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In my application, I am generating 7.5V from TPS5430 (Buck) regulator and to generate -5V, I am assigning inductor's one leg to input of 7905 for -5V o/p.

Circuit is shown. U14 is TPS5430 and U15 is 7905.

Is this a correct implementation to get -5V from pulsating DC?

I have seen this arrangement somewhere in design and trying to apply in my circuit.

enter image description here

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    \$\begingroup\$ It might not work well if the load on the 7.5V output is too light. \$\endgroup\$ – Spehro Pefhany Jun 11 '15 at 14:07
  • \$\begingroup\$ 7.5V is used as supply to L6203 and 7805 LDO. Will it work? And my question is voltage across inductor is pulsating above and below gnd reference, where as inductor waveform wrt gnd is unipolar pulse and unipolar pulses are applied to i/p to 7905,then how we can get -ve i/p to 7905? \$\endgroup\$ – Electroholic Jun 11 '15 at 15:32
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    \$\begingroup\$ The principle of using the capacitor is valid, you should get ~-11VDC into the 79L05, however the input waveform will depend on the load on the 7.5V supply. With a very light load you might lose the negative supply. Maybe you have to add a dummy load, I have not analyzed it in detail, which is why this is a comment and not an answer. \$\endgroup\$ – Spehro Pefhany Jun 11 '15 at 15:56
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    \$\begingroup\$ Spehro is correct, if your inductor current is discontinuous because of a light load on the primary output you will have problems getting the negative output to work. One solution would be to choose a different switcher that is synchronous and always operates in forced continuous conduction mode. That way the negative voltage rail will always work regardless of the load on the main supply. Also, the amour of current from the negative supply is limited by the amount of energy transferred by the cap in each cycle, so your negative load might be an issue as well. \$\endgroup\$ – John D Jun 11 '15 at 18:08
  • \$\begingroup\$ How to get this working for above circuit ? \$\endgroup\$ – Electroholic Jun 12 '15 at 2:12
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You could make your main inductor A coupled coil.Coupled coils are readily available off the shelf in SMD form from people like "coilcraft " .Keep the microhenries the same to avoid any other hassles and use the same current rating .You will find that the coupled coil wont be big or expensive .The 1st winding replaces the main inductor and the 2nd winding goes to a diode then a 10microfarad ceramic cap .The expected voltage is 7.5 because that's your output and the coupled coils that I have as samples are 1:1 .I recommend a LDO reg because the 7905 will struggle. If you don't want to do the coupled coil there is always the "GEEBIG " which uses separate coils the second or added coil having a high inductance compared to the main inductor .Remember that these methods work when the neg power needs are much lower than the pos power needs which would be the case for your diode pump anyway .

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  • \$\begingroup\$ If I want to test it with TPS5430 only.. can you suggest how to make it work with consistent regulation? \$\endgroup\$ – Electroholic Jun 10 '16 at 13:47

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