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In the description for the circuit below from this data sheet for TNY268 off-line SMPS switcher, it says

R5 is used to ensure that the Zener diode is biased at its test current and R6 centers the output voltage at 5 V.

SMPS using TNY268

The Zener current for the BZX79B3V9 is 5mA, the zener voltage is 3.9V and the forward voltage for the opto-coupler TLP181 is 1.0V Min.

How exactly does R6 'center' the output at 5V? How does R5 bias the zener at 5mA? I can't work out the math.

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  • \$\begingroup\$ Consider the case where there is no U2 ... How does the maths workout for you then? \$\endgroup\$ – Spoon Sep 9 '14 at 12:34
  • \$\begingroup\$ @Spoon - Well then, i = (Vo-Vz)/(R6+R5) = (5-3.9)/(59+680) = 1.5mA, way below 5mA. ...right? \$\endgroup\$ – Sohail Sep 10 '14 at 5:14
  • \$\begingroup\$ The LED starts dim at 1.5mA (a 10mA LED can produce light at 1.5mA, just very dim) ....then consider the voltage rising to 5.1V and the LED gets brighter ... what happens? \$\endgroup\$ – Spoon Sep 16 '14 at 11:37
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The knee of the curve is pretty mushy on Zeners < 5.1V, so this is important.

They've used the 'Y' bin opto, so the nominal CTR is about 100%. A quick look at the far side of the optocoupler should tell you that about 130uA comes from the 2M resistance and another 240uA from the TNY chip so 370uA. That means about 370uA at the LED.

Looking at the LED curve:

enter image description here

You can see that Vf is about 1.0V at 370uA. That means the current through the 680R resistor is 1.4mA, so the total current is 1.84mA. Then the voltage across the 59R resistor is 110mV. Looking at the zener curve:

http://pdf.datasheetcatalog.com/datasheets/400/114103_DS.pdf

At 1.84mA, the voltage will be about 3.6V.

So, the output voltage before the inductor is 3.6 + 1.0 + 0.11V = 4.71V nominal.

Conclusion: The resistor does not do what they said, and the nominal output voltage will be on the low side.

Compare this circuit, which uses the same zener, no series resistor, but 150R in parallel with the LED, so the zener current will be a bit higher than 5mA.

enter image description here

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  • \$\begingroup\$ That's great Spehro. (I've never used a zener less than `7 V.) \$\endgroup\$ – George Herold Sep 9 '14 at 23:13
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    \$\begingroup\$ That's great Spehro. (I've never used a zener less than ~7 V. (lm399)) I must admit the circuit with no R, but just two series diodes is a bit scary to me. (But then again I was scared by the double integrators in SV filters.) Is it OK because it's in a feed back loop. Does the temp coeff. of the zener and led cancel? (why no +1 votes? it's weird.) I always hit return at the start of my comment and then I only have 5 min for the edit... \$\endgroup\$ – George Herold Sep 9 '14 at 23:21
  • \$\begingroup\$ That makes sense. But Vf would be 1.0V (and not 1.0mV) at 370uA aye? I intend to skip the optional 2Mohm resistances; then @240uA at the optp-LED and only the 150ohm in parallel and using your charts and method, I get about 6.24mA through the Zener. \$\endgroup\$ – Sohail Sep 10 '14 at 5:45
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    \$\begingroup\$ Yes, typo, fixed now. That should get you fairly close to 5V. Of course a TL431 would get you a lot closer for a few cents more. \$\endgroup\$ – Spehro Pefhany Sep 10 '14 at 5:56
  • \$\begingroup\$ Few cents yes... But I'll keep it simple for now as a beginner. I'll have this in mind when accuracy becomes an issue. \$\endgroup\$ – Sohail Sep 11 '14 at 2:30
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The I-V for a zener is fairly steep where it is operated, so you don't need exactly 5mA to get the 3.9 volts. (2 mA will be fine.) Here's a spec sheet with some I-V curves.

http://www.onsemi.com/pub_link/Collateral/BZX84C2V4LT1-D.PDF

It doesn't show the 3.9V zener, but it won't be that much different than the 5 V one.

Edit: Well the 3.9 is a lot different than the 5.1V. See Spehro's answer.

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