Am I using the voltage reference correctly in this circuit? - Page 1
Forum section: electronics / projects

author: hkBattousai
author_rank: Regular Contributor
timestamp: November 09, 2014, 12:18:31 pm
content: Previously, I tried to understand the function of the circuit element called "voltage reference"here. Now I'm trying to use it in a practical hobby circuit. Is everything OK in this circuit? Am I using the voltage reference correctly?Planned circuit function:If the voltage per LED exceeds 3.5V, the voltage divider network (R21 and R22) outputs a voltage above 2.5V. When the input reference voltage of thevoltage referenceis above 2.5V, it starts draining current from its anode terminal, which turns the optocoupler on. The optocoupler feedbacks to a buck converter circuit and limits the LED current.

author: Dave
author_rank: Super Contributor
timestamp: November 09, 2014, 03:44:14 pm
content: You are using the reference correctly, however the circuit you want to make isn't going to achieve what you want.The forward voltage of a diode varies quite significantly with temperature, so regulating the current based on forward voltage makes no sense at all. If you want stable current, you need to make a current regulated supply. You should add a sensing resistor at the end of your diode chain and feed that signal back to the SMPS regulator.

author: dannyf
author_rank: Super Contributor
timestamp: November 09, 2014, 05:55:37 pm
content: The circuit is OK - the voltage reference here is really a comparator.However, it is far better to replace the last led with a voltage sampling resistor (and a v-ref with lower voltage reference).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 09, 2014, 10:36:00 pm
content: This is the, or one of the, correct way(s) of doing it:

author: hkBattousai
author_rank: Regular Contributor
timestamp: November 10, 2014, 04:57:00 am
content: The reason I didn't use a sensor resistor was to avoid power loss on it. If I were to use a small valued one, I had had to use a precision opamp and it would had add up to the cost. I came up with the idea of reading the LED current from the forward voltage drop on them. I know it won't be very precise, but I accept it. How much can the LED current vary in the worst case? I'm already using the LEDs very below the maximum allowed forward current; they can't be damaged without +50% current increase. Also, the whole circuit is designed to make low power loss; there won't be much temperature rise. How much difference can it do if the ambient temperature varies between -25oC and +50oC (in the worst case, outdoor usage)?

author: WarSim
author_rank: Frequent Contributor
timestamp: November 10, 2014, 05:39:25 am
content: More than what you would think. Even if there is a temperature derating charts still give head room. If there is no derating charts anything could happen.If you saved your money on the LEDs you should test and match the LEDs. I have seen some unexpected variances on some of the bargain LEDs I have come across.I really don't get the comment about sacrificing power for a current sense resistor that is big enough to avoid a precision opamp. Given the accuracy you have already given up. Giving up less that 1/8 W would be required for a current sense resistor while using a 0.50 OPAMP. With the specs you have accepted you could use a MOSFET to drive your feedback. Or how about a proportional current balance.These are just a few other options to use current sense instead of voltage sense.Every LED light I was asked to fix because it flickered used voltage control. We are just advising against it. In the end the project is yours, do what makes sense to you.Sent from my iPad using Tapatalk

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 10, 2014, 05:47:18 am
content: Well..Vf varies by current, temperature and manufacture. What gives you any idea that the LEDs will be at safe ratings at the voltage you've chosen?Especially at low current, where Vf is predominantly junction voltage rather than series resistance. The light output might vary by +/-10dB on temperature alone -- and that's if you've calibrated for the particular LED being sensed.My example spends less than 0.5W, while the LEDs produce up to 45W of heat and perhaps 5W of visible output. You can spend 5W on that damn resistor without making a lick of difference to the overall efficiency of the light.Tim

author: ejeffrey
author_rank: Super Contributor
timestamp: November 10, 2014, 06:18:41 am
content: Trying to save the cost of an opamp is false economy if you blow out the LEDs. You are also exaggerating the power waste issue. You could set the nominal operating point at 1V and still only waste <1% of the power used by the LED chain. Even a cheap opamp will not present any problem in this case. Consider that your proposed circuit will have far more than 1% variation in power usage due to the change in the voltage drop of the LED.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 10, 2014, 08:19:11 am
content: Or to put it another way:If you're going to do bad regulation of LEDs, just do it open loop on the switch side. With a current mode flyback controller, it's not as bad as it sounds, and way better than voltage regulation, and cheaper besides.Tim

author: dannyf
author_rank: Super Contributor
timestamp: November 10, 2014, 10:53:47 pm
content: :How much difference can it do if the ambient temperature varies between -25oC and +50oC (in the worst case, outdoor usage)? :it will depend on the particular less. At low current levels - 20ma or less for a power led - 4mv per C is not uncommon. A 75c swing would mean 300mv in vfwd reduction. For some less, that could mean life or death.basically, using a led as a current sensor defeats the purpose of having a constant current driver.if you are concerned about power loss, use a low voltage reference, like a npn transistor: at 150ma, the loss on the sense resistor is 0.1w.

author: AG6QR
author_rank: Frequent Contributor
timestamp: November 10, 2014, 11:44:34 pm
content: An issue is the way the I/V curve changes with temperature.If you hold voltage constant, then the higher the temperature, the more current the LED will draw (note that this is the opposite of the way most resistors behave under constant voltage with varying temperature). With LEDs, the variation is enormous. That produces an inherent instability, because higher temperature means higher current which leads to higher temperature, leading to still higher current. There's a positive feedback loop here, leading to huge current and burned out LEDs. That's why you drive LEDs with something approaching a constant current, never a constant voltage.If you sense the voltage across an LED, and use that as a proxy for the current, you are effectively regulating the voltage to that LED. You're effectively driving that LED with a constant voltage instead of a constant current.If that LED gets a little bit warm, it will draw far more current, which will heat it up, causing it to draw more current, and leading to a "death spiral".

author: hkBattousai
author_rank: Regular Contributor
timestamp: November 10, 2014, 11:48:23 pm
content: Thank you AG6QR. You made the best explanation. I will try to find a different feedback method.