1
\$\begingroup\$

I have two questions regarding the electrical characteristics of comparators:

  1. Is there a way to adjust the input offset voltage of a comparator to fit a custom application?
  2. Which comparator output type is the ideal choice for powering an LED circuit?

I am designing a circuit that will detect a failed LED (open or short circuit) by comparing two voltages and triggering a second LED, indicating a failure if the voltages ever differ. I am having trouble sourcing a comparator chip that will allow the correct amount of voltage fluctuation at it's inputs. For example, if the voltage fluctuates by 100mV this should not trigger the second LED but, most comparators have a very small input offset voltage. I am curious if there is a way to externally adjust the input offset voltage of the comparator to fit my needs.

Additionally, I'm having trouble understanding the best comparator output option for powering an LED. I believe the term "Output Short Circuit Current" is used to describe the amount of current that the chip is able to provide at it's output. Is there a specific output type that is preferred when powering something like an LED?

Please consider the following circuit: enter image description here

I plan on using a voltage reference (REF2933) rated at 3.3V as one input and the voltage across the LED as the second input. The TLV1701 will compare these two voltages and remain low until the voltage across the LED falls outside of a given range. If the voltage reference has a 2% accuracy (3.234 - 3.366V) and the voltage across the LED varies from 3.0 to 3.4V then it is possible for the comparator inputs to be 0.366V apart. I understand that the input offset voltage is used to determine when the comparator switches from high to low or low to high but, is there a way to change this value to allow more room for error?

\$\endgroup\$
0
\$\begingroup\$

Look at this circuit from the datasheet:

enter image description here

You can create Vth+ and Vth- with a voltage divider from a reference (or from the supply voltage if you don't care about accuracy to that degree).

Say you want the fault LED to come on at 2.8V and 3.6V (that's 200mV outside the range you stated- by the way, make sure that is the correct range for the current you are supplying).

You could use a 3.6V reference (such as an AN431 with two resistors), together with a 1.62K 1% resistor in series with a 5.62K 1% resistor to ground, to give you the 2.8V for Vth-. The LED goes in series with the Rpullup resistor in the above schematic, not to ground.

The (in)accuracy of the transitions will include the reference (in)accuracy, the resistor tolerance for the lower threshold, and the offset voltages of the two comparators.

As far as the maximum output current goes, you should not use the short-circuit current. That's a fault condition. The output is rated as follows:

enter image description here

So 4mA would be a reasonable output current. That should be plenty for a modern LED indicator. You can calculate the resistor value from the LED forward voltage at 4mA and supply voltage (the actual current will be a bit less because the output voltage will not be zero under fault conditions).

As far as actual part numbers that are suitable- you don't care about response time for this application, you do require that it will operate from a 5V supply, and you do require that the input common mode voltage includes the reference voltages. For the latter reason, the very cheap (< 10 cents) and common LM393 is unsuitable in this case. You care about offset voltages to the extent you want the fault indicator thresholds to be accurate, but there is no point using a part with uV accuracy if your reference is only accurate to tens of mV.

So, the circuit looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ Thank you for your explanation on how to use the TLV1702 as a window comparator. I am still concerned about the maximum output current of the comparator. I should've specified the type of LED I plan on using for my design since 4mA will not be enough to power them. The LEDs in this circuit will be mounted in a box behind a window and must attract the attention of an operator standing several feet away, especially if there is a fault detected. The LEDs I chose draw about 17mA.10mm LED \$\endgroup\$ – kprince Jul 21 '15 at 13:41
  • \$\begingroup\$ So you can simply add a buffer transistor to the output (eg. a single p-channel MOSFET as in the above edit) or use a different type of comparator- one that meets all your requirements, including ones such as offset voltage that are fuzzy at the moment. The circuit shown in the edit will handle more than 1A to the LED. If that's not enough... \$\endgroup\$ – Spehro Pefhany Jul 21 '15 at 13:56
  • \$\begingroup\$ How do you get 3.6V from the voltage reference? Shouldn't R3 and R17 be 10k? \$\endgroup\$ – Nick Gammon Dec 29 '16 at 1:16
  • \$\begingroup\$ Yes, 10K. Sorry for delay, in 'darkest' of areas for signals etc. \$\endgroup\$ – Spehro Pefhany Dec 31 '16 at 11:35
0
\$\begingroup\$

Offset voltage:
Forget the offset voltage (for your application). What is given in the datasheet usually is the max. value (wors case) or a range ranging from + to -. So you just can rely that its value is in the given range (it might be positive or negtive); you are not supposed to need to know the actual value.

Voltage reference (for threshold):
A reference IC for the theshold voltage is not a good idea. A simple voltage divider (two resistors) is better because it's more flexible (and cheaper): you can choose two values e.g. 3.9kΩ and 10kΩ which will give you ca. 3.6V.

OC output:
Because the comparator output is OC it only can sink current, i.e. you have to reverse the LED and connect one end to Vcc instead of GND (and swap + and - inputs of the comparator).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.