Question about the math calculations for the circuit below

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I have attached a URL for the circuit in quesiton. Here is the basics.
I am new to electroncis and was wanting a description of how to
calcluate the amount of volts that would be put out by the following
circuit and the reasoning for why this happens as it does (This mostly
applies to the prupose of R2 and R3, I know that with the values as
stated below I get a 10 time amplificaiton of the voltage that goes
through the positive of the op amp, but I dont know how this is
calculated and what the answers would be if I were to change the values
of the resistors). The light source will have a current of 1 microamp.
The resistors in the circuit below are as follows R1 = 1 megaohm, R2=
1kiloohm, R3= 9 kiloohm and R4 is 1 kiloohm. The purpose of this
circuit will be to use a light source to make an LED light. The light
source will be from a chemical reaction and if this reaction occurs
then I want to be able to see that due to the LED turning on.

Here is what I know so that maybe you wont have to do so much work in
answering my question:

Ohms law says that the Voltage applied to the + in the op amp will be
one volt. I understand this. What I am really struggling with is the
way in which that one amp is given an amplificaiton of 10 due to R2 and
R3.

If 10 Volts is the result of the circuit prior to R4 the the current
reaching the LED will be .01 amps. I am assuming this is enough to
light the LED but could be wrong.

The whole system is powered using a 9 volt battery.

HERE is the link to the circuit picture:


http://img207.imageshack.us/img207/8155/electronicsschematicforopoux0.jpg


Any help would be greatly appreciated.

Thanks in advance

Jeremy

 

[Lord Garth]

I have attached a URL for the circuit in quesiton. Here is the basics.
I am new to electroncis and was wanting a description of how to
calcluate the amount of volts that would be put out by the following
circuit and the reasoning for why this happens as it does (This mostly
applies to the prupose of R2 and R3, I know that with the values as
stated below I get a 10 time amplificaiton of the voltage that goes
through the positive of the op amp, but I dont know how this is
calculated and what the answers would be if I were to change the values
of the resistors). The light source will have a current of 1 microamp.
The resistors in the circuit below are as follows R1 = 1 megaohm, R2=
1kiloohm, R3= 9 kiloohm and R4 is 1 kiloohm. The purpose of this
circuit will be to use a light source to make an LED light. The light
source will be from a chemical reaction and if this reaction occurs
then I want to be able to see that due to the LED turning on.

Here is what I know so that maybe you wont have to do so much work in
answering my question:

Ohms law says that the Voltage applied to the + in the op amp will be
one volt. I understand this. What I am really struggling with is the
way in which that one amp is given an amplificaiton of 10 due to R2 and
R3.

If 10 Volts is the result of the circuit prior to R4 the the current
reaching the LED will be .01 amps. I am assuming this is enough to
light the LED but could be wrong.

The whole system is powered using a 9 volt battery.

HERE is the link to the circuit picture:


http://img207.imageshack.us/img207/8155/electronicsschematicforopoux0.jpg

Looks like your input can go higher than the supply rail...

The diode will clamp it when the input is higher than the cathode potential
plus the diode drop.

 

[Johnny Boy]

I have attached a URL for the circuit in quesiton. Here is the basics.
I am new to electroncis and was wanting a description of how to
calcluate the amount of volts that would be put out by the following
circuit and the reasoning for why this happens as it does (This mostly
applies to the prupose of R2 and R3, I know that with the values as
stated below I get a 10 time amplificaiton of the voltage that goes
through the positive of the op amp, but I dont know how this is
calculated and what the answers would be if I were to change the values
of the resistors). The light source will have a current of 1 microamp.
The resistors in the circuit below are as follows R1 = 1 megaohm, R2=
1kiloohm, R3= 9 kiloohm and R4 is 1 kiloohm. The purpose of this
circuit will be to use a light source to make an LED light. The light
source will be from a chemical reaction and if this reaction occurs
then I want to be able to see that due to the LED turning on.

Here is what I know so that maybe you wont have to do so much work in
answering my question:

Ohms law says that the Voltage applied to the + in the op amp will be
one volt. I understand this. What I am really struggling with is the
way in which that one amp is given an amplificaiton of 10 due to R2 and
R3.

If 10 Volts is the result of the circuit prior to R4 the the current
reaching the LED will be .01 amps. I am assuming this is enough to
light the LED but could be wrong.

The whole system is powered using a 9 volt battery.

HERE is the link to the circuit picture:


http://img207.imageshack.us/img207/8155/electronicsschematicforopoux0.jpg


Any help would be greatly appreciated.

Thanks in advance

Jeremy

The type of op-amp has a bearing on this. A typical op-amp, (e.g. 741),
will only go to within about 1.5V to 2V of the supply voltage. To get an
accurate gain of 10, you would be better with a supply voltage of about 12V.
Regarding gain, the gain is equal to (R2+R3)/R2. (1K+9K)/1K = 10.
Therefore, if you get 1V at the op-amp's +ve input, and if the supply is
high enough, then the output will be 10V. If we assume that the LED's
forward voltage is 1.6V, then the current through the LED will be (10V -
1.6V)/R4, which with an R4 of 1K will be about 8mA.
With the 9V supply, the op-amp will go into saturation, which doesn't
matter, and it's output will be in the neighbourhood of 7V to 7.5V. At 7V,
the LED current will be about (7V - 1.6V)/1K ; about 5.4mA. This will just
light the LED.

I hope I've made this as clear as you made the questions,
.... Johnny

 

[Johnny Boy]

The type of op-amp has a bearing on this. A typical op-amp, (e.g. 741),
will only go to within about 1.5V to 2V of the supply voltage. To get an
accurate gain of 10, you would be better with a supply voltage of about 12V.
Regarding gain, the gain is equal to (R2+R3)/R2. (1K+9K)/1K = 10.
Therefore, if you get 1V at the op-amp's +ve input, and if the supply is
high enough, then the output will be 10V. If we assume that the LED's
forward voltage is 1.6V, then the current through the LED will be (10V -
1.6V)/R4, which with an R4 of 1K will be about 8mA.
With the 9V supply, the op-amp will go into saturation, which doesn't
matter, and it's output will be in the neighbourhood of 7V to 7.5V. At 7V,
the LED current will be about (7V - 1.6V)/1K ; about 5.4mA. This will just
light the LED.

I hope I've made this as clear as you made the questions,
... Johnny

I should have added that my calculations rely on an input of 1V, which is
dependent on the light level hitting the photo-diode.
.... Johnny

 

[John Popelish]

how to
calcluate the amount of volts that would be put out by the following
circuit and the reasoning for why this happens as it does (This mostly
applies to the prupose of R2 and R3, I know that with the values as
stated below I get a 10 time amplificaiton of the voltage that goes
through the positive of the op amp, but I dont know how this is
calculated and what the answers would be if I were to change the values
of the resistors).

The voltage gain of the amplifier is 1+ R3/R2

The light source will have a current of 1 microamp.
The resistors in the circuit below are as follows R1 = 1 megaohm, R2=
1kiloohm, R3= 9 kiloohm and R4 is 1 kiloohm.

1 uA of photo diode current through R1 will produce a 1 volt drop.
That will be amplified to 1 V *(1 + 9000/1000)=10 volts.

The purpose of this
circuit will be to use a light source to make an LED light. The light
source will be from a chemical reaction and if this reaction occurs
then I want to be able to see that due to the LED turning on.

It takes only a couple volts to turn on an LED, but you have to limit
the current so you don't burn it up. If you want some precise turn on
point, add a comparator (you can use a second opamp for a comparator)
after the amplifier with the turn on set point voltage (from a pot
between the power supply and ground as a variable voltage source) into
the + input and the amplifier signal into the - input. When the
amplifier signal exceeds the setpoint voltage, the comparator output
will pull low. Connect an LED and series current limiting resistor
between the positive supply and the comparator output.

Here is what I know so that maybe you wont have to do so much work in
answering my question:

Ohms law says that the Voltage applied to the + in the op amp will be
one volt. I understand this. What I am really struggling with is the
way in which that one amp is given an amplificaiton of 10 due to R2 and
R3.

The concept is that the opamp has essentially infinite voltage gain,
so the only way the output can hold a steady voltage between
saturation limits is if the two inputs are essentially the same. So
the feedback network produces a matching 1 volt signal to the - input.
That requires a 10 volt output that gets divided by 10 by the two
resistors.

If 10 Volts is the result of the circuit prior to R4 the the current
reaching the LED will be .01 amps. I am assuming this is enough to
light the LED but could be wrong.

The 10 volts will also be available to drive other loads in parallel
with the feed back divider. You could add a 1k resistor and an led in
series between the output and ground. The LED will light dimly when
the output voltage is about 2 volts, and will brighten when the output
voltage rises. The comparator addition I mentioned allows the LED to
switch on cleanly at a precise light level.

The whole system is powered using a 9 volt battery.

Well, the output cannot reach 10 volts, then. Most opamps waste 1.5
to 2 volts to generate an output. So 7 volts positive is about all
most will produce. You could connect two 9 volt batteries in series
to provide an 18 volt supply that could produce up to 15 or 16 volts
out. You also need an opamp whose inputs work all the way to the
negative supply rail, like a dual LM358.

http://www.fairchildsemi.com/ds/LM/LM358.pdf#search="lm358"
The second section could perform the comparator function, if desired.

 

[Alan B]

Looks like your input can go higher than the supply rail...

The diode will clamp it when the input is higher than the cathode potential
plus the diode drop.

I think you've misinterpreted the circuit. This is how I see it: the input
device is a photo-diode. With no light, it will act as a typical diode; in
this case, reverse-biased +9V to zero. As light is applied, forward
current will be produced, varying according to the light intensity. This
forward current will vary the voltage at the non-inverting input via R1.
Here's a reference to photo-diode operation:

http://www.radio-electronics.com/info/data/semicond/photo_diode/photo_diode.php