| ??? 07/22/10 07:57 Read: times |
#177413 - Hold on Responding to: ???'s previous message |
Michael Karas said:
In the context of this MC34063A regulator part the Ipk value is not going to be just 90mA. Ipk is the peak current that the switching transistor in the chip can handle and is limited to 1.5A maximum. If you run the calculator I linked to before you will notice that Ipk gets to 1.5A when the load current on the 33V regulator gets to 90mA. (And this load of course has to include any current you intend to draw from the 7812 chip.
Once again I think you have missed an important detail here. I see you have said that you are obtaining the 5V power from a USB port. Be aware that your USB device will be overloading the USB ports legal current limit. USB ports are supposed to be designed to limit themselves to 500mA at 5VDC. This is equivalent to 2.5W for a device that draws power off the port. If you use the limit of output power out of the regulator at 33V @ 90mA be aware that this a total of 2.97W. This is well over the limit allowed for a USB device and does not even account for the inefficiency of the regulator itself or other load(s) you may have on the +5V. If one makes a brash guess that the regulator somehow approaches 70% efficiency and there are no other loads on the USB 5V then the total power from the USB port becomes 2.97 / 0.75 = 4.24 W. As you can see I think you'll have to re-think the idea of getting power from the a USB port. Michael Karas Thank Michael Hold on...I think that the Current didn't at 90mA.. The 33 VDC is only in 220uA and will investigated the +12 VDC.... Cause the schematics is very old...(load the 33VDC and 12 VDC). I knew that USB at 500mA. Thank you Jeckson Ben |
