ok so I'm working w/ this schematic.

http://www.uni-bonn.de/~uzs159/adsr2.png

i want to put an led on the other end of the last opamp. I want to have the voltage to the led to max out at the leds max brightness, i want it to get no higher, but i want the voltage to bottom out at the min brightness of the led and get no lower, how would i go about doing this, maybe put 1k resistor in parallel w/ the led and then a 1k in series w/ the cathode of led to ground?

idk help me please!

 

Have you read this?

LEDs, 555s, Flashers, and Light Chasers

This op amp is barely able to drive an LED, you'll probably want to use a transistor to drive the LED. The -15 won't get involved, you'll need to use ground since the op amp will only swing 0V to 13VDC.

Something like this perhaps.

I'll be honest, 555s are my specialty on this site, but I can't make heads or tails from that schematic. What is it for?

 

It's an adsr envelope generator (attack, decay, sustain, release)

well basically u put a voltage on the input marked gate, it charges up the capacitor, how quickly is determined by the pot marked a, as you know as soon as 2/3rds voltage is reached on the threshold it discharges how quickly it discharges is determined by the pot marked d, the voltage it discharges to is determined by the pot marked s, and finally when the voltage on the gate is realeased, the remaining charge in the cap is released, how quickly is determined by you guessed it, the pot marked r. I want to put an led on the output of the opamp, but the problem is, i dont want the voltage to surpass whatever is required to get the led at max brightness, and i don't want the voltage to dip below whatever is required for it to just be off, or at minimum brightness.

 

Then you need to control the current on it, not the voltage.

0 < LED current <= max rated current.

[eta]
The current through the LED needs to be a log function of the voltage envelope.

 

!!!!! how do i go about doing this???? 1k resistor in paralell w/ the led, then a 1k in series to ground?

 

A typical LED is going to require around 20ma for "bright"

You have to be sure the opamp (or whatever you use) can source that current.

The data sheet will have that information.

If it can, you use OHMS law to limit the max current to the LED. Say 20mA.

If you are getting 8v out of the opamp, and the led you want to use has a Vf of 2.2v
you subtract the two numbers and use Ohms law to find the resistor size for 20ma (.02A)

So, 8v - 2.2 = 5.8v
5.8v x .02 = R which is resistance needed in ohms
R = 116

If the opamp cannot supply enough current or voltage to meet the LEDs Vf and current, use the opamp output to turn on a transistor to send power from the rail to the led. You still use Ohms law to find the resistor to use, but with the rail voltage minus the LEDs Vf. The Vf is in the leds data sheet and can change from led to led.

If the number you get, in ohms, isnt a standard resistance value, round up to the next higher value.

 

I'm not sure if i read the datasheet correctly, but I'm thinking that I can't source enough current with that opamp, so after reading retched's response, these are the solutions I'm going to try.

1. jfet
2. Operational transconductance amplifier

hopefully one of these will get me where i need to be.

 

Or you could try the circuit I posted and see if it does what you want.

 

I agree with bill's response and circuit. Its job is to perform the task you want performed.

2 components and about 25 cents later, and your off and running. (not including your led) so 3 components and 30 cents

 

Following on what the Sargent said, I think you need a voltage-controlled current source.

Say we use a 100 Ohm resistor to limit and sense current.

The output from the 555 circuit will be zero to something below 15... so in the attached circuit, I have divided that down to a voltage from 0 to 2. This corresponds to 0-20mA on the 100 Ohm sense resistor. The MOSFET controls the current in the linear region, while the 500 Ohm resistor burns up all the excess current and limits against a blowup if the control malfunctions. At 20 mA, it burns up 10V, while the 100 Ohm burns up 2V, leaving only 3V from the 15V rail for the MOSFET and the LED. If you need some more voltage for your LEDs to shine full bright, then lower the 500 Ohms a bit. V/I = R, so (15 - 2 - Vf)/0.020 = R for a Vf for an LED. If the Vf for 20 mA is 3.2V then R = 490.