So If I put 14 diodes from an output to ground there would be no current flowing? Since all 10V is dropped across the diodes and the transistor would get 0V?
Not quite.
Look at the datasheet of a diode like the 1N4148. It has a voltage drop of 0.7V when its current is 5mA. But its voltage drop is 0.5V when its current is 0.1mA.
You must calculate the voltage drop of your 14 diodes and the current.
Not quite.
Look at the datasheet of a diode like the 1N4148. It has a voltage drop of 0.7V when its current is 5mA. But its voltage drop is 0.5V when its current is 0.1mA.
You must calculate the voltage drop of your 14 diodes and the current.
According to the chart as the voltage drop on the transistor nears 0v so does current. So my question is if you drop all the voltage on other components(like diodes), will there then be no current output? Again according the the chart, 0V Vdrop = 0 current.
The LED has a voltage drop. The voltage drop of the LED reduces the voltage across the resistance of the turned on Mosfet driver. Ohm's Law says that the current is less when there is less voltage across a resistance.
The LED has a voltage drop. The voltage drop of the LED reduces the voltage across the resistance of the turned on Mosfet driver. Ohm's Law says that the current is less when there is less voltage across a resistance.
Right, so then if you drop all 10V on diodes, making 0V on the transistor wouldn't current be 0mA? The chart says 0V Drain to Source = 0mA output. Drain to Source the voltage dropped across the transistor...correct?
Right, so then if you drop all 10V on diodes, making 0V on the transistor wouldn't current be 0mA? The chart says 0V Drain to Source = 0mA output. Drain to Source the voltage dropped across the transistor...correct?
Yes, yes and yes.
An active output of a CD4017 is high. With a 10V supply and an output shorted to ground the output current is typically 19mA. The graph shows the reduced current if the output load causes a voltage drop.
Yes, yes and yes.
An active output of a CD4017 is high. With a 10V supply and an output shorted to ground the output current is typically 19mA. The graph shows the reduced current if the output load causes a voltage drop.
Yes, the 4017 will short its output low, so that the maximum current can flow from the LED's cathode to ground. This goes back to the days of TTL, when the O/P was tied high through a high value resistor, but shorted to ground when the output transistors CE was saturated.
Yes, the 4017 will short its output low, so that the maximum current can flow from the LED's cathode to ground. This goes back to the days of TTL, when the O/P was tied high through a high value resistor, but shorted to ground when the output transistors CE was saturated.
Yes, the 4017 will short its output low, so that the maximum current can flow from the LED's cathode to ground. This goes back to the days of TTL, when the O/P was tied high through a high value resistor, but shorted to ground when the output transistors CE was saturated.