I designed an audio equalizer product using a TL074 (quad version of the dual TL072). Tens of thousands were sold and none was ever replaced.

My guess is that the system had a well designed power supply that never put out excessive voltage and never got the polarity wrong, And I also presume that all of those ICs were purchased from a reliable seller, or possibly direct from T.I.
AND, if those EQ packages were treated as well as mine has been, there has not been the abuse that the devices used by performers get, dragging to gigs and practice sessions.
Some of those ICs get a lot more abuse than others.

 

Is the V/2 supply a good spot to put a white LED, D1? Like this? Or probably be better in parallel with the 9v?

 

No, it’s a bad place. How much current is required for the LED? That current has to go through R3. What will happen to the voltage across R3?
A TL072 has a quiescent current of about 2mA, will your LED be consuming more power that the rest of the circuit? If so, you have just halved your battery life.

 

Interestingly enough, an LED can serve as a fair voltage reference, similar to a Zener diode, if operated at the flatter portion of the current/voltage curve. But in the circuit in post #142 it will not work at all as shown. For that circuit, to have the LED act as a power-on indicator, it could be in series with the 9 volt power source, but it would drop about 3 volts, probably less because of the lower current.

 

For reference, it will be a 9v power supply not a battery. Regardless, this seems better, yes? 3v and 20mA across white LED.

 

This circuit will work, but still there will be a larger current draw by the LED, plus there are two voltage dividers, where the one could be shared by both amplifiers.
R3 and R4 could be 47K instead of 10K, R10 an be eliminated and R11can be fed from the junction of R3 and R4. A common Vdd/2 rail is often used in multi-stage designs.

 

20mA through a white LED would be blinding, and would reduce your battery life to 27hours, compared to 270 hours with no LED (for a 550mAh alkaline PP3). You could put three LEDs in series, it would consume the same power, and you could also use it as a torch.

 

20mA through a white LED would be blinding, and would reduce your battery life to 27hours, compared to 270 hours with no LED (for a 550mAh alkaline PP3). You could put three LEDs in series, it would consume the same power, and you could also use it as a torch.

I was just going off the chart - 20mA is 3v on white. Yes I need to test the brightness of that and adjust the resistor, etc.

 

I never drive LEDs at 20mA. Quite often 1-2mA is enough.
Why white? Blue looks cool in the dark.

 

Personal preference I guess. When I see white, it makes me think clean, non-colored from a sound perspective.
Thanks for the suggestion.

 

......and you could also use it as a torch.

Yikes! A fire?
Oh, you must be from the UK and you douse a fire with wotah. We say a battery light is a flashlight and we swim in water.

 

Yikes! A fire?
Oh, you must be from the UK and you douse a fire with wotah. We say a battery light is a flashlight and we swim in water.

Prefer the torch. I found the flashing too distracting when I was trying to see in the dark!
water - rhymes with “daughter“, not with “flatter” as it would in Yorkshire, Lancashire and Cumbria, doesn’t sound like “warder”.

 

Getting back to the circuit, certainly a power-on indicator does not need to be obvious from 100 feet (30 M) away. And the brightness is almost the same, usually. I routinely drive LEDs from a (5V) CMOS gate and they are entirely adequate with only a 100 ohm series resistor.
That 20 mA was the maximum current for the claimed lifetime, not the required current specification.

 

I have a daughter but I do not say dotah like in the UK. I also drive on the ...., never mind.
Flash the LED occasionally to reduce its average current.

 

Wow you guys are right. I checked a white LED from 3.0 - 3.6V, and the current varied from 1-20mA... fairly linearly. And yes, anything more than 2-3 mA seemed quite bright!

So the problem is most 9v power supplies I have checked... including batteries, vary anywhere from 9.2 - 9.5 volts, with the exception being rechargeable batteries that are 8.2-8.4 volts. I'm not doing batteries. So let's assume I want 2 mA at the LED, and hit the middle on the voltage range and shoot for a voltage drop of 6.25 volts to get 3v at the LED. So I would need a 3.1k resistor. Is that the best way to approach a 9v power supply?

 

The simple approach is

R = (Vs - Vd) / Id

Hence R = (9V - 3V) / 2mA = 3kΩ

 

I started a new thread over here about PCB layout, prototyping, etc. for this circuit design. I'd love for y'all to chime in, you've all been so helpful!

 

Hi,

I've been learning about RC filters. I'm not sure a 2nd order Sallen-Key HPF is going to be good enough, steep enough for my application (80-1000 hz). As I understand it, getting to higher order filters may be easier to achieve with a universal filter IC.
UAF42, I think is only 2nd order.
LT1562, I think is only good for 10k-150kHz
Maxim 74xx series seems promising, but is a switched capacitor filter that requires a clock voltage. Probably beyond my ability.
NS AF100, MF5. Not sure.

What are my options to control the cutoff frequency of a higher order filter with an external pot?

Here is one of the best sources of information on Active filters I've ever read:

Active Filter Cookbook; 2nd Ed.
Author: Don Lancaster
ISBN-13: 978-0750629867
ISBN-10: 075062986X

 

Here’s another, but perhaps read Lancaster first.
https://www.analog.com/media/en/training-seminars/design-handbooks/basic-linear-design/chapter8.pdf

 

Why isn't R4 between R1 and R2? Then R2 and R3 would be a true 1/2 supply voltage divider. But in this schematic, it's not.