Will this circuit work

granddad · Mar 24, 2016

Lew Just had a look at HC-12 data sheet looks like good stuff , I worked recently with serial over 433Mhz, but the (Aurel RTX ) it has a power up sequence , this looks like a more useful module, 1000m is impressive have you tried it.
I would go for Mike's P MOSFET circuit.

lew247 · Mar 24, 2016

Maximum of 100mA it's actually more like 16mA
yes Its a brilliant module, works really well

I didn;t see Mike's Mosfet circuit at first, its a nice circuit but too many components, if the 3904 can be used as a switch why does it need a mosfet?

Switch the 5v line o the gnd line?
(last circuit on the previous page)

audioguru · Mar 24, 2016

lew247 said:
The transistor is working as a simple switch, to switch the 5v supply to the module on or off

Absolutely NOT! An emitter-follower is not a switch. 3.2V input gives 2.5V output.
But a switching transistor with 0V input gives +5V output. When its input is a little more than 0.7V with a resistor limiting the current from the PIC then the output is close to 0.1V so it makes a good switch.

lew247 · Mar 24, 2016

ok

HOW can I use a pic pin either 3.3v or 0v to work with a transistor to switch 5v to a module?

granddad · Mar 24, 2016

lew247 said:
if the 3904 can be used as a switch why does it need a mosfet?

What happens if the PIC 3.3 is off or ports are inputs after a reset , and still 5v at the 3904 transistor , the base is floating...

lew247 said:
but too many components

One more mosfet and resistor.
HC-12 Range 1000 meters ?

lew247 · Mar 24, 2016

How about

**broken link removed**

MikeMl · Mar 24, 2016

lew247 said:
Another options is this:

How is that any different than your post #1?

lew247 · Mar 24, 2016

its a mosfet

granddad · Mar 24, 2016

lew247 said:
How about

Better... what is the 3803 on resistance

lew247 · Mar 24, 2016

3803?

granddad · Mar 24, 2016

I think Mike referred to your post #20

granddad · Mar 24, 2016

the IRL3803 MOSFET on resistance...

Nigel Goodwin · Mar 24, 2016

lew247 said:
Another options is this

It's a poor option, using an emitter follower as an analogue amplifier - you need a 'switch'.

Assuming the PIC is run from 3.3V, then you can do it with an NPN transistor and a PNP one - check the 'Hardware Extras' section of my PIC tutorials.

Or if the PIC is 5V, just a PNP transistor and a resistor or two is all that's needed, with the PIC switching LOW to turn the transistor ON.

lew247 · Mar 24, 2016

Nigel
I did read your hardware extra bit and I thought I had got it right
post "5"

isn't my post 5 the same (or similar) to this?

The only difference I can see is I'm using 5v for the +ve and a module instead of a lamp

Would I be better off with a 1K resistor?

"The 'sinking' example is probably the easiest to understand, as the PIC pin goes high current flows through R1 turning the transistor ON, effectively connecting the bottom end of the lamp to ground. This method has a rather useful property, the +ve supply to the lamp isn't related in any way to the supply voltage of the PIC, so you could quite happily have the PIC fed from 5V, and a 12V supply feeding the lamp, you just need to make sure that they share a common 0V or Gnd connection."

Post 5's circuit

**broken link removed**

MikeMl · Mar 24, 2016

Still leaves the problem that the PIC port TX and RX pins will always be sinking current, even if they are tri-stated off!

Here is another way to do it: The Vf of the Red LED is used as a 2V level shifter. Note that the logic levels at pin 16 of the PIC must be inverted.

Nigel Goodwin · Mar 24, 2016

lew247 said:
Nigel
I did read your hardware extra bit and I thought I had got it right
post "5"

isn't my post 5 the same (or similar) to this?

The only difference I can see is I'm using 5v for the +ve and a module instead of a lamp

Would I be better off with a 1K resistor?

"The 'sinking' example is probably the easiest to understand, as the PIC pin goes high current flows through R1 turning the transistor ON, effectively connecting the bottom end of the lamp to ground. This method has a rather useful property, the +ve supply to the lamp isn't related in any way to the supply voltage of the PIC, so you could quite happily have the PIC fed from 5V, and a 12V supply feeding the lamp, you just need to make sure that they share a common 0V or Gnd connection."

That's fine if you're sinking current, but sourcing (with different supply voltages) requires two transistors. (or an LED like Mike has posted - or various other means as well).

From a personal point of view, I'd prefer to switch the positive

I'm currently doing a project with an FDTI USB/Serial board, and I was planning simply feeding it from a PIC pin to switch it ON and OFF. However, I haven't measured the current it takes yet

Plan 2 is to feed the FDTI from the USB socket, with a diode powering the PIC, and a PIC pin to detect when it's connected to USB.

lew247 · Mar 24, 2016

I don't get the "sinking current and sourcing"
but
Hopefully this will work?

but Why the LED? is it on or off when the switch is off?

The only thing I don't understand is does the pic pin go high or low to make the HC-12 work? (get power) I'm assuming low???

**broken link removed**

granddad · Mar 24, 2016

Lew ...#37 Always on ...

lew247 · Mar 24, 2016

That's no good then,
The idea of the switch is to conserve battery consumption not increase it

Although if the pin was high on power on (or told to go high) wouldn't that stop the led working and have no current draw?

Then if the pic pin went low, the transistor would conduct?

granddad · Mar 24, 2016

Did you not like post #35 .has my vote

Will this circuit work

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