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Understanding circuits

Hi Everyone,

I'm relatively new to studying electronics. I've purchased a few books (some college textbooks, a couple of math books, and the book titled "Electronic Circuits for the Evil Genius". I really like the last book because it helps me understand how various pieces of a circuit are used to do useful things. Again, the book is very high level - but at this point, I simply want to see the 'big picture'. By no means am I using this particular book as a sole guide - but I do find it very helpful.

I recently purchased a hit from Ramsey Electronics. Specifically, this one:

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I purchased this kit because it looked like something fun to build and my goal is to truly understand how it works. Unfortunately, there is only 2 short paragraphs written in the text as to how it works. This is a real bummer for me. I was hoping to see something like a block diagram showing each piece of the schematic and what its actually doing. I understand what a NPN, and PNP transistor is. I understand how a capacitor and a resistor can be used to form a timing circuit - but none of the circuit is really explained in the documentation (from the link above ). I've got a student edition of Electronics Workbench and tried building the small piece of the schematic (the section where Q1 and Q2 are ) all the way though to wear they connect to the LM-324 OPAmp. I can see how the capacitors 'fill' up but don't understand how the polarity reverses (there's a +9V DC supply this subcircuit is connected to do) .

Could someone who has some time, take a look at the schematic and get me started in understanding this? It would be great if someone could help me break this down into some sort of block diagram . I was thinking about just breadboarding everything before I actually solder and assemble the components - simply because I'm thinking I will learn more that way. If I put this thing together, I feel only my soldering skills will get better and I will leave this project not knowing anything more than I started.

Thanks in advance for any guidance.

Reply to
Drake
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I just noticed the schematic is not in the documentation online. Its supposed to be on page 8 but it looks like they removed it. :( Oh well...

Drake wrote:

Reply to
Drake

I found another place online that has the schematic I'm trying to understand:

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Its not exactly the same...but very close.

Thanks again and sorry for the back to back emails.

Drake wrote:

Reply to
Drake

There's no polarity reversal. Q1 and Q2 cut the +9V to about +6 to operate the LM324. The voltage divider R4 and R21 gives about 3V, which is used as the opamp virtual ground. As the documentation mentions, all the action occurs at Q3, the microwave oscillator, and D1. When any microwaves reflected from a moving object enter the circuit, they mix in D1 with the original frequency, producing a beat note of 10 to 40 Hz, as mentioned in the documentation. The first two stages of the 324, U1C and U1D, are a high gain amplifier and filter for that 10 to 40Hz signal. Then U1A is a buffer that keeps from loading the last part of the filter (the passive components connected to pin 3). The output of that buffer feeds an envelope detector (D2, C17 and the resistors). There a DC voltage develops if motion (a reflected, doppler-shifted signal) is being detected. That voltage is compared with about 0.3V that is developed across R20. If it is greater than 0.3, U1B's output goes high, turning on the MOSFET Q4, which turns on the LED. The circuit description calls this transistor Q8, and says it is driven by U1A, which is a little different than what is shown in the schematic. About the Q1-Q2 part- it's an unconventional thing, where they seem to be using Q2 as a zener diode or something. I got curious about this, and tried the circuit on the bench, and it seem to work, with several

2n3904s I tried as Q2 giving about 7 to 7.5V, relatively independent of the 9V supply. Then the diode drop at Q1 would give 0.7V less at the output, or about 6.8V. However, this didn't work in two simulator programs- LTSpice and Pspice, neither of which showed a zener breakdown for Q2. You may not be able to simulate this part of the circuit.

-- john

Reply to
John O'Flaherty

Q2 is connected so that the base emitter junction is reverse biased, to the point of zener break down. This produces a fairly constant voltage of a little more than 6 volts. Q1 is an emitter follower that copies this voltage (minus its base emitter diode drop) to act as a supply regulator that produces about 6 volts. That supply droves the LM324 and several voltage dividers that produce lower voltage references.

Reply to
John Popelish

Thanks John and John.

This is really great information - but I think i've got a long ways to go in terms of getting to the level where I wanna be in understanding electronics. I'm going to breadboard this circuit up and perform some testing around the circuit to make sure I understand what everything is doing. Your comments again are greatly appreciated.

One last question if you guys have a moment.

(concerning Q1 and Q2 ) does simple things. It essentially supplies the correct voltage to the respective areas.

the designer of the circuit not use a couple of series resistor in parallel to acheive the same thing ? For example:

If I have a 1k and a 2k resistor in series supplied by 9V , the voltage drop between them is 6 volts. If I have a 2k and a 1k resistor in series supplied by 9V, the voltage drop between them is 3 volts. If I connect each of these series in parallel, I think I get the same thing:

-9V+ -----R1 (1kOhm) ------(a) ----------R2(2kOhm) ------Ground | | ----R3(2kOhm) ------(b) --------R4 (1kOhm ) --------Ground

(these can be adjusted to acheive the desired current )

where :

(a) provides the voltage needed to operate the LM324 (b) gives the virtual opamp ground

To me, it seems like that whole area is over designed and my simple little voltage dividers is the same thing.

Is my thought process incorrect ?

John O'Flaherty wrote:

Reply to
Drake

"Drake" schreef in bericht news: snipped-for-privacy@75g2000cwc.googlegroups.com...

Yes and no. When using resistors, a voltage divider will do as you think. However, electronic parts like opamps are no resistors. They will not obey Ohms law. For the case of simplicity you can consider the opamp an variable resistor with an unpredictable and constantly variing value. So to make sure the opamp is powered by a (more or less) constant voltage some electronics are required.

(Beware! That "variable resistor" behavior is meant for the power consumption, not for its function as an amplifier.)

As for the regulator, most general purpose silicium transistors are known for the "zener" behavior of their reverse biased BE-junction. That's why the maximum reverse voltage for that junction is usually 5V. This zener behavior is not a part of the official specification of course and a transistor (mis)used this way may be internally damaged and not function as a good transistor anymore. But... A 2N3904 will much cheaper then a zener, especially when bought by (large) numbers.

petrus bitbyter

Reply to
petrus bitbyter

Drake wrote: (snip)

That is how it looks to me.

Because the two voltages connect to different loads that will change the voltage in different ways. They needed to be independent.

A better way to say it is that the 2 k resistor has a 6 volt drop across it. The 1k resistor has a 3 volt drop across it. Whether the node between them has 3 or 6 volts with respect to the zero volt node depends on which order they are connected.

Are you trying to show that the tow dividers are each connected between 9 volts and 0 volts?

Define "over designed".

Back to the schematic at:

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The 6 volt regulator supplies U1 (the quad opamp) shown at U1A.

It also supplies a voltage divider for U1B pin 6, made up of 3 resistors, R4, R21, and R20. This divider produces 2 voltages, one of 11/21 of the 6 volt supply, not exactly 3 volts, (11 k total on the ground side of the output, and 21 k total in the divider) and one of

1/21 of the 6 volt supply (about .286 volts) which is the drop across only the 1 k resistor. The .286 volt bias voltage is used by U1B, while the "3V" output goes to U1D.

It is used to produce a 3 volt bias with a pair of equal resistors at U1A, but this bias gets a signal added to it through R14. C2 is acts as a low pass filter to keep this 3 volt bias voltage quiet.

They couldn't use the same ~ 3 volt divider for both U1A and U1D, because the signal added to it at U1A would contaminate the signal at U1D.

Reply to
John Popelish

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