# Working out AT28C16 LED output from datasheet

I'm trying to work out how to connect some LEDs so I can view the value of the value stored in the AT28C16 (pdf) chip.

I'm actually following a youtube series. In the video he chose a 330Ω resistor but I'm not sure how he would work this out given we don't know the maximum VOH value.

My LED's are rated at

1.8v - 2.2v 20mA

Please could you explain how to read the datasheet to calculate the resistor value required.

Thanks.

Normally Voltage Output High will be a little less than VCC, as most ICs do not use internal boost circuits that could make it possible to be higher.

In this case, you could look at the Absolute Maximum rating:

Absolute Maximum Ratings*
Temperature Under Bias............... -55°C to +125°C
Storage Temperature ................. -65°C to +150°C
All Input Voltages (including NC Pins)
with Respect to Ground ...............-0.6V to +6.25V
All Output Voltages
with Respect to Ground ...............-0.6V to VCC + 0.6V
Voltage on OE and A9
with Respect to Ground ...............-0.6V to +13.5V


So the max you should ever see on this IC is VOH as (VCC - GND) + 0.6V. But it would likely be 5V.

So using Ohm's law for the led resistor, (5V - 2.2V) / 330 = 0.008A or 8 mA. But since the LED would not show 2.2 volts forward drop at 8 milliamps, then you should adjust the resistor up a bit. 470Ω or 500Ω may be better.

In any case, you are using the EEPROM in a non-standard way, which means it does not have specs for this type of operation. There is no way to know how well the outputs will deal with 8mA on each of the pins, if the voltage drops due to the higher current, if they will blow. The VOH spec is only tested at 0.4mA source.

• Wouldn't I want to reduce the resistor value if I'm aiming for 20ma? Something like (5V-2.2V)/160Ω=20mA Commented Jun 27, 2017 at 22:42
• @Jamie Is there some reason you actually need 20mA, times 8 so 160mA, through an IC not designed for anything like that? Most LEDs are quite visible at even a fraction of a milliamp, and fine for indicators at 5mA. 20mA is overkill, unless you are trying to light up the room. Commented Jun 27, 2017 at 22:48
• @jamie the video you link to, the 330Ω they are using, is not running the leds at 20mA. Commented Jun 27, 2017 at 22:49
• No reason for 20mA just trying to understand what "But since the LED would not show 2.2 volts forward drop at 8 milliamps" means? Commented Jun 28, 2017 at 2:13
• @Jamie a diode like an LED has an IV curve. It's nominal forward voltage of 2.2V is only seen at 20mA. At a lower current, it will have a lower forward voltage drop. Commented Jun 28, 2017 at 3:46

VoH is determined by Vcc (which I presume you would set at 3.3V or 5V but its not listed because you set it) and not much is given about it like the drop or what the outputs are. There also isn't any information on the current output of the chip because it was never intended to drive a heavy load. The only thing we do know about it is CMOS and TTL compliant and it can drive at least 400uA.

Reading the datasheet further suggests that it was tested with a 1.3k load at 5V which would be ~3mA. However, that is a far cry from 20mA. Since there is no information on the current output (probably because it doesn't have much current output) I would use an Octal Buffer/Driver (some support 24mA and list the Voh and IoH in the datasheet) on the output of the AT28C16 to drive the LED's.