CD4017 help.

[axro]

The current charges the capacitor, not the voltage.
When power is first applied to the circuit of a resistance in series with a capacitor then the charging current is the highest because the capacitor has no voltage yet. The resistance has the entire 9V across it.

When the capacitor is half charged then it has 4.5V across it and the resistance has 4.5v across it so the charging current is half of maximum.

So current can still flow into a capacitor(or other component) even if there is no voltage(because it's being dropped by resistance)?

When voltage is "dropped" it's actually gone from the circuit right? Lost as heat or w/e? Or is voltage drop just the amount of voltage being used in that section.

Like if I have 2 LEDS and 6V power supply. 1 LED is 4V one is 2V. Does the 1st LED "see" 6V or 4V?

 

[audioguru]

The voltage across the resistance creates current in the resistor. the current charges the capacitor.

If there is no voltage across the resistance then there is no charging current.

 

[axro]

I think I may need to just give up....I have an idea of how it works(a wrong one at that) and I can't shake it.

Voltage across = voltage drop?

voltage drop =?

 

[blueroomelectronics]

Get yourself a breadboard, bag of parts and a multimeter.

Or

Download the trial version of Electronic Workbench and try building some virtual circuits.

 

[axro]

I've done both.

The simulator I'm using is this

Circuit Simulator Applet

But doing that isn't answering the questions I'm asking.

OK how bout this. You have 2 resistors and 10v psu. A 100ohm resistor and a 1000 ohm resistor.

The 100 has 1 V across it and the 1000 has 9V. So does the 100 drop the 1V and pass the other 9V through it?

 

[blueroomelectronics]

I took electronics in school. You could try to find some old Forest M Mimms books which are an excellent intro to basic electricity and electronics.

 

[audioguru]

OK how bout this. You have 2 resistors and 10v psu. A 100ohm resistor and a 1000 ohm resistor in series.

The 100 has 1 V across it and the 1000 has 9V. So does the 100 drop the 1V and pass the other 9V through it?

The total resistance is 1100 ohms.
The current is 10V/1100 ohms= 9.09mA in both resistors because they are in series.
The voltage across the 100 ohms resistor is 100 x 9.09mA= 0.91V.
The voltage across the 1000 ohms resistor is 1000 x 9.09mA= 9.09V.
Ohm's Law plus simple arithmatic.

 

[axro]

Ok last question.

So that graph you posted. The number on the top is the voltage drop on the internal transistor?

Also I should be able to hook up a LED on an output with no resistor inline correct? Because the max current is less than 20mA.

 

[audioguru]

You must look at the power dissipation in the output transistor that causes it to heat. The max allowed dissipation is 100mW and the datasheet says over the full package temperature range.

As I said on page1, with a 10V supply the typical current into a 3.5V blue or white LED is 17mA so the output transistor dissipates 17mA x 6.5V= 111mW which is OK if the ambient temperature is not too hot.

If an output drives a 2V red LED with 19mA then the output transistor dissipates 19mA x 8V= 152mW which is too hot and a lower supply voltage or a current-limiting resistor is needed.

 

[Ross Craney]

The voltage supplied to a circuit will be automatically shared between any components in the circuit proportionally to their resistance.
If you have a 12V supply & two three ohm resistors in series then each resistor will drop 6 V . ie the current will be 2A. Now if you introduce a 6 ohm resistor into the circuit the current will decrease because of the higher total resistance. The current will now be 1A. The voltage drop across the three ohm resistors will halve to 3V & the remainder (6V) will be dropped by the by the six ohm resistor. The total voltage dropped around the circuit is still 12V but shared by all components in the circuit.

There are not many other ways of explaining it that haven't been put forward

 

[axro]

You must look at the power dissipation in the output transistor that causes it to heat. The max allowed dissipation is 100mW and the datasheet says over the full package temperature range.

As I said on page1, with a 10V supply the typical current into a 3.5V blue or white LED is 17mA so the output transistor dissipates 17mA x 6.5V= 111mW which is OK if the ambient temperature is not too hot.

If an output drives a 2V red LED with 19mA then the output transistor dissipates 19mA x 8V= 152mW which is too hot and a lower supply voltage or a current-limiting resistor is needed.

How would I calculate the resistance needed for that then? Do I need to take into account the resistance of the internal transistor, or do I just do the standard calculation? 10-2 = 8/.02 = 400ohm?

 

[audioguru]

How would I calculate the resistance needed for that then? Do I need to take into account the resistance of the internal transistor, or do I just do the standard calculation? 10-2 = 8/.02 = 400ohm?

You have 8V to divide between the output transistor and the resistor.
The output transistor can dissipate 90mW safely.
A 150 ohm resistor will have a voltage drop of 2.4V at 16mA then the output transistor will have a voltage drop of 10V - 4.4V= 5.6V and a dissipation of 5.6V x 16mA= 89.6mW.

 

[axro]

You have 8V to divide between the output transistor and the resistor.
The output transistor can dissipate 90mW safely.
A 150 ohm resistor will have a voltage drop of 2.4V at 16mA then the output transistor will have a voltage drop of 10V - 4.4V= 5.6V and a dissipation of 5.6V x 16mA= 89.6mW.

Where did you get the 16mA from? Did you just make a guess again and compare to the chart?

 

[audioguru]

I made a guess that was perfect.

 

[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?

 

[DEECEE]

IF A PROJECT (LED CIRCUIT) CALLS FOR A 4017 IC, IS IT THE SAME AS A CD40174BE CMOS Hex D Flip-Flop (RCA)?

 

[audioguru]

IF A PROJECT (LED CIRCUIT) CALLS FOR A 4017 IC, IS IT THE SAME AS A CD40174BE CMOS Hex D Flip-Flop (RCA)?

No.
A CD4017 is completely different to a CD40174.
Simply look at their datasheets to see.

Please do not type in CAPITALS. It is rude like screaming.

 

[axro]

Here is something I don't get. If you short a pin to ground and you have all 10V dropped across the transistor you get the max of 19mA output. But as voltage drop across the transistor decreases so does the current.

But with normal NPN transistors as the voltage drop of the transistor decreases(nearing saturation) the current output increases. Or is my logic flawed?

 

[audioguru]

Here is something I don't get. If you short a pin to ground and you have all 10V dropped across the transistor you get the max of 19mA output. But as voltage drop across the transistor decreases so does the current.

But with normal NPN transistors as the voltage drop of the transistor decreases(nearing saturation) the current output increases. Or is my logic flawed?

A CD4017 is made with weak tiny Mosfets, not powerful NPN transistors with an output of hundreds of mA.
The weak tiny mosfets limit the current to 19mA when they have a supply of 10V and drive a short. The voltage drop across the turned-on Mosfet decreases when the output current is less. The voltage drop across the turned-on Mosfet is zero when its output current is zero.

 

[axro]

A CD4017 is made with weak tiny Mosfets, not powerful NPN transistors with an output of hundreds of mA.
The weak tiny mosfets limit the current to 19mA when they have a supply of 10V and drive a short. The voltage drop across the turned-on Mosfet decreases when the output current is less. The voltage drop across the turned-on Mosfet is zero when its output current is zero.

Ok, Thank you.


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?