Amplifier design: how to connect a stage with the next one? - diyAudio
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Old 30th January 2005, 03:00 PM   #1
Bricolo is offline Bricolo  France
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Default Amplifier design: how to connect a stage with the next one?

Hi.

I know some theory and maths about diff pairs, common emitters, common collectors, miller... But nearly nothing to connect those stage together to make an amplifier.

Should we design all stages independently, but that stage N's output has a bias voltage of x volts, that matches the N+1's input bias point? Or can the stages have different bias voltages, and current will flow between them even without signal?
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Old 31st January 2005, 06:43 AM   #2
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One stage can output DC and, or AC into the next stage. The output impedance must match the input impedance of the next stage though. The voltage/current ratio must be the same in order for one stage to drive the next so the next stage will operate within it's specified range and so on. As for that specified range, it is determained in the design of the stage.

Say you have a VAS. This might be used to drive an emitter follower stage, with a current source bias. A 4Ohm EF stage has a beta of 250. If then the input Z of the EF stage is 1KOhms, and the peak voltage is 10V, then the VAS must be able to drive 10V at 10mA. Iq of the VAS must be greater than 10mA.
Usually you don't have to be exactly perfect in analog, altough there are perfectionists out there.
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Old 31st January 2005, 04:37 PM   #3
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usually output stage is in class AB and previous in A (SE or p-p)
it is nice when next stage has low influance on the previous (read:high input Z, low output). sorry if you find it cliche
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Old 31st January 2005, 05:08 PM   #4
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Regarding impedance matching: For optimum current transfer you want a high output impedance feeding a low input impedance. For optimum voltage transfer you want a low output impedance feeding a high input impedance.

The output impedance of a stage also affects what sort of voltage you can connect the output to. For instance a common-base stage has a very high output impedance, so you can connect it to any voltage and it won't mind (although you have to make sure to leave enough Vcb to avoid saturation). This also means you can't rely on the output voltage of such a stage to bias the next stage (which is why complementary differential amps can't have current mirror loading). On the other hand, a common-collector stage has a low output impedance, so if the bias voltage of the next stage is different from the output voltage, a large current will flow, which is not good.
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Old 26th February 2005, 02:39 PM   #5
Bricolo is offline Bricolo  France
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Ok here's the problem.

I'm trying to design an amp, a brief schematic is showed.
To have a proper bias on the VAS, R4 has to be 0.7V DC across it, right?

So I have the choice of
-very low bias for the diff pair, and a high value load resistor->high gain in the diff pair
-a few mA bias in the diff pair, but in order to have 0.7V DC on R4, R4 must be small -> low gain in the diff pair


Right? Or have I forgotten something?
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Old 27th February 2005, 08:35 AM   #6
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Old 27th February 2005, 08:49 AM   #7
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Your thinking is along the right lines and essential for open loop operation, but when overall feedback is applied by closing the loop from output to input, the LTP will do whatever it needs to maintain whatever is fed to its input, thus the VAS will be biased accordingly. This is especially true if you run a CCS on the VAS.
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Old 27th February 2005, 01:34 PM   #8
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Quote:
So I have the choice of
-very low bias for the diff pair, and a high value load resistor->high gain in the diff pair
-a few mA bias in the diff pair, but in order to have 0.7V DC on R4, R4 must be small -> low gain in the diff pair
Not necessarily so. The Diff Gain can be the same as it is a function of I and R.

Much of it depends on your VAS. If I2 is large and I1 is small, Q3 will "load" Q2 collector. With I1 small, you will also have more difficulty with C1.

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Old 27th February 2005, 07:05 PM   #9
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Quote:
To have a proper bias on the VAS, R4 has to be 0.7V DC across it, right?
Right.
Start by choosing I1 and I2. The current that must flow through R4 is the collector current of Q2 minus the base current of Q3. The voltage across R4 will be about 0.6V. Divide this by the current through R4 to give the optimum value of R4. Choose the nearest convenient value.
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