How do I calculate the resistors and the capacitor on this circuit, which has ground and virtual ground as you can see?
In the second image, how do I find the current I2?
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\$\begingroup\$ It depends on load and tolerance \$\endgroup\$ – Tony Stewart EE75 Dec 21 '17 at 18:40
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\$\begingroup\$ It is a circuit with 4 op-amps. So I calculate everything in order to find the current it needs and then I calculate the resistors and the capacitor of the above circuit right? \$\endgroup\$ – joe3489 Dec 21 '17 at 18:48
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\$\begingroup\$ 1) 50mA is too much for an Op Amp and feedback current has nothing to do with input bias current I2. If output is in linear range Vin+=Vin- \$\endgroup\$ – Tony Stewart EE75 Dec 21 '17 at 21:29
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\$\begingroup\$ Let's say instead of 50mA, it is 10mA. I suppose even in this case I cannot calculate I2, right? \$\endgroup\$ – joe3489 Dec 21 '17 at 21:35
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\$\begingroup\$ Correct I2 is a device dependent and often in uA range \$\endgroup\$ – Tony Stewart EE75 Dec 21 '17 at 21:57
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That is not a great method to use if you have more than one or two low current circuits attached to it.
The issue is any current passing back through that virtual ground has to make it through those resistors and changes the reference point. As such the resistors need to be small, which wastes a lot of current. The capacitors need to be fairly large regardless.
You would be better to use a more active circuit. Note the doubled up capacitors. This ensures a faster on time since they automatically charge up to half rail.
simulate this circuit – Schematic created using CircuitLab
As mentioned in the comments, op-amps are only good for a limited amount of current. If you need more than that there are specialized devices for this or you should be considering a proper dual-supply.
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1\$\begingroup\$ Better than the original circuit, but lots of op amps can't sink more than a couple of tens of mA. If you need more than that, best to go with a real "rail splitter", like the TLE2426, or boost the opamp with a transistor \$\endgroup\$ – Scott Seidman Dec 21 '17 at 20:04
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\$\begingroup\$ Why do you have the capacitor divider at the input/output? All AC noise will appear at the input and will be attenuated on by the factor of 2 \$\endgroup\$ – G36 Dec 21 '17 at 20:05
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1\$\begingroup\$ @ScottSeidman good point. And arguably if you have more than 20mA rail splitting might not be the best solution at all. \$\endgroup\$ – Trevor_G Dec 21 '17 at 20:06
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\$\begingroup\$ @G36 ya I know. it just feels wrong having them on one side only. C1 and C2 just keep the reference quiet. C2 and C4 are bypasses for the return currents in the rest of the circuit. \$\endgroup\$ – Trevor_G Dec 21 '17 at 20:08
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\$\begingroup\$ But how can C1, C2 keeps the reference quiet if they form a voltage divider? \$\endgroup\$ – G36 Dec 21 '17 at 20:24
