Confirm circuit design [Help]

Hi there, a newbie here regarding circuits.

I'm just about to mill myself a test-PCB for a transmitter module I've bought, and eventually for the receiver aswell.
But before I do that, I'd really appreciate if someone could confirm that my circuit is fine (See figure below).

Tx Module:
http://www.banggood.com/FPV-5_8G-200mW-Wireless-Audio-Video-Transmitter-Module-TX5823-p-84780.html

Voltage Regulator:
http://www.advanced-monolithic.com/pdf/ds1117.pdf
**broken link removed**



Cheers!

Edit:
I'm using two fixed voltage regulators, 5V & 3V3. In the datasheet I cannot find a typical circuit for the fixed ones. Thus I've looked at other places for typical "applications".
See the second link under "Voltage Regulator:"

That all looks good to me. You have the fixed regulators wired correctly.

Watch out for the antenna connection on the module, at 5.7GHz the track width and spacing are critical if you want most output to reach the antenna.

Brian.

betwixt said:

That all looks good to me. You have the fixed regulators wired correctly.

Watch out for the antenna connection on the module, at 5.7GHz the track width and spacing are critical if you want most output to reach the antenna.

Brian.

Click to expand...

Makes me glad to hear However, while we are on it I'd like to ask another question. For instance, if we look at the circuit C2 and C3 happens to be in parallell, thus the capacitance are added together (C2 + C3). Will this affect anything?

And regarding the trace going to the antenna signal, could I just make it big? Or does it have to be a required width?

Cheers!

There is a typo in your circuit: the capacitor C6 must be placed between the 3.3V pin and ground. The output of the last DS1117 (3.3V) has to be connected with the pin 3.3V of the module.
What about the input voltage to the first DS1117 ?
C3 can be eliminated, however its presence doesn't affect anything.
The RF trace have to be 50 ohm matched.

You have to know the dielectric constant "ε" and the thickness "H" of the substrate. The width "W" of the track will be given by:

W≈10*H/(ε+1)

If, for instance, you are using FR4 (ε=4.5) of thickness 1 mm then

W≈10*1/(4.5+1) = 1.8 mm

Note that if "H" was in mils then the resulting "W" was in mils too.

Under the track there must be a ground plane. Also keep the length of this line as short as possible.

An extended 8V battery will be used as a supply to the first regulator. Will be "hard" soldered to a pin on the board.

Lol I'm glad you cought my typo regarding C6! Will be fixed immediatly. And C3 will be eliminated, the less components on board the better
C4 & C6 will also be in parallell, maybe I can remove one of them?

I see, in that case I will have to change my width of the RF trace aswell.

Thank you!
Cheers