Help for understanding a Chinese U87 fake microphone schematics

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sunsplash

Member
Joined
Dec 27, 2013
Messages
8
Hello everyone,
I would need someone to help me understand the attached schematic.
I am discovering the microphone DIY and comparing with other schematics, this one seems very complicated.
I took a lot of care for the reverse engineering and I think there is no mistake.

Thank you for your help

U87 China General 1.jpg
 
It doesn't seem all that complicated, but it's definitely not a U87. The lower section generates the polarizing voltage (60V) from 15V. The two FETs are in a darlington configuration to lower the output impedance since there's no output transformer to step things down. The BC547B and BC557B drive the output, and probably further reduce the output impedance. It's definitely a little unusual, but if you have to design to hit a low price point (as many Chinese engineers are asked to do), you have to jump through some hoops. An output transformer could have made this circuit much simpler, but that costs money...
 
It's definitely a little unusual, but if you have to design to hit a low price point (as many Chinese engineers are asked to do), you have to jump through some hoops. An output transformer could have made this circuit much simpler, but that costs money...

... And yet, a Schoeps circuit would have needed one less JFET 😅
 
... And yet, a Schoeps circuit would have needed one less JFET 😅
Very true. I wonder if this was an attempt to make a circuit that had some more “color” than a Schoeps circuit (it’s supposed to be a “U87” after all). I haven’t simulated it, but it seems like this configuration might create some even-order harmonics due to slight asymmetry. Just a guess though…
 
Merci pour les réponses, c'est très gentil.
Je connais un peu l'électronique mais pas assez pour comprendre ce circuit que je trouve "tordu".
Désolé pour mon anglais, je parle français et je traduis avec Google. :)
 
The circuit as depicted in your first post doesn't seem to make much sense to me. If it were indeed a Schoeps, it would have had two PNP output transistors. I see only one PNP and one NPN. This is also not a darlington, because that would consist of two identical transitors, collectors tied together and emitter of the first connected to the base of the second. And what should most likely be a Hartley oscillator at the bottom of the circuit, is not drawn as such.

The two JFETs could be either put in parallel, or be configured as cascode circuit. The latter makes more sense, but is not drawn in the right way.

The whole circuit looks like spaghetti and hurts my eyes. It is common to have signal input(s) on the left of the schematic (i.e. capsule), output(s) on the right (i.e. the XLR), power rail on top, ground at the bottom of the circuit diagram. Then the schematic reads like a book for an Electronic Engineer.

In many cases, a complete schematic is the sum of sub-circuits that are commonly known circuits, such as Common Emitter (CE), CFP (aka Sziklai pair), long-tailed pair etc. But it may need some experience to recognize such standard circuit arrangements and draw appropriately. If drawn properly, it could also help others to better and more quickly understand the circuit.

If you are really interested in knowing the circuit, review and re-arrange the circuit diagram, such that it matches the actual circuit and becomes a more understandable circuit.

Jan
 
The circuit as depicted in your first post doesn't seem to make much sense to me. If it were indeed a Schoeps, it would have had two PNP output transistors. I see only one PNP and one NPN. This is also not a darlington, because that would consist of two identical transitors, collectors tied together and emitter of the first connected to the base of the second.
It's not a Schoeps circuit. The initial pair are JFETs, not BJT transistors, but they're still in a darlington configuration - the source of the first JFET is connected to the gate of the second, the drains are tied together. A similar circuit is discussed in this paper: https://www.ijareeie.com/upload/april/56_H_A New Circuit.pdf
 
Thanks for sharing the link. I didn't know that such a JFET darlington configuration was ever described. But I don't see any practical use in audio circuits either: why would I want an amp that would work only from 88.2074 Hz (what a nonse to use so many digits, haha!) and has several % THD, while a single JFET and Schoeps output circuit could give you something like 0.015%?

Anyway, even if it was intended to be like a JFET darlington in OP's circuit, it doesn't make sense to leave the two drains floating in mid-air. If this circuit is supposed to work as an amplifier, the drains would have to be connected to the Vdd rail through a resistor. As depicted, it will never-ever work.

Assuming the mike is actually working, the circuit diagram needs quite some rework before it makes sense. Whether it's supposed to be U87 style, Schoeps or whatever does not matter: like it's drawn here it will never work. But it would still be interesting to see the circuit drawn up correctly. You never know what interesting details the circuit has in store for us.

@sunsplash: could you make pictures from both sides of the PCBA, with a strong lamp behind the PCBA so the tracks become visible?

Jan
 
Hello and thanks for the answers.

@jp8: I really did my best to trace the circuit, as mentioned before, I am not a professional so I am posting the circuit traces (made with Photoshop) because the copper side is black. The white trace represents the copper on the component side.

It's up to you to tell me more... ;-)
 

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Here's a sketch of what might be going on

IMG_0742.jpeg

Q4's emitter & collector have been swapped from the original, for this to make sense.

Q1/Q2 form a cascode amp, not too unusual. Q4 drives the output with Q3 as its load, but it looks like the 1K/620K/220n network driving Q3's base is a White follower type of arrangement which varies the current in opposite phase to the Q4 current.

It would be useful to have DC voltage measurements around the FETs and transistors, to confirm whether or not this is right.
 
Aahhh, great! That makes sense. Great work Voyager10.

As I more or less expected, Q1 and Q2 are indeed a cascoded pair of JFETs. And with Q4 added, it creates a CFP (Sziklai) pair, but with a JFET as first active element, instead of a BJT. Interesting circuit. I'll drop this circuit in LTspice for further exploration of its properties.

Thanks @sunsplash and @Voyager10 !

Jan
 
@Voyager10: Why did you change the emitter and collector of Q4 and Q3?
I checked, they are indeed like that in reality. Note that the bias voltage is indeed - (minus)60VDC, which is obviously not common.
 
@Voyager10: Why did you change the emitter and collector of Q4 and Q3?
I checked, they are indeed like that in reality. Note that the bias voltage is indeed - (minus)60VDC, which is obviously not common.
It's possible for it to still work despite them being swapped, see e.g. https://electronics.stackexchange.c...istor-work-with-collector-and-emitter-swapped

If you'd like to discuss the circuit further:
  • Measure the voltages on each of Q3 and Q4's legs (with respect to ground), and mark them on the schematic (yours or mine)
  • Add reference designators (R1, C1, etc.) to all the components.
 
Q2: E = 354mV; B = 1V; C = 1.29V

Q3: E = 19.3V; B = 18.67V; C = 9.5V

Q4: D = 10.16V; S = 9.73V; G = 18.88V

Q5: D = 10.16V; S = 135mV; G = 9.73V
 
My slow brain realized now that the designer of the circuit uses the Drain and Source connections on the JFETs interchangeably. With many JFETs, that can be done, indeed. This is why in the circuit diagram you'll see the two Drains connected, causing me to not recognize the cascode configuration that Voyager10 had in his circuit diagram.

Jan
 
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