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It doesn't make a difference. If i tie it to ground the VT LED is constantly on as it is constantly transmitting, hence why i have put it through the switch. I was trying to get some sort of control..One thing I noticed, your diagram for the HT12E does not match the breadboard layout. In the diagram you have 'TE' grounded. On the breadboard it is tied to 'D11'.
Im not sure if my comment posted on the other website...Like I said on the other website, the capacitors for the 5V regulator are not ceramic and are too far away so I think the regulator is oscillating.
This is my regulator, so i will try this and hopefully you were right!All datasheets for the 7805 regulator are spec'd with and show a 0.33uF input capacitor and a 0.1uF output capacitor. They are not electrolytic because electrolytic caps have too much inductance. Regulators are stable with capacitance at high frequencies, not inductance which is the opposite.
Your multimeter will not show high frequency oscillation and the capacitance of your scope' cable might stop oscillation.
it's not v stable without it....From what I have seen of these two ICs, they have a wide operating voltage 2.4V~12V. So there is no need for a 5v regulator in your circuit.
You have a low dropout 5V regulator. It is completely different to a normal 7805 regulator because to achieve low dropout its series pass transistor is a PNP that has some voltage gain, instead of an NPN emitter-follower without any voltage gain used in a 7805 regulator. Because the PNP has voltage gain then a pretty big output capacitor is needed, and tantalum is spec'd because it has much lower inductance than an electrolytic.
National Semi (not the Taiwan copy-cat company) invented the LM2940 low dropout regulator a long time ago, and it has problems. It needs an expensive tantalum output capacitor and it needs a load current.
For all my 9V battery operated 5V regulators I use the National Semi (they have been purchased by Texas Instruments) LM2931 low dropout 5V regulator. It uses an inexpensive 100uF electrolytic output capacitor and does not have a minimum load current.
The input and output capacitors should be mounted VERY CLOSE to the pins of the regulator IC. Yours are too far away.
You are using a 9V battery. Its voltage drops to 6V during its life but a 7805 voltage regulator does not regulate when its input voltage drops below about 7V. You should use a low dropout 5V regulator when you use a 9V battery.
The minimum output current from an output of the HT12D is only 1mA which is barely enough to dimly light an LED but you are shorting its outputs with much more than 1mA. An output goes high when a button for it is pressed so to turn on a motor then an output feeds a series 1k resistor to the base of an NPN transistor with its emitter at ground. The collector will go to ground when a corresponding button is pressed and can drive a motor that has its other wire at +5V or +9V. The LED and its current-limiting resistor must be disconnected from the output and instead connected to the emitter of a PNP emitter-follower with its collector connected to 0V that has its base directly driven from the output.
Originally i was going to power the transistor from my output where the LEDs are, but as i said the output goes low when i press the button.
I am just trying to apply what you just said to this. Maybe im just missing something
Im not sure if i was clear before. The outputs go LOW when you press the data input switch. The LEDs energise when power is applied to the circuit...
You can use a relay or a Mosfet if the motor draws a very high current. But then the little 9V battery will die very soon.
You said before the LEDs de-energize (turn off) when the button is pushed, now you say an output goes low (which will turn on an LED) when the button is pushed. Which way is it?
Yes, the decoder outputs are low when the power supply is turned on which lights the LEDs.
When a button is pushed then an output goes high which turns off its LED.