To achieve a duty cycle of less than 50% a diode can be added in parallel with R2 as shown in the diagram. This bypasses R2 during the charging (mark) part of the cycle so that Tm depends only on R1 and C1:
Hello there tbr75,
The equation for the time period (high or low) is:
t=ln(5/3)*R*C
where R is either R8 or R5 and C is the capacitor C1.
Because you have diodes in the feedback and each resistor works on only one part of the cycle, that equation has to be used twice, once for R8 and once for R5, and the two times have to be added to get the entire time period for one cycle.
This means we have:
t1=ln(5/3)*R8*C1
and
t2=ln(5/3)*R5*C1
and the total time period is:
tp=t1+t2
and that's how the two time periods work together.
The way we get the first equation above is by starting with the equation for a charging capacitor and equate that to the two voltages that the other resistors produce with an op amp output of either high or low, then solve for t.
That makes sense. Thank you MrAl. However when I plug in the resistor and capacitor values I get t1=.235s and t2=1.175s and tp=1.41s. And the Lt. Spice simulation gives the correct t1 and t2. Any idea why this is?
Hello again,
Yes. You have to read my previous post again because i added something to it while you were making your reply. It said that the real life times could be lower because the output of the LM324 does not go completely up to +5v, and in fact at 5v it would only go up to around 4v or so. You can verify this with your simulator.
The effects are that the charge time with the output high changes to:
t=ln(625/406)*R*C
and the discharge time will be slightly decreased also:
t=ln(198/125)*R*C
My original post was not meant to be super accurate, only to represent how things in general work. If the output went up to the full 5v then those equations would apply, but it doesnt go that high so we have to modify the equations a little.
Hi MrAl,
I have found with this configuration that the input bias and offset currents must be considered when using a general purpose op amp like the LM324. Note the impact this may have on the charge and discharge rate of the capacitor. The typical Ib of the LM324 is ~45nA with an Ios of ~100na, and will yield a lower output frequency that is not easily predictable due to its impact on the capacitors charge rate. A Fet input amp would be the one of choice with an Ib in the pA range, but that is not within the bounds of the OP, of course. Just something to consider.
Cheers
Merv
Hello again,
Yes. You have to read my previous post again because i added something to it while you were making your reply. It said that the real life times could be lower because the output of the LM324 does not go completely up to +5v, and in fact at 5v it would only go up to around 4v or so. You can verify this with your simulator.
The effects are that the charge time with the output high changes to:
t=ln(625/406)*R*C
and the discharge time will be slightly decreased also:
t=ln(198/125)*R*C
My original post was not meant to be super accurate, only to represent how things in general work. If the output went up to the full 5v then those equations would apply, but it doesnt go that high so we have to modify the equations a little.
Okay, so I'm guessing you are using this equation: Vc/Vmax = 1-e^(-t/RC). I have tried to figure out how you are getting your values but with no success. I was thinking along the right path as far as realistic outputs before reading your post, but even with that I can't figure it out. For my purposes, 4V is a close enough estimate for the high voltage and 1V is the low voltage.
You say to equate the Capacitor equation to the voltages across the other resistors. So does that mean when calculating the high time I should set Vc equal to the voltage across the 1meg resistor? (which shows as 1.86V peak in Lt. Spice).
Sorry for the questions, but I'd like to know how this works so I know for the future, not just turn in the correct values without knowing how I got there.
I did calculate the 2 volt hysteresis first. The calculations led to R4 being 3/2 of R3. Also, given only a 5v power supply I needed to add a voltage divider, thus the R2 resistor.
My frequency calculations did not include the impedance of the diodes.
That would get you close, but what of the calculations required for the project; are those not required to be presented and explained in your writeup? Ones approach to the solution is more important than the accuracy of the solution. The wrong approach with a close answer by accident was always graded down over the correct approach with a wrong answer. Anyone can drop a decimal point, for instance.
Here is a suggestion. Since the voltages at all points in both states are known, to calculate the values quickly, first set R4 at 10k or some other convenient value. Then take the voltage ratios of the drops across R3 & R4 in both states, and average the two ratios for the scalar of R2. That leaves summing the currents of R2 & R4 in one state and dividing that into the drop of R3 in the same state to get its resistance being mindful that the current of R4 reverses in the change of state. Sketching it out and filling in the known voltage drops helps me keep things straight being in my dotage.
Your configuration of presents a problem with two diodes in the feedback; one for each of the periods t1 & t2. Given the complexity of the dynamic resistance of both diodes in such a low current range, would it not be a good idea to eliminate the one with the lowest average current and the greatest swing in its dynamic resistance range? Doing that allows t1 to be calculated with just the RC components and a swamping effect for the t2 timing with R5 in parallel with R8 and the diode, mitigating the impact greatly.
Now to come up with the timing for t1, the 0.8s period. The capacitance can be selected based on a few factors like the diode impact. Period t1 is 0.8s and lets put the capacitance at 10uf and are the knowns so Rt1 must be determined. The charge on the cap over that period can be determined as (1+Ln t1)*C. here is a start on the equation:
R = X/((1+ln t1)*1e-5)
EDIT: OOPS!
After checking my math, I discovered a transposition error. However, you should get the idea and the path to follow. Sorry for the posting an error.
I'll leave X for you to puzzle out. Here's another hint...take a look at the LM555 datasheet and this app. note: **broken link removed**
I'll add a few attachments also. One plot will display the plot of a diode's dynamic resistance with the cursor at 1uA to display the issues noted above. The second will of the circuit I came up with just in case I wasn't clear in my explaination with the plot of the circuit with all parameters at less than 1% error. The only the know values are displayed in the schematic. Note the hysteresis and cap charging between the trigger level of 1.5V and the threshold level of 3.5V.
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