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Question about the math calculations for the circuit below

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:

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Any help would be greatly appreciated.

Thanks in advance

Jeremy

Reply to
JJKarr
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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.

Reply to
Lord Garth

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

Reply to
Johnny Boy

741),
12V.

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

Reply to
Johnny Boy

snipped-for-privacy@gmail.com wrote: (snip) > how to

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

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.

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.

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.

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.

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.

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The second section could perform the comparator function, if desired.

Reply to
John Popelish

On Sat, 07 Oct 2006 05:44:23 GMT, in message , "Lord Garth" scribed:

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:

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Reply to
Alan B

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