Full m/lab SD2000 schematics

[ZED]

teacher does anyone have a good and correct fcb layout without any errors. and it's been tested thank you

Nop ! not that i know off.

[liekwang]

teacher does anyone have a good and correct fcb layout without any errors. and it's been tested thank you

[GoldenKiwi]

Many thanks for taking the time to work this through to explain it!

[Carl-NC]

Wow, that's a really good app note. Ree is the ohmic resistance of the emitter and adds to the ac impedance of the emitter. So in my last post, "everything in the emitter" is actually (re + Ree) = 6.9+0.4 = 7.3 ohms. So the gain reduces to 1350/7.3 = 185. There are other terms I'm ignoring: Rbb/beta, Rs/beta, and ro. So maybe the actual gain is 180. Close enough.

This gain is part of the open loop gain. The total open loop gain AOL is the gain of the diffpair times the open loop gain of the opamp (which is about 100,000) because they are both inside the feedback loop. So the total AOL = 185*100,000 = 18,500,000. But feedback reduces this to (1 + R118/R119) = 46 or so.

[GoldenKiwi]

Q13 and R112 are not directly part of the gain. They are the current source for the diffpair; the collector current of Q13 is (5 - 0.75)/560 = 7.59mA. Let's call it 7.5mA. This current gets split in half in the LM394 diffpair so each leg sees 3.75mA, and that is the current through the collector resistors, which is 1.35k each side. Thus the DC outputs are sitting at about 0V.

The gain of the diffpair is "everything in the collector divided by everything in the emitter." The collector side is easy, it's just 1.35k. The emitter side has no physical resistors, just the impedance of the emitter junction itself. That is called "re" and is VT/I where VT is the thermal voltage kT/q and I is the DC current. We know I -- it is 3.75mA -- and VT is 26mV at room temp. Therefore re = 26mV/3.75mA = 6.9 ohms. So the overall gain of the first stage is 1350/6.9 = 195. It is fully differential.

Hi Carl, Many thanks! Apologies for the delayed reply, it took me a while to find the schematic version with components numbered, then follow along with your explanation and maths.
I note that the App note AN-222 reference by waltr above mentions on page 5 that the Ree of the LM194 is 0.4R. Does this change the calculation or is this value of 0.4R Ree insignificant and we are talking about the effective in circuit value due to the constant current formed by Q13?

Is the resultant gain of the compound LM394 Diff Pair + NE5534 = 195 * (1000/22) = ~ 8863 ?

Still waiting for my SD2000 to arrive!

[liekwang]

ok I have begun to understand what's in the scheme thank you sir for the quick reply

[6666]

[ATTACH]50832[/ATTACH]
how can this land symbol be accessed by the scheme sent by zed and kt instead of this positive path batrai

What they are drawing is the positive plate of the battery is ground, where you see the ground symbol all those points are connected together on the PCB, also to make drawing the schematic easier and to reduce the amount of lines they connect the component pins to the ground symbol.

[waltr]

[ATTACH]50842[/ATTACH]
I do not understand what type of capacitor symbol is used and why there is a GND logo on the negative battery bar that I circled

The 'plus' symbol on the cap indicates a polarized cap, typically either electrolytic or tantalum.

the "GRND" Name is a local 'ground' reference for the circuits on that page.

[liekwang]

I do not understand what type of capacitor symbol is used and why there is a GND logo on the negative battery bar that I circled

[liekwang]

how can this land symbol be accessed by the scheme sent by zed and kt instead of this positive path batrai

[liekwang]

[QUOTE=Carl-NC;271306]Yes, should work fine.[/QUOTE

OK
Thank you for the info.

[Carl-NC]

what is the BAS16 diode replacement can I use diode 4148

Yes, should work fine.

[liekwang]

what is the BAS16 diode replacement can I use diode 4148

[Carl-NC]

I'm still scratching my head over this input stage.
When I first looked at it, I assumed it was a composite of a single-ended differential BJT input providing gain at very low noise, followed by an op-amp as second stage of gain. Having built a Hammerhead years ago I recalled reading something to effect that splitting the gain (default x1000) into two stages provides better settling through maintaining the gain bandwidth product - and assumed this was essentially the same principle.

I assumed the gain of the LM394 was essentially 1350/560, while not understanding the current source arrangement, multiplied by the opamp gain, which I wasn't sure how to calculate in this arrangement(positive feedback?). This is clearly not the case!

What am I missing? I now guess the LM394 is not singled ended and the feedback loop is including the LM394. Is this practically including the BJT LTP as the input of the opamp?

Apologies to ask what are probably basic questions!

Q13 and R112 are not directly part of the gain. They are the current source for the diffpair; the collector current of Q13 is (5 - 0.75)/560 = 7.59mA. Let's call it 7.5mA. This current gets split in half in the LM394 diffpair so each leg sees 3.75mA, and that is the current through the collector resistors, which is 1.35k each side. Thus the DC outputs are sitting at about 0V.

The gain of the diffpair is "everything in the collector divided by everything in the emitter." The collector side is easy, it's just 1.35k. The emitter side has no physical resistors, just the impedance of the emitter junction itself. That is called "re" and is VT/I where VT is the thermal voltage kT/q and I is the DC current. We know I -- it is 3.75mA -- and VT is 26mV at room temp. Therefore re = 26mV/3.75mA = 6.9 ohms. So the overall gain of the first stage is 1350/6.9 = 195. It is fully differential.

I expect the diffpair was added to the NE5534 to reduce noise. The LM394 has about half the input-referred noise of the 5534 (3.5nV/rtHz). With today's opamps (down to 0.9nV/rtHz) it doesn't make much sense, but there are other discrete front-ends that can beat the noise down to 0.5nV/rtHz, so it's still a valid technique.

[GoldenKiwi]

I'm still scratching my head over this input stage.
When I first looked at it, I assumed it was a composite of a single-ended differential BJT input providing gain at very low noise, followed by an op-amp as second stage of gain. Having built a Hammerhead years ago I recalled reading something to effect that splitting the gain (default x1000) into two stages provides better settling through maintaining the gain bandwidth product - and assumed this was essentially the same principle.

I assumed the gain of the LM394 was essentially 1350/560, while not understanding the current source arrangement, multiplied by the opamp gain, which I wasn't sure how to calculate in this arrangement(positive feedback?). This is clearly not the case!

What am I missing? I now guess the LM394 is not singled ended and the feedback loop is including the LM394. Is this practically including the BJT LTP as the input of the opamp?

Apologies to ask what are probably basic questions!