Need help understanding this circuit

[VILeninDM]

Hi everyone,

I am relatively new to the whole field. Been trying to teach myself this stuff on and off, but for some reason it just not sinking in. I've read few basic books and found them extremely simple, but none of them actually taught how to use transistors (2 did explain what they are but gave very vague details in only 10 pages or so).

Then I found 'Art of Electronics' by Horowitz and Hill. Very good book, but now the problem is that I feel it is just slightly over my head. It can take me about half an hour to read a single page because I always try to figure out what they are actually saying, and authors make way too many assumptions (IMO) that reader just understands all this, so their explanations are lacking at times. At the same time, I don't keep going until I figure out what I just read, so everything just seems to take forever.

Every circuit that I didn't get right away, I put in pspice and run the simulation and play with the numbers. So far PSpice has been a great tool in helping me understand what is actually going on, and this way I got through 89 pages so far.

Now attached is a circuit. According to pspice it works as designed, but I am still not sure why. Can someone provide a verbal analysis in their own words (in case you know what section in the book it came from) of what goes on here?

 

[david90]

i don't think it does anything in particular. I think it is made for viewers to do analysis on. It's like a drill problem to test u on your circuit analysis skill.

 

[eziggurat]

Check this website https://www.4qdtec.com/csm.html I've seen someone use the current source for a power supply design but not the current mirror. You can possibly find out more about it in the internet.

 

[VILeninDM]

k, I'll be a little more specific. This circuit is a current source where R2 is actually the load. R1 sets the current where I = ( Vcc - 2Vbe ) / R1. Theorically current flowing through R2 should always be equavalent to current through R1 as long as transistors stay in the active range.

What I was wondering is if someone could explain to me how this circuit actually works. I know it does because I've seen it in pspice, but pspice doesn't provide explanations. It just showns voltages and current values.

 

[armsgroup]

look in razavi anloge for similar current mirrors or you can see www.rfic.co.uk

 

[jiangwp]

There is the junction voltage and resistor voltage summing.

So, it is the negative temperature coeffecient and postive temperature coeffecient summing. To some degree, it is like a bangap circuit. Only, it has some temperature compensation function.

You can do some temprature simulation to evaluate its temperature feature.

I think the author is the meaning.

 

[Engineer_Bob]

Hi VILeninDM,

I have some advice for you that many may disagree with,

The book your using may be one of the worst books that a beginner can come across!

In my opinion I’d throw it in the bin (even now as a professional engineer I think its poor).

I think that the art of electronics is useful only to professional engineers as a memory aid.

I have read this book and I can’t understand why it’s so popular, the explanations are poor and an iee reference is a better memory aid.

I recall (many years ago) an excellent book (when I started out) Duncan ISBN: 0 7195 4449 1.


Hope that helps Bob

 

[vale]

I agree with Bob.
The art of electronics covers many quite different areas, such as basic circuit, rf, and mcu. It's not a textbook, just a reference. For beginner, I think neamen's 'electronic circuit and design' is a easy book to read. Then Allen's, Razavi's or Grey's classic books can be continued.

 

[Kral]

As you observed, the voltage across R1 = V1 - Vbe(Q1) -Vbe(Q3). The contribution of the Q3 base current to the current (I1) thru R1 is minimal if the Beta of Q3 is reasonably high. So let's assume the current I1 = (V-2Vbe)/R1. Since the base current of Q3 is minimal, the bulk of the current thru R1 is the collector current of Q1. The emitter-base voltages of Q1, Q2 are equal. If Q1, Q2 are matched, then the emitter currents will also be equal. If we ignore the base current of Q3, then the emitter current of Q3 will equal the emitter current of Q1, which also equals the emitter current of Q1, which equals I1.
.
So, essentially, we are forcing the emitter current of Q3 to eqaul I1. This is a convenient method of setting the operating current of Q3.
.
Regards,
Kral

 

[dkace]

It is definatelly a courent sourse utilizing a simple current mirror. Check basic electronic books with BJT's for current mirrors. We use this kind of circuits for:
producing a current from a voltage source
mirroring it to the output (R2), without loosing much of it in the process. It can also multiply a current etc, if you choose the right BJTs or the right dimension of FETs - if FETs are involved- . For the shake of exersise, take the basic equations of BJT and apply them to each and every device of your circuit. Then confirm with spice. It will help a lot.
After that check other current mirror topologies that have to do with temp coefficient, β of the transistor etc. Equations are leading the way my friend.

D.

 

[checkmate]

The top half of the circuit is the familiar current mirror configuration, so one can make a good guess that this is some sort of current mirror.
The mathematical way of analyzing common mirror circuits are to find the relationship between the currents through R1 and R2, using the basic transistor rules. For a circuit with identical beta values for all transistors, the relation I derived is shown below. You can see that for high beta values, I1≈I2.

 

[VVV]

The goal of this circuit is to equalize the output currents.
Look at the circuit at the left. The current through Q2 should reflect the current through the current-setting resistor, R1. But the transistors need base currents, which are supplied by R1, too. So Q2 current is less than R1 current.

The improved version is in the picture to the right. Here the base currents and Q3 current are supplied by Q5. Its emitter current contains its base current, which is now added to the mirrored current. The overall effect is that the mirrored current (Icq4+Ibq5) is almost equal to the programmed current (flowing through R2. There is a small error, associated with the fact that the collector current of Q5 is really Ic+Ib, not Ic. Thus, its base current is slightly higher than the other Ib currents, but the error is small, since the gain of the transistor is high: Ib/βq5.

Try simulating both versions to see the difference.