Understand driver strength of digital IC

[farsi]

Hi,

I have a shift register that switches Vout from 0 to 5V.
How can I measure the driver strength? I have put a 10nF cap on one of the pins, and can measure the rise time.

But how to translate this to an output current remains unclear.

Anyone has a pointer?

Thanks!

[jahonen]

Neglecting capacitance voltage dependency, you can use i = C*du/dt to calculate the driving current. Just determine the derivative of the capacitor voltage for each voltage in V/s, and multiply that by the capacitance. That gives the output current for any given output voltage.

Regards,
Janne

[farsi]

Thanks!
Ah... that was the formula I needed, I was somehow getting lost with formulas about RC time.

[retrolefty]

Wouldn't the device datasheet specify maximum sinking and sourcing current capacity?

[Richard Crowley]

The data sheet for the (unidentified) shift register will state the manufacturer's rated output drive specification. 
This published value is what we use to design circuits.
Attempting to measure this is a very poor and error-prone substitute for simply using the published specification.  NOT RECOMMENDED!

Furthermore, using the C*du/dt method is even worse because the rise-time of the shift register output is an important (and UNKNOWN) factor.

[Gall]

Thanks!
Ah... that was the formula I needed, I was somehow getting lost with formulas about RC time.

Look, that is quite simple.

A resistor is a component with constant voltage/current ratio: I = U/R
A capacitor is a component with constant voltage speed/current ratio: dU/dt = I/C
An inductor is a component with constant current speed/voltage ratio: dI/dt = -U/L

Here, dU/dt and dI/dt mean derivatives (as in calculus), that is, they represent momentary speed. dt should be very small (infinitely small).

If we want to calculate the voltage change for a finite time, let's do the following. Consider the capacitor C is charged from source Us via resistor R. The voltage on the capacitor depends upon time and is U(t). Consider that the capacitor is initially discharged, U(0)=0.

It's obvious that I = (Us - U)/R in this circuit. Substituting the capacitor equation C*dU/dt for I, we get
dU/dt = (Us - U)/(RC).
Rewrite as follows:
dU/(U-Us) = -dt/RC
Now integrate both parts of the equation:
ln(|U-Us|) = const - t/(RC)
Since U < Us,
ln(Us-U) = const - t/(RC)
We have to find const. Note that for t=0
ln(Us-0) = const
and thus
ln(Us-U) = ln(Us) - t/(RC)
or
ln(Us-U)-ln(Us) = -t/(RC)

Finally,
ln((Us-U)/Us) = -t/(RC)
or, the same
Us-U = Us*exp(-t/(RC)).

Here RC is called "time constant".

For discharging it's done in a similar way.

[T3sl4co1l]

If it's a CMOS device, RC time constants are just what you want.

The peak current output occurs at the instant of switching, which sets the initial rising slope dV/dt.  You can calculate it from either direction (current and slope, or time constant and RC), it is of course Ipk = V/R = (5V) / R.

Typical strength for 74HC class logic is around 50 ohms pull-down, 80 ohms pull-up (NMOS is stronger than PMOS), or a short-circuit current under 100mA.  Or for a typical 0.4V drop condition (V_OH = 4.6V or V_OL = 0.4V), around 10-20mA (I_OH / I_OL -- typical, not the minimum).

Tim

[bingo600]

Some packages says they can deliver ie. 100ma per pin , but remember to observe the max package draw.

You could have a package with 8 pins , each 100ma , but a max package current of 500ma.

/Bingo

[jlmoon]

Thanks!
Ah... that was the formula I needed, I was somehow getting lost with formulas about RC time.

Look, that is quite simple.

A resistor is a component with constant voltage/current ratio: I = U/R
A capacitor is a component with constant voltage speed/current ratio: dU/dt = I/C
An inductor is a component with constant current speed/voltage ratio: dI/dt = -U/L

Here, dU/dt and dI/dt mean derivatives (as in calculus), that is, they represent momentary speed. dt should be very small (infinitely small).

If we want to calculate the voltage change for a finite time, let's do the following. Consider the capacitor C is charged from source Us via resistor R. The voltage on the capacitor depends upon time and is U(t). Consider that the capacitor is initially discharged, U(0)=0.

It's obvious that I = (Us - U)/R in this circuit. Substituting the capacitor equation C*dU/dt for I, we get
dU/dt = (Us - U)/(RC).
Rewrite as follows:
dU/(U-Us) = -dt/RC
Now integrate both parts of the equation:
ln(|U-Us|) = const - t/(RC)
Since U < Us,
ln(Us-U) = const - t/(RC)
We have to find const. Note that for t=0
ln(Us-0) = const
and thus
ln(Us-U) = ln(Us) - t/(RC)
or
ln(Us-U)-ln(Us) = -t/(RC)

Finally,
ln((Us-U)/Us) = -t/(RC)
or, the same
Us-U = Us*exp(-t/(RC)).

Here RC is called "time constant".

For discharging it's done in a similar way.

in the above.. aren't  the the terms "U" really meant to be "V".. am I correct?    such as in I = C(dv/dt) ?

[T3sl4co1l]

U is traditionally used for potential (voltage) in Europe, I think.

American physicists typically use it more generally as a field potential (electric, gravitational, ...), using U or V for electric potential specifically.

E is also common (electromotive force), especially in regards to induced voltage (scalar, and usually a cursive "E") or electric field (vector).

Tim

[David Hess]

For CMOS I just measure the static output resistance in both states using a load resistor.  The maximum output current ratings will be part of the specifications.

[farsi]

Thanks for all the discussion.
I can't see why measuring the rise/fall times of a toggling pin, and applying dV/dt * C would be a problem. With a rise/fall time of 5V/us and an external cap of 10nF --> Iout would be 50 mA

If this is is a problem. it would be great to see an example with numbers. Thanks a lot!

[Richard Crowley]

1) Because there are too many unknowns by that method.
2) Because measuring a single unit (or even a handful of units) will NEVER reveal what the manufacturer's specifications were.
3) Because designing to the performance of a single unit is living on the edge and just asking for trouble.
4) Because measuring the slew rate does not necessarily correlate to the steady-state performance.
5) Because measuring the performance of a single pin (even if it was 100% accurate) doesn't tell you anything about the dissipation of the entire package rating.

I still don't even understand the question.  If we want to know the output current capacity of an integrated circuit, actually looking it up in the published data sheet is both EASIER as well as being MORE RELIABLE. Please remind me why we are trying to measure this for ourselves. AND why using such a roundabout method?  Sorry I just don't get it on a very fundamental level.

[jahonen]

Another way is that if you can get an IBIS model for the chip you are using, it contains pretty accurate output current vs output voltage data among with other things. Not just a single number.

Regards,
Janne

[farsi]

Thanks - sorry if the question was answered, and I missed it.

Background is that if you assemble some device, e.g. a shield for Arduino, more of a prototype.

[Richard Crowley]

Background is that if you assemble some device, e.g. a shield for Arduino, more of a prototype.

Are you not aware that there are very detailed published technical specifications for any chip that you can buy?
You never answered the question why measure it instead of just getting the official answer from the data sheet.

[farsi]

Sorry, so here is the real background: I can't use the datasheet since the devices will be self-made for a prototype device.

[Richard Crowley]

Sorry, so here is the real background: I can't use the datasheet since the devices will be self-made for a prototype device.

OK, but that still doesn't answer the question of "extrapolating" the stead-state characteristics from the dv/dt vs. simply measuring it directly?
I am always suspicious when people discard the straightforward and direct way of doing something, and try the "exotic" method.  Why?

By self-made, are you saying that you are are making your own integrated circuit? 
Do you not have the design-rules that tell you the current capacity of a transistor of the size and shape you have plotted?

[David Hess]

Sorry, so here is the real background: I can't use the datasheet since the devices will be self-made for a prototype device.

Going back to your original question:

I have a shift register that switches Vout from 0 to 5V.
How can I measure the driver strength? I have put a 10nF cap on one of the pins, and can measure the rise time.

But how to translate this to an output current remains unclear.

If you know the load capacitance, then the current is proportional to slope or derivative of the voltage with respect to time.  You can just calculate it.  Many DSOs can calculate the derivative directly but it is just as easy to measure the slope as dV/dT in the area of interest and multiply by the capacitance to find the current in that area.  For a 5 volt output with a 5 nanosecond transition time into 200 picofarads and looking at the whole edge, that would be (10^9 volts/second) * (200*10^-12 farads) = 200 milliamps approximately.  Typical loads are more like 20 picofarads making the output current 20 milliamps which is pretty close to the average real value in typical CMOS outputs.

In practice however this is just specified as output resistance in the form of worse case output voltage into some load and switching time into a typical or worse case load capacitance.  For CMOS outputs, I just measure the change in output voltage for a change in load resistance to determine the effective output resistance which has nothing to do with the rated output current.  Then you can use the RC formula.

This may be tricky with bipolar totem-pole output stages because they usually have very asymmetrical drive combined with a change in output resistance do to saturation effects.  That is not normally a problem for them since it goes along with their asymmetrical input characteristics.

[rob77]

let me support Richard here..

look at the datasheet and design your circuit accordingly.
if you measure the "what it can do" and you'll operate your circuit under such a circumstances... then it might fail in a short time.
furthermore... if you measure the "what it can do" on one pin and you'll load all output pins in such a way - your IC  will blow up for sure - it's just a matter of time.

if you'll do it anyways.. then wear safety glasses while breadboarding... see the attached picture what a OVERLOADED IC can do.... you don't want that little shrapnel in your eyes or someone elses eyes, right ?

[David Hess]

The output resistance and drive current that you measure have nothing to do with the absolute maximum ratings shown in datasheets and the absolute maximum ratings themselves are not operating parameters you can use and expect reliable operation without derating.