What this circuit does?

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leonel

New Member

• Jun 8, 2005

• #1

What´s the use of putting two transistor collectors shunted like the circuit in attachement! I don´t understand what this circuit does! Can anybody explain me? Thank you!

 

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Nigel Goodwin

Super Moderator

Most Helpful Member

• Jun 8, 2005

• #2

leonel said:

What´s the use of putting two transistor collectors shunted like the circuit in attachement! I don´t understand what this circuit does! Can anybody explain me? Thank you!

Click to expand...

It's a simple logic gate, either OR or NAND, depending on the logic polarity used.

 

L

ljcox

Well-Known Member

• Jun 8, 2005

• #3

leonel said:

What´s the use of putting two transistor collectors shunted like the circuit in attachement! I don´t understand what this circuit does! Can anybody explain me? Thank you!

Click to expand...

It looks like a "wired OR gate".

The collectors of Q2 & Q3 are low when one or the other (or both) transistor is on.

It can only go high when both transistors are off.

As I can't see the rest of the circuit, I don't know what the objective is.

len

 

L

leonel

New Member

• Jun 8, 2005

• #4

Here it is the all circuit...
Can you give me the circuits basic ideia?
Thank you...

 

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ljcox

Well-Known Member

• Jun 8, 2005

• #5

The letters are too small to read.

Can you convert it to a .gif ?

Or partition it into larger sections.

Len

 

L

leonel

New Member

• Jun 9, 2005

• #6

Hello.
Here´s the *.gif. If you can´t read please say I´ll convert into different designs.
Thank you

 

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Nigel Goodwin

Super Moderator

Most Helpful Member

• Jun 9, 2005

• #7

leonel said:

Hello.
Here´s the *.gif. If you can´t read please say I´ll convert into different designs.
Thank you

Click to expand...

It doesn't help at all, because you've only converted the previous JPG to GIF, the JPG (as well as being far larger than a GIF) has already lost all the detail of the diagram, so converting the JPG to a GIF doesn't help at all. You need to start with a decent quality format, THEN convert it to GIF, any package should be able to generate a BMP (which is HUGE!), then convert that to GIF. I would also suggest you make the picture a bit bigger, even at good quality it's a bit small to see.

 

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ljcox

Well-Known Member

• Jun 9, 2005

• #8

I agree with Nigel.

I draw circuits using PowerPoint and then convert them to .gif or .png

Len

 

L

leonel

New Member

• Jun 9, 2005

• #9

New schema

Here´s a new schema. My voltage regulator is 18V, and, almost, my resistances values are 490K and 5K6.
Thank you.
Regards
Leonel Monteiro

 

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• esquema_total_616.gif

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ljcox

Well-Known Member

• Jun 9, 2005

• #10

It is still too dificult to read.

Len

 

Nigel Goodwin

Super Moderator

Most Helpful Member

• Jun 9, 2005

• #11

ljcox said:

It is still too dificult to read.

Len

Click to expand...

Still too small!.

 

L

leonel

New Member

• Jun 9, 2005

• #12

I put the resistances values... My transistor is BC547. If this helps... If it doesn't helps, that´s ok, i'll try other solution.
Regards

 

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ljcox

Well-Known Member

• Jun 9, 2005

• #13

That may help. Its 8:15 pm here so I'll look at it in the morning

Len

 

L

leonel

New Member

• Jun 9, 2005

• #14

Thank you

Ok thank you...
In Portugal it´s 11h15 am

 

audioguru

Well-Known Member

Most Helpful Member

• Jun 9, 2005

• #15

Hi,
With the bases of a few transistors being fed from a 490k to 5.6k voltage divider, the input signal must be at least 57.5V for the transistors to begin to turn on. Where is such a high voltage coming from?

 

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ljcox

Well-Known Member

• Jun 9, 2005

• #16

I agree with Audioguru. The "490k" resistors must be much lower.

I have extracted the essential elements of the circuit and have redrawn it in a recognisable configuration. I could not read the transistor numbers, so I labelled them myself.

Q1 and Q2 are configured as a Set/Reset Flip Flop. it is set and reset by transistors Q3 and Q4.

Assume that A and B are low, and Q1 is on, therefore Q2 is off. If a high is applied to A, Q4 is turned on, but no change will occur since Q1 is on. But if a high is applied to B (with A low), Q3 will turn on, this will turn Q1 off and so Q2 will turn on.

The FF is now in its other state Q2 on Q1 off. Now if a high is applied to A (with B low), Q4 will turn on, this will turn Q2 off and so Q1 will turn on.

The FF is now in its initial state.

Len

 

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• circuit_extract.gif

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L

leonel

New Member

• Jun 13, 2005

• #17

Resistor value

Your right!
The correct value is 39K. Thank you for your help!

 

L

leonel

New Member

• Jun 15, 2005

• #18

Just one stupid question...
I don´t understand if my inputs A and B are high??

 

L

leonel

New Member

• Jun 15, 2005

• #19

Confused

In the image i have 4 input possibilities
0 0
0 1
1 0
1 1

What´s the state of my output in each possibility?
I don´t understand if for example A=1 and B=0. With A my output=0 and for B my output=1. There´s a conflict between each transistor! It´s possible having this configuration?
Please help me... i´m confused.

 

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• transistores.jpg

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audioguru

Well-Known Member

Most Helpful Member

• Jun 15, 2005

• #20

Hi Leonel,
It's an OR gate with an inverted output. The output is low if A or B or both are high.