Dear all,

I am learning digital electronics and for a long time I have played with logic IC's but there's something that I really need to know.

Firstly, I am working with CMOS chips, specifically the 74HC series of chips.

What I want to know is, should I add resistors at the Vcc or the logic Inputs of these chips? I have no idea what the current should be at these places. The IC's seem to work without them. The datasheet says that the "Input diode current" should be 20mA, but what is Diode Input ? is it just the inputs? Why call it diode? Also it says that at 6V, the "Quiescent Supply Current" is 4uA at 25C and 40uA from -40 to 85C. This information seems crazy.

Anyway. how do I calculate the resistors to use, in case they're needed? It seems also that the inputs have very high resistance and so maybe resistors are not needed at inputs.

Please let me know

Much appreciated

Thank you,

Paulo

 

The datasheet says that the "Input diode current" should be 20mA

I bet it doesn't say "should be". That rating is the maximum allowable current (let's call it Imax), beyond which damage to the chip might occur.
CMOS chips have diodes connected from input pins to the Vdd and Vss rails to help protect the chips from excessive input pin voltages.
However, these diodes can withstand only a certain amount of current without overheating, so input resistors are sometimes needed to limit the current. For an input voltage Vin> Vdd, the resistor value would be greater than (Vin-Vdd)/Imax.

Also it says that at 6V, the "Quiescent Supply Current" is 4uA at 25C and 40uA from -40 to 85C.

Look again carefully. 4uA is a "typical" value. 40uA is the maximum over that big temperature range.

 

You should never use a resistor on the Vcc or GND pins. Connect them directly to the power supply. Inputs which are routed to external circuits can sometimes use resistors to limit input current, or you can use another kind of interface device besides CMOS. If teh input to a chip is connected to another chip on the same board then using resistors to limit current is not required.

 

You don't need anything to limit the supply current to the chip's power or inputs as long as they are within the voltage limits of the device.
The chips will take whatever current they need.
These currents can vary with temperature and the data sheet is just telling you what those currents may be.

It's a common newbie error to think that the current rating of a device means you must somehow supply that current.
In most cases that current is either a limit or just what it normally draws when operating.

 

Thanks everyone.

Ok I understand now that I shouldn't add resistors to Vcc.

Inputs are still not clear however. Usually these chips can take up to 6V at inputs. I am currently using 3.5V from my power source, but it can supply up to 1.5 Amps of current. Of course this is a lot compared to my chip's 20mA maximum. So what to do here? Should I not add resistors?? I have noticed one of my chips taking 23mA current at one of its "Parameter" inputs, like Enable, etc.

Please help!

 

I am currently using 3.5V from my power source, but it can supply up to 1.5 Amps of current.

I am currently using 240 volts from the local power company, but they can supply thousands of amps. Should I put a resistor in series with my house?

No. As long as the input voltage is within the proper range, the power supply will not shove all its amps into your chip.

I have noticed one of my chips taking 23mA current at one of its "Parameter" inputs, like Enable, etc.

Did you measure 23 milliamps going into or out of the enable pin or did you read that 23 ma is the maximum allowed when you apply too much voltage to an input pin? If you choose to apply too much voltage to an input pin, you are supposed to use a resistor to limit the current to 23 ma...or you can choose not to apply too much voltage.

 

Let's not say NEVER add resistors to the supply lines. I have seen it done with commercial analog circuits and this http://www.designers-guide.org/design/bypassing.pdf note explains why.

"Input diode current" should be 20mA, but what is Diode Input ?

There is a "parasitic" diode that's formed and often you will see specs of -.3+Vdd to Vcc+0.3. these are the potentials of the Schotkey diodes that exist. ESD diodes may be on chip as well to protect against the human body discharge model. They can also be added externally.

The REAL importance of this rating is what happens when Vcc and Vdd are both zero. i.e. the device is unpowered, but it's fed a live input.
This is when you have to limit the input current. The parasitic diodes are sometimes shown and sometimes not. This goes for OP-amps too.

 

The answer is NO.
No resistors are required on the inputs.
Why?

Read the data sheet. The maximum leakage current is 1μA. That means that the actual leakage current is less than 1μA. Hence, with a Vcc supply of 6V, the input resistance looks like about 10MΩ.

You are not going to draw much current through a 10MΩ load.

BUT, you never leave an input pin floating. So, if you want to connect a switch across the input for input to GND (i.e. pull-down to logic LOW), or from input to Vcc (i.e. pull-up to logic HIGH) you need to do so through a load that takes more than 1μA. That means that a pull-up or pull-down resistor should be at least 10-times lower than the input resistance, i.e. 1MΩ or lower, 100kΩ is a good answer. The switch does the opposite of the pull-up or pull-down, i.e. if the switch is to GND, you need a pull-up resistor. If the switch is to Vcc, you need a pull-down resistor.

Grounding the input pin or shorting to Vcc does no harm to the chip.

 

The answer is NO.
No resistors are required on the inputs.
Why?

Read the data sheet. The maximum leakage current is 1μA. That means that the actual leakage current is less than 1μA. Hence, with a Vcc supply of 6V, the input resistance looks like about 10MΩ.

You are not going to draw much current through a 10MΩ load.

BUT, you never leave an input pin floating. So, if you want to connect a switch across the input for input to GND (i.e. pull-down to logic LOW), or from input to Vcc (i.e. pull-up to logic HIGH) you need to do so through a load that takes more than 1μA. That means that a pull-up or pull-down resistor should be at least 10-times lower than the input resistance, i.e. 1MΩ or lower, 100kΩ is a good answer. The switch does the opposite of the pull-up or pull-down, i.e. if the switch is to GND, you need a pull-up resistor. If the switch is to Vcc, you need a pull-down resistor.

Grounding the input pin or shorting to Vcc does no harm to the chip.

I don't even know what leakage current is. Now I googled it, and it seems to have to do with when a device is OFF. How does it come in when the IC is ON ?

 

Let's not say NEVER add resistors to the supply lines. I have seen it done with commercial analog circuits and this http://www.designers-guide.org/design/bypassing.pdf note explains why.



There is a "parasitic" diode that's formed and often you will see specs of -.3+Vdd to Vcc+0.3. these are the potentials of the Schotkey diodes that exist. ESD diodes may be on chip as well to protect against the human body discharge model. They can also be added externally.

The REAL importance of this rating is what happens when Vcc and Vdd are both zero. i.e. the device is unpowered, but it's fed a live input.
This is when you have to limit the input current. The parasitic diodes are sometimes shown and sometimes not. This goes for OP-amps too.

I've built and released a large number of products in the last half century. I've seen not a single instance where a damping resistor was required on a supply line (Vcc or GND). The one exception for inputs was on the address lines for a 64K byte array of 16Kx1 DRAMs. The actual problem was the distributed inductance of the traces on the PCB. For random logic on even a two layer board this has never been a problem up to several tens of megahertz. It is true that I've never done an ECL design.

 

Since we're here, let me please ask another related question.

If I have an "input wire", that is a logic level that I will bring either from Vcc or GND, and I want this level to go to various IC inputs, can I just divide the wire at a junction and send each to its IC? Will this cause problems? I know the voltage will be the same, but the current might divide too much and not be able to drive all of the inputs ?

 

what are you saying ? I didn't even know what leakage current was. Now I googled it, and it seems to have to do with when a device is OFF. How does it come in when the IC is ON ?

Leakage current, from an input when the device is ON but undriven, will put a voltage on a pullup or pulldown resistor. If the resistor is large the voltage will be large, Maybe enough to make a low look like a high or a high look like a low. What you never want to do is leave CMOS inputs unconnected. If you do their value can be anything and the parts will misbehave.

Since we're here, let me please ask another related question.

If I have an "input wire", that is a logic level that I will bring either from Vcc or GND, and I want this level to go to various IC inputs, can I just divide the wire at a junction and send each to its IC? Will this cause problems? I know the voltage will be the same, but the current might divide too much and not be able to drive all of the inputs ?

You can do exactly that. CMOS inputs are very high impedance and require only a very small amount of current. Look again at the datasheet and look for the specification for typical input currents for an Input Low and an Input High.

 

..............If I have an "input wire", that is a logic level that I will bring either from Vcc or GND, and I want this level to go to various IC inputs, can I just divide the wire at a junction and send each to its IC? Will this cause problems? I know the voltage will be the same, but the current might divide too much and not be able to drive all of the inputs ?

Let me say again.
The device will take whatever current it requires as long as you supply the correct voltage.
At a wire junction the currents will just divide as needed, nothing to worry about.
The wire is such a low impedance as compared to the small currents, that it has no significant effect on the signal.

The inputs of CMOS circuits are a very high impedance (they look like a small capacitor) but there is some slight leakage current through the input protection diodes which go to ground and Vcc.
This leakage is there for either high or low inputs (although it will likely vary some between high and low).

 

I've built and released a large number of products in the last half century. I've seen not a single instance where a damping resistor was required on a supply line (Vcc or GND). The one exception for inputs was on the address lines for a 64K byte array of 16Kx1 DRAMs. The actual problem was the distributed inductance of the traces on the PCB. For random logic on even a two layer board this has never been a problem up to several tens of megahertz. It is true that I've never done an ECL design.

I haven't seen it in digital designs either. Analog, yes.

 

The inputs of CMOS circuits are a very high impedance (they look like a small capacitor) but there is some slight leakage current through the input protection diodes which go to ground and Vcc.
This leakage is there for either high or low inputs (although it will likely vary some between high and low).

You mean the diodes go to ground and vcc, or the current goes to ground or vcc? Confused here because as far as I know currents go only to ground not vcc.

Would a comma help there or am I missing something ?

 

Incredible how I could forget Ohms law here!

I was forgetting these chips have internal resistance and so as long as the voltage is the required one, the current will not burn the chip. How can I forget this. But then again, if my power supply can supply 1.5 Amps at the normal 3.5V, how can I be sure that the resistance inside my chip not not cause it to get burned ? This seems strange...........Does the manufacturer add enough internal resistance to the Chip so that the mosfets inside them dont get toasted?

But when I put all my chips together with a clock, of course, most inputs were varying voltage quickly, and because I was investigating currents, I saw that at the "Clear" input of a Flip Flop, the current was varying from 0 to 23mA, and 23mA > 20mA which was the said maximum, what could this mean?

 

Also my dear friends,

I've heard many times about not leaving unused inputs floating. Fine! But does this mean strictly all unused inputs ? Because suppose I am using an octal D-Flip flop chip which has 8 D-Inputs, and if I'm not using some of the bits, it doesn't seem harmful to leave these unused inputs floating since their outputs are not going anywhere in the circuit. But, might these inputs build up charge and then suddenly shot out charge into nearby inputs changing logic levels?

So my question is, in case flip flop case say, will it matter ?

 

You mean the diodes go to ground and vcc, or the current goes to ground or vcc? Confused here because as far as I know currents go only to ground not vcc.

Would a comma help there or am I missing something ?

Current can go anywhere. Current can go to GND. Current can go to Vcc.

CMOS inputs have protection diodes.
One diode is from the input pin to Vcc. If the input voltage exceeds Vcc + Vdiode, the current goes to Vcc.

Another diode goes from GND to the input pin.
If the input voltage goes more negative than -Vdiode, the negative current goes to GND.

 

Incredible how I could forget Ohms law here!

I was forgetting these chips have internal resistance and so as long as the voltage is the required one, the current will not burn the chip. How can I forget this. But then again, if my power supply can supply 1.5 Amps at the normal 3.5V, how can I be sure that the resistance inside my chip not not cause it to get burned ? This seems strange...........Does the manufacturer add enough internal resistance to the Chip so that the mosfets inside them dont get toasted?

But when I put all my chips together with a clock, of course, most inputs were varying voltage quickly, and because I was investigating currents, I saw that at the "Clear" input of a Flip Flop, the current was varying from 0 to 23mA, and 23mA > 20mA which was the said maximum, what could this mean?

The manufacturer DOES NOT add internal resistance to the chip.
The input resistance is already very high, about 10MΩ.

Your power supply could deliver 1.5A @ 3.5V
Connect a 10MΩ resistor across the power supply. Ohm's Law still applies.
The current through the resistor is I = V/R = 3.5V/10MΩ = 0.35μA
The power supply supplies 0.35μA, no more, no less.

 

Also my dear friends,

I've heard many times about not leaving unused inputs floating. Fine! But does this mean strictly all unused inputs ? Because suppose I am using an octal D-Flip flop chip which has 8 D-Inputs, and if I'm not using some of the bits, it doesn't seem harmful to leave these unused inputs floating since their outputs are not going anywhere in the circuit. But, might these inputs build up charge and then suddenly shot out charge into nearby inputs changing logic levels?

So my question is, in case flip flop case say, will it matter ?

NEVER leave inputs not connected.
Yes, it does matter with a flip-flop and every other CMOS chip.
ALL unused inputs must be connected to VDD, VCC, VSS or GND.