This is the circuit I have to create a negative rail for a local op-amp which needs the rail in order to manage to get low on the output rail. I've tried running a single supply but that did not perform well enough. The circuit works and with my diodes I get an output of -4.4V from 5V positive which is enough for my needs.

The problem is that the output is a bit noisy and I can see some low freq. pickup as well. How can I filter the output of the circuit? I've tried to increase the C4 capacitor which reduced the noise however the did not change the low freq. pickup.

The circuit will supply an op-amp used in analog circuit so low noise is important.

UPDATE: The schematic was wrong (did it in a hurry before leaving duh), however the circuit was hooked up correctly.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ As drawn your "cheap circuit" will only generate heat in D2 and could very well trend to producing a failure of the MCU output that supplies the 31.25kHz drive signal. To get closer to intentions you need to put C3 in series with the signal source and the connection between the two diodes D1 and D2. \$\endgroup\$ – Michael Karas Dec 23 '14 at 14:15
  • \$\begingroup\$ @MichaelKaras Please see update in the original post, there was an error in the schematic however the circuit was hooked up as should. \$\endgroup\$ – user34920 Dec 23 '14 at 15:49

Your circuit:

Can't work as you describe. If this is supposed to be a charge pump, then C3 is in the wrong place. A proper charge pump is:

Issues of noise are nonsensical until you actually have a charge pump. Try the above circuit. Then you can ask another question if you still have problems with that.


Are you using the MCU output directly as the source of the square wave? The MCU output impedance (probably low 2 digits of ohms range) transfers to the charge pump output and needs to be taken into consideration.

If the low frequency noise is indeed resulted from low frequency load variation and it is somewhat repetitive and can be estimated with a low frequency limit, along with the estimate of the charge pump output impedance, you can calculate a capacitor value to meet the noise requirement.

Alternatively, to reduce the output capacitor size, lower the impedance of the square wave source with a buffer of some sort.

The LDO suggestion in the other answer may give you the cleanest output if there is some voltage to spare.

  • \$\begingroup\$ I had this formula for the capacitor size: C = 1 / (10*f*R) where R is the output resistance of the source, f is the freq. and C is the capacitor size(s). I did not have R so I used R=V/I so this works out to 5 / 20m = 250 Ohm. f is 31.25KHz and the cap size is about 0.01uF. \$\endgroup\$ – user34920 Dec 23 '14 at 23:40
  • \$\begingroup\$ So 0.01uF is going to limit the ripple due to the switching frequency 31.25KHz. But the load may have dynamically changing element that is much slower (for example, the opamp is driving something at audio frequency of 300Hz, that is 100 times slower). Then, at first glance, the 0.1uF could be 100 times too little to take care of the ripple from that effect. \$\endgroup\$ – rioraxe Dec 24 '14 at 2:23
  • \$\begingroup\$ I added a RC LPF with a 680 Ohm resistor + 470uF cap, the noise is much better but still not acceptable. I think I might have to used an LDO, I'm not using a differential probe to check this but the noise is not common mode. \$\endgroup\$ – user34920 Dec 24 '14 at 9:51

This low frequency noise can be generated by changes in the current consumption (negative rail) of the op amp. Maybe a LDO could do the trick. For the high frequency, use any low pass filter.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.