Author Topic: Help me understand parts of a schematic  (Read 2678 times)

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Offline EvyTopic starter

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Help me understand parts of a schematic
« on: August 12, 2017, 08:04:31 pm »
Hello!
I have got a schematic of a negative feedback amplifier which I am trying to decipher. I am basically trying to understand the functionality of some parts of the circuit.



I am trying to understand the purpose of T3, R5, R6, R4. It looks like the transistor with R4 is an voltage-current amplifier and that the R6 and R5 are voltage dividers for the input into the voltage-current amplifier I mentioned earlier.

Is this reasonable?
 

Online Benta

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Re: Help me understand parts of a schematic
« Reply #1 on: August 12, 2017, 08:18:32 pm »
The part you are asking about is a negative constant current source ("current sink"). It is the lower leg of the class A output stage, the T2 being the upper leg.
 
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Offline EvyTopic starter

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Re: Help me understand parts of a schematic
« Reply #2 on: August 12, 2017, 08:21:44 pm »
Thank you for your reply!!!

But how does the current source operate? Are my assumptions about the voltage-current amplifier true?

I think that T3 and R4 is a voltage to current amplifier (transconductance amplifier which outputs current) and then there is input into that amplifier which comes from the voltage divider of R5,6 if I am correct.
« Last Edit: August 12, 2017, 08:24:27 pm by Evy »
 

Online Benta

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Re: Help me understand parts of a schematic
« Reply #3 on: August 12, 2017, 08:26:11 pm »
Evy, this is exactly the same amplifier you asked about in your other thread, only with real components instead of current source symbols.

 
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Offline EvyTopic starter

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Re: Help me understand parts of a schematic
« Reply #4 on: August 12, 2017, 08:28:38 pm »
Evy, this is exactly the same amplifier you asked about in your other thread, only with real components instead of current source symbols.

Oh yes, I realize that I was just curious how they implemented it. Thank you for your replies once again.
 

Online Benta

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Re: Help me understand parts of a schematic
« Reply #5 on: August 12, 2017, 09:36:37 pm »
Quote
I was just curious how they implemented it.

The analysis is simple. R5 and R6 form a voltage divider between ground and VEE with a voltage we can call U56.

The voltage over R4, which sets the constant current, is U56 - VBE (referenced to VEE). Again, as beta is reasonably large, we can ignore the base current into T3.

« Last Edit: August 12, 2017, 09:47:42 pm by Benta »
 
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Offline T3sl4co1l

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Re: Help me understand parts of a schematic
« Reply #6 on: August 13, 2017, 03:31:56 pm »
Why ask about T3 (and associated components) only, and not the unlabeled components in the top left?

The astute reader will note that these are complementary sub-circuits (change NPN for PNP). :) Except for that diode, why would they add that?

(It is actually a more interesting question: why they wouldn't add the diode in series with R5?  The inconsistency is probably just an accident by the author.)

And yes, these sub-circuits can be called transconductance amplifiers (voltage in, current out; gain is out/in == current/voltage == conductivity (siemens), hence the name), though because their inputs are fixed (static resistor divider!), the outputs also do not change, therefore they are simply fixed current source/sinks. :)

Tim
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Offline EvyTopic starter

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Re: Help me understand parts of a schematic
« Reply #7 on: August 13, 2017, 09:51:47 pm »
Why ask about T3 (and associated components) only, and not the unlabeled components in the top left?

That part of the circuit is too complicated for me   :o

But the part where there is an PNP instad of an NPN is very interesting too since there is an NPN at that same spot in the unbiased version of this circuit. The first time they swap the two  in the book they mention that its done to "adress the voltage level difference" which I didn't understand.
 

Online Zero999

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Re: Help me understand parts of a schematic
« Reply #8 on: August 13, 2017, 10:38:59 pm »
Why ask about T3 (and associated components) only, and not the unlabeled components in the top left?

That part of the circuit is too complicated for me   :o
The unlabelled components form a constant current source. The diode is just makes it less sensitive to changes in temperature because its forward voltage changes to compensate for changes in the base-emitter voltage of the transistor.
 
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Offline EvyTopic starter

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Re: Help me understand parts of a schematic
« Reply #9 on: August 13, 2017, 11:15:50 pm »
Why ask about T3 (and associated components) only, and not the unlabeled components in the top left?

That part of the circuit is too complicated for me   :o
The unlabelled components form a constant current source. The diode is just makes it less sensitive to changes in temperature because its forward voltage changes to compensate for changes in the base-emitter voltage of the transistor.

Is that what's called a current mirror?
 

Offline T3sl4co1l

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Re: Help me understand parts of a schematic
« Reply #10 on: August 13, 2017, 11:16:00 pm »
That part of the circuit is too complicated for me   :o

Surely that cannot be!  Consider the analogy:



Spotting these symmetries is one of the very useful skills to learn in electronics.  :)

Tim
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Offline T3sl4co1l

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Re: Help me understand parts of a schematic
« Reply #11 on: August 13, 2017, 11:19:39 pm »
Is that what's called a current mirror?

Yes, the collector current equals the input current times a gain factor, with very little offset.

In practice, you'd use a "diode strapped transistor", because the Vf of such a "diode" will be nearly equal to that of the ordinary transistor.  Whereas if you take a random, say, 1N914 and 2N3904, the Vf's will be very different, and you won't get a good mirror (the current will be equal to the input plus an offset).  But you will still get much better results than without the diode at all (where the base-emitter voltage gives a full ~0.6V offset, and whatever current through the resistors it corresponds to).

Tim
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Offline EvyTopic starter

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Re: Help me understand parts of a schematic
« Reply #12 on: August 13, 2017, 11:59:04 pm »
Woah, what is this sorcery  :clap:  :popcorn: .

Thank you for your illustration!
Unfortunately I get stuck in the first step where you swap a PNP for an NPN, reverse the diode and switch Vcc for -Vcc. What property of the BJT transistor are you using for the circuits to be equivalent?
 

Offline T3sl4co1l

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Re: Help me understand parts of a schematic
« Reply #13 on: August 14, 2017, 12:08:17 am »
A diode is just a P-N junction, and a transistor is just a pair of junctions (very close together, so that the two diodes, er... transist).

Swap P for N, N for P, and the polarity reverses.  NPN becomes PNP.  Nothing more than that. :)

If you check out old circuits -- back when germanium PNP were all the rage, you'll see a lot of emitters pointing "up", and "positive ground" circuits.  You look at these seemingly-odd circuits, blink once or twice, then realize: they've still put ground at the bottom, emitters at the bottom (for common-emitter circuits), and the collectors point up towards the supply.  It's just upside-down, because the supply is negative and the transistors are PNP.  It's actually nothing strange at all, it just looks odd today!

Tim
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Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Online Zero999

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Re: Help me understand parts of a schematic
« Reply #14 on: August 14, 2017, 07:48:05 am »
When configured in this manner, the transistors are behaving as voltage controlled current sources, with the collector current being dependant on the base-emitter voltage. A small base current flows, but that can be ignored at this point and is more dependant on the temperature, than collector current: Hfe varies widely.

A diode strapped transistor is just a BJT, with its base connected to the collector, so you're only using the base-emitter diode junction. It's better than an ordinary diode, in this case, because if the transistors have the same part number, they'll pass a very similar current, when given the same base-emitter voltage. The collector currents would match more closely, if the two transistors were on the same piece of silicon, as there's a higher chance they'll share the same characteristics and perhaps more importantly, they'll be at the same temperature.
 


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