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Treez,
Now it was very clear.
This image is the noise sensed by the lead in the nearest ambient of the shunt resistor. The GND clip was not connected:
The noise does not seem to be a problem I think and the 50Hz signal is very noticeable. The oscilloscope scale for the voltage is 20mV.
No. So now I think that I did not understand you.
You suggest that I must put the lead in the same place when I sense the current (first image of this thread). So I do not understand where I must connect the GND clip. Sorry.
Treez,
I think that I understood what you were asking me. I have connected the GND clip to the probe to see the noise in the oscilloscope and I have realized that the oscilloscope noise is 10mVpp.
See the attached picture:
KlausST,
Yes, the signal frequency is 20Hz (the signal of the first post).
I´m working with Processor Experts, it let configure the conversion time (I set it to 23uS) and number of conversions (I set it to 20). So the result of the ADC is an average between 23uS x 20 samples. What frequency do...
I do not understand this.
This is the testing circuit:
- - - Updated - - -
Klauss,
20Hz.
Conversion time 23uS.
I need Vrms over the shunt resistor to know the current. After I could know this value I check the voltage over a resistor that is in the center of the HBridge, so I can...
Dear,
I am sensing current over a shunt resistor, to achieve this I sense the voltage over the 1ohm shunt resistor.
As you can see in this image:
This is the voltage at the ADC input. Before the ADC the shunt resistor signal is connected to a AO INA118.
As simple as this, the voltage is...
The Vin of the HBridge is 400Vdc, I need to have a 10Hz semi sinusoidal signal and the possibility to change the duty cycle to change the average voltage at the output.
The idea is correct? Can I do that with a HBridge?
Thanks.
Dear,
I need to change the duty cycle of a HBridge, with fix frequency (10Hz). The Vin is 400V. Has someone some good document to read and understand? I will control it with a microcontroller.
Thanks.
Guys,
Well, I will give more details about the projects to received from you, the possibilities of success.
The generator is the inverter that we are talking from the beginning, as I need to rectify the signal to obtain a DC voltage, I connect the capacitors and the limiter resistance.
The...
Sunny,
The charge will not be necessarily a lamp in the future. I'm using it because I have not got a resistance with such power capacity.
I connected a resistor of 10ohm in series with the relay and the circuit is working.
Thanks all of you, this forum is real helpful.
BR.
Booboo1,
The gain formula is wrong. The correct one is (1+Rfeed/Rin) as FvM told you.
Try with bigger values, for instance, if Rfeedback is 12K and Rin is 3.3K the gain will be:
Gain = 1 + (12K/3.3K) = 4.6 (Similar to the one you calculated).
FvM,
I connected a circuit with the rectifier diodes and 2 capacitors of 220uF/250V in series (~110uF/500V) with two series resistors of 1kohm. After two cycles TAU (ResxCap) a control circuit shorted the resistors through a relay. Once the control circuit short the resistors inverter´s voltage...
FvM, sorry the AN is not so clear for me. If I connect the resistor in parallel with the capacitor how will the resistor decrease the inrush current generated by the capacitors?
Can you help me with the approximate values of the the resistor and the capacitors if the output voltage is 400V and...
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