Need to build an interface between an audiometer and a set of LED lights

[xXAmyXx]

Hi Guys!
,
Could one of you help me I need some advice, I need to build an interface between an audiometer and a set of LED lights. The audiometer outputs a dc signal that correlates to the sound reading it registers. To make it nice and easy it outputs 60dB as 0.6v, 70dB as 0.7v and so on.
What I need to build is something that can take that signal and turn on some LEDs in accordance to the output. The LEDs are part of a tower we have already bought and run on 24v and 500mA per set. I need the first set of LEDs to come on at 0.6v the second at 0.7 and the third at 0.8.

Using my rather limited electronics knowledge my first thought was to have a seperate 24V battery for the leds that connect through three seperate transistors and then have the audiometer signal trigger the transistors by having 3 zener diodes block the signal untill the breakdown voltage is achieved.
A few questions on this, is this the best idea? can you get zener diodes that have that low a breakdown?
I have built the rather basic circuit on multisim in an attempt to test the circuit replacing the leds with lights at 12v and while the premise of the circuit works its only at 1.3v 1.4v and 1.5v that the lights come on even though i set the Zeners at 0.6v 0.7v 0.8v. Any ideas why thats the case?
Also how will I calculate what resistors i need?

All help will be much appreciated, im new to electronics

Click this image to show the full-size version.

Thanks, Amy.

 

[Gryd3]

Hi Guys!
,
Could one of you help me I need some advice, I need to build an interface between an audiometer and a set of LED lights. The audiometer outputs a dc signal that correlates to the sound reading it registers. To make it nice and easy it outputs 60dB as 0.6v, 70dB as 0.7v and so on.
What I need to build is something that can take that signal and turn on some LEDs in accordance to the output. The LEDs are part of a tower we have already bought and run on 24v and 500mA per set. I need the first set of LEDs to come on at 0.6v the second at 0.7 and the third at 0.8.

Using my rather limited electronics knowledge my first thought was to have a seperate 24V battery for the leds that connect through three seperate transistors and then have the audiometer signal trigger the transistors by having 3 zener diodes block the signal untill the breakdown voltage is achieved.
A few questions on this, is this the best idea? can you get zener diodes that have that low a breakdown?
I have built the rather basic circuit on multisim in an attempt to test the circuit replacing the leds with lights at 12v and while the premise of the circuit works its only at 1.3v 1.4v and 1.5v that the lights come on even though i set the Zeners at 0.6v 0.7v 0.8v. Any ideas why thats the case?
Also how will I calculate what resistors i need?

All help will be much appreciated, im new to electronics

Click this image to show the full-size version.

Thanks, Amy.

If I were tackling this project, I'd probably use an LM 3914. It's a bar-graph driver, you can use the output to drive the transistors to fire the LED modules (or other outputs) and it's fairly easy to adjust with a couple potentiometers. (Or some correctly sized resistors)
http://www.ti.com/lit/ds/symlink/lm3914.pdf

Now, as far as the simulation you have worked on is concerned, notice the difference between what you set and what you got is 0.7V? That would be a property of the transistors you used. They don't turn on as soon as they exceed 0V... they usually turn on at 0.7V, so your source voltage needs to be 0.7V + your zener to trigger it.

I would suggest using a comparator or other digital component that can almost instantaneously toggle from off to on once the target voltage is reached. The other unfortunate side-affect of the transistor idea, is that they may end up operating in a 'linear region' which isn't fully ON. This will result in a voltage drop across the transistor making the LEDs dimmer until it's switched completely on. This voltage drop will also create a LOT of heat... Hence why I suggest something that will slap it fully on.
(You can make a comparator with an opamp, and the reference voltages of 0.6 , 0.7 and 0.8V can be supplied with the zener diodes, or voltage dividers.)

 

[davenn]

All help will be much appreciated, im new to electronics

Click this image to show the full-size version.

I cannot see the image

 

[Arouse1973]

Hi Amy, welcome to EP. The LM3914 seems a good option. Do you want the LED strips to stay on as the sound level increases or only light one strip at a time dependant on the sound level?
Thanks
Adam

 

[Gryd3]

Hi Amy, welcome to EP. The LM3914 seems a good option. Do you want the LED strips to stay on as the sound level increases or only light one strip at a time dependant on the sound level?
Thanks
Adam

Good question. Luckily, toggling the state of pin 9 on the LM3914 can do both
I'm sure it will be sensitive enough. I've got a few tucked away for a project I never got to so we can test before-hand if need be.

 

[Arouse1973]

Good question. Luckily, toggling the state of pin 9 on the LM3914 can do both
I'm sure it will be sensitive enough. I've got a few tucked away for a project I never got to so we can test before-hand if need be.

Oh that must be what bar-graph and dot mode is then .

 

[xXAmyXx]

If I were tackling this project, I'd probably use an LM 3914. It's a bar-graph driver, you can use the output to drive the transistors to fire the LED modules (or other outputs) and it's fairly easy to adjust with a couple potentiometers. (Or some correctly sized resistors)
http://www.ti.com/lit/ds/symlink/lm3914.pdf

Now, as far as the simulation you have worked on is concerned, notice the difference between what you set and what you got is 0.7V? That would be a property of the transistors you used. They don't turn on as soon as they exceed 0V... they usually turn on at 0.7V, so your source voltage needs to be 0.7V + your zener to trigger it.

I would suggest using a comparator or other digital component that can almost instantaneously toggle from off to on once the target voltage is reached. The other unfortunate side-affect of the transistor idea, is that they may end up operating in a 'linear region' which isn't fully ON. This will result in a voltage drop across the transistor making the LEDs dimmer until it's switched completely on. This voltage drop will also create a LOT of heat... Hence why I suggest something that will slap it fully on.
(You can make a comparator with an opamp, and the reference voltages of 0.6 , 0.7 and 0.8V can be supplied with the zener diodes, or voltage dividers.)

Oh that must be what bar-graph and dot mode is then .

Thanks for the replies!
Im looking at the datasheet for the 3914 and it seems to be exactly what im looking for but im not 100% sure how it works, would one of you mind briefly explaining it to me
Because my LEDs need so much voltage do I still need the transistors connected to the desperate power supply?

 

[Gryd3]

Thanks for the replies!
Im looking at the datasheet for the 3914 and it seems to be exactly what im looking for but im not 100% sure how it works, would one of you mind briefly explaining it to me
Because my LEDs need so much voltage do I still need the transistors connected to the desperate power supply?

Take a look at page 8 on the datasheet here : http://www.ti.com/lit/ds/symlink/lm3914.pdf
There is an internal voltage divider between pins 6 and 4 that has 10 steps.
So... if you put pin 4 on ground, and pin 6 at 1V, then each step would be 0.1V.
You can simply use pins 14, 13, and 12 for your 0.6v, 0.7v and 0.8v outputs and ignore the rest.

How man 'segments' do you want to be able to control? In theory, you could set this up to turn on one segment at 0.6V, and an additional segment every additional 0.03V, or 0.05V. You can have 10 segments with one LM3914, and can daisy chain them together if need be.

To control a high power output, grab a PNP transistor, because the LED 'Outputs' on this chip are active Low.
The Transistor Base pin to the specific LED output you want. (outputs 6 - 8 in this example)
Transistor Emitter to the higher voltage source for your LED modules.
Transistor Collector to the LED modules.
You can use a separate power source for the LM3914 and the LED modules, but you need to make sure the negative is tied together. If not, the transistors will misbehave and either not light or behave erratically.

 

[xXAmyXx]

Take a look at page 8 on the datasheet here : http://www.ti.com/lit/ds/symlink/lm3914.pdf
There is an internal voltage divider between pins 6 and 4 that has 10 steps.
So... if you put pin 4 on ground, and pin 6 at 1V, then each step would be 0.1V.
You can simply use pins 14, 13, and 12 for your 0.6v, 0.7v and 0.8v outputs and ignore the rest.

How man 'segments' do you want to be able to control? In theory, you could set this up to turn on one segment at 0.6V, and an additional segment every additional 0.03V, or 0.05V. You can have 10 segments with one LM3914, and can daisy chain them together if need be.

To control a high power output, grab a PNP transistor, because the LED 'Outputs' on this chip are active Low.
The Transistor Base pin to the specific LED output you want. (outputs 6 - 8 in this example)
Transistor Emitter to the higher voltage source for your LED modules.
Transistor Collector to the LED modules.
You can use a separate power source for the LM3914 and the LED modules, but you need to make sure the negative is tied together. If not, the transistors will misbehave and either not light or behave erratically.

Wow thats awesome thankyou so much!
I just need three segments like you put, the only other issue im having is with the LEDs they need high voltage at 24V but low current if I use some resistors would it not cause a lot of heat?
Also how will I limit the power supply to 1V for the LM3914?

 

[Gryd3]

Wow thats awesome thankyou so much!
I just need three segments like you put, the only other issue im having is with the LEDs they need high voltage at 24V but low current if I use some resistors would it not cause a lot of heat?
Also how will I limit the power supply to 1V for the LM3914?

In the example I used, pin 6 would need the 1V 'reference'. You can generate this 'reference' a number of ways.
Pin 7 for example is a 'reference' pin that doubles as a 'brightness' adjustment for the LEDs. This output pins provides 1.25V as-is, but if you connect it to a potentiometer wired as a voltage divider, you can use it to make adjustments to this reference value. The data-sheet has some examples, but does not go into great detail about adjusting this reference.
There is literally 10x 1K resistors in series between pins 6 and 4, so you could technically just put a resistor between V+ and pin 6. If V+ is a 5V source, you would need a 40K resistor on pin 6. (4V would drop across the 40K resistor, and the remaining 1V would be present on pin 6) It's strongly suggested to use a potentiometer here depending on what kind of accuracy you require.
Remember this is simply a reference and does not require or draw much current at all. As long as you get 1V to it you'll be fine.

Now as far as the LEDs are concerned... and the resistors... heat will be generated based on the 'power' they are handling. Power is a combination of voltage and current of course, so if the current is low, the resistors will remain cool even with higher voltages.
If you have specifications on the LED modules we can provide details for that as well.
The fact that you quote 24V almost sounds like they have built-in current limiting... in which case you would NOT need a resistor unless you planned to feed it more than 24V.

 

[xXAmyXx]

In the example I used, pin 6 would need the 1V 'reference'. You can generate this 'reference' a number of ways.
Pin 7 for example is a 'reference' pin that doubles as a 'brightness' adjustment for the LEDs. This output pins provides 1.25V as-is, but if you connect it to a potentiometer wired as a voltage divider, you can use it to make adjustments to this reference value. The data-sheet has some examples, but does not go into great detail about adjusting this reference.
There is literally 10x 1K resistors in series between pins 6 and 4, so you could technically just put a resistor between V+ and pin 6. If V+ is a 5V source, you would need a 40K resistor on pin 6. (4V would drop across the 40K resistor, and the remaining 1V would be present on pin 6) It's strongly suggested to use a potentiometer here depending on what kind of accuracy you require.
Remember this is simply a reference and does not require or draw much current at all. As long as you get 1V to it you'll be fine.

Now as far as the LEDs are concerned... and the resistors... heat will be generated based on the 'power' they are handling. Power is a combination of voltage and current of course, so if the current is low, the resistors will remain cool even with higher voltages.
If you have specifications on the LED modules we can provide details for that as well.
The fact that you quote 24V almost sounds like they have built-in current limiting... in which case you would NOT need a resistor unless you planned to feed it more than 24V.

Certainly not looking to feed it more then 24V
The LED tower we have is this one http://uk.rs-online.com/web/p/pre-configured-beacon-towers/8495383/

 

[Gryd3]

Certainly not looking to feed it more then 24V
The LED tower we have is this one http://uk.rs-online.com/web/p/pre-configured-beacon-towers/8495383/

Awesome. Well, it would appear to have built in current limiting for each 'tier' so you don't need a resistor. You can buy a 12V or a 24V version.
According to the data sheet:
** Current consumption: 50 mA per tier / buzzer **
That makes it pretty damn easy. The voltage is higher than most beginning projects, but the current is really low. You won't need a very 'strong' transistor to drive these things.
There is one question now that must be addressed to be able to determine the best method to hook this up...

Does this device have a single 'Common' (Ground) and a +24V input wire for each teir?
Or does this device have a single 24V input, and a separate Ground for each teir?

It's not uncommon for automotive and industrial devices to switch on the negative side, if this is the case, we will need to make sure the power supply and transistors are connected properly.

 

[cjdelphi]

To be honest, an arduino board (or mini version with usb)

And then a bit of code to read voltage and act accordingly ...

 

[xXAmyXx]

Awesome. Well, it would appear to have built in current limiting for each 'tier' so you don't need a resistor. You can buy a 12V or a 24V version.
According to the data sheet:
** Current consumption: 50 mA per tier / buzzer **
That makes it pretty damn easy. The voltage is higher than most beginning projects, but the current is really low. You won't need a very 'strong' transistor to drive these things.
There is one question now that must be addressed to be able to determine the best method to hook this up...

Does this device have a single 'Common' (Ground) and a +24V input wire for each teir?
Or does this device have a single 24V input, and a separate Ground for each teir?

It's not uncommon for automotive and industrial devices to switch on the negative side, if this is the case, we will need to make sure the power supply and transistors are connected properly.

Wow, thats awesome to hear. The device has a common ground and a pin for each LED set.
Any recommendation for the transistor I do have a look but there seems to be about a bazillion different ones to choose from.

 

[Gryd3]

Wow, thats awesome to hear. The device has a common ground and a pin for each LED set.
Any recommendation for the transistor I do have a look but there seems to be about a bazillion different ones to choose from.

The current draw is pretty low, so the heat will be as well.
PNP-Type. They must handle 24V, and you shouldn't have a problem finding one that handles 50mA
If you have problems finding one let us know and we can track one down.

 

[xXAmyXx]

The current draw is pretty low, so the heat will be as well.
PNP-Type. They must handle 24V, and you shouldn't have a problem finding one that handles 50mA
If you have problems finding one let us know and we can track one down.

Would something like this be ok?
http://uk.rs-online.com/web/p/bipolar-transistors/0263908/

 

[Arouse1973]

Would something like this be ok?
http://uk.rs-online.com/web/p/bipolar-transistors/0263908/

Yes that should work Amy. Now you also need to have a think about some resistors to protect the base/emitter junction from over current. Also a pull-up resistor will be required from the base terminal to the + V supply terminal (emitter), this ensures the transistor is off when it should be. Looking at the data sheet for the LM3914 device it says you can programme the output current from 2 mA to 30 mA.

I would set this at maximum current so you don't have to worry about setting it again if you need to adjust the output current for some reason in the future. So you just need to work out some resistor values for base current limiting and pull up. Do you want to have a go at working that out? I don't know how much you know about electronics so I am not going insult you by telling you how to do it, unless you are unsure.

Adam

 

[xXAmyXx]

Yes that should work Amy. Now you also need to have a think about some resistors to protect the base/emitter junction from over current. Also a pull-up resistor will be required from the base terminal to the + V supply terminal (emitter), this ensures the transistor is off when it should be. Looking at the data sheet for the LM3914 device it says you can programme the output current from 2 mA to 30 mA.

I would set this at maximum current so you don't have to worry about setting it again if you need to adjust the output current for some reason in the future. So you just need to work out some resistor values for base current limiting and pull up. Do you want to have a go at working that out? I don't know how much you know about electronics so I am not going insult you by telling you how to do it, unless you are unsure.

Adam

Dont worry about insulting me Im pretty new to electronics so please explain away.

 

[Arouse1973]

Ok, it’s generally good practice to include a current limiting resistor to the base of a BJT that’s operating in common emitter mode. All though not always necessary it’s good practice to always think first base “Resistor!” Do I need one? In this case I think we need one.

If you look at the data sheet for the ZTX550 you will see something called static forward current transfer ratio. (hFE). This is one of the h-parameters for the transistor in common emitter mode. Which is the dc collector current divided by the dc base current. People also call this the gain of the transistor or “Beta”.

If you look at the graph on the data sheet that corresponds to VBE(on) v IC. You will notice that at 50 mA collector current the base emitter voltage drop is approx... 0.7 Volts. So let’s double this maximum current to 100 mA to ensure when the transistor is on its collector emitter junction can supply 50 mA. We now notice the VBE is only slightly higher at approx...0.75 Volts (probably closer to 0.78V Volts, difficult to see on graph).

So we know we need a voltage of 0.75 Volts across the base emitter terminal, so the rest of the supply voltage needs to drop across the resistor. But we need to work out what base current we need first. How do we do that? Well we don’t have any information to hand that allows us to work out the voltage drop of the base emitter diode versus current, this would be easy. You could use the formula IB=AE*(qni^2/GE)*((e^qVBE/kT)-1). But I don’t know anyone that does it this way, I certainly don’t.

With the limited data available to us, all we can do easily is take a gain figure that we think might be reasonable. If you look on the data sheet it states a minimum gain of 100 for the test conditions Ic=150 mA @ VCE = 10 Volts. So we will go with that. So if we now divide our collector current by the gain we get the base current required.

So that’s 1 mA. Now drop 24 Volts-0.75 Volts = 23.25 Volts across a base resistor which has a current of 1 mA. So 23.25 Volts / 1 mA = 23.25 K. We know we are over in our current requirements @ 100 mA and we need 50 mA. So we can use the standard value of 22 K for the base resistor, a ¼ W will be fine. Now the pull up resistor, I normally use the *10 rule of thumb, so 10 times the base resistor will be 220 K which is another standard value. Again ¼ W will be fine.

Due to the complexity of predicting the gain off the transistor at a certain currents I wouldn't bother trying, just ensure you obey the maximum ratings of the device and that you have enough minimum gain for what you want to do.


Thanks
Adam

 

[xXAmyXx]

Ok, it’s generally good practice to include a current limiting resistor to the base of a BJT that’s operating in common emitter mode. All though not always necessary it’s good practice to always think first base “Resistor!” Do I need one? In this case I think we need one.

If you look at the data sheet for the ZTX550 you will see something called static forward current transfer ratio. (hFE). This is one of the h-parameters for the transistor in common emitter mode. Which is the dc collector current divided by the dc base current. People also call this the gain of the transistor or “Beta”.

If you look at the graph on the data sheet that corresponds to VBE(on) v IC. You will notice that at 50 mA collector current the base emitter voltage drop is approx... 0.7 Volts. So let’s double this maximum current to 100 mA to ensure when the transistor is on its collector emitter junction can supply 50 mA. We now notice the VBE is only slightly higher at approx...0.75 Volts (probably closer to 0.78V Volts, difficult to see on graph).

So we know we need a voltage of 0.75 Volts across the base emitter terminal, so the rest of the supply voltage needs to drop across the resistor. But we need to work out what base current we need first. How do we do that? Well we don’t have any information to hand that allows us to work out the voltage drop of the base emitter diode versus current, this would be easy. You could use the formula IB=AE*(qni^2/GE)*((e^qVBE/kT)-1). But I don’t know anyone that does it this way, I certainly don’t.

With the limited data available to us, all we can do easily is take a gain figure that we think might be reasonable. If you look on the data sheet it states a minimum gain of 100 for the test conditions Ic=150 mA @ VCE = 10 Volts. So we will go with that. So if we now divide our collector current by the gain we get the base current required.

So that’s 1 mA. Now drop 24 Volts-0.75 Volts = 23.25 Volts across a base resistor which has a current of 1 mA. So 23.25 Volts / 1 mA = 23.25 K. We know we are over in our current requirements @ 100 mA and we need 50 mA. So we can use the standard value of 22 K for the base resistor, a ¼ W will be fine. Now the pull up resistor, I normally use the *10 rule of thumb, so 10 times the base resistor will be 220 K which is another standard value. Again ¼ W will be fine.

Due to the complexity of predicting the gain off the transistor at a certain currents I wouldn't bother trying, just ensure you obey the maximum ratings of the device and that you have enough minimum gain for what you want to do.


Thanks
Adam

Hi Adam,
Thanks for that breakdown, I think im following along for the most point, Im only really familiar with NPN transistors. Is the base not connected to the LED output on my LM3914? Therefore not getting 24V to begin with?
I may be being a complete ditz here so ill appologise in advance.