0
\$\begingroup\$

I have an LDO TPS7B69 which takes in up to 40V DC and regulates it to 5V. This LDO typically takes in DC, but I would also like it to be powered by AC 40 volts peak max at 50Hz or greater.

Rectification would first come to mind, and I would prefer using a full bridge rectifier since it will utilize both positive and negative cycle thus not needing huge caps values, which also help against Reverse Input protection (?) when using DC. But I do not know if a DC input on the FBR would also yield a DC output. I personally think it would work since it would just be like a peak voltage on an AC.

Is my approach acceptable? Are there any better ways to do this?

\$\endgroup\$
  • \$\begingroup\$ Yes, FBR will output DC, but the voltage drops of the diodes will fluctuate with the current, and you will have a diode between your circuits negative rail and ground, not cool. I would actually have a DC input separately and through the diode of DCIN to the main power line of your project to make sure the rectified power doesn't go into your DCIN if you connect them at the same time by accident. In fact, DC from FBR and DC from DCIN should both meet together with common cathode diodes, I would say. Sounds like the safest option and will give stable circuit \$\endgroup\$ – Ilya Jul 29 at 22:02
  • \$\begingroup\$ I was hoping that i would only have one power connector, to remove the possibly of as you have said having the 2 inputs connected at the same time. but i see your point that the voltage drop changes with current,.. but wouldnt that be very little and the regulator should be able to filter that out (my max current draw is 50mA) \$\endgroup\$ – DrakeJest Jul 29 at 22:09
  • \$\begingroup\$ "40V peak" AC, so about 26VAC? \$\endgroup\$ – Jasen Jul 29 at 22:44
  • \$\begingroup\$ @Ilya if the diode is a problem, would changing to half bridge solve the problem? \$\endgroup\$ – DrakeJest Jul 29 at 22:44
  • \$\begingroup\$ @Jasen yes, 40v is the highest point, but after rectification the DC voltage is probably lower. That value is not really what im going to typically use but the max \$\endgroup\$ – DrakeJest Jul 29 at 22:47
2
\$\begingroup\$

It is a standard practice to make linear power supplies by rectifying AC into a reservoir capacitor and use that unregulated DC voltage to create regulated DC voltage.

However, with such a high input voltage, the regulator has to dissipate a lot of power as heat. So depending on which regulator IC package you have and how much current you would like to draw, it might or might not be possible depending on how much the regulator temperature goes up.

Use the formula provided by analogsystemsrf to calculate how large capacitance you need.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ According to the formula of the datasheet my power dissipation is 1.75 watts, i got confused though how to calculate my max operating ambient. TAmax= TJmax– (ZθJA× PD) where pd = 1.75w, ZθJA = 64.2, i dont know how to get Tjmax \$\endgroup\$ – DrakeJest Jul 29 at 22:00
  • \$\begingroup\$ My main concern is still present though would the FBR work with a DC input? \$\endgroup\$ – DrakeJest Jul 29 at 22:04
  • \$\begingroup\$ Great, that's about 50mA load. Now it just depends on chip package type, heat sinking and ambient temperature how fast the silicon die heats up to 150 °C and melts. That is quite unrealistic power dissipation for a linear regulator. The FBR has no problems with DC input. \$\endgroup\$ – Justme Jul 29 at 22:09
  • \$\begingroup\$ @DrakeJest as I said, you would have a diode between your device's negative rail and ground dc connection. FBR diode. I wouldn't rely on it in terms of stability, but I'm not an expert \$\endgroup\$ – Ilya Jul 29 at 22:35
  • \$\begingroup\$ yes 50mA is the worst case load and im assuming 40v. I tried simulating a DC input im getting mixed results. On the oscilloscope im getting 0v while on the voltmeter is 11.2v \$\endgroup\$ – DrakeJest Jul 29 at 22:35
-3
\$\begingroup\$

Using Q = C * V, taking the derivative with respect to dT,

you find

I = C * dV/dT

And by re-arranging, yyou get

C = I * dT/dV

which for 1 amp, 1/100 second (50Hz fullwave bridge), 1 volt ripple, you get

C = 1 * (1/100)/1 = 0.01farad == 10,000 uF

| improve this answer | |
\$\endgroup\$

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.