Re: Re: short circuit protection with lme49810
It takes 50uA (max 200uA) into pin 2 to turn the chip on.
so set the time constant with R1/C1, and R1+R2 limit the current.
jackinnj said:
It takes 50uA (max 200uA) into pin 2 to turn the chip on.
so set the time constant with R1/C1, and R1+R2 limit the current.
An externally hosted image should be here but it was not working when we last tested it.
and a transistor between anode of C1 and earth, driven by the over-current/temperature/ugly shirt detector at the output of the amplifier, immediately shutting down the drive to the output.
cant get much simpler than this.
the output transistors probably survive a complete shortcircuit of say 75-80V for the few microseconds it takes for the circuit to react and shutdown drive (check SOA!) to the output.
cant get much simpler than this.
the output transistors probably survive a complete shortcircuit of say 75-80V for the few microseconds it takes for the circuit to react and shutdown drive (check SOA!) to the output.
'Ugly shirt detector'???
That is an interesting phrase that I haven't heard before. Do you have an example?
That is an interesting phrase that I haven't heard before. Do you have an example?
that's an ugly "Princeton detector". I thought Princeton guys wore Brooks Bros blazers and spats.
Sent this out for prototyping last evening:
An externally hosted image should be here but it was not working when we last tested it.
How did you get LME49600's? I've heard what was probably these a month ago in a headphone amplifier at a National tech seminar around here.
Here's the prototype just back and soldered up -- a piece of PCB material serving as a heat sink:
An externally hosted image should be here but it was not working when we last tested it.
Interested to hear reports of how it works out... Hopefully will be able pick up some 49810 driver boards and lateral fet output boards off tech-diy in the future.
Hello everyone,
I am new to power amp design and need a little guidence. I have currenly taken up the task of leading a group of fellow students in designing a new power amp for my school's recording studio. We have decided to base the amp off of the LME49810 because of its simplistic design and the opportunity to explore different types of output configurations. I have worked with audio equpiment a good bit before but for some reason designing this is given me a bit of trouble (damn EE majors for being too much theory, not enough practical application).
We are starting with the basic design supplied in the application notes so some questions I have are how do you determine what supply voltage you want and how much gain do you want with the output transistors.
If it matters at all:
The amp will be driven by a DMX R-100, typ output +4dBu into 10k load
The speakers being driven are Alesis Monitor Twos (4 ohm impedance, 150W Continuous (200W peak)
Heres what I have been thinking:
I want to provide 150W into a 4ohm load, so to find the voltage and current I just use the following equations:
P=V^2 / R
P=I^2 * R
Giving me a desired nominal voltage of 24.5V and nominal current of 2.5A. So is all I need is a 30 V supply(to account for peaks and transistor loss) and a driver/output transistor gain of about 100?
I am new to power amp design and need a little guidence. I have currenly taken up the task of leading a group of fellow students in designing a new power amp for my school's recording studio. We have decided to base the amp off of the LME49810 because of its simplistic design and the opportunity to explore different types of output configurations. I have worked with audio equpiment a good bit before but for some reason designing this is given me a bit of trouble (damn EE majors for being too much theory, not enough practical application).
We are starting with the basic design supplied in the application notes so some questions I have are how do you determine what supply voltage you want and how much gain do you want with the output transistors.
If it matters at all:
The amp will be driven by a DMX R-100, typ output +4dBu into 10k load
The speakers being driven are Alesis Monitor Twos (4 ohm impedance, 150W Continuous (200W peak)
Heres what I have been thinking:
I want to provide 150W into a 4ohm load, so to find the voltage and current I just use the following equations:
P=V^2 / R
P=I^2 * R
Giving me a desired nominal voltage of 24.5V and nominal current of 2.5A. So is all I need is a 30 V supply(to account for peaks and transistor loss) and a driver/output transistor gain of about 100?
Hi Martin,
you'll need a lot more than +-30Vdc to give 150W into 4r0.
150W & 4r0 implies a maximum voltage of 34.6Vpk=24.4Vac.
The PSU will sag at maximum power so allow about 5Vdc for that.
There will also be about 2 to 4Volts lost through the amplifying chain requiring supply rails of about 34.6+5+2=+-42Vdc.
You can get this from a 30Vac transformer. Two channels will require 500VA to give 150+150W into 4ohm
For PA work in a big hall, I would be tempted to design for a bit more tolerance to abuse.
200W into 3ohm (reactive) and testing into 1r5 (resistive).
you'll need a lot more than +-30Vdc to give 150W into 4r0.
150W & 4r0 implies a maximum voltage of 34.6Vpk=24.4Vac.
The PSU will sag at maximum power so allow about 5Vdc for that.
There will also be about 2 to 4Volts lost through the amplifying chain requiring supply rails of about 34.6+5+2=+-42Vdc.
You can get this from a 30Vac transformer. Two channels will require 500VA to give 150+150W into 4ohm
For PA work in a big hall, I would be tempted to design for a bit more tolerance to abuse.
200W into 3ohm (reactive) and testing into 1r5 (resistive).
Hello
Let me give you a few tips as they come out of my head.
For what you are saying ( EE students), the design around LME49810 is a great choice. You can concentrate on the main aspect of power amplifiers: the output stage.
This is not a trivial task.
I would first evaluate the constraints.
If possible I would avoid the packaging constraints by making the device a large open frame with for example a plexiglas protection plate.
This will allow you to reach without constraints:
optimal cooling, optimal LF magnetic shielding ( by distance and orientation of transformer).
Then I would start from the heatsink. If money permits, Fisher elektronic makes great extruded compact forced air cooled heatsinks with machined faces to allow good thermal contacts. The heatsink is a rectangular block with integrated ventilator and two opposit mounting faces for the power transistors. Very good theta ( 0.1°/watt). You can mount the pcb's with the LM on the other 2 sides, one per channel.
Then I would estimate the power dissipated by the output transistors on a reactive load of 45°.
A good rule of thumb is : average power dissipated in transistors = 60% of power delivered. Another good rule is keep average T° of heatsink lower than 70°C or better 60°C.
Then You have to make the choice of output transistors and their number in parallel. I would use BGT's from Onsemi ( thermaltrack with diode integrated)
Then you need to estimated the +-Vcc's and the SOA aspect's related to output protection and number of transistors. and peak power in reactive load
The Vcc is an estimate because voltage drops at full power (so clipping) depends on a lot of unknows: ac dips, transformer efficiency, rectifier and fuses voltage drops, capacitor filtering sizes and output headroom at full power before clipping transistors. Because it is an internal project and because of LME flexibility I would start for your 150w/4ohms with a 45Volts dc. Yes this is a lot more than the simple calculation of : average power ( 150W) = Vcc^2/2R ( Vcc is peak amplitude!!!).
Now that Vcc is estimated you can find the voltage of the ac transformer ( see some rule of tumbs on toroidal transformer vendors site like powertronix). Tweaking later can be done by adding or removing some turns on the power transformer with enameled pair of wires of right AGW.
Now you need to estimate the VA rating of the transformer. A good rule of thumb is minimum twice the delivered average power. In this case 300VA or more if money permits.
Now you need to estimate the peak power and peak current in load therefore in output transistor. This will allow you to estimate the type and number of output transistors in //. In your case I would guess 4NPN's and 4PNP's (NJL 3281, 1302).
Peak power dissipated on reactive load can be a lot more than on resistive. A good estimate would be: take a worse case minimum load impedance module ( of complex impedance) half of the rated 4 ohms and a worse case of 45° phase shift that the load could present at a certain frequency. These are estimates because music is transient ( see Benjamin in the AES papers for a good overview). Estimate the current per single transistor ( divide by the number of transistors)
With this load, draw on the SOA spec the elliptic load line and see how many transistors you need to accomodate a non crossing of the SOA curve ( you will be surprise). If necessary, put more transistors.
Do not forget to derate the curve at maximum case temperature that you can now estimated ( if your heatsink remains at 60°C worse case) knowing the peak power per transistor and the theta of case to heatsink ( about 1°c/watt if very good mounting).
This procees requires some interation.
You are now able to design a good non intrusive SOA protection device ( see Leach papers and book or very good paper from Mikek diy member.
You still need to estimate a minimum capacitor filtering size. Some rule of thumb is 2 joules of power stored per 10 watt ouput power. Energy = Vcc^2 C/2 joules
Now you need a topology and to tune the biasing of your output transistors. I would suggest to use the triple output stage ( Locanthi) that is well described by Leach. The first emitter follower of this triple stage is the output transistor of the LME.
Use a 0.1 ohm or 0.15 ohm in each emitter of the output transistors. They help in thermal stability ( thermal runaway avoidance) and current equilisation in // transistors). These are power resistors !!! With 4 ohm load try to match the ouput transistors for beta if you can or use same batch from supplier and then go for 0.1 ohm. This will give you better crossover distortion in case of optimal biasing but weaker thermal stability.
Very good heatsink and thermaltrack helps there but you must try. With the resistance chosen, ajust optimal bias current in output transistors for minimal crossover distortion: this is gmRe =1 where Re is emitter resistor and gm is transconductance of output transistor. this is equal to IcRe=Vt=26mV, so ajust current ( Vbe multiplier or thermaltrack diodes with some potentiometer) until V on Re is 26mV
This is theoretical because Re should include the internal resistance of the transistor + Rb/Hfe where Rb is the resistance in the base divided by Hfe of transistor. On top of that the 26mV is Vt = kt/q which is temp dependant.
The best is to allow the transistor to reach average operating temperature and bias then for minimum distortion using a spectrum analyzer. The formula gives you a good approach.
Your SOA protection device is also a short circuit protection. An handy way of using it is to activate the mute pin of LME via optocoupler and some delay before restart.
You need perhaps a DC servo to avoid using coupling capacitors and a DC protection that activates an ouput relay to protect your speakers in case of power transistor failure or any other in the signal chain. A good DC protection circuit can be found in SELF's book. I would switch the relay with an optocoupler.
The relay contacts should be well investigated for no distortion.
Even then contact protection is mandatory. Because you will switch a reactive load , spikes are generated. The usual two clamping diodes seen in // with collector emitter or output transistors should be placed after the contact( speaker side) between +Vcc/ speaker ouput and -Vcc speaker output. This will clamp the surge to VCC. Now you know that the worse case voltage on the contacts will be less than 300V wich is the start of glow arcing.
You are still left with metallic arc discharge on your contact. A good rule is that the voltage gradient ( dV/dt) between contacts remains under 1V/microsecond to avoid this arc. Using a large bipolar capacitor as contact protection in // with them will do it. C dV/dt=I peak load. Then, C greater than I load /1.000.000 so for 100A peak C> 100microF !!
With this capacitor you still need to protect the contatcts when closing them. This is usually done via a resistor in serie with Cbut this resistor has a detrimental effect on opening because a voltage equal to R times I peak will be present and arcing will happen at the begining.. In the particular case of switching speakers we can switch speakers on with mute of LME still engaged therefore there is no voltage on the contacts on closing and they will never be degraded. This is very important and overlooked or avoided by not using relays.
Finally, you get a state of the art amplifier !!
A final hint/ I would use BJT's because of very accurate model made by Andy_c , diy member for onsemi thermaltrack's. Then your guys should then play with LTspice and estimate operation of the amplifier. They will be able to show and use their theoretical knowlege before learning practical through this excellent project🙂
Cheers
Jean-Pierre
Let me give you a few tips as they come out of my head.
For what you are saying ( EE students), the design around LME49810 is a great choice. You can concentrate on the main aspect of power amplifiers: the output stage.
This is not a trivial task.
I would first evaluate the constraints.
If possible I would avoid the packaging constraints by making the device a large open frame with for example a plexiglas protection plate.
This will allow you to reach without constraints:
optimal cooling, optimal LF magnetic shielding ( by distance and orientation of transformer).
Then I would start from the heatsink. If money permits, Fisher elektronic makes great extruded compact forced air cooled heatsinks with machined faces to allow good thermal contacts. The heatsink is a rectangular block with integrated ventilator and two opposit mounting faces for the power transistors. Very good theta ( 0.1°/watt). You can mount the pcb's with the LM on the other 2 sides, one per channel.
Then I would estimate the power dissipated by the output transistors on a reactive load of 45°.
A good rule of thumb is : average power dissipated in transistors = 60% of power delivered. Another good rule is keep average T° of heatsink lower than 70°C or better 60°C.
Then You have to make the choice of output transistors and their number in parallel. I would use BGT's from Onsemi ( thermaltrack with diode integrated)
Then you need to estimated the +-Vcc's and the SOA aspect's related to output protection and number of transistors. and peak power in reactive load
The Vcc is an estimate because voltage drops at full power (so clipping) depends on a lot of unknows: ac dips, transformer efficiency, rectifier and fuses voltage drops, capacitor filtering sizes and output headroom at full power before clipping transistors. Because it is an internal project and because of LME flexibility I would start for your 150w/4ohms with a 45Volts dc. Yes this is a lot more than the simple calculation of : average power ( 150W) = Vcc^2/2R ( Vcc is peak amplitude!!!).
Now that Vcc is estimated you can find the voltage of the ac transformer ( see some rule of tumbs on toroidal transformer vendors site like powertronix). Tweaking later can be done by adding or removing some turns on the power transformer with enameled pair of wires of right AGW.
Now you need to estimate the VA rating of the transformer. A good rule of thumb is minimum twice the delivered average power. In this case 300VA or more if money permits.
Now you need to estimate the peak power and peak current in load therefore in output transistor. This will allow you to estimate the type and number of output transistors in //. In your case I would guess 4NPN's and 4PNP's (NJL 3281, 1302).
Peak power dissipated on reactive load can be a lot more than on resistive. A good estimate would be: take a worse case minimum load impedance module ( of complex impedance) half of the rated 4 ohms and a worse case of 45° phase shift that the load could present at a certain frequency. These are estimates because music is transient ( see Benjamin in the AES papers for a good overview). Estimate the current per single transistor ( divide by the number of transistors)
With this load, draw on the SOA spec the elliptic load line and see how many transistors you need to accomodate a non crossing of the SOA curve ( you will be surprise). If necessary, put more transistors.
Do not forget to derate the curve at maximum case temperature that you can now estimated ( if your heatsink remains at 60°C worse case) knowing the peak power per transistor and the theta of case to heatsink ( about 1°c/watt if very good mounting).
This procees requires some interation.
You are now able to design a good non intrusive SOA protection device ( see Leach papers and book or very good paper from Mikek diy member.
You still need to estimate a minimum capacitor filtering size. Some rule of thumb is 2 joules of power stored per 10 watt ouput power. Energy = Vcc^2 C/2 joules
Now you need a topology and to tune the biasing of your output transistors. I would suggest to use the triple output stage ( Locanthi) that is well described by Leach. The first emitter follower of this triple stage is the output transistor of the LME.
Use a 0.1 ohm or 0.15 ohm in each emitter of the output transistors. They help in thermal stability ( thermal runaway avoidance) and current equilisation in // transistors). These are power resistors !!! With 4 ohm load try to match the ouput transistors for beta if you can or use same batch from supplier and then go for 0.1 ohm. This will give you better crossover distortion in case of optimal biasing but weaker thermal stability.
Very good heatsink and thermaltrack helps there but you must try. With the resistance chosen, ajust optimal bias current in output transistors for minimal crossover distortion: this is gmRe =1 where Re is emitter resistor and gm is transconductance of output transistor. this is equal to IcRe=Vt=26mV, so ajust current ( Vbe multiplier or thermaltrack diodes with some potentiometer) until V on Re is 26mV
This is theoretical because Re should include the internal resistance of the transistor + Rb/Hfe where Rb is the resistance in the base divided by Hfe of transistor. On top of that the 26mV is Vt = kt/q which is temp dependant.
The best is to allow the transistor to reach average operating temperature and bias then for minimum distortion using a spectrum analyzer. The formula gives you a good approach.
Your SOA protection device is also a short circuit protection. An handy way of using it is to activate the mute pin of LME via optocoupler and some delay before restart.
You need perhaps a DC servo to avoid using coupling capacitors and a DC protection that activates an ouput relay to protect your speakers in case of power transistor failure or any other in the signal chain. A good DC protection circuit can be found in SELF's book. I would switch the relay with an optocoupler.
The relay contacts should be well investigated for no distortion.
Even then contact protection is mandatory. Because you will switch a reactive load , spikes are generated. The usual two clamping diodes seen in // with collector emitter or output transistors should be placed after the contact( speaker side) between +Vcc/ speaker ouput and -Vcc speaker output. This will clamp the surge to VCC. Now you know that the worse case voltage on the contacts will be less than 300V wich is the start of glow arcing.
You are still left with metallic arc discharge on your contact. A good rule is that the voltage gradient ( dV/dt) between contacts remains under 1V/microsecond to avoid this arc. Using a large bipolar capacitor as contact protection in // with them will do it. C dV/dt=I peak load. Then, C greater than I load /1.000.000 so for 100A peak C> 100microF !!
With this capacitor you still need to protect the contatcts when closing them. This is usually done via a resistor in serie with Cbut this resistor has a detrimental effect on opening because a voltage equal to R times I peak will be present and arcing will happen at the begining.. In the particular case of switching speakers we can switch speakers on with mute of LME still engaged therefore there is no voltage on the contacts on closing and they will never be degraded. This is very important and overlooked or avoided by not using relays.
Finally, you get a state of the art amplifier !!
A final hint/ I would use BJT's because of very accurate model made by Andy_c , diy member for onsemi thermaltrack's. Then your guys should then play with LTspice and estimate operation of the amplifier. They will be able to show and use their theoretical knowlege before learning practical through this excellent project🙂
Cheers
Jean-Pierre
Or start smaller if the above scares you. Get some LM3886 chips or LM4780s. Run two channels in parallel with a good heat sink (1.5C/W or lower). Build a simple unregulated supply with a torriod transformer, fully integrated bridge, and at least 6,800uF per rail and see how you like it. The set up will not get you a full 150W all day long with sine waves but it will be close enough. Check out the datasheets for the parts listed reading them completely then download the design guide National offers to do some calculations. Also read AN-1192 for good info. Start with this first and what you learn from thermals, power supply, GND'ing connections, hum and noise issues, case design, PCB design and modularity, interconnects and hardware will be a huge help without breaking the bank or getting in over you head. Then you might be ready for the LME49810 or one of it's cousins to build what you really want. You'd be surprised how loud a good solid 120W of power can sound and the LM chips are reasonably good sound quality. Here is the link to the design guide. Down load the instructions as well. Links are in the middle of the page.
http://www.national.com/appinfo/audio/
-SL
http://www.national.com/appinfo/audio/
-SL
To be complete, some practical tips on layout and grounding.
You should avoid problems (coupling) within the amplifier and coupling from outside.
The rules to follow are :
1 understand where currents are flowing,
then minimize the size of the loop formed by a current and its return current. The return current will follow the path of least impedance (not least resistance) towards its origin . Therefore many currents will follow the path of minimum loop ( minimum inductance) and then radiate less (mutual magnetic coupling) in other loops. This is very important for the loop made from postive current flowing from decoupling cap power supply at output transistor level to NPN power transistor to load and back to ground of decoupling cap. This current has a lot of high frequencies and this loop should be minimized and twisted. The same for the negative part.
2 Use a single point grounding at audio frequencies per group of currents : one group for load output currents and decoupling caps at power transistor level, one group for low level audio ( currents for the LME and its decoupling caps.
Single point grounding means each current has its own return to ground point and back to origin.
3 put a small resistance between Vcc of power transistors and Vcc's decoupling caps of LME. This will force the currents to stay in their original loops. This will avoid ground loop coupling.
4 connect each ground to main power supply smoothing caps central point. This point should be individually connected to the central tap of transformer and to the chassis safety ground. A small ( one ohm) resistance can be put in the ground connection coming from the ouput transistors ground point. The usefulness of this resistor should be tested by measurements. Ac noise filters should be connected to the chassis at the entrance.
5 make a clean ground piece (large isolated pcb strip) where the external shield of of incoming signal and input filtering caps should be connected at the entrance. This clean ground way not carry internal currents and should be connected with a strip to the chassis safety ground. This connection will divert any external ground loop currents or shield picked up noise from entering the circuit.
two more tips:
avoid proximity of ground to feedback node of LME (- INPUT). This would create a parasitic capacitance with oscillation problems.
Be carefull with the layout of the feedback resistor. The current comes from the LME output transistor, flows in the bases of the emitter followers to the next one, through the power resistors and then back via the feedback resistor. This loop should be minimized and the connection of the feedback resistor track to the output should be after the connection point of the power resistors.
Comments are welcome but this is a classical approach.
Jean-Pierre
You should avoid problems (coupling) within the amplifier and coupling from outside.
The rules to follow are :
1 understand where currents are flowing,
then minimize the size of the loop formed by a current and its return current. The return current will follow the path of least impedance (not least resistance) towards its origin . Therefore many currents will follow the path of minimum loop ( minimum inductance) and then radiate less (mutual magnetic coupling) in other loops. This is very important for the loop made from postive current flowing from decoupling cap power supply at output transistor level to NPN power transistor to load and back to ground of decoupling cap. This current has a lot of high frequencies and this loop should be minimized and twisted. The same for the negative part.
2 Use a single point grounding at audio frequencies per group of currents : one group for load output currents and decoupling caps at power transistor level, one group for low level audio ( currents for the LME and its decoupling caps.
Single point grounding means each current has its own return to ground point and back to origin.
3 put a small resistance between Vcc of power transistors and Vcc's decoupling caps of LME. This will force the currents to stay in their original loops. This will avoid ground loop coupling.
4 connect each ground to main power supply smoothing caps central point. This point should be individually connected to the central tap of transformer and to the chassis safety ground. A small ( one ohm) resistance can be put in the ground connection coming from the ouput transistors ground point. The usefulness of this resistor should be tested by measurements. Ac noise filters should be connected to the chassis at the entrance.
5 make a clean ground piece (large isolated pcb strip) where the external shield of of incoming signal and input filtering caps should be connected at the entrance. This clean ground way not carry internal currents and should be connected with a strip to the chassis safety ground. This connection will divert any external ground loop currents or shield picked up noise from entering the circuit.
two more tips:
avoid proximity of ground to feedback node of LME (- INPUT). This would create a parasitic capacitance with oscillation problems.
Be carefull with the layout of the feedback resistor. The current comes from the LME output transistor, flows in the bases of the emitter followers to the next one, through the power resistors and then back via the feedback resistor. This loop should be minimized and the connection of the feedback resistor track to the output should be after the connection point of the power resistors.
Comments are welcome but this is a classical approach.
Jean-Pierre
SpittinLLama said:
Hi SpittinLLama
I fully agree with what you are saying.
But if these guys are EE's wanting to experiment with output stage design they should not be afraid by my comments.
If they are, then your approach is the best but forget about experimenting with output stage.
Your comments are welcome anyway.
Cheers
JP
Thanks everyone for the comments, especially Jean-Pierre for the extremely detailed answer! It looks like I have a bunch of studying to do this weekend with your responses and Im sure I will be back with some questions 🙂 .
After quickly reading through I have a simple question.
I have often seen around here and audio sites that P = Vcc^2/2R. Where does the '2R' come from? I thought P = V^2/R? Im sure i am missing something obvious.
-Jamie
After quickly reading through I have a simple question.
I have often seen around here and audio sites that P = Vcc^2/2R. Where does the '2R' come from? I thought P = V^2/R? Im sure i am missing something obvious.
-Jamie