# Choosing Component Values for DCO Design

This is a sawtooth DCO, the comparator U1A resets the ramp at the desired frequency (thus controlling the frequency), and the voltage V7 applied to integrator U2A determines the rate of change of the the ramp (determines the amplitude). I am trying to find the values of RS2 and C1 of the integrator circuit.

Known:

1. Maximum Integrator Input Voltage V7 = 5V
2. Desired integrator Output Voltage after 500 microseconds = around 5V
• (500us since a frequency of 2000Hz as used in equation below)
3. Op amp: TL084

Unknown:

1. RS2
2. C1
3. Input current at maximum voltage

I understand that the current flows straight through (virtual ground), so the equations below should apply. My question is, which value of C1, RS2 or max current is chosen first, and why?

• You haven't figured out the "list item" markdown syntax yet, Jack. Put a blank line before the list and the -  (hyphen space) before each list item line. Do capitalise and punctuate properly. No need for the work "question" in the title. All questions are questions. Just state the topic. Apr 3, 2018 at 21:02
• Sorry I was still editing it, I didn't realise that you could see the change below. how is that? Apr 3, 2018 at 21:05

My question is, which value of C1, RS2 or max current is chosen first, and why?

Generally we'll pick the capacitor first. They come in fewer values per decade (E3 or E6 series) than resistors which are available in E12 or E24 series. Pick the capacitor and then calculate the resistor based on that. But the two are intertwined ...

We want to pick a mid-range resistor value. Too low and we'll load the previous stage and need a large capacitor. Too high and we'll suffer bias-current offsets and integration of the bias current. 10k might be a nice value. It would limit the current to 0.5 mA on a 5 V input.

From $Q = CV$ we can differentiate and get $I = \frac{dQ}{dt} = C \frac {dV}{dt}$. Rearranging this we get $C = \frac {I}{\frac {dV}{dt}}$. Using your values of 5 V / 500 µs = 10000 V/s we get $C = \frac {0.5m}{10k} = 0.05 \ \text {µF}$.

Let's increase this to 0.1 µF for convenience and recalculate. $I = C \frac {dV}{dt} = 0.1µ \times 10000 = 1 \ \text {mA}$.

For 1 mA from a 5 V supply we can calculate $R_{S2} = \frac {V}{I} = \frac {5}{1m} = 5 \ \mathrm {k\Omega }$.

• Thank you, that was very clear! The previous design used a capacitor value of 1nF and resistor of 270Kohm, giving very low currents of around 18uA. I was surprised that such values were used. is there benefits to having such low currents? As the 270Kohm could cause probems no? input bias current is 30pA and offset is 5pA Apr 3, 2018 at 21:47
• Sure. Maybe for a battery powered application the low current might be an advantage or if V7 was a high impedance source. 30 pA into 1 nF would give a drift of $\frac {dV}{dt} = \frac {I}{C} = \frac {30p}{1n} = 30 \text {mV/s}$. This is negligible compared with the 10 kV/s of your spec so that RC value should work well too. Thanks for accepting my answer but I recommend you unaccept for a day or two to let the world spin around at least once and give the whole of humanity a chance to answer your question. You might attract much better answers. Apr 3, 2018 at 22:00
• Actually the original design did use a 9V battery, but even so surely this is capable of supplying about 50mA of current no ? and V7 is from a DAC, are they typically high impedance sources?? Apr 4, 2018 at 9:08
• Thank, I have unaccepted it to see if anything is added :). Apr 8, 2018 at 12:21
• I am just experimenting with some different RC vales to see how drift error changes. what would be the maximum acceptable drift amount? Before any difference would be a problem Apr 8, 2018 at 13:13