RC in Parallel

I'm trying to figure something out and so far no one has come up with an answer.

If I have a voltage divider network (Just 2 resistors) and I bypass the bottom one with a cap with an AC source in. How do I figure out the impedance?

My goal is the following: put a sine wave in with say 4v p-p in @1kHz, have an R1 and R2 (equal value) with a bypass cap on R2 (the other side of R2 is connected to ground) and say I get 1v p-p out (the middle of R1 and R2 is my output). I want to take that circuit and replace it with just R1 and R2 (keeping R1 the same value) and get the same amplitude out. I don't care about phase right now.

Anyone have any ideas?

Reply to
Peter
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I've had no need to figure anything remotely close to this in years but; Don't you have to calculate XC or capacity reactance to get Z impedance in an RC network ?

good luck !

Roy ~ E.E.Technician

Reply to
Roy Q.T.

If you already know what you want for Vout (1 volt) and you know that Vin is 4 volts, why do you care what the cap is doing, since it won't be in the new circuit? You know its presence in the existing circuit with equal Rs causes the the thing to divide by 4.

Anyway: Vout = Vin*NewR2/(R1+NewR2) where Vout appears across the NewR2

If you want to know the impedance of your existing R2 and cap, solve the above equation for the new R2. Then solve for Zc with R2new = R1*Zc/(R1+Zc). Or figure the capacitave reactance at 1 khZ Xc = 1/2*pi*f*C. In this case, Xc and Zc are the same.

Ed

Reply to
ehsjr

Do you mean the impedance as seen by the supply? In which case, simply work out the impedance of one resistor in series with a parallel combination of the capacitor, the other resistor (and what ever load you might want to work it out for).

Do you mean the impedance as seen by a load? In which case it is the impedance of the parallel combination of the resistor and capacitor, in series with the resistor (and the impedance of ths source, if not considered negligable).

The voltage source will see R1 in series with the parallel combination of R2b and your load impedance. Assuming that your load impedance is high enough to be insignificant, this reduces to R1 in series with R2b. If you can't work out the value of R2b - bearing in mind that the same current flows through R1 and R2b but that the voltage across R1 is three times the voltage across R2b, then you want to take up needlework and/or cookery and leave electrickery to us girls.

Reply to
Palindr☻me

Why do you have the resistor bypassed if you are trying to make a voltage divider? In a high frequency attenuator, both resistors are in parallel with capacitors and have their time constants equal like in a scope. But, I don't understand what you are trying to do. Bob

Reply to
Bob Eldred

presented to the previous stage or the next stage?

"Bypass" means AC short, so the impeadance presented to the next stage will be low (looking mostly at the cap impeadance)

You have 1/2 voltage out R2/(R1+R2) (no cap)

Now your divider is 1/4 so Z of the cap is about 1/3 of R (estimate)

So take the cap out? and have a voltage division of 1/4?

1/4 = R2/(R1+R2) so 4*R2=R1+R2 or R1= 3*R2

So if R1 is 1k then R2= 330 ohms and presented Z to previous stage is about 1.3k and Z presented to next stage is about 250 or so. (need Z's of the stages to have real answers)

Reply to
meqshesher

The reason I'm trying this circuit is because I was trying to understand bypass caps in a common emitter transistor circuit. Everytime I tried figuring out the impedance looking into the base, I didn't get what I expected.

So I built this circuit to learn from it then use what I learn and apply that back into a common emitter circuit.

Reply to
Peter

That's a different kettle of fish. Try this site to see if it helps:

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Ed

Reply to
ehsjr

Temperature stability of transister turn on threshold sys you must have some R in E leg to ground, then bypass it with cap to increase gain for AC amp

tiz anodder fish entirely!

Reply to
Schweinkolben

How far off are you calculation? Remember to include the base /emitter junction resistance. Mostly this value is insignificant unless the the total emitter resistance is by passed. If the emitter resistor is completely bipassed this may have the most significant effect on the input impedance depending on the circuit design.

Reply to
Jimmie

Well that's why I'm using THIS circuit so I don't have to worry about all the factors of the transistor.

If I can figure out how to go from a 2 resistor series circuit with the bottom resistor bypassed by a cap and the middle of the two resistors as my Vout to just a 2 resistor series circuit with ONLY changing the bottom value resistor to the equivilent of the R and C combiation and get the SAME Vout, then I will know for sure that my emitter is all set and any variations I get in calculations will be due to temp or whatever.

This is why this circuit is important. I don't care about phase. I ONLY care about how to go from 1kHz 4Vp-p with both resistors equal to 1k and the bottom resistor bypassed with a 1uf cap while I get 610mVp-p and then take the R and C to get a total impedance, remove the cap and install my new calculated value reistor so I can get the 610mVp-p output with JUST two resistors.

when I calculate the bottom resistor, I get 157ohms, but that does not equal

610mVp-p on the output. The value that does give me the same amplitude is 178ohms, but I have NO idea how to calculate that value.

Thanks again

Reply to
Peter

If I recall (its been 25 years since I had this class) the magnitude of the capacitive reactance is 1/2pi fC in ohms. If you run the numbers this is 159 ohms for your capacitor at 1kHz. in parallel with the 1kohm you get 137 ohms. you then have a voltage divider with 1k and 137 ohms. I think this would gove you a theoretical output of 480mV with 4 V in. (check my math; I could easily have a dumb error here). you say you were getting 610 mV out. This would suggest that the internal resistance of your cap is high making the divider different than if you assume your cap to be purely capacitive. you might try this without the lower resistor of the divider and see if you can determine if the cap value is correct (another possibility) and if you are perhaps getting a high ESR in the cap.

Reply to
no_one

is there a source impedance of the 4 vp-p generator in your sim?

Reply to
no_one

What's the big problem? E=I*R

Vr1=(4-0.610) voltage across r1 r1=1k a given V/R=I (0.00339) find current through r1 Vr2=(0.610) a given V/I=R find the resistance that will drop 0.61V at 3.39mA (0.610)/(0.00339)=179.9410029499

you said it works with a 178 ohm

Reply to
TheTechnician

OK. figure out the resistance of the R and C without using any voltages

Reply to
Peter

This has been answered before. In case you missed it: Xc = 1/2*pi*F*C That gives you the capacitve reactance in ohms. If you already know the total resistance of R and C in parallel: Rtot = (R*Xc)/(R + Xc) Solve for R

If you know R but do not know Rtot, the equation is the same: Rtot = (R*Xc)/(R + Xc) Solve for Rtot

Ed

Reply to
ehsjr

Sorry, this doesn't give the correct results. Xc is displaced 90 degrees from R. A 10 ohm resistor and a capacitive reactance of 10 ohms in parallel with it does *not* give you (10*10)/(10+10) = 5 ohms impedance.

With parallel components, to combine them it is useful to calculate the individual conductance and susceptance, then add them (being sure to account for resistance vs. reactance), then convert the resulting admittance back to impedance (you just got to love all these terms ;-)

For example, a 10 ohm resistor has a conductance of 0.1 mho. Similarly, the capacitor with 10 ohm reactance has a susceptance of 0.1 mho. These combine to give the complex admittance

For parallel circuits.... admittance = sqrt (conductance^2 + susceptance^2) = sqrt (0.1^2 + 0.1^2) = 0.1414 mho

So the impedance is 1/0.1414 = 7.07 ohm

Lumped all together, for a resistance R and capacitive reactance Xc....

Z = 1 / (sqrt( (1/R)^2 + (1/Xc)^2))

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If the resistance or reactance is significantly larger than the other, your formula will give you *approximately* correct results. But if the ohmic values are near each other, the proper calculation is what I just described.

daestrom

Reply to
daestrom

Right you are (I knew 25 years lack of use had me forgetting something!). What is troubling is that OP seems to have gotten numbers from both a sim as well as a breadboard circuit that don't match what the analysis says he should get. His 157 ohms matches what I calculate (using the magnitude of the suseptances) but I get a calculated output of 543 mV rather than his

610. I think the discrepancy is with the voltages that he is using as his basis for the replacement of the bottom leg of his voltage divider.

Reply to
no_one

Thanks - you are correct and my answer was wrong. Started thinking DC (obviously) when its AC. I hate it when I do that. Ed

Reply to
ehsjr

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