For a rough estimate on power in Watts you'll need to dissipate, it is (Vin-Vout) x Iout.

 

just out of curiosity, in an effort to cut down on the heat dissipation, could i put a diode right at the input to make sure 1) protection of backward connection and 2) the V drop across it so the regulator will have to dissipate less V? the datasheet says 1V drop - so if i have 12V coming in, would the regulator then see 11V to drop down to 8V?

fwiw, the unaltered circuit w/ 11.44V coming and 7.94 going out is ambient + 35F, so even in the middle of summer, although while warm, the specs will be within tolerance of even the 85C ceramic caps as max temp should be 160F on the regulator.

please advise.

thanks in advance,
bob

 

That would work, but you've now migrated your heating issue to another component. It might be best to heatsink the voltage regulator instead.

 

heatsink is on regulator (albeit the heatsink is a bit small so i could pick up a larger one too), was just hoping to spread around the heat dissipation - have multiple items cutting a bit of V before it gets to regulator. it is just a temporary setup to see how this project turns out too.

 

just to verify, the diode on the input side of this circuit:

will drop the V 1V, so if the datasheet calls for the forward V on the 4n33 says 1.3V for its led, i should do the resistor value so that it brings the V down to 2.3 and then the diode will will drop it the other 1V? is 1V an approximate value for the 1n4004 diode or approximate because i measured the V from gnd to pin 1 w/out the diode and got 1.3V, w/ the diode it was 120mV, so it seems a bit more than 1V, but the 4n33 still worked in the fact that it triggered the input.

i just happen to have some 470 and 560Ω 1% 1/2W resistors that i think will be the perfect fit for this, w/ that close, does it matter which one i use? fwiw, they are metal film resistors if that makes any difference.

thanks in advance,
bob

 

just to verify, the diode on the input side of this circuit:

will drop the V 1V, so if the datasheet calls for the forward V on the 4n33 says 1.3V for its led, i should do the resistor value so that it brings the V down to 2.3 and then the diode will will drop it the other 1V? is 1V an approximate value for the 1n4004 diode or approximate because i measured the V from gnd to pin 1 w/out the diode and got 1.3V, w/ the diode it was 120mV, so it seems a bit more than 1V, but the 4n33 still worked in the fact that it triggered the input.

i just happen to have some 470 and 560Ω 1% 1/2W resistors that i think will be the perfect fit for this, w/ that close, does it matter which one i use? fwiw, they are metal film resistors if that makes any difference.

thanks in advance,
bob

Well, the diode is normally reverse biased, so does not conduct anything (apart from leakage current but you can probably ignore that.)

Also, why are the grounds from the isolated sides connected? That breaks any isolation barrier. Might want to fix that!

 

the graphic is not accurate, i have removed the trace between pin 2 & 4

 

ok,

i rec'vd the boards and i made a mistake on the opto one. due to time constraints i submitted a board that had 1 groundplane that connected both sides of the isolator. i understand the issues now i was going to fix that on the next order.

i have put a couple parts on the board - the 12V input on the left and then also a resistor in the R2 spot - 2.6KΩ, i thought i would get ~1.3V & ~5mA if i measured between pins 1 & 2 of a 6pin socket, but i am not, i am getting the full V as if there is no resistor in place.

so, where did i screw up?

thanks in advance,
bob

this is the board i sent in to be made:

fwiw, the power one worked flawless, albeit a bit tight like the render showed, but it measured 7.95 coming out of it and no smoke

 

You didn't screw up. Try placing a couple of diodes (like 1N4001-1N4007 or perhaps 1N4148/1N914) in series across pins 1 and 2, and measure the voltage then.

 

thanks SgtWookie, i actually had a 5mm led sitting here and put it in between pins 1 & 2 and measured the V - i had a 2600Ω resistor in (i think, i hadn't soldered anything in R2 yet) and it read 1.8V on the led pins - i was reading the pins w/out the 4n33 in it, but i thought it would only show the 1.3V - what am i missing here? i have been up for a long time, but i think i am still missing something...

so i guess i measured it wrong , but it looks like my math is ok and will solder one in tomorrow since i have now starred at it so long thinking WTF??

just out of curiosity, when i had it breadbarded, i would put the dmm +/- across pins 1 & 2 of the 4n33 and it would read basically input V but still worked fine. am i just reading at the wrong spot on the pins or is the resistor dropping that much V?

 

Your multimeter has a very high input impedance; 10s or hundreds of megohms. By comparison, the 2.6k resistance is very small. Thus, nearly all of the voltage was being dropped across the meter instead of the resistor.

When you insert an LED, your IR emitter, or a couple of diodes in series, the Vf of the LED/IR emitter/diodes causes most of the voltage to be dropped across the current limiting resistor.

 

thank you for the explanation and helping me keep my sanity

 

You're welcome.

 

thanks again to all who helped . i think i cooked a 4n33 but i have a lot of those sitting here. when they go, do they just quit working? or do they smoke? i ended up w/ a 1.2KΩ resistor and that has been triggering all day w/ 11.44V input. measured V is 1.15V across pins 1 & 2 on the 4n33 itself in the socket. i was trying for 1.3V but the resistors are 5% and i am using 1/2V less so i think i am pretty close. this gives me supposedly 10mA at the led in the 4n33, so it should be reliable but still on the low side.

also, for future reference, is there a calculator for trace width for power circuits? i am sure mine are more than adequate on both boards, but it would be nice to know when i make a board for a 2-5A switching power supply.

last, is there a mtbf for the led in the 4n33? or is it like any regular led as far as longevity is concerned?

 

When optocouplers go due to abuse, they generally just quit working. When properly fed in a well-designed circuit, they can last a long time.

11.44v in, Vf is 1.15, so I(IR emitter)= (11.44-1.15)/1.2k = 10.29/1200 = 8.575mA.
10.29v/10mA=1029 Ohms.
If you put a 7.5k resistor in parallel with the 1.2k resistor, you'll be close to 10mA.

Roman Black hosted PCBTemp on his website: http://romanblack.com/pcbtemp.htm
Very useful utility. Ultracad no longer supports the freeware version, but it's plenty good enough for hobbyist use. The thing to watch is the thermal rise for a given trace width and copper weight (1 oz, 2 oz, etc.) vs the current in the trace.

Something else to consider is the component cooling that wide traces provide. People forget that diodes have a Vf; if you're running 1A through a 1N400x diode, you'll have about 1v-1.1v Vf on the diode, so you'll be dissipating over 1W power. If your traces are wide, that'll give the heat someplace to go rather than toasting the diode.

As far as the 4N33 MTBF - it's kind of a mixed bag. What happens is that as the device ages, the CTR (current transfer ratio) decreases due to the lower emissions from the IR emitter. If you didn't plan on this in your design, it'll stop working a lot sooner than you'd expect it to.

For example, if the CTR when new is 1, and you're running 10mA through the emitter, if you design requires sourcing or sinking 10mA from/to the output, it's going to break pretty quickly as that CTR lessens. However, if your design only requires 1mA sink/source from the output, it will last a very long time.

 

thanks for the info SgtWookie.

i came across the program PCBToolkit and the power board looks like i did ok, at least for the trace width, next time i will make it w/ a ground plane which i should have done in the first place, but i am learning as i go.

i decided on the ~10mA for the 4n33 because that is what spec sheets seems to use for a lot of their tests. i have a couple more boards and i can change that to 5mA just for longevity if failure arises. they claim 80mA max, so i am quite far from that.

again, thanks for taking the time to explain this all to me, appreciate it greatly.

bob