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42 Threads found on Level Follower
As mentioned by AMS012, the part marked replica buffer is just used to generate the required bias voltage for the main pseudo differential buffer. Basically you need to set the output common mode voltage to a comfortable level as supported by the ADC which the buffer drives. The high gain opamp sets the bias voltage so that the output DC level is s
I want to design an non-inverting OP,with output DC level be 900mV to have largest output swing . (The supply is 1.8V and 0V). Can I add a 900mV-DCvoltage source to Vin- so that the output DC level can be fixed at 900mV because of negative feedback?
I had a store-bought line follower robot called Movit. It had two light detectors held snug against the floor. A single op amp (maybe 2) was all the logic it had. If one sensor dropped in light level, it caused the op amp to send an 'On' signal to the opposite motor. When the other sensor detected dark, the other motor was energized. The end resul
A transistor with a high output level and no negative feedback has high distortion with an output that is not symmetrical:
Using a complementary BJT voltage follower raises respectively reduces low and high level by about 0.7 V which involves no problem with standard MOSFETs.
Hello. please help me to solve my problem. I want to run my test bench on gate level Verilog netlist in modelsim. but I am getting the follower error: #**Error: /home/abbas/Desktop/pomyprfirst/tsmc18.v(7947): $hold( posedge CK:1170 ns, posedge RN:1170 ns, 500 ps ); # Time: 1170 ns Iteration: 1 Instance: /test_tbmpe/uut/c1/\c_reg # ** Er
I can't imagine any good reason for a 0.1V level shifter. But if Vcc is higher than the Vh level, simply use a pullup resistor of appropriate value. Compute the ESR of the driver from the VI curve and apply ohms law to pull up 0.1V or more.
Did you try to connect the output directly to the 50 ohm input? If you fear to overload the output, place a series resistor. You didn't tell about required level of the 50 ohm signal, but I expect that you will able to use a passive resistor network and don't need a transistor amplifier stage. Otherwise use a common collector follower as buffer amp
hey i am a beginner level learner. i use mikroc compiler. i use adc before but with a single channel,do not know how to multiplex adc coding..i want to build PIC based line follower robot using ADC multiplexing.please help me
Hi, Can somebody tell me how the output DC level of a source follower is established? Suppose the load is resistive and since there is no current flowing from gate to source, the gate to source terminal can be viewed as a capacitor then the circuit be visualized as an RC circuit. In such a case, the capacitor would be charged to the input DC le
Hi, Use a source follower as a level shift of the input signal whose CM voltage is beyond your comp.. By the way, may I know the bandwidth or frequency of your input signal. I feel that you have not got the best resolution. LEE
The simple analog way is to use an op amp (OTA) to drive a source follower that has a resistor tail, closing the loop on the resistor such that I=Vin/R. That I can be shipped down to the ground referred side through as many cascode levels as you need. You push that current into an identical resistor (presuming your drawn resistor at two backg
You're free to design charge pump (switched capacitor) converters at any current level. The question is, what approximately means for you? A witchd capacitor converter als well as e.g. a push-pull forward converter can provide approximately 1:1 conversion, but with some internal resistance. To avoid any load dependant voltage drop, a conver
pretty much to lower signal from the level it was for another Vt. Output from the diffamp is VDD-R*i so after SF you will have another 500mV or so down.
Dear friends I wanted to share some images of my line follower robot here.. This Robot secured second place in state level mini project competition.. Here are the images... Feedbacks are welcome..
Hi, every body, I have a question that is about the source follower (common-drain amplifier) and body effect. Why does increasing the output bias level can reduce the percentage of source to bulk voltage due to the signal? Could everybody explain the reason?
Hi, I need a cmos voltage level shifter circuit which shifts *exactly* 400 mV on a DC analog signal. I thought about voltage adder/substractor circuit but they are not applicable because of the resistors. Does anybody have an idea how to shift exactly 400 mV on-chip? Thx, Krivan
First check output voltage when input is 2V. If output is not 2V then you introduced some offset and attenuation to the "follower". When follower is working correctly up to certain input level and then goes wrong then check stability at this level.
I agree with leehying. By the way, level shift is also a comparator or amplifier during the changing process! Larger gain, larger resolution and faster speed.
It depends on the configuration of the IR leds you use. For instance to follow a line using only a single IR LED you will have to pass the signal from the photo transistor or photo diode in order to determine the level of reflection and as such determine if the robot is still on the desired path. A much simpler method is to use more than one IR