Input Voltage to Sawtooth oscillator Confusion

Thread Starter

Jack Tranckle

Joined Jan 20, 2016
73
I have attached a schematic of a project that I am working on, it is following a design that has already been done and works well. My problem is regarding V7. now the schematic below works perfectly when V7 is a DC source of the required voltage, in this case 0.039V. But the actual source of V7 is from a microcontroller, which sends 12 bits to a DAC, which then outputs a voltage which is what V7 is simulating. My question is, wouldn't the output from the DAC be an AC waveform? However when I replace V7 with an AC waveform it doesn't function properly at all. Actually, the AC signal can work but only when the frequency of it is equal to 1Hz. what am i missing?
 

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Hymie

Joined Mar 30, 2018
721
View attachment 149450 I have attached a schematic of a project that I am working on, it is following a design that has already been done and works well. My problem is regarding V7. now the schematic below works perfectly when V7 is a DC source of the required voltage, in this case 0.039V. But the actual source of V7 is from a microcontroller, which sends 12 bits to a DAC, which then outputs a voltage which is what V7 is simulating. My question is, wouldn't the output from the DAC be an AC waveform? However when I replace V7 with an AC waveform it doesn't function properly at all. Actually, the AC signal can work but only when the frequency of it is equal to 1Hz. what am i missing?
Rather than just copying a circuit, you need to understand the principle of operation before you make changes.

From basic circuit analysis IC U2A is configured as an integrator (with a capacitor in the op-amp feedback path). This gives the ramp output waveform; fet Q1 is short circuiting the op-amp feedback – controlled by the input voltage V1.

So increasing the voltage V7 will result in a faster ramp change rate, conversely reducing this voltage will decrease the ramp change rate. Varying the high/low period of voltage V1 will determine the allowed ramp time, and the time the output voltage is held at 0V.

Based on the above analysis, I would guess the reason your circuit appears to operate correctly with a 1Hz input, is that the timing and voltage of the 1Hz wave just happen to be correct for the output you want.
 

ebp

Joined Feb 8, 2018
2,332
Jack, your confusion may have arisen because you have seen "multiplying" DACs used in synthesizer circuitry. A multiplying DAC can accept an AC signal as an input and produce an output that has been multiplied by a digitally-set value. Unless the DAC has an amplifier built it, the output is usually a current proportional to the product of the input voltage and the digital setting. They can also be used to produce DC output if the input is DC. With appropriate amplifiers at the output they can do full four-quadrant multiplication.

a couple of docs:
http://www.analog.com/media/en/news-marketing-collateral/solutions-bulletins-brochures/AnalogMultiplyingDACs.pdf
http://www.linear.com/docs/4081

From what I've seen (and it isn't that much) most DACs used on boards sold primarily to the hobby market are simple voltage-output types that produce a unipolar DC voltage that is some fraction of a reference voltage (e.g. a 12 bit DAC might output 0 to 4.095 volts in 1 millivolt steps). You can produce all sorts of waveforms under program control. A DAC built into a microcontroller would be the same type, but they aren't very common.
 

Thread Starter

Jack Tranckle

Joined Jan 20, 2016
73
Thank you, I understand now. Since I am sending a constant value, a constant DC value will be outputted. I have another question regarding this schematic, I am not sure if the forum would prefer me to do this in another post. I will update this one and see that works :)
 
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