Voltage Divider Question

[digi2t]

Ummm, going in a slightly different direction here...

How does voltage dividing, urrr, reducing I guess would be a better term here, work with Zener diodes? I've really gotten into the Colorsound Vocalizer and Dipthonizer lately, and I see that they use Zeners to reduce the voltage. What's the deal here? Just bleeding off the excess voltage to ground, with a resistor to make up for the little bit extra, and a cap for filtering?




Like I said, not really a voltage divider question, but I was just wondering.

[jaapie]

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.

You've got the right idea. I'm grossly oversimplifying here, but basically the transistor acts like a valve. The current going into the base controls how much current flows through the path between the emitter and collector... kind of like squeezing a garden hose to control how much water flows through it.

read the first few pages here (http://www.allaboutcircuits.com/) on DC and AC circuits and semiconductors and it should be a lot easier to figure out what's going on in this (or any other) circuit.

[jaapie]

Ummm, going in a slightly different direction here...

How does voltage dividing, urrr, reducing I guess would be a better term here, work with Zener diodes? I've really gotten into the Colorsound Vocalizer and Dipthonizer lately, and I see that they use Zeners to reduce the voltage. What's the deal here? Just bleeding off the excess voltage to ground, with a resistor to make up for the little bit extra, and a cap for filtering?

using a diode in place of the lower resistor helps regulate the output voltage if the supply voltage changes. for instance, if you have a 9v supply and a divider made up of two 100k resistors, you'll get 4.5 volts out as long as the supply stays at 9v. If the supply sags down to, say, 7v under a heavy load, though, you'll only be getting 3.5 volts out of the divider. Depending on what comes after the divider, 3.5v might still work out ok. but if whatever comes next has to have exactly 4.5v to operate correctly, you've got a problem.

if you replace the lower resistor with a 4.5v zener, the output of the divider will always be 4.5v because the voltage dropped by the diode doesn't change with the supply voltage like it does with a resistor.

[lopsided]

ok, let's try to sort it out. It helps me to understand the stuff better myself.

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.   

the last paragraph seems to be wrong. I wasn't sure before, but i am pretty sure that we have it wrong.
We can not take this as a simple Voltage divider. I think it is because the 440K resistor isn't grounded and
is connected to the base.

There is java circuit simulator: http://www.falstad.com/circuit/
best just paste this link into your browser

Code Select Expand

http://www.falstad.com/circuit/#%24+1+5.0E-6+10.634267539816555+43+2.0+50%0Aw+384+256+384+304+1%0Aw+384+224+384+176+1%0A172+384+112+384+80+0+6+9.0+9.0+0.0+0.0+0.5+Collector+Voltage%0At+336+240+384+240+0+1+-7.983092873805362+0.09853954517627035+100.0%0Aw+272+240+336+240+1%0Ar+384+112+384+176+0+50000.0%0Ar+384+176+272+176+0+440000.0%0Ar+384+304+384+352+0+1000.0%0Ag+384+352+384+384+0%0Ag+272+176+272+208+0%0A


in this case we have grounded the 440K resistor and we really have 8v at the collector (just point your mouse over the part). But our base has no voltage - the transistor does not conduct.

now compare to our actual DC circuit:

Code Select Expand

http://www.falstad.com/circuit/#%24+1+5.0E-6+10.634267539816555+43+2.0+50%0Aw+384+256+384+304+1%0Aw+384+224+384+176+1%0A172+384+112+384+80+0+6+9.0+9.0+0.0+0.0+0.5+Collector+Voltage%0At+336+240+384+240+0+1+-0.6666721065460202+0.5287184951300044+100.0%0Aw+272+240+336+240+1%0Ar+384+112+384+176+0+50000.0%0Ar+384+176+272+176+0+440000.0%0Ar+384+304+384+352+0+1000.0%0Ag+384+352+384+384+0%0Aw+272+176+272+240+0%0A


we have just 1,3 V at collector, and 0,68 at the base.

Conclusion - we can not apply the simple voltage divider rule for the 50K , 440K part of this circuit.
Because the 440K is not grounded and the transistor makes it more difficult (still trying to understand this).

I will try to simulate the AC part, but I recommend checking the Simulator yourself, there are various pre-made circuits (including the filters), which can help understand

if you have a breadboard or a possibility to build the circuit, try it and then check the stuff with a voltmeter. It always helps when you see it with your own eyes.

[lopsided]

One more thing. From where do you have the schematic in the original post? I can not seem to find it. And I do not understand why the volume pot is the way it is. A standard way would be to put a volume pot with one lug to ground after the last cap. and this is also the way it is suggested on the original schematic here http://www.diystompboxes.com/analogalchemy/sch/easydrive.html

[Gurner]

One more thing. From where do you have the schematic in the original post? I can not seem to find it. And I do not understand why the volume pot is the way it is. A standard way would be to put a volume pot with one lug to ground after the last cap. and this is also the way it is suggested on the original schematic here http://www.diystompboxes.com/analogalchemy/sch/easydrive.html

It may not be standard, but it's a completely valid way of using it. The voltage rail is fixed at 9V....very low impedance - it ain't budging! There will be less & less AC signal present on the pot wiper as it approaches the 9V lug end of the travel (just like with a standard setup, there's less & less AC as the wiper travels towards the ground lug end of the travel). Of course there'll be more and more DC, but the follow on cap removes that. Also going the way of the OP's posted schematic, you save one component, because the transistors load resistor also acts as the volume control pot (whereas with the more conventional way you need the collector load resistor & a volume pot!)

[PRR]

> DC travels out the TRANSISTOR and down to the GAIN pot too (I guess). I have no idea what the voltage would be here.

You can put limits on it.

Assume transistor is open-circuit. That blocks all DC to the 1K resistance. Zero current times 1K is zero voltage.

Assume transistor is short-circuit. We have 50K and 1K in series. This is a voltage divider, output about 1/50 (actually 1/51) of supply. With 9V supply, the most we could have is 0.176V, a very small value.

In fact so small that it hardly matters for DC. And because the cap is large, it hardly matters for small AC voltage. IMHO, that whole emitter network may be neglected for linear analysis. It may matter for overload analysis (and since this is an overload box, it may matter; but overload analysis is too-too-tedious for hand computation).

For DC you first remove all the caps, 'cuz they are open-circuit for DC. With caps gone, the diodes can't do anything, erase them. The two 220K act as one 440K. This 440K leaks base current "down" into transistor, so draw it vertical. The 1K is teeny compared to 50K. Unless we find some reason why 0.18V matters, we may ignore it.

Assume the Base voltage is "small" compared to 9V.



What is collector current? Depends on base current times hFE. What is base current? Depends on voltage across R2. Which depends on voltage dropped by R1. Which depends on collector current.

Is it circular analysis? No. Note that if collector current increases, R1 drop increases, R2 drop decreases, base current decreases, collector current decreases. The system tends to find a balance.

To estimate the balance, we need a number for hFE. 2N5089 at 0.5*9V/50K or 100uA has hFE of 300 to 900. We could take 520 as the average value, but "440" is also valid and makes the arithmetic simpler.

440K, divided by hFE of 440, acts-like a 1K from collector to emitter. The low base current is multiplied by hFe to give collector current. The collector-emitter voltage is nearly equal to collector-base voltage. So a collector-base resistor can be approximated by a collector-emitter resistor hFE times smaller.

Now we have a voltage-divider 50K+1K. You should be able to work this out.

What is hFE is not 440? Try 100: 440K/100= 4.4K, divider is 50K+4.4K. Try 1000: divider is 50K+0.4K.

Note that for any hFE 100 to 1000 the R2 drop is very small compared to the R1 drop. R1 drops 91%-99% of supply voltage. For 9V supply nearly-all dropped in 50K, the collector current is 0.18mA.

Now add-back some things we skipped. The 1K in the emitter adds 1K*0.18mA or 0.18V. Base-emitter really drops 0.55V. R2 drops 0.8V to 0.08V. Base-ground voltage will be 1.5V to 0.8V. R1 voltage will be 9V-1.5V= 7.5V to 9V-0.8V= 8.2V. R1 current is really 0.164mA to 0.15mA.

We initially assumed 0.1mA then refined to 0.18mA. Checking that gave 0.164mA-0.15mA. We could go around the computation loop again with the revised values for a more exact solution. However the 3:1 variance in hFE and 10% variance in all resistors (5% fixed but 20% in pots) overwhelms the small uncertainty due to simplifying the DC analysis.

Collector stands at 0.8V to 1.5V. R1 current is 0.16mA. Emitter voltage is 0.16V.

Since collector is about 1V and emitter is about 0.2V, a perfect transistor could only pull-down 1V-0.2V or 0.8V negative peak. Even without the diodes, this thing is easy to overload.

[fuzzy645]

One more thing. From where do you have the schematic in the original post? I can not seem to find it. And I do not understand why the volume pot is the way it is. A standard way would be to put a volume pot with one lug to ground after the last cap. and this is also the way it is suggested on the original schematic here http://www.diystompboxes.com/analogalchemy/sch/easydrive.html

I ended up trying modify it again (on paper anyway) and the last version (I am calling ver B) Mark Hammer took a look at and indicated he thinks it will work. Here is the diagram:


As far as your point that we may not have a voltage divider at all because the resistors are not grounded, well there is one ground point between the pair of 220K resistors (that flows through a cap) so I wonder if that is the reason for that ground point. 

I will now hook all of this all up and measure my voltages and report back to the forum once I have it hooked up!!

[lopsided]

It may not be standard, but it's a completely valid way of using it. The voltage rail is fixed at 9V....very low impedance - it ain't gudging! There will be less & less AC signal as the pot wiper approaches the 9V lug end of the travel (just like with a standard setup, there's less & less AC as the wiper travels towards the ground lug end of the travel). Of course there'll me mor and more DC, but the follow on cap removes that.

I understand. Thanks for clarifying. I was a little unsure about the setup because the simulation gave me positive voltage on the other side of the cap and I could not understand it:

Code Select Expand

http://www.falstad.com/circuit/#%24+1+5.0E-6+1.500424758475255+66+2.0+50%0Aw+368+256+368+304+1%0At+320+240+368+240+0+1+-0.036049604844298644+0.5310215613364145+100.0%0A174+368+304+368+336+0+1000.0+0.5495000000000001+Gain%0Ag+368+336+368+384+0%0Aw+368+224+368+160+0%0Ar+368+160+304+160+0+220000.0%0Ar+304+160+240+160+0+220000.0%0Ac+304+160+304+192+0+1.0E-8+0.8608784517681266%0Ag+304+192+304+208+0%0Aw+320+240+240+240+0%0Aw+240+240+240+160+0%0Aw+240+160+240+112+0%0Ad+240+112+288+112+1+0.805904783%0Aw+288+112+288+96+0%0Aw+240+112+240+96+0%0Ad+288+96+240+96+1+0.805904783%0Ac+288+96+368+96+0+1.0E-8+-0.19290836672051148%0Aw+368+112+368+160+0%0Aw+368+112+368+96+0%0A174+368+96+400+32+0+50000.0+0.33170000000000005+Volume%0AR+368+32+320+16+0+0+40.0+9.0+0.0+0.0+0.5%0Ac+240+240+176+240+0+1.0E-8+0.1891228767214826%0Aw+384+320+432+320+0%0Ac+432+320+432+384+0+2.2E-5+0.06490913296238507%0Ag+432+384+432+416+0%0Av+176+240+176+336+0+1+400.0+0.5+0.0+0.0+0.5%0Ag+176+336+176+368+0%0Ac+400+64+448+64+0+1.0E-8+-5.664801960847399E-12%0AO+448+64+480+64+0%0A

the other way worked for me fine

Code Select Expand

http://www.falstad.com/circuit/#%24+1+4.9999999999999996E-6+1.7725424121461644+67+2.0+50%0Aw+336+256+336+304+1%0At+288+240+336+240+0+1+-0.8709305928290256+0.5160014826884474+100.0%0A174+336+304+336+336+0+1000.0+0.0050+Gain%0Ag+336+336+336+384+0%0Aw+336+224+336+160+0%0Ar+336+160+272+160+0+220000.0%0Ar+272+160+208+160+0+220000.0%0Ac+272+160+272+192+0+1.0E-8+0.7932800820835654%0Ag+272+192+272+208+0%0Aw+288+240+208+240+0%0Aw+208+240+208+160+0%0Aw+208+160+208+112+0%0Ad+208+112+256+112+1+0.805904783%0Aw+256+112+256+96+0%0Aw+208+112+208+96+0%0Ad+256+96+208+96+1+0.805904783%0Ac+256+96+336+96+0+1.0E-8+-0.3497435695792195%0Aw+336+112+336+160+0%0Aw+336+112+336+96+0%0AR+336+32+288+16+0+0+40.0+9.0+0.0+0.0+0.5%0Ac+208+240+144+240+0+1.0E-8+0.3935012586649378%0Aw+352+320+400+320+0%0Ac+400+320+400+384+0+2.2E-5+0.08992934491019093%0Ag+400+384+400+416+0%0Av+144+240+144+336+0+1+400.0+0.5+0.0+0.0+0.5%0Ag+144+336+144+368+0%0Ac+336+112+384+112+0+1.0E-8+0.9682140398296556%0Aw+400+144+480+144+0%0AO+480+144+528+144+0%0A174+384+112+384+176+0+100000.0+0.0050+VOlume%0Ag+384+176+384+192+0%0Ar+336+32+336+96+0+68000.0%0Ao+24+32+0+34+0.625+9.765625E-5+0+-1%0Ao+28+64+0+34+1.1692013098647223+1.52587890625E-56+1+-1%0A

never mind I am still figuring the program out.

[lopsided]

As far as your point that we may not have a voltage divider at all because the resistors are not grounded, well there is one ground point between the pair of 220K resistors (that flows through a cap) so I wonder if that is the reason for that ground point. 

no, the ground point is after a cap - therefore not significant for a voltage divider of Direct Current.
The cap and the ground is there probably for filtering of some frequencies.
But let's put this aside for a while and move to the bench to see if you will make it work.
Good luck

[Gurner]

Thanks for clarifying. I was a little unsure about the setup because the simulation gave me positive voltage on the other side of the cap and I could not understand it

Perhaps it's my PC, but I'm not seeing your code/simulations...but fwiw, you'll need a pull down resistor to ground on the other side of the cap...else that right hand side cap leg will just float.

no, the ground point is after a cap - therefore not significant for a voltage divider of Direct Current.The cap and the ground is there probably for filtering of some frequencies.

exactly, here's what I said earlier...

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)

[Gurner]

Thanks for clarifying. I was a little unsure about the setup because the simulation gave me positive voltage on the other side of the cap and I could not understand it

Perhaps it's my PC, but I'm not seeing your code/simulations...but fwiw, you'll need a pull down resistor to ground on the other side of the cap...else that output cap right hand side leg will just float.

[Quackzed]

on those 2 220k resistors with the cap to ground in the middle.
if you omit the cap, you have 440k from collector to base,  negative feedback. a high resistance of 440k only lets a small amount of signal back into the base. being out of phase with the signal at the base, when you add a bit of this 'opposite phase' signal to the normal 'in phase' signal -you reduce the level of the input signal by a little.  does that make sense?
if we added 2 signals of the same phase, they would add together and be twice as large- or loud!
if we add 2 signals of opposite phase they would cancell each other out and be silence!
so adding a tiny bit of opposite phase signal to the input reduces its level a little.
if we made that 440k resistance -say 20k we would add more out of phase signal tothe input and reduce its level a lot!
now having that cap there between the 220k resistors is  sort of a low pass filter.
from the collector through the right side 220k resistor then through a .01uf cap to ground, you have a basic low pass filter that would roll off treble at @ 72hz. (simplification)
BUT!!! its rolling off those frequencies in the negative feedback path, so whatever you send through the negative feedback path REDUCES the input volume... if some treble gets grounded on the way, then it cant make it back to the base to REDUCE those treble frequencies and consequently it only reduces the lows below 72hz fully, then reduces volume at the input less and less (3db per octave) for frequencies higher than 72hz.
...thats my take on it. i'm just a hack but that may help you to get an idea of whats happening there.

[digi2t]

Ummm, going in a slightly different direction here...

How does voltage dividing, urrr, reducing I guess would be a better term here, work with Zener diodes? I've really gotten into the Colorsound Vocalizer and Dipthonizer lately, and I see that they use Zeners to reduce the voltage. What's the deal here? Just bleeding off the excess voltage to ground, with a resistor to make up for the little bit extra, and a cap for filtering?

using a diode in place of the lower resistor helps regulate the output voltage if the supply voltage changes. for instance, if you have a 9v supply and a divider made up of two 100k resistors, you'll get 4.5 volts out as long as the supply stays at 9v. If the supply sags down to, say, 7v under a heavy load, though, you'll only be getting 3.5 volts out of the divider. Depending on what comes after the divider, 3.5v might still work out ok. but if whatever comes next has to have exactly 4.5v to operate correctly, you've got a problem.


if you replace the lower resistor with a 4.5v zener, the output of the divider will always be 4.5v because the voltage dropped by the diode doesn't change with the supply voltage like it does with a resistor.

Thanks a bunch. Goin' to bed less stoopid, again. God I love this place 

[Gurner]

...thats my take on it. i'm just a hack but that may help you to get an idea of whats happening there.

which was a more elegant explanation of what I said earlier today ...

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)

If you cut the lows, it has the overall vibe of accentuating the highs.