Hi friends
The Leach amp has 2 transistors (Q8,Q9) for VAS stage protection that should conduct when VAS (Q12,Q13) current is more than 20mA (for example when protection is active).
But with diff. stage current of 3.5mA and selected VAS current gain (R23/R11 ratio, 1,2k/360) , VAS current will never more than 14mA, and Q8,Q9 will never conduct! If even Q1 or Q3 will shorted, as long as Q5, Q6 are OK, current via R11, R12 will not more 3.5mAx2 and VAS will not overloaded.
I simulated this and found that even if VAS output is shorted to ground and amplifier input stage is heavy overloaded by 10V sin. signal, VAS current never goes above 12mA and Vbe of Q8, Q9 is always less than 360mv - exact as I thought...
But if M.Leach added these transistors - they are required. Where is my mistake ?
http://users.ece.gatech.edu/~mleach/lowtim/graphics/ckt.pdf
The Leach amp has 2 transistors (Q8,Q9) for VAS stage protection that should conduct when VAS (Q12,Q13) current is more than 20mA (for example when protection is active).
But with diff. stage current of 3.5mA and selected VAS current gain (R23/R11 ratio, 1,2k/360) , VAS current will never more than 14mA, and Q8,Q9 will never conduct! If even Q1 or Q3 will shorted, as long as Q5, Q6 are OK, current via R11, R12 will not more 3.5mAx2 and VAS will not overloaded.
I simulated this and found that even if VAS output is shorted to ground and amplifier input stage is heavy overloaded by 10V sin. signal, VAS current never goes above 12mA and Vbe of Q8, Q9 is always less than 360mv - exact as I thought...
But if M.Leach added these transistors - they are required. Where is my mistake ?
http://users.ece.gatech.edu/~mleach/lowtim/graphics/ckt.pdf
Hi,
the VAS protection is required. Your simulation must be wrong.
The output protection shorts the driver base to the output rail.
The VAS when driven by the input signal still tries to vary the output voltage to where it thinks it should be. The VAS sends increasing current to the shorted base trying in vain to lift the voltage back up to where it should be. It kills itself!
Most schematics omit this protection. Unfortunately VAS protection does not save the drivers. How do we achieve that?
Theoretically the maximum VAS current is twice the VAS bias current. I changed the protection resistors to allow a peak VAS current of three times VAS bias current to ensure the VAS never entered self protection mode with any valid audio signal.
The Leach version of the IV output protection has omitted the rail to base resistor. The Leach protection locus is seriously flawed as a result. Add the missing resistor trace & pads to your PCB.
the VAS protection is required. Your simulation must be wrong.
The output protection shorts the driver base to the output rail.
The VAS when driven by the input signal still tries to vary the output voltage to where it thinks it should be. The VAS sends increasing current to the shorted base trying in vain to lift the voltage back up to where it should be. It kills itself!
Most schematics omit this protection. Unfortunately VAS protection does not save the drivers. How do we achieve that?
Theoretically the maximum VAS current is twice the VAS bias current. I changed the protection resistors to allow a peak VAS current of three times VAS bias current to ensure the VAS never entered self protection mode with any valid audio signal.
The Leach version of the IV output protection has omitted the rail to base resistor. The Leach protection locus is seriously flawed as a result. Add the missing resistor trace & pads to your PCB.
I agree...
I still agree..
Why? How?
It's right if there no emitter resistor, or it's too small - 10...47Ohm.
But in Leach amp diff stage can't deliver more 3.5mA and thanks to feedback via R23 or R24, VAS cant amplify it to more than 12-14mA, and power dissipation on VAS tr-r will less than 1W. I Don't see a way how it can kill itself!
Hi,
the signal voltage on the base of the VAS goes low (current sunk to LTP) opening up the VAS to pass more current.
The current route is: Rail thro first protection monitor R, thro Re, thro VAS, thro forward biased diode, through open output protection Q, to output rail.
The Vce of the prot Q could be 0.1V, Vf of diode coud be 0.7V.
Output rail connected to 0r1 and flowing 20Apk is sitting at 2V.
The supply rail has sagged to 50V (20pk flowing to shorting load).
The voltage across [VAS+resistors] is about 50-[0.1+0.7+2.0]~=47V.
Calculate the VAS dissipation for a variety of unprotected emitter currents. (65mA through this assembly is about 1.5W through each of Re and Q)
If the front end is supplied from an independant supply at 58 to 60Vdc the situation is slightly worse.
BTW,
it's the NFB trying to match input voltage conditions on it's two inputs that drives the VAS base lower and lower to send more and more current through the VAS. If the VAS current limits with 3mA extraction from it's base and a current gain of 100, that's still 300mA at 47V across the emitter resistors+Vce. When the VAS protection triggers it connects the Rail to VAS base and LTP sinks current through the prot Q. No harm done.
the signal voltage on the base of the VAS goes low (current sunk to LTP) opening up the VAS to pass more current.
The current route is: Rail thro first protection monitor R, thro Re, thro VAS, thro forward biased diode, through open output protection Q, to output rail.
The Vce of the prot Q could be 0.1V, Vf of diode coud be 0.7V.
Output rail connected to 0r1 and flowing 20Apk is sitting at 2V.
The supply rail has sagged to 50V (20pk flowing to shorting load).
The voltage across [VAS+resistors] is about 50-[0.1+0.7+2.0]~=47V.
Calculate the VAS dissipation for a variety of unprotected emitter currents. (65mA through this assembly is about 1.5W through each of Re and Q)
If the front end is supplied from an independant supply at 58 to 60Vdc the situation is slightly worse.
BTW,
it's the NFB trying to match input voltage conditions on it's two inputs that drives the VAS base lower and lower to send more and more current through the VAS. If the VAS current limits with 3mA extraction from it's base and a current gain of 100, that's still 300mA at 47V across the emitter resistors+Vce. When the VAS protection triggers it connects the Rail to VAS base and LTP sinks current through the prot Q. No harm done.
AndrewT,
Sorry, but some Your assumptions are incorrect, because You forgot Rc of diff. stage (R11 or R12). You should look on VAS tr-r "in circuit" and not like it is on the table.
Incorrect! In this case voltage across R11 should be 47+0.7=47.7V. And to obtain it, current thru R11 must be 47.7V/1.2k~=40mA! It's impossible, it's limited by 3.5mA.
Twice incorrect!
1. In this circuit current gain is not 100! It's equal to (Diff. stage Rc)/(VAS Re) = 1.2k/(360+30) = 3 ONLY!!! And max. VAS current is 3.5*3 = 10.5mA.
2. VAS base current is much less than R11 current! Almost all Q1 current flow via R11 and not via VAS base!
Look on it in other way: let's VAS base current is negligible. Max VAS input voltage will 3.5mA*1.2k=4.2V. And with 2.5mA/V transconductanse of VAS, we still get 10.5mA! And voltage drop across VAS Re is always less than 4.2-0.7 = 3.5v!
Thank You anyway - during this discussion I get more and more sure that VAS protection in Leach amp is needless.
Sorry, but some Your assumptions are incorrect, because You forgot Rc of diff. stage (R11 or R12). You should look on VAS tr-r "in circuit" and not like it is on the table.
Incorrect! In this case voltage across R11 should be 47+0.7=47.7V. And to obtain it, current thru R11 must be 47.7V/1.2k~=40mA! It's impossible, it's limited by 3.5mA.
Twice incorrect!
1. In this circuit current gain is not 100! It's equal to (Diff. stage Rc)/(VAS Re) = 1.2k/(360+30) = 3 ONLY!!! And max. VAS current is 3.5*3 = 10.5mA.
2. VAS base current is much less than R11 current! Almost all Q1 current flow via R11 and not via VAS base!
Look on it in other way: let's VAS base current is negligible. Max VAS input voltage will 3.5mA*1.2k=4.2V. And with 2.5mA/V transconductanse of VAS, we still get 10.5mA! And voltage drop across VAS Re is always less than 4.2-0.7 = 3.5v!
Thank You anyway - during this discussion I get more and more sure that VAS protection in Leach amp is needless.
Hi,
I hope you're right.
Did you use a sim to get to these earlier conclusions or did it come about by thinking through the failure mechanism?
What is the limiting value of Vre that provides this automatic protection? Leach's 360r+30r drops a very high Vre in normal bias operation.
Yes, I have experienced exactly what you describe, talking through a problem forces one to describe it coherently for the listener. That process clears the mind and opens it, often to find it's own conclusions without the listener even replying. I use it at work.
I hope you're right.
Did you use a sim to get to these earlier conclusions or did it come about by thinking through the failure mechanism?
What is the limiting value of Vre that provides this automatic protection? Leach's 360r+30r drops a very high Vre in normal bias operation.
Yes, I have experienced exactly what you describe, talking through a problem forces one to describe it coherently for the listener. That process clears the mind and opens it, often to find it's own conclusions without the listener even replying. I use it at work.
Adjust the protection and you will listen better sound
It is there to protect...needed.. but has a sub product very interesting.... sound!
As amplifiers are made for audio, for sound reproduction, and not exactly to calculator machines satisfaction...then.....
It is needed, cannot remove?.... maybe you can adjust.... you will reduce the transistor protection and increase your listening pleasure protection.
Remember that in the first World War...people use to attack walking slowly into the enemy.... this was the technique use those early days....they received shots.... thousands were killed to respect the traditional technical rules that existed.
regards,
Carlos
It is there to protect...needed.. but has a sub product very interesting.... sound!
As amplifiers are made for audio, for sound reproduction, and not exactly to calculator machines satisfaction...then.....
It is needed, cannot remove?.... maybe you can adjust.... you will reduce the transistor protection and increase your listening pleasure protection.
Remember that in the first World War...people use to attack walking slowly into the enemy.... this was the technique use those early days....they received shots.... thousands were killed to respect the traditional technical rules that existed.
regards,
Carlos
Re: Adjust the protection and you will listen better sound
I think you have omitted a condition to your header.
Protection is audible if it interupts or contaminates a valid audio signal.
If the protection NEVER contaminates a valid audio signal then it cannot be audible for any valid audio outputs.
There should be no need to adjust the protection to be able to listen to a better sound when it is properly designed.
Protection should only interupt non valid audio signals. eg, signal clipping, and/or shorted output, and/or wrong speaker load, and/or loss of mains supply, and/or DC input, etc.
Hi Carlos,destroyer X said:
I think you have omitted a condition to your header.
Protection is audible if it interupts or contaminates a valid audio signal.
If the protection NEVER contaminates a valid audio signal then it cannot be audible for any valid audio outputs.
There should be no need to adjust the protection to be able to listen to a better sound when it is properly designed.
Protection should only interupt non valid audio signals. eg, signal clipping, and/or shorted output, and/or wrong speaker load, and/or loss of mains supply, and/or DC input, etc.
Hi Lumba,
let's go to an extreme as an example.
The output relay is normally triggered after start-up to give the amp time to settle down.
Now use this same relay controlled by a monitoring circuit that looks for excessive (non valid audio signal) output current. The relay ONLY opens when a fault causes the excessive output current.
Now tell me the monitoring circuit is altering the signal quality.
let's go to an extreme as an example.
The output relay is normally triggered after start-up to give the amp time to settle down.
Now use this same relay controlled by a monitoring circuit that looks for excessive (non valid audio signal) output current. The relay ONLY opens when a fault causes the excessive output current.
Now tell me the monitoring circuit is altering the signal quality.
Hi Andrew,
it was my instant reaction to the contents of your post(#9).
Many of those protection circuits have a considerable impact on the sound.
You see, all components have some impact. We can not fool physics, can we. Just ourselves.
I can not say you are logically wrong.
Something else: I have been thinking too - about the base current issue...
it was my instant reaction to the contents of your post(#9).
Many of those protection circuits have a considerable impact on the sound.
You see, all components have some impact. We can not fool physics, can we. Just ourselves.
I can not say you are logically wrong.
Something else: I have been thinking too - about the base current issue...
To AndrewT :
I'm desining a high-power amplifier now, and I double and triple check every point in my design before I go to PCB production. Initially "it come about by thinking through the failure mechanism", then I didn't believe and tested it with Spice - results were the same.
VAS transconductanse is equal to 1/Re, so in Leach amp it's set to 1/390r~=2.5ma/V. Such transconductanse value makes VAS behavior predictable and almost independent from transistor parameters deviations, but there is no free cheese exept mousetrap - You will pay by voltage drop.
But my analyzing shows that this voltage drop is not major factor that reduces max. output voltage - for instance 0.33r output stage resistors "contribute" much more voltage drop in real amplifier.
So IMO There is no limiting value, in +/-56V rails version I'll use 360r, in +/-112V version I'll use 200r and max VAS current will be ~20mA.
To Lumba Ogir:
I'm desining a high-power amplifier now, and I double and triple check every point in my design before I go to PCB production. Initially "it come about by thinking through the failure mechanism", then I didn't believe and tested it with Spice - results were the same.
VAS transconductanse is equal to 1/Re, so in Leach amp it's set to 1/390r~=2.5ma/V. Such transconductanse value makes VAS behavior predictable and almost independent from transistor parameters deviations, but there is no free cheese exept mousetrap - You will pay by voltage drop.
But my analyzing shows that this voltage drop is not major factor that reduces max. output voltage - for instance 0.33r output stage resistors "contribute" much more voltage drop in real amplifier.
So IMO There is no limiting value, in +/-56V rails version I'll use 360r, in +/-112V version I'll use 200r and max VAS current will be ~20mA.
To Lumba Ogir:
I dont think that prof. M.Leach added protection circuits to improve sound quality, and without it the amp will sound worse
I apologize for bringing this old thread up, but it seems to me Prof. Leach's VAS protection does indeed serves purpose when the amp is heavily over-driven.
It is true that under certain over driving input level the tail current of the input pair limits the maximum VAS emitter current and the VAS protection will never engage, despite the fact that increased over-driving voltage increases the tail current in a dynamic fashion when the pair is driven out of the linear zone.
However when the input pair receives a very high driving level that the b-e junction in the inverting side transistor gets reversely biased to breakdown like a 6v or so zener, the tail current may no long be a limiting factor to the output voltage of the input stage, and the VAS can see an emitter current that results the protection to engage.
My calculation indicates, assuming a 6V reverse breakdown voltage at the emitter junction of the input transistors, an instantaneous voltage of 14.4V at the input can result in about 7mA collector current at the non-inverting input transistor, and that's the point the VAS reaches 20mA that the protection circuit was designed to watch for.
14.4V is a horrendous voltage to feed an amp, it is not impossible, however.
It is true that under certain over driving input level the tail current of the input pair limits the maximum VAS emitter current and the VAS protection will never engage, despite the fact that increased over-driving voltage increases the tail current in a dynamic fashion when the pair is driven out of the linear zone.
However when the input pair receives a very high driving level that the b-e junction in the inverting side transistor gets reversely biased to breakdown like a 6v or so zener, the tail current may no long be a limiting factor to the output voltage of the input stage, and the VAS can see an emitter current that results the protection to engage.
My calculation indicates, assuming a 6V reverse breakdown voltage at the emitter junction of the input transistors, an instantaneous voltage of 14.4V at the input can result in about 7mA collector current at the non-inverting input transistor, and that's the point the VAS reaches 20mA that the protection circuit was designed to watch for.
14.4V is a horrendous voltage to feed an amp, it is not impossible, however.
None of these discussions are about the behaviors of the amp with the input stage being driven far beyond its linear range, or +1V ~ -1V differential unless I missed something. Within such a "normal" driving level the VAS needs no protection as there is no chance for the VAS transistors to see a collector current anywhere near 20mA that might poke its SOA envelope.
Therefore, IMHO, the VAS protection was designed for much more severe, abnormal scenarios. One of the scenarios that I can think of is that two bad things take place at the same time: 1) a high level input that can cause input transistors' emitter junction to reversely breakdown hence allowing much larger output at the differential pair to drive the VAS, and 2) meanwhile the output stage over-current protection engages and will sink/source the excessive current from/to the VAS transistors when they are so driven.
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