 2023 at 4:43:21 PM UTC-4, Ricky wrote: > > On Tuesday, March 28, 2023 at 4:13:02 PM UTC-4, Fred Bloggs wrote: > > > On Tuesday, March 28, 2023 at 3:42:11 PM UTC-4, Ricky wrote: > > > > On Tuesday, March 28, 2023 at 3:39:12 PM UTC-4, Fred Bloggs wrote: > > > > > On Tuesday, March 28, 2023 at 2:33:35 PM UTC-4, Ricky wrote: > > > > > > The circuit has to operate in both differential input and single ended input modes. When in single ended input mode, there's no way to ground one input, so it will float. > > > > > > > > > > > > In this circuit, the two inputs are tied through C9. So I can't see how this circuit would allow single ended operation. If, however, C9 is split in two, as a pair of 1 uF capacitors to ground, that would seem to solve that particular issue. But I'm not certain it would still operate the same way. > > > > > > > > > > > > It seems to me the split capacitor will operate both in common mode and in differential mode, while the single capacitor is only effective in differential mode. > > > > > > > > > > > > Any comments? > > > > > Care to link the circuit diagram image? > > > > Sorry about that. > > > > > > > > https://www.eevblog.com/forum/beginners/high-pass-filtering-for-differential-opamp/?action=dlattach;attach=492248;image > > > That circuit is intended for a bipolar supply on the OA, then you can GND an input. Common mode gain is 0. > > You don't seem to understand the question. > > > > 1) I can't ground an input, as there is no circuitry to do that. There is no room on the board to add a switch. One input will have to float when used single ended. > > > > 2) This will run on a single power supply voltage, but you can consider the terminology of "ground" to refer to a reference voltage that is half the power supply. > > > > 3) The inputs will be coupled to the signal source through capacitors, so no DC path. > > > > 4) The concern is that when using the circuit with a single ended input, and one input floating, that C9 will cause signal to drive the floating input, impacting both the gain, and the filter response. Actually, the filter response might not be so important. The single ended mode does not use the full 20Hz to 20kHz bandwidth. It's a 1 kHz sine wave with 100 Hz amplitude modulation. > > > > Does this make more sense? > I suspected that's what you were saying but wasn't sure. All that amplifier is a standard differential amplifier with some RCs hung off the intermediate nodes to get the 2-pole roll off. > > If you leave one input open you lose the DC common mode rejection. Whatever side, (+) or (-), you choose to drive, leaving the other open, the signal average will be multiplied by a gain.Hmmm... I'm not sure you understand. If the non-inverting input is grounded, what you say will be useful. The non-inverting input will be held at an AC constant value and the circuit will work as intended for the single ended input. But, leave the non-inverting input open, it will be dragged about by C9, the input will not be an AC constant value, and the resulting output will not be the same as if the non-inverting input were AC grounded. When running with a single ended input, there is no DC common mode rejection, because it's single ended and the concept of common mode does not apply. DC is not an issue anyway, as the source is coupled to the circuit input through capacitors. I'm more interested in finding a similar circuit that will work effectively with differential inputs, as well as with a single ended input. That's why I asked about replacing C9, with a pair of capacitors, each to AC ground Then there's no direct path through the capacitors to the other half of the circuit. The amplifier + input will be connected to the DC reference through the network and only the inverting half of the circuit will operate, as in this drawing. http://i.stack.imgur.com/yykyQ.png I'm not sure connecting two capacitors in place of one, will not degrade any performance in the two use cases. I am currently drawing up the circuit to simulate. I'm pretty convinced this will work fine, I just thought it might be interesting to discuss. I had to do a lot of searching to find any low pass, active filter, that was designed for a differential input and single ended output. It's funny that Larkin says the circuit is "bizarre". But then he's never claimed to be well versed in filter design, has he? -- Rick C. -- Get 1,000 miles of free Supercharging -- Tesla referral code - https://ts.la/richard11209
Reply by ●March 28, 20232023-03-28
On Tuesday, March 28, 2023 at 8:21:45 PM UTC-4, Ricky wrote:> On Tuesday, March 28, 2023 at 6:28:01 PM UTC-4, Fred Bloggs wrote: > > On Tuesday, March 28, 2023 at 4:43:21 PM UTC-4, Ricky wrote: > > > On Tuesday, March 28, 2023 at 4:13:02 PM UTC-4, Fred Bloggs wrote: > > > > On Tuesday, March 28, 2023 at 3:42:11 PM UTC-4, Ricky wrote: > > > > > On Tuesday, March 28, 2023 at 3:39:12 PM UTC-4, Fred Bloggs wrote: > > > > > > On Tuesday, March 28, 2023 at 2:33:35 PM UTC-4, Ricky wrote: > > > > > > > The circuit has to operate in both differential input and single ended input modes. When in single ended input mode, there's no way to ground one input, so it will float. > > > > > > > > > > > > > > In this circuit, the two inputs are tied through C9. So I can't see how this circuit would allow single ended operation. If, however, C9 is split in two, as a pair of 1 uF capacitors to ground, that would seem to solve that particular issue. But I'm not certain it would still operate the same way. > > > > > > > > > > > > > > It seems to me the split capacitor will operate both in common mode and in differential mode, while the single capacitor is only effective in differential mode. > > > > > > > > > > > > > > Any comments? > > > > > > Care to link the circuit diagram image? > > > > > Sorry about that. > > > > > > > > > > https://www.eevblog.com/forum/beginners/high-pass-filtering-for-differential-opamp/?action=dlattach;attach=492248;image > > > > That circuit is intended for a bipolar supply on the OA, then you can GND an input. Common mode gain is 0. > > > You don't seem to understand the question. > > > > > > 1) I can't ground an input, as there is no circuitry to do that. There is no room on the board to add a switch. One input will have to float when used single ended. > > > > > > 2) This will run on a single power supply voltage, but you can consider the terminology of "ground" to refer to a reference voltage that is half the power supply. > > > > > > 3) The inputs will be coupled to the signal source through capacitors, so no DC path. > > > > > > 4) The concern is that when using the circuit with a single ended input, and one input floating, that C9 will cause signal to drive the floating input, impacting both the gain, and the filter response. Actually, the filter response might not be so important. The single ended mode does not use the full 20Hz to 20kHz bandwidth. It's a 1 kHz sine wave with 100 Hz amplitude modulation. > > > > > > Does this make more sense? > > I suspected that's what you were saying but wasn't sure. All that amplifier is a standard differential amplifier with some RCs hung off the intermediate nodes to get the 2-pole roll off. > > > > If you leave one input open you lose the DC common mode rejection. Whatever side, (+) or (-), you choose to drive, leaving the other open, the signal average will be multiplied by a gain. > Hmmm... I'm not sure you understand. If the non-inverting input is grounded, what you say will be useful. The non-inverting input will be held at an AC constant value and the circuit will work as intended for the single ended input. But, leave the non-inverting input open, it will be dragged about by C9, the input will not be an AC constant value, and the resulting output will not be the same as if the non-inverting input were AC grounded. > > When running with a single ended input, there is no DC common mode rejection, because it's single ended and the concept of common mode does not apply. DC is not an issue anyway, as the source is coupled to the circuit input through capacitors. > > I'm more interested in finding a similar circuit that will work effectively with differential inputs, as well as with a single ended input. That's why I asked about replacing C9, with a pair of capacitors, each to AC ground Then there's no direct path through the capacitors to the other half of the circuit. The amplifier + input will be connected to the DC reference through the network and only the inverting half of the circuit will operate, as in this drawing. > > http://i.stack.imgur.com/yykyQ.png > > I'm not sure connecting two capacitors in place of one, will not degrade any performance in the two use cases. I am currently drawing up the circuit to simulate. I'm pretty convinced this will work fine, I just thought it might be interesting to discuss. I had to do a lot of searching to find any low pass, active filter, that was designed for a differential input and single ended output.At F= 1kHz, the 470n has an impedance of 1/(2piE3 x 470E-9)= 340 ohms.> > It's funny that Larkin says the circuit is "bizarre". But then he's never claimed to be well versed in filter design, has he? > -- > > Rick C. > > -- Get 1,000 miles of free Supercharging > -- Tesla referral code - https://ts.la/richard11209
Reply by ●March 28, 20232023-03-28
On Tue, 28 Mar 2023 17:44:40 -0700 (PDT), Fred Bloggs <bloggs.fredbloggs.fred@gmail.com> wrote:>On Tuesday, March 28, 2023 at 8:21:45?PM UTC-4, Ricky wrote: >> On Tuesday, March 28, 2023 at 6:28:01?PM UTC-4, Fred Bloggs wrote: >> > On Tuesday, March 28, 2023 at 4:43:21?PM UTC-4, Ricky wrote: >> > > On Tuesday, March 28, 2023 at 4:13:02?PM UTC-4, Fred Bloggs wrote: >> > > > On Tuesday, March 28, 2023 at 3:42:11?PM UTC-4, Ricky wrote: >> > > > > On Tuesday, March 28, 2023 at 3:39:12?PM UTC-4, Fred Bloggs wrote: >> > > > > > On Tuesday, March 28, 2023 at 2:33:35?PM UTC-4, Ricky wrote: >> > > > > > > The circuit has to operate in both differential input and single ended input modes. When in single ended input mode, there's no way to ground one input, so it will float. >> > > > > > > >> > > > > > > In this circuit, the two inputs are tied through C9. So I can't see how this circuit would allow single ended operation. If, however, C9 is split in two, as a pair of 1 uF capacitors to ground, that would seem to solve that particular issue. But I'm not certain it would still operate the same way. >> > > > > > > >> > > > > > > It seems to me the split capacitor will operate both in common mode and in differential mode, while the single capacitor is only effective in differential mode. >> > > > > > > >> > > > > > > Any comments? >> > > > > > Care to link the circuit diagram image? >> > > > > Sorry about that. >> > > > > >> > > > > https://www.eevblog.com/forum/beginners/high-pass-filtering-for-differential-opamp/?action=dlattach;attach=492248;image >> > > > That circuit is intended for a bipolar supply on the OA, then you can GND an input. Common mode gain is 0. >> > > You don't seem to understand the question. >> > > >> > > 1) I can't ground an input, as there is no circuitry to do that. There is no room on the board to add a switch. One input will have to float when used single ended. >> > > >> > > 2) This will run on a single power supply voltage, but you can consider the terminology of "ground" to refer to a reference voltage that is half the power supply. >> > > >> > > 3) The inputs will be coupled to the signal source through capacitors, so no DC path. >> > > >> > > 4) The concern is that when using the circuit with a single ended input, and one input floating, that C9 will cause signal to drive the floating input, impacting both the gain, and the filter response. Actually, the filter response might not be so important. The single ended mode does not use the full 20Hz to 20kHz bandwidth. It's a 1 kHz sine wave with 100 Hz amplitude modulation. >> > > >> > > Does this make more sense? >> > I suspected that's what you were saying but wasn't sure. All that amplifier is a standard differential amplifier with some RCs hung off the intermediate nodes to get the 2-pole roll off. >> > >> > If you leave one input open you lose the DC common mode rejection. Whatever side, (+) or (-), you choose to drive, leaving the other open, the signal average will be multiplied by a gain. >> Hmmm... I'm not sure you understand. If the non-inverting input is grounded, what you say will be useful. The non-inverting input will be held at an AC constant value and the circuit will work as intended for the single ended input. But, leave the non-inverting input open, it will be dragged about by C9, the input will not be an AC constant value, and the resulting output will not be the same as if the non-inverting input were AC grounded. >> >> When running with a single ended input, there is no DC common mode rejection, because it's single ended and the concept of common mode does not apply. DC is not an issue anyway, as the source is coupled to the circuit input through capacitors. >> >> I'm more interested in finding a similar circuit that will work effectively with differential inputs, as well as with a single ended input. That's why I asked about replacing C9, with a pair of capacitors, each to AC ground Then there's no direct path through the capacitors to the other half of the circuit. The amplifier + input will be connected to the DC reference through the network and only the inverting half of the circuit will operate, as in this drawing. >> >> http://i.stack.imgur.com/yykyQ.png >> >> I'm not sure connecting two capacitors in place of one, will not degrade any performance in the two use cases. I am currently drawing up the circuit to simulate. I'm pretty convinced this will work fine, I just thought it might be interesting to discuss. I had to do a lot of searching to find any low pass, active filter, that was designed for a differential input and single ended output. > >At F= 1kHz, the 470n has an impedance of 1/(2piE3 x 470E-9)= 340 ohms. > > >> >> It's funny that Larkin says the circuit is "bizarre". But then he's never claimed to be well versed in filter design, has he?I'd donate a nice circuit if he wasn't such a consistent jerk.
