Help with circuit diagram

ericgibbs · Jul 8, 2009

axro said:
Ok so let me see if I have this right. R2 is connected to one end and R3 is connected to the other. There is always 10k ohms through the pot in this configuration. The wiper terminal is connected to the op amp, and moving it ONLY adjusts voltage beween max and min.

If this is correct, does the op amp see the 10kohms from the pot?

You are almost there!.

The opa dosnt see the resistance, it sees the voltage at the wiper of the pot.

The voltage on the wiper can be varied fro +0.5V thru +11.5V.

axro · Jul 8, 2009

ericgibbs said:
You are almost there!.

The opa dosnt see the resistance, it sees the voltage at the wiper of the pot.

The voltage on the wiper can be varied fro +0.5V thru +11.5V.

The opamp can "see" the 940 from R2/R3 though correct?

Also I see how you get the 11.5. But I don't understand the .5. I get that if the pot is wide open, the opamp won't see the whole 12 because of the resistors. But how does it get the 0.5? If you "close" the pot completely shouldn't it see 0 not .5?

MikeMl · Jul 8, 2009

axro said:
...If this is correct, does the op amp see the 10kohms from the pot?

Yes to the first question. Let me paraphrase your second question:

If I were standing on the non-inv input of the opamp, looking back at the pot wiper, what is the source resistance of the voltage that I see as the wiper is moved from top to bottom?

The answer can be answered with some algebra which I don't have time to solve for you for all positions of the wiper. The solution is called the "Thevenin Equivalent Source Resistance".

Lets just take a single position of the wiper; where the wiper is in the center. The resistance above the wiper is 5K (half the POT) + 470 = 5470Ω. So is the resistance below the wiper. For computing the source resistance, you find the parallel combination of the two 5470Ω resistances, which is 5470/2= 2735Ω.

The open-circuit voltage with the wiper centered is 12/2=6V.

If you were standing on the non-inv input looking back(electrically speaking), you cant tell if the non-inv node is being driven with the resistors/pot or a 6V source in series with a 2735Ω resistor.

Now, knowing the Thevenin Equivalent, you can ask the following very important question. If the input bias current into the non-inverting opamp input is e.g. 10uA, what is the actual voltage there?

Using the Thevenin Equivalent resistance of 2735Ω, what is the voltage drop across the resistor at 10uA? E=I×R=10e-6×2735=27.35mv, so the actual voltage at the non-inv input would be 6-0.02735= 5.973V. This is can be important (or insignificant) depending on what the opamp circuit is asked to do.

Now, using the Thevenin Equiv above, compute what the output voltage at the pot wiper would be if you were trying to draw 10mA off it? This goes to your suggestion that a pot is like a voltage regulator.

ericgibbs · Jul 8, 2009

axro said:
The opamp can "see" the 940 from R2/R3 though correct?

Also I see how you get the 11.5. But I don't understand the .5. I get that if the pot is wide open, the opamp won't see the whole 12 because of the resistors. But how does it get the 0.5? If you "close" the pot completely shouldn't it see 0 not .5?

hi,
I would suggest you enrol on a study course for basic electronics.

axro · Jul 8, 2009

MikeMl said:
Yes to the first question. Let me paraphrase your second question:

If I were standing on the non-inv input of the opamp, looking back at the pot wiper, what is the source resistance of the voltage that I see as the wiper is moved from top to bottom?

The answer can be answered with some algebra which I don't have time to solve for you for all positions of the wiper. The solution is called the "Thevenin Equivalent Source Resistance".

Lets just take a single position of the wiper; where the wiper is in the center. The resistance above the wiper is 5K (half the POT) + 470 = 5470Ω. So is the resistance below the wiper. For computing the source resistance, you find the parallel combination of the two 5470Ω resistances, which is 5470/2= 2735Ω.

The open-circuit voltage with the wiper centered is 12/2=6V.

If you were standing on the non-inv input looking back(electrically speaking), you cant tell if the non-inv node is being driven with the resistors/pot or a 6V source in series with a 2735Ω resistor.

Now, knowing the Thevenin Equivalent, you can ask the following very important question. If the input bias current into the non-inverting opamp input is e.g. 10uA, what is the actual voltage there?

Using the Thevenin Equivalent resistance of 2735Ω, what is the voltage drop across the resistor at 10uA? E=I×R=10e-6×2735=27.35mv, so the actual voltage at the non-inv input would be 6-0.02735= 5.973V. This is can be important (or insignificant) depending on what the opamp circuit is asked to do.

Now, using the Thevenin Equiv above, compute what the output voltage at the pot wiper would be if you were trying to draw 10mA off it? This goes to your suggestion that a pot is like a voltage regulator.

27.35 volts?

ericgibbs said:
hi,
I would suggest you enrol on a study course for basic electronics.

Haha I was waiting for that. I'm not as dumb as I sound. Basically I'm just trying to confirm what I "think" I know. I've been reading everything I can online, and I'd like to get a good book to go through. Taking a course is not really an option. I'm out of college already, and I really just want to do this as a fun hobby.

MikeMl · Jul 8, 2009

axro said:
MikeMl said:

...
Now, using the Thevenin Equiv above, compute what the output voltage at the pot wiper would be if you were trying to draw 10mA off it? This goes to your suggestion that a pot is like a voltage regulator.

Click to expand...

27.35 volts?
...

Yes, that would be the drop across the 2735Ω resistor if 10mA were flowing through it, but notice that the Thevenin Equiv voltage is only 6V, so there is no way you can draw 10ma. It was a trick question.

In fact, if you grounded the wiper, the maximum current (with the wiper at zero volts) would be I=E/R= 12/(470+5000)= 2.2mA max! So much for it being a substitute for a regulated voltage.

Stare at the attached simulation. Can you tell the difference between the behavior of the Original Circuit on the left compared to the Thevenin Equivalent on the right? The horizontal axis is the Load Resistor value plotted logarithmically. Note that as the load resistor get closer to zero Ω, the voltage is near zero, and the maximum current is 2.2mA, as calculated above.

Hero999 · Jul 8, 2009

Mark68 said:
What program makes that circuit diagram?

I don't know, I just edited it with MS paint.

I don't think anyone's noted my point that the 741 isn't a very good comparator. The LM311 is designed for the job and can directly driver a relay.

ericgibbs · Jul 9, 2009

axro said:
Haha I was waiting for that. I'm not as dumb as I sound. Basically I'm just trying to confirm what I "think" I know.... I really just want to do this as a fun hobby.

hi,
I never suggested you were dumb.
Its just some of the questions you are asking, are absolutely basic to understanding circuits.
I think its good you do things for fun, so do I.

You should get some of the basic bits of test gear, a multimeter, solderless project boards, a simple power supply etc. If you have the basic hardware you can do your own measurements on simple circuits and learn that way.

The problem in trying to explain some of the basics using a Forum is that we can only guess at your technical expertise and so often we have to dumb an explanation down.

Its a pity we just cant look over your shoulder, we could have answered the 10K pot problem in a few seconds.

You are obviously keen to learn, good luck with your new hobby.

axro · Jul 9, 2009

MikeMl said:
Yes, that would be the drop across the 2735Ω resistor if 10mA were flowing through it, but notice that the Thevenin Equiv voltage is only 6V, so there is no way you can draw 10ma. It was a trick question.

In fact, if you grounded the wiper, the maximum current (with the wiper at zero volts) would be I=E/R= 12/(470+5000)= 2.2mA max! So much for it being a substitute for a regulated voltage.

Stare at the attached simulation. Can you tell the difference between the behavior of the Original Circuit on the left compared to the Thevenin Equivalent on the right? The horizontal axis is the Load Resistor value plotted logarithmically. Note that as the load resistor get closer to zero Ω, the voltage is near zero, and the maximum current is 2.2mA, as calculated above.

I think I'm starting to understand. Thanks for all the help. I think I may need to study the Thevenin equation a little more to really get a good handle on it.

Just to make sure I have some of this right. In this pic(555timer astable circuit) :

**broken link removed**

If you had a 9volt power supply would the voltage at pin 7(I know thats the discharge pin, but just for learning sake) be 8.96?

ericgibbs · Jul 9, 2009

axro said:
If you had a 9volt power supply would the voltage at pin 7(I know thats the discharge pin, but just for learning sake) be 8.96?

hi,
If the astable is running it will not be at 8.96V.

How are you measuring it.?

EDIT:
Read the text at the bottom of this image.

its 2/3 and 1/3 Vsupply.

axro · Jul 9, 2009

ericgibbs said:
hi,
If the astable is running it will not be at 8.96V.

How are you measuring it.?

EDIT:
Read the text at the bottom of this image.

its 2/3 and 1/3 Vsupply on the discharge pin

I guess it was kind of a dumb question. More theoretical than anything.

But If I am understanding right. When output is high, the discharge pin is disconnected. The capacitor is then charged through R1/R2. So I was just asking if thats what would be seen there.

When the Cap gets to 2/3 Voltage then the discharge pin is connected and my numbers would be diff. But I would think my 8.96 would be correct or am I missing something?

ericgibbs · Jul 9, 2009

axro said:
I guess it was kind of a dumb question. More theoretical than anything.

But If I am understanding right. When output is high, the discharge pin is disconnected. The capacitor is then charged through R1/R2. So I was just asking if thats what would be seen there.

When the Cap gets to 2/3 Voltage then the discharge pin is connected and my numbers would be diff. But I would think my 8.96 would be correct or am I missing something?

hi,
Look at these waveforms for the threshold and discharge pins.
Ive changed the resistor values to make the wave more clear.

axro · Jul 9, 2009

Are the reasons my numbers are wrong because the Capacitor is in series with the 2 resistors and not parallel?

I can't find what I was reading before, but I seem to recall when connect a capacitor in series with a circuit and not parallel acts differently. Like it blocks the flow of electricity when it is full.....or something.

ericgibbs · Jul 9, 2009

axro said:
Are the reasons my numbers are wrong because the Capacitor is in series with the 2 resistors and not parallel?

I can't find what I was reading before, but I seem to recall when connect a capacitor in series with a circuit and not parallel acts differently. Like it blocks the flow of electricity when it is full.....or something.

hi,
The use of a capacitor in a 'coupling' circuit , when its used to block the dc component of a signal is different from the 555 circuit.

In the 555 astable, the cap charges up and down, so the voltage on the cap rises and falls.
Its used as timing capacitor and the two resistors control the current into the cap [charge] and flowing out of the cap [discharge]

axro · Jul 9, 2009

ericgibbs said:
hi,
The use of a capacitor in a 'coupling' circuit , when its used to block the dc component of a signal is different from the 555 circuit.

In the 555 astable, the cap charges up and down, so the voltage on the cap rises and falls.
Its used as timing capacitor and the two resistors control the current into the cap [charge] and flowing out of the cap [discharge]

Right, so then why would the Charging of the capacitor affect the voltage on the discharge pin? Or even say the discharge pin was replaced with the + lead of an LED. Shouldn't the discharge pin see the voltage from the battery?

Sorry again, I'm sure my noobness is showing

ericgibbs · Jul 9, 2009

axro said:
Right, so then why would the Charging of the capacitor affect the voltage on the discharge pin? Or even say the discharge pin was replaced with the + lead of an LED. Shouldn't the discharge pin see the voltage from the battery?

Sorry again, I'm sure my noobness is showing

hi,
You shouldnt consider the Res and Caps without knowing how the 555 works.

The 555's internal circuit detects when the cap voltage is 1/3 and 2/3 of the supply voltage and switches the two resistors in and out of the circuit to charge and discharge the cap.

Do you have the 555 datasheet.?

axro · Jul 9, 2009

ericgibbs said:
hi,
You shouldnt consider the Res and Caps without knowing how the 555 works.

The 555's internal circuit detects when the cap voltage is 1/3 and 2/3 of the supply voltage and switches the two resistors in and out of the circuit to charge and discharge the cap.

Do you have the 555 datasheet.?

I'm pretty sure I know how the 555 works. It detects when it gets to 2/3 on the threshold. When it gets to 2/3 Discharge connects to ground and cap discharges through R2. When Trigger sees 1/3 voltage discharge disconnects and cap charges again through both R1/R2. Right?

ericgibbs · Jul 9, 2009

axro said:
I'm pretty sure I know how the 555 works. It detects when it gets to 2/3 on the threshold. When it gets to 2/3 Discharge connects to ground and cap discharges through R2. When Trigger sees 1/3 voltage discharge disconnects and cap charges again through both R1/R2. Right?

hi,
Looking good.

Check this image, tell me what you think.. RA is the top resistor and RB the bottom.

axro · Jul 9, 2009

ericgibbs said:
hi,
Looking good.

Check this image, tell me what you think.. RA is the top resistor and RB the bottom.

I've actually used that before It tells you how long it takes to Charge and discharge to the 2/3 and 1/3 based on the value of the resistors and the capacitance.

ericgibbs · Jul 9, 2009

axro said:
I've actually used that before It tells you how long it takes to Charge and discharge to the 2/3 and 1/3 based on the value of the resistors and the capacitance.

Look at the actual threshold waveform thats specified.

Help with circuit diagram

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