5
\$\begingroup\$

I am prototyping this voltage doubler circuit based off the datasheet of the TC962. Capacitors are GRM21BZ71E106KE15L. When applying 9V at the input, the output is 17.96V at no load. When drawing around 67mA using a 215Ohm resistor across the output, the voltage output drops to 14.5V.

At this stage I'm trying to understand whether this voltage drop occurs from the diodes alone or whether I can improve my circuit in general.

enter image description here

\$\endgroup\$

1 Answer 1

11
\$\begingroup\$

The output resistance of the chip is typically 35Ω at 80mA output so that accounts for 2.3V of the drop with a 67mA load.

There are also two diodes and an inductor and the external MOSFET.

The diode has typically 500mV drop at 70mA, so that's 3.3V.

If we assume the inductor, the ripple and the reverse protection MOSFET drops are insignificant, you would expect 14.7V at the output with 9V in, which is essentially what you are seeing.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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