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Pure Class A Single End Amplifier Idea!
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Old 10th June 2008, 01:55 AM   #71
woody is offline woody
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Pure Class A Single End Amplifier Idea!
That IRF530 with it's 78W rating and wimpy 220 case will be
rather hot disipating 48W !
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Old 10th June 2008, 06:11 AM   #72
AKSA is offline AKSA  Australia
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If current through each active device of the LTP is identical, it is difficult to see how AC currents are imbalanced, particularly when, with matched devices, bias currents would be identical.
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Old 10th June 2008, 06:23 AM   #73
GK is offline GK  Australia
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Quote:
Originally posted by AKSA
If current through each active device of the LTP is identical, it is difficult to see how AC currents are imbalanced, particularly when, with matched devices, bias currents would be identical.

No, it isn't difficult to see. The current mirror provides a complementary drive current to the VAS. If the current mirror has disimilar emitter degeneration then it will not act a a 1:1 current "mirror". The AC current in the leg of the LTP that drives the current mirror will then differ from that of the other leg by the modified ratio of the current mirror.
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Old 10th June 2008, 06:56 AM   #74
AKSA is offline AKSA  Australia
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Therefore the sins of unbalanced active devices are about the same as the AC unbalance of the current mirror... and what of the fact that one device in the CM is a diode, the other is a transistor, that too will introduce unbalance.

Six to one, half a dozen to the other it seems to me.
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Old 10th June 2008, 07:20 AM   #75
GK is offline GK  Australia
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Well, by attempting to fix the DC imbalance by unbalancing the emitter degeneration of the current mirror you are just shifting the problem somewhere else.

Below is a sim of what happens when the current mirror is unbalanced to compensate to the VAS bias current. In this case I deliberately made a bad VAS with a bias current of 550uA to make the degree of error visible on the graph.

The green trace is the current provided to the VAS by the current mirror leg, the blue trace is complementary current provided to the VAS from the LTP leg.

The former has a DC value 550uA less due to the 550uA sourced from the VAS. The current mirror emitter degeneration resistors are 100 ohms and 73 ohms to make the collector currents of each LTP transistor equal.

The amplitude of the green trace is equal to the amplitude of the blue trace * (73/100).

The circuit is therefore incapable of providing an accurate complementary push-pull drive current to the VAS, a common-mode AC difference signal then circulates and the THD rises somewhat proportionally.
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Old 10th June 2008, 07:32 AM   #76
AKSA is offline AKSA  Australia
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Thank you for the curves; I see your point, the math is clear.

I wonder if you would hear this in practice.........
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Old 10th June 2008, 08:55 AM   #77
AndrewT is offline AndrewT  Scotland
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Default Re: Balancing an LTP

Quote:
Originally posted by AKSA
If LTP stage current is 2mA and beta of an 8mA VAS is 100, then Ib will be 80uA, and assuming a perfect current mirror this gives active collector at 1040uA and inactive collector at 960uA. The percentage imbalance this represents is +4%, -4%, 8% in total.
But, Glen would assert this is bad design.

I accept that the effect of the base current and of the capacitive loading should be minimised/optimised to get best performance from the LTP.

I cannot accept that a Mirror load automatically balances the LTP for best performance.

I see two factions, those that claim mirror loading of the LTP sounds best and they probably also use Cdom around the VAS, the other group who minimise Cdom or even eliminate it AND use resistor loading of the input LTP to obtain best sound.
Somehow I can never see these two groups of designers agreeing with each other. Glen appears to be in the Mirror+Cdom group but at least Glen recognises where the shortfall of this technique is and addresses it in his way.
I prefer an accurately set up LTP (balanced) with resistor loading and if possible eliminate Cdom. But my skills are lacking and I often have to add some Cdom (with a very low capacitance VAS) to stabilse the amp for a wide range of reactive loads.

However, when Symasym came along and others before it, I can see that loading the LTP symetrically has big advantages. Maybe that's why I bought a couple of PCBs to listen to the effectiveness of the topology.
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Old 10th June 2008, 10:32 AM   #78
AKSA is offline AKSA  Australia
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Andrew,

You've introduced quite a few other issues into this, but why is my example bad design? I'm curious..... and I do not agree that there are two design camps here, I have use both, but have found the stock CM is flawed for reasons I alluded to earlier.

Hugh
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Old 10th June 2008, 11:10 AM   #79
destroyer X is offline destroyer X  Brazil
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Default Take an amplifier fellows, and put a resistance as load


and then install a mirror.... those two transistors as load..... and then revert them to a resistance...... revert them to the mirror once more.... listen carefully.... you will conclude a lot of interesting things.

Audio..... do that listening.

What i think, has no interest....the important is your own conclusion into the real world... into listening tests... so easy to make... so conclusive and so definitive.... the rest is beliefs... alike Religion...not scientific.

But if you prefer to discuss, instead to have the experience...well.... good pain into the stomach for you.

regards,

Carlos
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Old 10th June 2008, 12:35 PM   #80
GK is offline GK  Australia
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Default Re: Re: Balancing an LTP

Quote:
Originally posted by AndrewT
I cannot accept that a Mirror load automatically balances the LTP........

Then you simply do not understand how the circuit works, particularly the way in which the global negative feedback steers the quiescent DC operating point of the amplifier.

If a design is made without an LTP current mirror, particular attention has to be taken WRT to the value of the LTP load resistor in order to ensure that the DC currents in each leg of the LTP are balanced, right?

This concern doesn’t exist when a current mirror is used because the current mirror ensures that the balancing act is done "automatically". There are of course sources of error (as there is in every circuit), but these can be easily made negligible for all practical purposes.
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