Dx Blame MKIII-Hx - Builder's thread

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Post your pictures here please.

Group buy for this amplifier is in it's last days, so, i have opened this thread in order to post some final instructions, updated schematic, adjustments and some tips and tricks.

First of all, the schematic:

It has lifted ground (low noise ground) in the input..that 10 ohms resistance (1 watt or more) lifts the input ground.

I have removed that...but Alex though i have missed and he has installed in his layout... in the reality it was not missed...i have decided to avoid the lifted ground because this demands special connectors to the input jack, insulated connectors not to short the chassis ground with the lifted ground.... because this deactivates the low noise ground producing a bypass in that input lifting ground resistor.

Also if you use a metalic potentiometer into your input, and also a metalic knob, then you will produce noises (from mains captured by your body) when touched...this demands you ground the potentiometer...and you cannot use the ground comes from the shielded cables because they are lifted ground and will make electrical contact (continuity) with the chassis.... and our chassis is grounded..the secondary center tap goes attached to ground... so.... once again you deactivate.....because of that i have removed..and Alex mm installed....well...this is not bad...can be this way too... it is all right.

But that resistance was not in the schematic.....now it is in the schematic.

regards,

Carlos
 

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I will show you the previous schematic ground

and the new ground scheme that matches the layout made by Alex mm.

regards,

Carlos
 

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How to monitor your current, the stand by current.

My suggestion is to adjust your trimpot (preset is 500 ohms...or midway) in order to measure 1 milivolt above the power transistor's emitter resistor.

But for protection, as you can have mistakes in your board (errors and mistakes while building) install 100 ohms resistance - 5 watts, in series with your rail voltages..one resistance to each rail..so, they interrupt the connection and forces supply current to cross them...when supply current crosses develops a voltage drop over these resistances...and this can be measured in volts DC....you should read 4.4 volts... and this is 44 miliamperes to the positive rail...if you have different readings, then adjust your trimpot in order to obtain this voltage.

In the image you see a voltimeter and an amperometer together this resistance...you will use only a DC voltimeter...avoid amperometer.

Current is the voltage divided by the resistance...so....4.4 volts divided by 100 (100ohms) is 0.044 and this is 0.044A or 44 miliamperes.

Why we do this?

If you have mistakes in your board..or if your trimpot is misadjusted (the preset for first moment of power on is 500 ohms), then you may have huge current (max. 350 miliamperes with your trimpot in maximum resistance that is 1K).or much more if you have mistakes..current can be 10 amperes or more if you have shorts in your board caused by bad soldering method... solder balls shorting....melted solder connecting leads and so on....this will overheat and may burn your protective resistance......and it is better this resistor to be burned then you face board damaged with traces melted and board carbonized....the resistance goes first..as 10A represents 1000 watts and we will not install a 1000 watts resistance there because of that.

With normal current of 44 miliamps, then your resistor will be dissipating 3 watts...so...your resistor will be hot..more than warm...but if something goes wrong, the voltage will be higher and this resistor will smoke.

Check off set...it must be small... 1 to 25 milivolts will be fine.

Both.... the offset and the bias trimpot adjusting...operating..shows your amplifier is fine.... if offset is strange (high) and trimpot does not adjust properly, then check your construction because you made mistakes (this happens... a lot....humans does that)

regards,

Carlos
 

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Byron made the kindness to produce parts list.

I have checked but i use to make mistakes (eyes not that good anymore)... help me to check this.

I have removed supplier name, but you may have this making contact with Byron if you want....i want correct parts, not matter what the supplier is.

regards,

Carlos
 

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You can use 1 watt, 2 watts, 3 watts or 5 watts

This is another fashion thing that is used since the dawn of the electronic civilization and no one discusses....and everybody think it must be 5 watts.

In normal operation you will never have high watt dissipation there...... the only condition you may have high wattage there is if your power transistor blow out and produces a short from colector to emitter...then you will have 64 volts from the supply crossing it...136A or something alike that will flow...and now you will have the need of a 870 watts resistor..... 5 watts resistor will smoke anyway...so.... the use of 1 watt will smoke under this conditions... 2 watts will smoke also, and 3 watts will go on flames...... 5 watts will be destroyed..... and so on.... only a 870 watts resistor will survive, but will overheat and will burn your pcboard anyway.

What is the normal?....it is the worst case scenario when someone uses 8 ohms crazy speakers that may have 4 ohms in some frequency of the audio spectrum... and will be operating full undistorted power that will be reached from time to time...this is 400 watt average... this is normal high power for this amplifier...the way people may use to have clean sound...playing standard music (not PA use or continuous tone amplification)

So, if someone wants to have this resistor surviving after a long therm short in some power output transistor...and when the supply can feed 136 amperes (no supply can do that..even the mains cannot do that) while keeping 64 volts over this resistor... and this does not happens in real world, but if the one decides to calculate this way...then must install a 870 watt resistor to each power transistor..imagine the size of this thing?

You see, 1 ampere will cross these resistances while your amplifier will be producing 1 Kilowatt of sound (10 percent distortion)..... well....first i doubt you will have a supply that can do this job, secondly you will not tollerate the sound at this level of distortion continuously...will transform your ears in a garbage can and you will not tolerate unless you have it as a peak, as a transient of sound...short therm distorted peak during high level dinamic transient..... under this condition, 10 amperes will be shared by 10 output transistors in each rail...each one of them will manage 1 ampere... and this will be 0.47 watt dissipation in that resistor (power is current multiplied by voltage..... and voltage is 0.47V.... current is one ampere)... and this is not a realistic way to use it.

I do believe you understood... the ones does not understand use 5 watts...i do think some designers use this value because they have never think about the stuff.... this is automatic alike a rule:

- "Power emitter resistors are 5 watts units"...and then everybody goes doing this way...and they feel angry when someone has the arrogance to use a different value....ahahahaha!:)

regards,

Carlos
 
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Be aware that high power amplifiers are dangerous

Not only because supply voltage is high (64 volts X 2) but also because you have a lot of current available to start some fire.

When we have huge amount of energy available in an audio amplifier .... energy from the power supply, if we make a mistake creating an unbalanced circuit, then we will have all this power available to be redirected to the junctions of semiconductors ... and this will burn them .... it is basically what differentiates one high-power unit from a low power unit .... The fact that we have available energy from the power source will be the determining factor that will trigger the destruction of several components.

‪Be carefull... check everything prior to power on your amplifier‬‏ - YouTube

regards,

Carlos
 
numbered schematic & board layout

Greetings,

I put together this numbered schematic and board layout to help myself make a parts list. Note that these have not been carefully scrutinized by either Carlos or Alex. There may be mistakes. The point of posting them is to make it easier for others to put together their parts list. In addition, in the process, people will (hopefully) post my mistakes or make suggestions, so everyone benefits.
 

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Comment on use of 3W Resistors

Hello Carlos,

Thanks for getting this builder's thread going.

I would just like to comment on your explanation regarding the use of the 3W resistors back on post #7. If you don't mind, I'll try to explain this as simple as possible, so please bear with me.

Below is the upper half (positive-going) output section. Let us do a DC analysis to see what the power is across those 0.47 ohm resistors. The speaker load is RL.

A1.jpg

To make the explanation simple, let's assume that for all transistors the Collector-Emitter voltage drop (Vce) is 5V when all the transistors are fully saturated (e.g. full output). With a 5V Vce drop on the transistors, we can simplify the circuit. This is an equivalent circuit:

A2.jpg

Notice that we now have a voltage divider circuit where the speaker load is the lower resistance; and the upper resistance is formed by (5x) 0.47ohm degeneration resistors. Note that five 0.47 ohm resistors in parallel is 0.094 ohms.

Therefore, the voltage across the (5x) 0.47 ohms is calculated as:

V = (60 V * 0.094 ohms) / (0.094 ohms + RL)

First correction:

.... the only condition you may have high wattage there is if your power transistor blow out and produces a short from colector to emitter...then you will have 64 volts from the supply crossing it...136A or something alike that will flow...and now you will have the need of a 870 watts resistor...

Not TRUE --- just looking at the equation above, RL (speaker load) actually determines the maximum voltage across the five 0.47 ohm resistors. The lower the RL, the more voltage those resistors will have across them.

Correct answer is worst case condition for those resistors occur when the output load (RL) is shorted to GND. With this condition, those resistors will see the 60V (on this example). However, a shorted power transistor will remove the effect of the Collector-Emitter voltage drop. :)

Now going back to the main topic, regarding the use of 3W resistors:

Assuming the speaker load is 2 ohm resistance (we are still in DC analysis). Then the actual voltage and power across all five 0.47 ohm resistors is:

V = (60 V * 0.094 ohms)/(0.094 ohms + 2 ohms) = 2.6934 V <--- Very small, right? ;)

However:

P(total) = (2.6934 V * 2.6934 V) / 0.094 ohms = 77.175 Watts

This total power of 77.175W is spread around 5 resistors. Therefore, each resistor will dissipate about 15.4W.

Don't dispair, since this is only the positive half of the signal, the resistors are powered for only 50% of the time. Also, the scenario above occurs only when the amp is clipping. When it clips, chances are (depending on the signal), it only clips for say about 10-20% out of the complete period.

The scenario above is an extreme case (imagine connecting 2ohm load) so at normal operations (typical home application), those 3W resistor might be fine.

However, for professional applications, those 3W resistors will burn at continous high output with critical speaker loads.

Also, I noticed that on the schematic, the circuit doesn't have an output protection where the voltage across the 0.47 ohm resistors are monitored. Some designs have this protection where it turns off the output when a certain voltage threshold is present across those resistors.

Other things to consider:

1. The above calculations are based off on nominal calculations. It might be wise to factor in the resistor's tolerance rating as well as other parameters (e.g. Vce, supply ripple and actual power supply rating)

2. As explained, the speaker's impedance is really critical on this issue. All speaker impedance rating is somewhat an "average resistance" on the frequencies it was designed for. The actual "resistance" is determined on a particular frequency that is being passed (e.g. some 8-ohm speakers will be 4-ohm at certain frequencies).

3. The number of output pairs will help in dissipating the power. Therefore, when you plan to reduce the output pairs, you need to increase those resistor's power rating (say from 5 pairs to 2 pairs) to handle difficult loads.
 
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They can work there.... of course dimensions will be wrong

And also you will have lower gain in these power units used in the driver position..they work...but not that good.

I have used in many amplifiers.... i really could not perceive difference in sonics.... maybe for low impedances we gonna need the original transistor gain.

You see the current gain is the multiplication of driver and power output gain.... voltage gain there is 1....means no gain as multiplication by 1 does not increase value.... gain of current is the one is very high....output just add current..voltage comes from the Voltage Amplifier Stage (VAS).

Imagine that we can have gain 40 in the output power transistor(s) while high current will be crossing these output units...and maybe we gonna have gain 90 to the power transistor while working as driver....multiplication result is 3600..

Now imagine we can have gain 40 in the output power transistor(s) and 200 in the original MJEXXXX driver.... result is 8000..... so...original will help the amplifier to drain much more current from the power supply...and transfer much more current to the speaker...and this means power..... using 8 ohms we will not perceive..but using 2 ohms, for sure you will have losses...got it?

regards,

Carlos
 
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