Voltage Divider Question

[fuzzy645]

OK, I want to see if I am understanding the concept of a voltage divider.

I think I found the correct formula as:



It seems to me if the value of the 2 resistors is the same it will cut the voltage by 50%. I actually tested this with a multi-meter.  Here is a diagram of what I tested:



First I used a pair of 220K resistors for both R1 and R2, and measured the voltage at the "hanging" wire (presumably that would be V out), and lo and behold it measured approx. 4.5 V.

I then swapped out both resistors for a pair of 1.1M resistors and since the resulting fraction still reduces to 1/2, I got the same voltage, namely 4.5 V.

OK, here are my questions:

1. In the Davisson Easy drive schematic, he has one of these voltage dividers, but what I would be considering the V Out wire (like my hanging wire) passes through a cap then to ground. That seems like a waste to me. Why would he bother to go through the effort of cutting the voltage in 1/2, and then proceed to send the halved voltage straight to ground (do not pass go, do not collect $200).  I have a diagram of Davisson's circuit below.

2. Is the purpose of the voltage divider to send less juice to the transistor so we don't blow it out?

3.  Who decides that 4.5 is the magic number for volts?

4. In the Davisson circuit below, which was does the current flow?  It looks like to me (and my NOOB eyes) that the battery is connecting into both the collector and bass of the transistor.

[arawn]

I can tell you why 4.5v it's half of 9. Basically your Vref determines how much you tranny/opamp can swing voltage wise. half of supply gives you the most swing. no matter what V= those equal dividers will always give you half.

[deadastronaut]

hmmm interesting question...

so does that mean that any value resistors in a voltage divider would give 4.5v?.......as long as they are the same value?....hmmmmm...

[Gurner]

Here's how to work out junction voltage.

Assume two resistors in the divider chain ...a top resistor called R1 (let's say 1k) & lower resistor R2 (let's say 5k) ....and the supply voltage 'V' (let's say 9V)

First work out the total current through both resistors....

total current  = 9V / (1k + 5k)   ...or 9/6000 , which results in 0.0015A ...or 15mA

now just multiply the lower resistor by the total current ....so 5,000 (ie 5k) * 0.0015 = 7.5V

The voltage at the junction is 7.5V.

Or if you're lazy, just use a potential divider calculator  http://www.electronics2000.co.uk/calc/potential-divider-calculator.php  

(assuming no other 'reactive' component in the chain the above works for DC  or AC voltage)

so does that mean that any value resistors in a voltage divider would give 4.5v?.......as long as they are the same value?

Assuming that the voltage at the top of the chain is 9V & the bottom of the chain is 0V (ground) - Yes.

[harmonic]

My question (and i think the OP's as well) is why bother to split the voltage and then pass it to ground. And, why through the cap?

[deadastronaut]

ahhh gotcha...(sound of penny dropping)...cheers G..

[lopsided]

somebody knowing sure will elaborate this. just my two cents: this part of Easy drive is not a DC voltage divider, because the cap will block DC.
It's more likely some kind of frequency filter for the feedback loop. Or so I would guess...

J.

[Gurner]

To my eyes, that looks like -ve feedback, with a LPF incorporated....therefore higher frequencies will not be as attenuated as much as lower frequencies as a result of the -ve feedback (ie therefore high end accentuation)

So while it looks like a potential divider...imho I don't think that its purpose here.

[fuzzy645]

My question (and i think the OP's as well) is why bother to split the voltage and then pass it to ground. And, why through the cap?

Yes, that is one of my  main questions. It seems to me the transistor will see the full 9 volts.

[fuzzy645]

To my eyes, that looks like -ve feedback, with a LPF incorporated....therefore higher frequencies will not be as attenuated as much as lower frequencies as a result of the -ve feedback (ie therefore high end accentuation)

So while it looks like a potential divider...imho I don't think that its purpose here.

Thank you.

What exactly is -ve feedback?

[lopsided]

My question (and i think the OP's as well) is why bother to split the voltage and then pass it to ground. And, why through the cap?

Yes, that is one of my  main questions. It seems to me the transistor will see the full 9 volts.

remember that you can not pass DC through a cap.
when you consider the DC aspect of the circuit you have: 50K from V+ to collector, 440K (220K and 220K in series) from collector to base and 1K from emitter to ground.
the other components affect the AC (your guitar) signal

so the way I see it (and I might be wrong) you have a 50/440 divider which gives 8,1 to the collector and 0,9 to the base.

"What exactly is -ve feedback?"

it's negative feedback: you take a little from the output signal (at the collector) - which is phase inverted to the original signal - and pass it back to the input.  In this case it is clipped by the diodes and passed through the RC filter. Because it's phase inverted it is "subtracted" from the original signal - hence negative feedback.

[fuzzy645]

... you have: 50K from V+ to collector,
440K (220K and 220K in series) from collector to base
.....
so the way I see it (and I might be wrong) you have a 50/440 divider which gives 8.1 to the collector

Now THAT I understand perfectly.  Makes sense 

....... and 0,9 to the base.

...and now you lost me again.  Why .9 to the base?  Personally, I see 2 paths into the base.  One path is through those diodes, and the other path is through the pair of 220K resistors in series.  Where do you get the .9 volts into the base?

I must say the multiple paths in some of these circuits confuses me more than anything.

"What exactly is -ve feedback?"
it's negative feedback: you take a little from the output signal (at the collector) - which is phase inverted to the original signal - and pass it back to the input.  In this case it is clipped by the diodes and passed through the RC filter. Because it's phase inverted it is "subtracted" from the original signal - hence negative feedback.

Is this phase inverted business a "rule of thumb" - in other words any signal that goes through a transistor gets phase inverted when it comes out the other end of the sausage machine, simply because it passed through a transistor?  I'm having a rough time visualizing this one and understanding what its doing for us?

Thanks you!

[kurtlives]

hmmm interesting question...

so does that mean that any value resistors in a voltage divider would give 4.5v?.......as long as they are the same value?....hmmmmm...

Yep

Though if you try with really high value resistors you might see funky looking numbers on your DMM. That's just your DMM being loaded down.

[jaapie]

...and now you lost me again.  Why .9 to the base?  Personally, I see 2 paths into the base.  One path is through those diodes, and the other path is through the pair of 220K resistors in series.  Where do you get the .9 volts into the base?

There's a capacitor between the diodes and the battery, so DC can't take that path.

[fuzzy645]

...and now you lost me again.  Why .9 to the base?  Personally, I see 2 paths into the base.  One path is through those diodes, and the other path is through the pair of 220K resistors in series.  Where do you get the .9 volts into the base?

There's a capacitor between the diodes and the battery, so DC can't from taking that path.

Thank you...that makes sense!!!  Now stuff is really starting to make a bit more sense to me (so I think anyway)

So it seems to me that caps are used for 2 main reasons in this case:

1). to be some kind of "filter" - which will be the case when the cap heads to ground (bleeds signal there).

2). as a way to block DC from some part of the circuit

That leads to more questions:

a). Why the 22 uf elecrolytic cap at the end?  That one seems to have a different purpose (although it is going to ground)? 

b). I notice most of these circuits take care to put an input cap and output cap (usually .1 uf) to block DC on input and output. What would happen if we didn't do that, would the guitar player get a "shock" from the 9 Volts, or would it just sound like garbage?

c). I notice these caps that block DC are all .1 uf.  How do we determine that value (.1 uf)?  Is there some formula to determine the cap value to block DC given a max voltage of 9V??

[lopsided]

I see 2 paths into the base.  One path is through those diodes, and the other path is through the pair of 220K resistors in series.  Where do you get the .9 volts into the base?

First of all. Keep in mind I am quite a newbie myself and am to say just one step ahead of you in these things.

One think I have learned: when you a DC analysis of a circuit , try to redraw it , and change all caps into resistors of infinite resistance or just cut the circuit open there.  This way the diodes aren't a part of the circuit because they are connected just at one side to the circuit - no DC can pass through them

Is this phase inverted business a "rule of thumb" - in other words any signal that goes through a transistor gets phase inverted when it comes out the other end of the sausage machine, simply because it passed through a transistor?  I'm having a rough time visualizing this one and understanding what its doing for us?

It is not so simple there are more transistor configuration. Not every inverts the signal. One of them is called "common emitter", it is the most common in the stompbox circuits, and it inverts the signal phase. You can usually identify a transistor amplifier in common emitter configuration , when 1) signal comes to the base, 2) emitter is grounded (sometimes through a resistor), 3) output signal is at the collector.

[lopsided]

1). to be some kind of "filter" - which will be the case when the cap heads to ground (bleeds signal there).

correct, but keep in mind they never work as a filter, alone but always in configuration with a resistor. They are called RC filters (look http://upload.wikimedia.org/wikipedia/commons/e/e0/RC_Series_Filter_(with_V%26I_Labels).svg  or search high-pass and low-pass filter)

a). Why the 22 uf elecrolytic cap at the end?  That one seems to have a different purpose (although it is going to ground)? 

It gets a little more complicated again. As I have said when we do a DC analysis we can leave this cap out completely. But when we do AC/guitar signal analysis the cap works as a bypass of the emitor resistor. As much as I know this makes the gain bigger. But I am not hundred percent sure how this works myself.

[ashcat_lt]

a). Why the 22 uf elecrolytic cap at the end?  That one seems to have a different purpose (although it is going to ground)?  

This one mostly blocks DC on the wiper of the Gain pot.  As you turn the pot, the AC gain changes, but the DC gain and therefore the bias of the transistor remain the same.

b). I notice most of these circuits take care to put an input cap and output cap (usually .1 uf) to block DC on input and output. What would happen if we didn't do that, would the guitar player get a "shock" from the 9 Volts, or would it just sound like garbage?

Without the "coupling" caps at the input and output you'll have DC across input and output, which might damage or cause unpredictable performance of other circuits to which it is connected.  Those circuits in turn might mess up the somewhat delicate biasing of this one.

c). I notice these caps that block DC are all .1 uf.  How do we determine that value (.1 uf)?  Is there some formula to determine the cap value to block DC given a max voltage of 9V??

All caps block DC.  More specifically, a capacitor's impedance increases with decreasing frequency, until it becomes infinite at 0 Hz.  If all you care about is blocking DC you can use any cap with a Voltage rating somewhat greater than what you expect to see across it.  The cap value is chosen to allow the low frequency AC signals through while blocking the DC.  They can be adjusted to change the bass response of the circuit.  

[lopsided]

1). to be some kind of "filter" - which will be the case when the cap heads to ground (bleeds signal there).

correct, but keep in mind they never work as a filter, alone but always in configuration with a resistor. They are called RC filters (look http://upload.wikimedia.org/wikipedia/commons/e/e0/RC_Series_Filter_(with_V%26I_Labels).svg  or search high-pass and low-pass filter)

to make it clear, you're sentence in not hundred percent correct, because they work as filters also when the the cap doesn't head to ground, but the resistor after the cap does.
reading through the wikipedia entries for high-pass and low-pass filter can make it a little clearer.

[fuzzy645]

OK, so here is my analysis based on what I've learned from your EXCELLENT responses (thank you to all who have responded).

DC ANALYSIS (by a simpleton)
------------------------------------
9V DC heads in at the top and immediately encounters a 50K resistor.   

DC would try to make a left turn towards the output, but it can't due to the .1 uf cap. 

DC current then travels straight and encounters yet another intersection. It tries to make a right turn here but can't again due to yet another pesky .1  uf cap.   

OK, DC current continues straight down the schematic and encounters yet another intersection. This is a very interesting intersection.  9V DC will continue to travel straight down to the COLLECTOR of the transistor.  However, upon making a left turn a voltage divider is formed (R1 = 50K and R2 = 440K) so DC measuring at approx 8V then travels around entering the BASE of the transistor.   

Then DC travels out the TRANSISTOR and down to the GAIN pot too (I guess). I have no idea what the voltage would be here.  The 22 uf cap here blocks more DC to ground (still fuzzy on this entire part for sure)?

Can anyone create a corresponding AC analysis now???  I would guess that AC heads into the base of the transistor and comes out louder the other end, and then out to the output (ok, that is too simple, please help LOL!!!)

..also,

please feel free to correct my DC analysis.