I am a newbie and was hoping someone could assist me. Given the following schematic which takes a voltage of 12v and considers it off and a voltage of 2v and considers it on.
Can someone tell me what changes would need to take place regarding capacitors and resistors (if any) to account for 20v instead of 12v? Meaning that 20v is considered off and 2v is considered on.
Duplicate post deleted. Please do not post the same question in multiple places.
With 12V, Vbe will be 12*4700/(1000+10000+4700) = 3.59V
With 2V at the bottom of the 1K resistor, Vbe will be 2*4700/(10000+4700) = 0.64V
That should be too low to turn on the inherent Vbe diode and bring the Q1 emitter down low enough to turn off Q2.
Change 12V to 20V, and the first voltage just goes up: 20*4700/(1000+10000+4700) = 5.99V
With the switch closed, the bottom of the 1K is 0V, so you wouldn't see 2V no matter what.
Not possible to calculate if you don't know the LED's forward voltage.
A calculated guess is ~6-7mA through the 1k resistor connected to +12volt. Most of it goes to the LED.
That means that you have to add ~150ohms per volt to the 1k resistor.
8 volts more, so change the 1k resistor that connects to 12volt to 2k2.
I don't see any problems if you power the original circuit directly from 20volts. Only the LED would be a bit brighter.
Does the +5volt also change?
on/off output voltages are meaningless. The transistor is just a switch to ground.
Those voltages appear when you connect something to the output.
Leo..
Can someone tell me what changes would need to take place regarding capacitors and resistors (if any) to account for 20v instead of 12v? Meaning that 20v is considered off and 2v is considered on.
It's a digital switch. If it was OFF for 12V, it will certainly be OFF for 20V. The difference in current is not enough to damage the transistors. It should actually work better with 20V than with 12V, (as Crossroads implied, (I think))
The 4.7k resistor is wrong. Should be 47k.
This is a doorbell, isn't it?
Hi,
What is the application that needs the outcome inverted and changed to 20V logic?
There may be a simpler solution, especially if we know what you are going to do with the 20 to 2 V signal.
Tom...... 