DC-DC Converter/Regulator Question

[Dinnin]

I Have a AP1086 three prong regulator. I want to use it to convert a 9V battery to 5V, for my lcd/pic and every thing else.

The Equation on the data sheet reads like Vo=Vref * ( 1 +R2/R1).

So my question is, if I want my Vo to be 5V then I plug in 5V as Vo and find what ratio i need R2/R1. The problem i am having is what is Vref?? I think i read it was 1.25V, but then elsewhere in the data sheet it said 1.5V i am not sure.

So if i need 5V to be Vo and i use 1.25 as Vref i get R2/R1= 3, meaning i should use 300/100 Ohms resistors to get the desired effect?

I am new to this so not sure.... I would have thought that your Vin would have some affect on how you choose your resistors but i guess i would be wrong

 

[Hero999]

The reference voltage for the AP1086-Adj is 1.25V, see page 5 of the datasheet.
https://www.electro-tech-online.com/custompdfs/2010/04/AP1086.pdf

This site has an L317 calculator which uses exactly the same formula as your regulator.
LM317 Calculator

You got the correct ratio, if R1 is 120R, R2 can be 360R or R1 = 100R and R2 = 300R like you said.

 

[Dinnin]

cool, does the value of the resistors matter as much as the ratio? For example could i use 3k and 1k resistors, as long as its a 3:1 ratio?

 

[Hero999]

There's a minimum load current of 10mA which means R1 can't be above 125R unless there's a load permanently connected to the regulator.

The LM317 has a current that flows through the adj pin which can also compromise the accuracy of the setting when the value of R2 is high but no mention is made of it in this regulator's datasheet so assume it's negligible.

My advise is keep R1 above 62R otherwise the quiescent current becomes excessive (>20mA) and don't choose too high values either, keep below 2k4 in case there is an adj pin current which messes things up.

 

[Dinnin]

Well i have a small problem, I set it up with the resistor values of 120 and 360 but it does not work as its supose to. When i put ion 5V i get 4V.. Also when i put in 9V i get 8V. So i tried the standard setting with 2x 120R and i checked 5V and it outputs 4V..... I had a friend check it for me and he has no idea why it wont work... i also tested 2 others that i had same result

The label ob the bag says AP1086T33L- U.... The actual device says 1086-33 0743AL... Can any one help i dont know what the heck is going on

 

[Hero999]

Something itsn' right.

Are you sure you have the AP1086-ADJ, not the AP1086-3.3 or AP1086-5?

You need to have the ADJ version.

If you have the 1.5V to 2.5V versions you can still use it for 3V, you just have to replace Vref with the voltage, for example, if it's 1.8V replace 1.25 in the formula with 1.8.

If it's the 3.3V version, you can probably just use it as is but accept the output will be 3.3V, not 3V. If it's a higher voltage then you're stuck.

 

[bychon]

One of my favorite tricks is to use a diode (1n4001 to 1n4007) as a "spacer". I call it that because it always uses up about the same voltage instead of the voltage across it changing according to current like a resistor does. Adjust the output voltage to 3.6 and put a diode in series with the output to use up 6 tenths of a volt. That will get you back down to 3 volts if you have a 3.3 volt regulator.

There is also the chance that you connected the prongs wrong or you left out the capacitors or the capacitors are not right next to the regulator. Regulators sometimes oscillate if you dont have the capacitors right next to them, then the output wanders all over the place.

 

[Dinnin]

I actually read the data sheet more closely and realized the exact part number is AP1083T33 and the 33 means its a fixed 3.3V out not an adjustable regulator i felt so stupid, but i ordered the correct one now. Thanks for the help1

 

[Hero999]

Well you could've used the spacer diode idea mentioned above, the trouble is the forward voltage of diodes isn't very well controlled so it will drift a bit.

If you really need 3V, then ordering the correct part, is the best solution.

 

[indulis]

I call it that because it always uses up about the same voltage instead of the voltage across it changing according to current like a resistor does.

Not true... the diode forward voltage will change with current and temperature.

 

[bychon]

So, if the current changes from a tenth of an amp to one amp, a 6 ohm resistor will use up 6 tenths of a volt at the low current and 6 volts at the high current, rendering the power supply useless. How bad will it be if a diode is used?

 

[indulis]

Per the diode equation... ~2mV/*°C for temperature... that's also true for the B-E juction in a BJT. As for change over current, that depends on the diode... the curve is on the data sheet for most diodes. A 100°C temperature change for power circuit components isn't all that odd (200mV) and big current changes could push upwards of a few hundred millivolts. For a high voltage output not a big deal, but for lower voltages it becomes a big percentage.

 

[bychon]

So, as the current rises from a tenth of an amp to a whole amp, the diode voltage will increase by about .12 volts, and if the diode heats up to 100 C, the voltage across the diode will decrease to about .04 volts at one amp. (1.333% to 4% of 3 volts) I'd call that "about the same voltage", especially compared to the 1000% voltage change across a 6 ohm resistor.

 

[Hero999]

It depends on what he needs his 3V for.

If it's just a general purpose 3V supply, he could easily get away with the 3V spacer diode, but if the 3V is also being used as a reference for say an ADC then it's a bad idea.

 

[bychon]

The 1086 chip has a 2% initial guarantee and 1/2 % drift with temperature. Is that good enough to be the reference for an ADC?

 

[indulis]

So, as the current rises from a tenth of an amp to a whole amp, the diode voltage will increase by about .12 volts, and if the diode heats up to 100 C, the voltage across the diode will decrease to about .04 volts at one amp. (1.333% to 4% of 3 volts) I'd call that "about the same voltage", especially compared to the 1000% voltage change across a 6 ohm resistor.

In the industry, line and load regulation numbers above ±1% of Vout are considered poor.

 

[bychon]

So...the regulator chip isn't good enough to be an ADC reference in the first place.

 

[Hero999]

I suppose it all depends on the accuracy required.

The problem with the diode is about the load regulation, the regulator has a much better load regulation than 2%, it's 0.2%

 

[Electronworks]

depends on the resolution of the ADC. At a guess, a reference/regulator with a '%' drift with temperature is not good enough for any moderately precise ADC. generally you cannot use a linear regulator as a reference for an ADC - it has to be a proper reference IC with its drift measured in ppm/degC.

Now for the maths:

For an 8 bit resolution ADC, this measures to 256 steps. With a 2.5V reference, your ADC will measure from 0 - 2.5V in steps of 9.76mV. Your ADC needs to drift no more than 0.5LSB over temperature. If your equipment operates from 10 degC to 40 degC, your refernce must not drift more than 4.88mV over 30 degrees = 162uV per deg C.

162uV/2.5 = 65 x 10^-6 = 65ppm/deg C.

65ppm/deg C = 0.0065%

OK, that handles the drift which you cannot control. the initial offset is calibrated out at production stage.

 

[bychon]

I guess that ends the idea of using a diode as a spacer.