CD4017 help.

[audioguru]

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.

 

[axro]

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.

OK, well if you get enough diodes to drop all the voltage, then there would be no current, because there would be no drop on the transistor?

 

[audioguru]

If there is no output current then the output Mosfet will not have a voltage drop. So what?

 

[axro]

If there is no output current then the output Mosfet will not have a voltage drop. So what?

Thats not what I meant.

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.

 

[audioguru]

It is the current in the output transistor that causes it to have a voltage drop because of its on-resistance.

 

[axro]

It is the current in the output transistor that causes it to have a voltage drop because of its on-resistance.

So why then when you add a LED the current goes down?

 

[audioguru]

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.

 

[axro]

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?

Help me understand where I am wrong.

 

[audioguru]

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.

 

[axro]

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.

Thank you for all the help.

 

[axro]

If you used an output to directly power the base of an NPN transistor, how would you determine the current into that?

If it was 10v output would you just account for the 0.7V Vdrop across the Base-Emitter Connection and the use the chart. Giving almost the full 19mA?

Is this correct?

 

[audioguru]

Is this correct?

Yes.
But the output Mosfet will have a voltage of 9.2V and a current of 19mA which is a power dissipation of 175mW which will destroy it.

You must add a resistor in between to reduce the current and share the power dissipation.

 

[axro]

Say you were going to use an ouput to directly power a relay. How would you determine the current ouput for that case? Coil resistance and the chart?

 

[q5101997]

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.

 

[axro]

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.

I'm sorry I'm a little confused. Which post are you referring to

 

[audioguru]

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.

No.
The outputs of a CD4017 are high when they are active. They are all low when the IC is reset.

No again.
The output low and output high currents of a Cmos logic IC like the CD4017 are symmetrical.

No again.
The outputs of most old TTL ICs were push-pull and did not use and did not need a pullup resistor.