Power LED from 48 volt phantom power

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#61
 
#62
Earlier I thought the function of this thing would be "mute while holding" so the button would have to be held down for as long as one wanted the mic to be muted. But it appears you actually want a "push (and release) to mute" and the light to come on, and then "push (and release) to unmute" and the light to go off.

There are mechanical pushbutton switches that latch, that are push-on and push-off. That's what I would use to get that function. Here's one:

http://www.radioshack.com/product/index.jsp?productId=2049717
 
#63
No, you got it right benb. My primary use is a momentary mute. And the design for version 1 is all but finalized.

The rest of the discussion was thinking about more advanced function, such as a switch to convert the main switch into a latching function.

After I understand what the opto-isolater vs. opto-mosfet decision turns out to be, I'll build the version 1 design. And that may be all I need, just the momentary switch powering an indicator led and activating the mosfet to short pins 2 & 3.

I've been learning lots through this thread.

thanks again
 
#64
The opto mosfet will be fast i.e. the change of R will occur in an instant, and u will hear the click whereas an LDR in an optic coupler(it is not called opto isolator and the term is reserved for ICs) is slow.

The idea with the opto-isolator is power hungry...and for that matter both, as u will have to supply 20mA for driving the LED in both cases.
...but on second thought, the LED will draw power when the mike is muted and so will not affect the performance.

In an optic coupler, LDR is the sensing element and it has a slow response to the change of light.

For more understanding, read Lightspeed attenuator thread on this forum.

http://www.diyaudio.com/forums/part...lightspeed-attenuator-new-passive-preamp.html

Gajanan Phadte
 
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#65
If you look at an actual data sheet you'll find direct set and reset pins which must be dealt with. CMOS must have all inputs tied to something whether it be power, ground or the output of another stage. Input to input does not count. Inputs of unused sections must also be tied down. The Power On Reset would drive either the Set or Reset with the other one tied low.

http://www.onsemi.com/pub_link/Collateral/MC14013B-D.PDF

The 4013 is the venerable old slow as mud (a gazillion times faster then you need) super low power CMOS device that can run directly from a 9 Volt battery. Look at the output drive current and you might think 'that's OK' but you'll be wrong. It's not made for that kind of load so you use the MOSFET (amplifier) on the output.

The MCU solution would fix nearly all your problems except for the MOSFET driver. I'm partial to Freescale while lots of folks use PICs. If you've never worked with MCUs there is a learning curve that might be steep depending on your programming skills. For little projects like this I work in assembler which is about as deep into the innards as you'll ever get. It can be the most compact code you'll ever see. Small numbers of bytes of RAM and way less than a 1K of code for this. Debouncing done in software. Power on reset in software. Even test modes if you like. Mission creep is very easy with software - but fun.

G²
 
#66
Ok, thanks for the explanation gmphadte.
So the optocoupler would have a gradual decline of resistance compared to the optomosfet.
I'm not sure why the speed of change would cause a click. Will this happen even though the circuit will have a capacitor?

About power consumption. I had understood that the leds could work on as little as ~2 mas. What would require 20mas to function?

@stratus: Yep, that clinched it. version 2 will be mcu based. My only question about that is a clarification of the hookup for the mosfet driver.
I'd guess I'd have the mcu put a pin high to activate the mute...that high pin would go to the gate, which would connect the source and drain?
What am I missing here?
 
#67
Sudden short or open e.g. in a mosfet will lead to harmonics which will be heard.

With 2 mA, the led will light, but if u use the opto-coupler that I mentioned, u need to pass 20mA (or whatever mentioned by the mfr) through the LED so that it lights bright enough to lower the LDR resistance.

Gajanan Phadte
 
#68
I continue to learn.

I took a moment to google up some optocoupler datasheets. If I'm reading this one correctly, the input current can be in the range of 40 microamps to 1.5 milliamps.

I don't see the "on-resistance" listed...it must be, but I'm missing it.

Granted, special low current optocouplers will be more expensive, $2 for this one.

So would something like this work for me?
 
#72
The opto coupler u attached has a low LED drive(40µA). That is fine, but the transistors will draw a leakage current from your phantom power, which will form a voltage divider and affect the mike even when the LED is not driven.

Gajanan Phadte
 
#73
So the muting circuit will draw say, 2-3ma of current from phantom power when ever it is connected, regardless of the mute status. This means the mic has that much less current to work with.

Could you explain a bit about the voltage divider?

In this case the power source is 48v and the muting circuit will be connected to ground on pin 1. So the circuit will have some resistance. This would be equivalent to the bottom resistor in a voltage divider?

48v------mute circuit---gnd
.........|
.........|
.power to mic

The amount of voltage dropped will depend on the resistance in the mute circuit?

So attaching this thing to the phantom power will incur both a current drop, and a voltage drop.

However, the mics I'm thinking of using all have large input voltage tolerances. Often down to 12v.

Did I understand your point properly?
 
#74
What I meant was that the leakage in the transistor will be resulting in a resistance across the mic that u r using. Find out what voltage it (mic) has across it. This R along with the phantom biasing Rs will form a voltage divider.

Gajanan Phadte
 
#75
gmphadte said:
What I meant was that the leakage in the transistor will be resulting in a resistance across the mic that u r using. Find out what voltage it (mic) has across it. This R along with the phantom biasing Rs will form a voltage divider.

Gajanan Phadte
Click to expand...
Look at the schematic in post #56. The transistor is in series with a capacitor (and has a high value resistor across the transistor to charge the capacitor to whatever voltage imbalance there might be while the transistor is off), and they connect to pins 2 and 3, so it will make no DC change in pins 2 and 3 as the transistor switches on and off.

Unfortunately, the circuit in #56 uses/needs a photo mosfet, and in more recent posts TheFixer is listing an optoisolator using a Darlington bipolar phototransistor. This will not work well, as the transistor will be very nonlinear, and may not conduct well at all for low-level signals. Go back to using a photo mosfet.
 
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