I see what you _want_ to do regarding the light. How do you simultaneously power the mic and _not_ the LED? What parameter does 'mute' change that turns on the LED while killing the audio?
G²
@stratus: Well, power is tapped off of the phantom pins to light the led and power the muting circuit. So when the mute is not active, a switch will be open, preventing both the led and the muting circuit from getting power.
To mute the mic, the switch allows some phantom power to light the led and activate the muting circuit.
That was my thinking anyway.
@benb: I'm still learning about the current/power relationship.
I think I might have figured out this bit.
So, both of the above examples have the same heat generated, the same "power" even though one current is 10 times the other.
Is that the point?
To mute the mic, the switch allows some phantom power to light the led and activate the muting circuit.
That was my thinking anyway.
@benb: I'm still learning about the current/power relationship.
I think I might have figured out this bit.
But also, if a flashlight uses 0.01 amps at 30 volts, the total power is 0.01 x 30 = 0.3 watts.
So, both of the above examples have the same heat generated, the same "power" even though one current is 10 times the other.
Is that the point?
Yes, your calculations and understanding are exactly correct.
But I can also understand stratus46's question and concern about muting the mic. There are many to do it that will cause a big pop in the audio, and there are ways that will mute the mic quietly. I'd think powering the LED with the mute switch on would complicate this. We can judge better from a schematic than from a general description.
But I can also understand stratus46's question and concern about muting the mic. There are many to do it that will cause a big pop in the audio, and there are ways that will mute the mic quietly. I'd think powering the LED with the mute switch on would complicate this. We can judge better from a schematic than from a general description.
I use high-efficiency LED's and run them at 0.5-2mA. These are GaInN greens or whites with 3-3.6V for Vf.
If putting the indicator on XLR (balanced line), you would use two resistors to keep things balanced.
I always put a reverse diode across the LED to prevent static electricity or voltage spikes from ruining them, especially high-eff LEDS that can't take any reverse voltage.
If putting the indicator on XLR (balanced line), you would use two resistors to keep things balanced.
I always put a reverse diode across the LED to prevent static electricity or voltage spikes from ruining them, especially high-eff LEDS that can't take any reverse voltage.
Notes:
The value of the large resistor (47k) and the capacitor were taken fro the pro co cough drop tech sheet. I have seen other values in other schematics such as 100k resistor and 1000 microfarad cap.
Popless Mic On/Off Switch
When pressed, the switch sends current through the led and then to ground (pin 1)... And also provides a path from phantom power to the gate.
If I understand the concept correctly, the capacitor will prevent a dc offset from existing, thus stopping the pop that a straight switch would cause.
Questions:
Will I need to drop the voltage applied to the mosfet gate? in this image the full phantom voltage is applied, will 48v damage the mosfet?
Will the mosfet allow bi-directional current flow? or will I need to use 2 mosfets, one in each direction?
Any thoughts?
Thanks!
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I see what you are trying to do... mute a balanced mic by shorting the + and - signals using phantom power to activate a mosfet switch.
To shunt AC signals you need two MOSFETS. Also the MOSFETS need voltage limiting on the gate, as 48V drive is too much. Usually a 10-15V zener clamp is used.
You can use a photo-MOSFET solid-state relay; they have on-resistances 1-200 ohms, the higher vales will not guarantee 100% muting, depending on the mic preamp's gain.
Data Sheet - ASSR-1218, ASSR-1219, ASSR-1228, Form A, Solid State Relay (Photo MOSFET) (60V/0.2A 380 KB)
Datasheet - Vishay LH1500
You can wire the opto's LED in series with your indicator LED.
To shunt AC signals you need two MOSFETS. Also the MOSFETS need voltage limiting on the gate, as 48V drive is too much. Usually a 10-15V zener clamp is used.
You can use a photo-MOSFET solid-state relay; they have on-resistances 1-200 ohms, the higher vales will not guarantee 100% muting, depending on the mic preamp's gain.
Data Sheet - ASSR-1218, ASSR-1219, ASSR-1228, Form A, Solid State Relay (Photo MOSFET) (60V/0.2A 380 KB)
Datasheet - Vishay LH1500
You can wire the opto's LED in series with your indicator LED.
I predict some massive 'thumps' in your future. If you're lucky there won't be any preamp failures.
G²
G²
My thanks prariemystic, I did not know there was such a thing as a photo-mosfet. Looking over the data sheets, they look perfect. One question, I'm not positive I'm reading these right, do those photo-mosfets work in both directions... passing ac current?
If so, my search for a solution is almost over.
@stratus: Would you care to provide a reason why I'll be getting thumps? Or perhaps a different way of doing it to prevent thumps? I've looked high and low on the internet and this capacitor low pass filter is the only technique for silent muting I've found. Do you have a better way in mind?
thanks
If so, my search for a solution is almost over.
@stratus: Would you care to provide a reason why I'll be getting thumps? Or perhaps a different way of doing it to prevent thumps? I've looked high and low on the internet and this capacitor low pass filter is the only technique for silent muting I've found. Do you have a better way in mind?
thanks
You have an audio signal in the milliVolt range riding on top of the DC power in the many Volts range. When you flip your mute switch you're going to change the DC loading of the signal lines which will drop the DC level several Volts - over a thousand time the level of the audio. If both lines don't change at the same rate the differential will show up as audio. After all that's what balanced audio is. Since the change isn't desired audio, it will show up as a 'thump' or 'pop'. Even if the lines change exactly the same the mic itself may have problems with the changing power supply.
The way you're handling the actual audio should be good. It's the power to do it that will cause the problems. I'd be delighted to be proven wrong.
I take it the end goal is to have a phantom powered mic with a box/switch placed in the cable near the mic end of the cable. You want to press a switch on the box and have it mute the audio and light an LED reminding you to mic is in mute. Is that correct? The only thing I can think of in this case it to have an extra power supply to trip your mosfet and light the light. A battery could do that for a long time but would eventually die.
When you started the thread I got the impression the mute function was originating in a mixing console and you were trying to send the mute status on the mic cable which was why I was asking 'how' you intended to make it work.
G²
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Thanks for the helpful reply Stratus.
I think I see what you mean. I posted another thread on this forum to try to figure out what components are used in an electret mic capsule, unfortunately that thread died.
Could you clarify what DC loading is exactly?
I had hoped, perhaps naively, that since the mics I'm dealing with are all electret, they should only be using a fraction of the power available since phantom power was desgn to power old full condenser mics.
Are you concerned that tapping off enough power to run the 2 leds will cause enough of a voltage drop on the main line that the mic will throw a fit?
Or, is it more about the two signal lines not changing dc (phantom power) voltage at the same instant. If the later is the concern, why would they not change at the same instant? The preferred idea so far is to tap off equal power from both lines through identical resistors. I suppose if those resistors are not of high tolerance one may draw more than the other, but hypothetically they should draw equal power and at the same moment...or did I miss something?
Sorry for any confusion on the purpose of this build, I'm not necessarily the best at describing technology.
I'm appriciative of all the help I've gotten here. So many smart people have contributed. I can already say I've learned here.
Thanks
I think I see what you mean. I posted another thread on this forum to try to figure out what components are used in an electret mic capsule, unfortunately that thread died.
Could you clarify what DC loading is exactly?
I had hoped, perhaps naively, that since the mics I'm dealing with are all electret, they should only be using a fraction of the power available since phantom power was desgn to power old full condenser mics.
Are you concerned that tapping off enough power to run the 2 leds will cause enough of a voltage drop on the main line that the mic will throw a fit?
Or, is it more about the two signal lines not changing dc (phantom power) voltage at the same instant. If the later is the concern, why would they not change at the same instant? The preferred idea so far is to tap off equal power from both lines through identical resistors. I suppose if those resistors are not of high tolerance one may draw more than the other, but hypothetically they should draw equal power and at the same moment...or did I miss something?
That's it precisely. I would add that I intend to locate the switch a few feet away from the mute box to make it fit my use.
Sorry for any confusion on the purpose of this build, I'm not necessarily the best at describing technology.
I'm appriciative of all the help I've gotten here. So many smart people have contributed. I can already say I've learned here.
Thanks
One method of supplying phantom power. A microphone or other device can obtain DC power from either signal line to ground terminal, and two capacitors block this DC from appearing at the output. R1 and R2 should be 6,8k ohms for 48 volt phantom, and R3 should not be used. >>>>>
Inside your mic preamp or mixer the phantom power is applied to the 2 signal leads with a 'matched' pair of usually 6.8K resistors. This will cause a voltage drop due to the power required by the mic. The DC load is the couple mA for the LED. You propose to draw some current to light your LED which will necessarily reduce the phantom voltage and there is no way around it. Since those are the signal lines I'm virtually certain it WILL pop. Again, I'd be happy to be wrong.
G²
There have been one or two diversions in the course of this thread - but it is going generally in the right direction.
Remember that it CAN be done:
Welcome to Orchid Electronics
Stratus is quite correct in that any change in current fed to the LED will cause a click or bang. As the above item is my design, which took considerable development time - and I believe to be fairly unique, regard that as a clue to the next step you need to take!
Remember that it CAN be done:
Welcome to Orchid Electronics
Stratus is quite correct in that any change in current fed to the LED will cause a click or bang. As the above item is my design, which took considerable development time - and I believe to be fairly unique, regard that as a clue to the next step you need to take!
@ stratus: So the issue I now face is that when the mic in unmuted, the electret will be supplied with 48v, and when the mic is muted it will be supplied with, say, 45v. This being due to the power siphoned off to run the leds?
Why would the mic see less voltage? the mixer puts out 48v...and the wires run straight to the mic. I know there will be less current available to the mic since the leds require power...but why would that change the voltage seen by the mic?
The most obvious solution would be to eliminate the "change" in voltage by switching between the led load in the mute condition, and a dummy load (perhaps 2 hidden leds?) in the unmute condition. This would cause the mic to "see" only one voltage all the time, regardless of mute condition. Would this fix the "change in dc loading" issue?
Question: can photo-mosfets pass current in both directions? this datasheet indicates yes. http://www.toshiba-components.com/docs/opto/TLP222G_en_datasheet.pdf
@John: I am quite certain the it can be done. I expect the most expensive component involved will be the enclosure.
@amco: Thanks for posting...could you talk a little bit about your schematic? I can't see it too well, but it looks like you have pins 2 & 3 tied together with a pair of resistors. Then you pull both lines to ground through capacitors. ? I'm not sure what you mean to do.
Why would the mic see less voltage? the mixer puts out 48v...and the wires run straight to the mic. I know there will be less current available to the mic since the leds require power...but why would that change the voltage seen by the mic?
The most obvious solution would be to eliminate the "change" in voltage by switching between the led load in the mute condition, and a dummy load (perhaps 2 hidden leds?) in the unmute condition. This would cause the mic to "see" only one voltage all the time, regardless of mute condition. Would this fix the "change in dc loading" issue?
Question: can photo-mosfets pass current in both directions? this datasheet indicates yes. http://www.toshiba-components.com/docs/opto/TLP222G_en_datasheet.pdf
@John: I am quite certain the it can be done. I expect the most expensive component involved will be the enclosure.
Did you mean... any change in current fed to the MIC ... ? Your hint, if I understand it, is that any change in power supplied to the mic will cause audible issues. The solution being to prevent the mic from seeing any change in power.
@amco: Thanks for posting...could you talk a little bit about your schematic? I can't see it too well, but it looks like you have pins 2 & 3 tied together with a pair of resistors. Then you pull both lines to ground through capacitors. ? I'm not sure what you mean to do.
The circuit shown is what is found inside the mixing desk - the resistor R3 is there to offer a little extra decoupling (together with the capacitor to ground) - providing some isolation between channels. Remember that a big mixer may have 32 or more channels, each with a network as shown.
The purpose of this drawing is to show how the DC will flow through each of the 6K8 resistors and return to ground.
You can draw current from either wire to ground - remember that the current is limited by the resistors.
Unplug an XLR from a microphone - and connect an LED directly from pin 2 or pin 3 (+ or long leg of LED) and pin 1 ground. It will illuminate if phantom is present.
You can now make a simple cable tester with a couple of LED's - which will show if either audio leg or the screen is open circuit (disconnected).
For the purpose of the muting discussion - ignore the microphone, it will draw whatever it needs to operate.
Time to try out some of the schemes discussed - remember that you cannot do any damage to the mixer input whatever you connect (or damage the microphone). Just keep the gain down so you can't blow speakers or deafen yourself!
The purpose of this drawing is to show how the DC will flow through each of the 6K8 resistors and return to ground.
You can draw current from either wire to ground - remember that the current is limited by the resistors.
Unplug an XLR from a microphone - and connect an LED directly from pin 2 or pin 3 (+ or long leg of LED) and pin 1 ground. It will illuminate if phantom is present.
You can now make a simple cable tester with a couple of LED's - which will show if either audio leg or the screen is open circuit (disconnected).
For the purpose of the muting discussion - ignore the microphone, it will draw whatever it needs to operate.
Time to try out some of the schemes discussed - remember that you cannot do any damage to the mixer input whatever you connect (or damage the microphone). Just keep the gain down so you can't blow speakers or deafen yourself!
Yes, the photo-MOSFETs are bi-directional, they pass AC.
I thought LED activation will cause a common-mode drop since it is powered from both + and - lines... just like microphones... that use phantom power... as power...
I thought LED activation will cause a common-mode drop since it is powered from both + and - lines... just like microphones... that use phantom power... as power...
Yes and no - in a perfect world the two 6.8k resistors inside the preamp have the exact same value. In the real world, they don't.
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