# How to determine the max. frequency of the 74HC93 IC at a specific voltage?

I'm trying to use a 74HC93 counter. It was working fine when the clock was 4 MHz, but when it is increased to 10 MHz the output is not producing what I expect it to be, so I was thinking that maybe the 74HC93 doesn't work at that frequency.

I have been looking at the datasheet to check for the max. frequency of the 74HC93, and from what I've gathered I need to look at both the propagation delay time and the output transition time. The problem is the voltages in the datasheet aren't the same as mine.

The datasheet's propagation delay time and output transition time are specified for 2, 4.5, and 6 V, but my microcontroller's voltage is 3 V, so how am I going to get the right data for my voltage?

Also, am I right in thinking that the max. frequency that the 74HC93 counter can take can be calculated by 1/(propagation delay time + output transition time) at the right ambient temperature?

• Iterate between specified values in the data sheet. Commented Nov 18, 2022 at 8:10
• – CL.
Commented Nov 18, 2022 at 8:58
• Max frequencies are not "proportional" to Vcc. You should need something as this digchip.com/datasheets/parts/datasheet/364/74LVC161-pdf.php google.be/… Commented Nov 18, 2022 at 9:28

## 1 Answer

To find values in between those given on a datasheet, use interpolation: the spreadsheet below shows how.

It looks from the 74HC93 datasheet that 10 MHz should be in range, as fMAX is about 17 MHz at 3 V at 25°C, (but drops to about 12 MHz under some extreme temperature conditions). So perhaps it's another parameter that you're exceeding; any if you're anywhere near a device's limits you need to take a view about how much headroom you want.

To do the interpolation, you put your values into a spreadsheet and add a trend line and make the graph big enough to read the value you need.

If you don't know what kind of trend line is appropriate, use linear interpolation and make a judgement about how conservative you need to be. To interpolate between the two nearest points, use y = mx + c, which in your case is m = (y1-y0)/(x1-x0) = (30-6)/(4.5-2) = 9.6, and c = y0 - m x0 = 6 - 2 m) = -13.2, which for x = 3 V gives an estimate of 15.5 MHz as the minimum fMAX at 25°C.