How do I interpret the maximum Output voltage Vo on the STP16CPC26

[HABHAB]

I want to connect 7 LED's on one output in series, using a current of 20 mA. Each LED can have a forward voltage of 3V, which means 7 x 3 = 21 V. I want to use a power supply of 24V.

I read in the specifications that the max Vo (Output Voltage) is rated 20V. But how do I interpret this? If I drive my LED's, each taking 3V, the voltage on the output pin will only be 24V - (7 x 3V) = 3V, which is far below the max rated Output Votage.

With other words, powering my LED's with a 24 V power rail is not a problem?

[Peter BENSCH · Accepted solution]

Welcome, @HABHAB​, to the community!

The data sheet of the STP16CPC26 specifies an absolute maximum rating of 20V for the output voltage, because the output transistors are guaranteed to withstand this voltage. If this voltage is exceeded at the output, the LED driver can be destroyed.

For LED drivers, however, two states have to be considered:

• Switch-on state, LED driven with constant current
• Switch-off state, LED current = zero

It is precisely in this switch-off state that the full operating voltage occurs at the LED output, so that you must not operate the LEDs at the STP16CPC26 with 24V. Please do not use a higher LED supply voltage than 20V to avoid destroying the driver.

To determine the LED supply voltage, it is helpful to take a look at AN3981, section 5, where this point is dealt with in detail. 

If your LEDs have (e.g. by tolerance) a forward voltage of 3.2V instead of 3.0V and you connect 6 of them in series, the total voltage will be 19.2V. The difference to 20V is too close to ensure a safe operation at 90mA together with the drop voltage (see data sheet, section 9.3).

Does it answer your question?

Regards

/Peter

[HABHAB]

Hello ​@Peter BENSCH​ ,

I really appreciate the time you took to give me such a detailed answer. The link to the application note is really helpful!

I get your point... I guess. When in "Switch-off state", there is no voltage drop over the LED's. That means that the voltage on the Output (Vo) is VLed - [sum of voltage drops over LED's] which means VLed - 0 = VLed. Hence, if my VLed is 24V, the Vo will be 24V as well, which exceeds the maximum specification.

But I was in the impression that, because of the fact that the current is 0 as well, there is no power dissipated, hence that 24V wouldn't be considered as harmful. But reading your explanation, that is a wrong assumption, right?

I have also tried to measure the Vo when the output is in the "Off-state". But that is also very unreliable, because the resistance of my Volt-meter seems already to "play around" with the numbers. In the "Off-state", using a VLed of 12V, I'm reading a Vo of about 5 Volt. But that is a false measurement of course.

Thanks a lot!

Hans

[Peter BENSCH]

An LED driver can be destroyed not only by too much power dissipation (while driving), but also by too much voltage.

A comparison: if you hold a door shut and someone from the other side pushes harder than you, they will be able to open it. With the LED driver, however, this can lead to complete destruction because a very large current can flow suddenly, resulting in very high power loss.

To stay with the comparison: you then stumble backwards and get knocked flat by the door crashing in (and all the people outside pushing).

It also makes little sense to measure the voltage Vo in the driving, i.e. switched-on state: LEDs are not voltage-driven but current-driven components. The operating state of the LEDs is therefore determined by the flowing current. If necessary, however, you can check the forward voltage of the LED(s) when they are switched on.

Regards

/Peter

[HABHAB]

Thanks again Peter. Great explanation! I will reduce my VLed to 18V, which is now nicely within specifications.

I'm also using a resistor in series, to reduce the voltage (hence power) on the chip. I'm only using 20 mA, so a few 100 ohms doesn't need a high power resistor.

[Peter BENSCH]

If you reduce Vled even further, you can possibly save the additional resistors.

OK, if the problem is resolved, please mark this topic as answered by selecting Select as best under your preferred answer. This will help other users find that answer faster.

Good luck!

/Peter