Means what exactly? dv/dt proportional to Vsupply?the "ramp" will vary linearly according to the supply voltage
Yes. Current source has to be proportional to Vsupply to achieve the intended characteristic. PWM input controls an analog switch discharging the capacitor, a buffer can provide a low impedance "copy" of the ramp voltage. That's the basic design.for generating a sawtooth you usually need a cpapcitor oand a constant current source.
So more information and a sketch:Hi,
please give some examples or sketches of the waveform ... regarding duty cycle amd supply voltage.
for generating a sawtooth you usually need a cpapcitor oand a constant current source. The capacitor then can´t charge up the supply voltage, because the constant current source needs some voltage headroom to work.
So we nned some definitions how close the PWM should go to the supply rails... and what "fast" means. Please: values with units.
Klaus
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... and dv/dt proportional to 1/duty_cycle?
Klaus
The maximum voltage is not a problem, because I'll need no more than 1V, I only need to have a linear dependency to the supply voltage.Hi,
So it does not need to go to supply voltage.
And dv/dt does not depend on duty cycle.
So what maximum voltage do you want? Let's say at 15V supply and 95% duty cycle. (100% makes no sense in your case)
And what minimum voltage do you need? (Zero is hard to achieve. Please don't say "as low as possible")
And how fast do you need it go from maximum to minimum voltage?
For good quality:
I'm thinking about
* a reference voltage
* an inverting Opamp circuit that generates: Vi = V_ref - a × V_supply
* an (inverting) integrator referenced to V_ref
* an analog switch for zeroing (PWM)
Klaus
It "charges" when the PWM is high, so for 10% it's 1us and for 80% it's 8us.If PWM is 10% duty cycle what is ramp time beginning to end ?
Same but for PWM 80% ?
Regards, Dana.
The PWM is not the issue, the sawtooth generator is.So the ramp rate is changing, dV/dT ?
Stated another way the PWM is used to set supply voltage in a control loop,
and you want this sawtooth wave to linearly track its ramp rate to the supply
which is directly related to PWM duty cycle ?
Or am I missing something......?
What is PWM size ? 8 or 16 bits ?
Regards, Dana.
Absolutely feasible. Use a MOSFET as discharge switch. Regarding current source, you should talk about required accuracy. An OP based current source can achieve high control linearity, but may be a simpler approach, e.g. a current mirror is sufficient.The maximum voltage is not a problem, because I'll need no more than 1V, I only need to have a linear dependency to the supply voltage.
Minimum voltage - Less than 0.1V.
Fall time from 1V to "zero" - less than 0.5us.
This is probably my solution.OK, thats what i showed you with icky processor solution.
For a discrete approach you could take PWM output, convert it to DC with a LPF,
and use that to control a current source that charges a cap. Reset of sawtooth simply
a MOSFET with a time element to guarantee MOSFET fully discharges cap. Thats all
run off PWM as well.
Note PWM LPF has a latency associated with it, how much time delay can you tolerate ?
Voltage controlled current source, single opamp type solution easy, just google
"voltage controlled current source", lots of hits on web.
View attachment 169286
Regards, Dana.
If yes, then ramp rate (dv/dt) does not depend on duty cycle.Example:
for 15V and 17% I'll get a sawtooth that goes up to 0.7V
for 3V and 83% I'll also get 0.7V max in the sawtooth
You are right, it was a specific example of some measurements that I had, but assuming that everything is perfectly linear, what you wrote will be correct.Hi,
is this still valid?
If yes, then ramp rate (dv/dt) does not depend on duty cycle.
(more precise 15V and 17% gives the same as 3V and 85% (not 83%), since 15 x 17 = 3 x 85 )
Klaus
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