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Analog Isolation help

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matth

New Member
Hello

I have a problem with an analog isolation, I want to isolate a DC signal various between 0 - 60mV.
I use a HCNR201 and the schematic of the datasheet (page 9, figure 12B)
http://www.datasheetcatalog.org/datasheet/hp/HCNR200.pdf
When I check the voltage at the end of the circuits a have always the same voltage.
I want to check the component step by step to make sure it's not damaged. What's the easiest way to do it?
If I undestand right when I force some current through LED (i.e 12mA) the output diodes should conduct the current if are properly supplied, am I right?
Any help would be appierciated as I've spent lot of time and still have nothing.
 

ericgibbs

Well-Known Member
Most Helpful Member
Hello

I have a problem with an analog isolation, I want to isolate a DC signal various between 0 - 60mV.
I use a HCNR201 and the schematic of the datasheet (page 9, figure 12B)
http://www.datasheetcatalog.org/datasheet/hp/HCNR200.pdf
When I check the voltage at the end of the circuits a have always the same voltage.
I want to check the component step by step to make sure it's not damaged. What's the easiest way to do it?
If I undestand right when I force some current through LED (i.e 12mA) the output diodes should conduct the current if are properly supplied, am I right?
Any help would be appierciated as I've spent lot of time and still have nothing.
hi,:)
Which of the 'basic' configurations shown in the datasheet are you using.?

EDIT:
OK, Ive found it.

EDIT2:

What are the values of the 2 resistors in your circuit of 12B..??
 
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matth

New Member
thank you for you answers

the power supply is 5V and the values of resistance are
12Kohm for the resistor connected to the "-" of the first Op-Amps
330ohm for the resistor between the Anode of the LED and the output of the first Op-Amps
12Kohm between the "-" and the output of the second Op-Amps

the values of capacitor are 0.33uF for the first Op-Amps and 10nF for the second Opamp.
 

ericgibbs

Well-Known Member
Most Helpful Member
thank you for you answers

the power supply is 5V and the values of resistance are
12Kohm for the resistor connected to the "-" of the first Op-Amps
330ohm for the resistor between the Anode of the LED and the output of the first Op-Amps
12Kohm between the "-" and the output of the second Op-Amps

the values of capacitor are 0.33uF for the first Op-Amps and 10nF for the second Opamp.
Ok, I'll look thru the data and your values, get back to you.:)
I was looking at these devices a couple of years ago for project I was considering.
 

ericgibbs

Well-Known Member
Most Helpful Member
ok, thank you very much,
I hope that you find the solutions because I have no idea now.
hi,
If I read your first post correctly you are trying to drive the LED.??

You should connect your 0 to 60mV to Vin on A1 and read a 0 to 60mV on A2, Vout.

As you have a 12K on A1 and A2 you have an overall gain of unity [1]

The LED is controlled by A1 via the 330R, you do not control the LED.
 

matth

New Member
You mean I don't need to control LED or there is no need to do it?
In my mind current flowing thru PD2 and PD1 is proportional to current for LED. So if I provide supply for LED (let's say 5V and 300ohm resistor I should generate decent current for LED), then if I connect PD2 or PD1 to 5V thru 330ohm resistor I shoul have some current. If I change voltage (current) for LED, voltage drop on resistor for PD2 or PD1 should change (indicating that current is different). That's how I wanted to test the component in the first place...
but have no current in PDa or PD2...

Is it something wrong with my thinking?
 

ericgibbs

Well-Known Member
Most Helpful Member
You mean I don't need to control LED or there is no need to do it?
In my mind current flowing thru PD2 and PD1 is proportional to current for LED. So if I provide supply for LED (let's say 5V and 300ohm resistor I should generate decent current for LED), then if I connect PD2 or PD1 to 5V thru 330ohm resistor I shoul have some current. If I change voltage (current) for LED, voltage drop on resistor for PD2 or PD1 should change (indicating that current is different). That's how I wanted to test the component in the first place...
but have no current in PDa or PD2...

Is it something wrong with my thinking?
hi,
Look at this image.
The A1 controls the LED drive, which is seen by both PD1 and PD2, you dont change the LED drive.
 

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matth

New Member
Hi

This image is exactly my circuit.
I see that the LED with PD1 create something likes a feedback no ? so in this case, maybe I have a problem with the gain of A1 ?

If i understand my input voltage can't change the LED drive when i change the input voltage ?
 

ericgibbs

Well-Known Member
Most Helpful Member
Hi

This image is exactly my circuit.
I see that the LED with PD1 create something likes a feedback no ? so in this case, maybe I have a problem with the gain of A1 ?

If i understand my input voltage can't change the LED drive when i change the input voltage ?
hi,
No image in view yet, Admin will clear it first.

Forget about the LED drive, leave it to the A1 amp via the 330R.
It uses the LED to optically couple the two halves of the isolator.
There is feed back [LED] to PD1 and PD2 in order to make the coupler a LINEAR analog coupler.

You feed your input to Vin

How much gain do you want.?

At the moment the 12K with a 60mV signal gives 5uA current into A1, look at the datasheet graph.

If you look at the GAIN formula in the sheet you can adjust the two resistors to set the gain.

OK.?:)
 
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matth

New Member
Hi,

sorry, when i said "this image is exactly my circuit", I spoke about your schematic.

I want a Gain of 1 or of 83.3. If i take a gain of 1, after i will use a no-inverter op-amps. Because, I have a signal of 0-60mV in the input and i have to transform this signal to O-5V DC with an isolation.

For the current in pd1, I think understand why i have 50uA (Ipd1=60mV / 12k),

in the datasheet, the gain of the formula is Vout/Vin = K * (R2/R1) ?
After this formula, the datasheet says that K=K3. And in the page 6, it says that the typical gain of K3 is 1.

So, I have take two resistors of 12kohm in the order to have a gain of 1, but maybe with this i don't have enough current , Do you think that it is better to use a resistor like 1kohm or 100ohm ?

and thank you again for your help.
 

ericgibbs

Well-Known Member
Most Helpful Member
hi,
The IPD1 in your example, 60mV and 12K gives a current of 5microAmps

I would suggest that you decrease the R1 value from 12K to 1.2K this will give a IPD1 of 50microAmp.
It will also increase the available span usage of PD1 and hence the resolution.
Note: at very low levels of IPD1 < 3microAmp the K3 value is no longer linear [Ref: Figure #2]

Make the value of R2 say 12K, this will give a gain of 10, assuming that K3=1 [which I would suggest as a starting point]

With a gain of 10 in the opto circuit, it will give an output signal of 600mV, which could amplified again
with a following amplifier by [*8.33] to the +5V that you require.

What type of OPA's are you using in the circuit.?
 
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matth

New Member
Ok if i understand, currently my current Ipd1 is to weak, because in the figure 2, we can see the linearity of K3 range from about 5uA to 50uA. And to have the best linearity, we need to have 50uA for 60mV no ?

I will try say with the gain of 10 this monday. Maybe it's fine, and i after i will say you if it works or not.

I use a Lm324 for the OPA, I have forget to says taht maybe the Voltage drop by the is too big, in this case, i will use a supply of 6V.

Thanks you again for your time :)
 

ericgibbs

Well-Known Member
Most Helpful Member
Ok if i understand, currently my current Ipd1 is to weak, because in the figure 2, we can see the linearity of K3 range from about 5uA to 50uA. And to have the best linearity, we need to have 50uA for 60mV no ?

I will try say with the gain of 10 this monday. Maybe it's fine, and i after i will say you if it works or not.

I use a Lm324 for the OPA, I have forget to says taht maybe the Voltage drop by the is too big, in this case, i will use a supply of 6V.

Thanks you again for your time :)
hi,
Yes.

The LM324 requires at least a supply voltage of +1.5V greater than the maximum output required.
A 6V supply will only give a Vout of 6-1.5 = 4.5V
 
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matth

New Member
thanks you, for your helps

i know that the drop of the Lm324 is important, maybe later i will take a another OPA, but at the moment i juste have this OPA.

Monday, morning I will try to change the first resistance, and normally after the circuit should be fine ? I hope really.

have a good week end.
 

matth

New Member
hi

sorry for the double post,
I have test the circuit with a resistor of 1.2Kohm for R1, and my circuit don't works again. I have always about 2mV at the output.

I woul like to know if it's important or not, if i have invert PD1 and PD2 ?
or maybe the component it's broken, i don't know
 
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ericgibbs

Well-Known Member
Most Helpful Member
hi

sorry for the double post,
I have test the circuit with a resistor of 1.2Kohm for R1, and my circuit don't works again. I have always about 2mV at the output.

I woul like to know if it's important or not, if i have invert PD1 and PD2 ?
or maybe the component it's broken, i don't know
hi,
Can you post how you had the opto-islolator wired initially when you were trying to drive the LED.?
What do you mean by invert the PD1 and PD2, I assume that the inputs are 0mV thru +60mVdc .????

There is a chance the ic maybe damaged.
 
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matth

New Member
Hi, thank you for your answer.

When i said, i have invert PD1 and PD2, i said that, i use pd2 for the first OPA and PD1 for the second OPA. for example, where in the schematic we can see diode PD1, I have put the diode PD2.
i don't invert cathod-anod.

When i had try the optocoupleur, I the schematic in attach fils, and i saw the current in R2 and R3, I had nothing.
 

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ericgibbs

Well-Known Member
Most Helpful Member
Hi, thank you for your answer.

When i said, i have invert PD1 and PD2, i said that, i use pd2 for the first OPA and PD1 for the second OPA. for example, where in the schematic we can see diode PD1, I have put the diode PD2.
i don't invert cathod-anod.

When i had try the optocoupleur, I the schematic in attach fils, and i saw the current in R2 and R3, I had nothing.
hi,
The LED is shown the wrong way around in your diagram.

You must keep PD1 as the input to A1 and PD2 to the input of PD2.

Whats your location.
 
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matth

New Member
ok i will try to invert PD1 and PD2.

what do you mean when you say "The LED is shown the wrong way around in your diagram." I have invert anod and cathod of the LED ? because in the datasheet i see this schematic (in attach files)

my location :confused: ? I'm in south of UK :)
 

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