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Old 27th March 2011, 02:58 PM   #1
rjm is offline rjm  Japan
Richard Murdey
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Default Diamond Buffer Emitter Resistor Question

Attached schema is from National LH0002.

Would anyone care to lecture me as to why resistors R3 and R4 are required? At only 2 Ohms, I'm wondering if one might as well not bother.
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Old 27th March 2011, 03:30 PM   #2
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They limit the current in output.
Without them maybe is too much current and HOT HOT
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Old 27th March 2011, 04:47 PM   #3
AndrewT is offline AndrewT  Scotland
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R3 & R4 aid Thermal Stability of the stage.
Take Vbe of Q1 @ 10mA ~620mV and Vbe of Q3 @ 100mA ~620mV.
Reduce R3 and R4 to 0r0.
When the input diamond is biased to 10mA the output is biased to 100mA.
Now let the input bias and/or Vbe change slightly to 10.5mA @ 618mV.
In the meantime the output has warmed up and the Vbe for 100mA has become 610mV.
The input diamond applies 618mV but the output only needs 610mV.
What happens to the excess of 8mV?
It turns on the output harder. The output bias current increases. The output transconductance determines the increase in output bias. Let's assume the output transconductance @ 100mA is 8S, i.e. 8A /V = 8mA /mV. For a 8mV cgange we would expect about 64mA of extra current to flow. The output bias has changed from 100mA to ~ 164mA. How hot do the output Tj become? How low does the Vbe become?
How high does the output bias current become?
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Old 28th March 2011, 03:02 AM   #4
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Thanks Andrew. If I understand you correctly they are a safety measure to guard against thermal runaway resulting from mismatched transistor characteristics.

I'm still missing something. Why is the current though the output (Q3,4) 10 times larger than the current through the input drivers (Q1,2)?

The transistors are the same.

In the example, R1,2 are 5k. Say the circuit runs on bipolar 5 V supplies. The voltage drop across R1 is about 4.4 V, so Q1 (and Q2) is biased to just under 1 mA. If the Vbe for 1 mA is 0.66 V, this voltage is applied to the base of Q3 and Q4, so, with R3,4 at 0R, Q3 and Q4 also run at 1 mA, assuming they are the same model of transistor as Q1,2.

If R3,4 are 2R, the Vbe of Q3,4 is reduced by a relatively trivial 2 mV, reducing the current through Q3,4 to something slightly less than Q1,2.

My problem is if I want to run the driver and output stage at 30 mA, the resistance value of R3,4 for an equivalent Vbe drop would have to be 15 times smaller than the value listed, about 0.15 ohm, to avoid quenching the output bias current.

Alright, not a big problem: those resistors do exist, just not something I have on hand. My question though is for the case where driver and output run at the same current, how necessary are they? (The transconductance for BD135/6 I'm using appears to be about 1 S.)
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Old 28th March 2011, 03:32 AM   #5
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Perhaps a better, certainly more concise way as asking the question:

Is there a rule of thumb for estimating how large R3,4 should be?

R=1/S for example?
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Old 28th March 2011, 09:06 AM   #6
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The diagram does not state the transistors are the same.
Now set the current cold and Vbe cold for all the trannsistors and start from there as the circuit operates.

The emitter resistors are a form of feedback. Not just helping to adjust for imbalances in devices parameters but setting the stage amplifier characteristics.
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Old 28th March 2011, 10:46 AM   #7
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If the transistors are the same then the resistors guarantee that the output stage runs at lower current than the input stage. I assume this is not what is intended, so probably the output stage uses bigger transistors. The resistors then help regulate the current, reduce thermal runaway, and may also smooth out the crossover region by reducing effective gm while both output transistors are contributing to the signal. R3/4=1/S would smooth the crossover, approximately.

My gut reaction is that this is a poor design, as you don't really have much control over quiescent current. It could easily be that the AC and DC requirements call for different values of the resistors, so you have to adopt a compromise. However, there could be some clever trick in it which I have missed.
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Old 28th March 2011, 11:17 AM   #8
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Remember that the original design called for very low bias current, 2 mA, so all the transistors could be same without blowing the power budget. And with a low current and only 2 ohm emitter resistors, the driver and output transistor currents would be almost the same, though as you say, the output (Q3,4) will always be slightly lower than the driver (Q1,2).

As a line driver it's not an especially poor design. As a power amp ... not a good idea. I am hoping to nudge it up a little bit from line driver to headphone amp.
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Old 28th March 2011, 12:04 PM   #9
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Quote:
Originally Posted by DF96 View Post
If the transistors are the same then the resistors guarantee that the output stage runs at lower current than the input stage. I assume this is not what is intended, so probably the output stage uses bigger transistors.
You are correct.
If we put bigger transistors like a TO-126 pair such as BD139,BD140,
then there is a larger effect of these emitter resistors.

See diagram.
Left has no resistance 0 Ohm, Right has 2.2 Ohm resistors.
This makes current much lower in output pair.
25mA vs 10mA
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Old 28th March 2011, 12:34 PM   #10
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Or, if all the transistors are BD139/140, and R1,2 is 150 ohms for +/- 5 V, for 30mA through Q1,2 then the current through Q3,4 is reduced to about 12 mA when R3,4 is 1 ohm.

(I've just tried this out on the bench, and it is so.)
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