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17 Threads found on edaboard.com: Low Current Clock
One issue with manually powering off the LCD, when it is powered up again it will require a delay in the order of 30ms to 50ms for the display controller chipset (HD44780 compatible) to fully initialize and be ready to accept input from the microcontroller. A better approach would be to utilize a couple of BJTs or MOSFETs in between the power r
Hi, 1) the 0.4mA Active current is @ 1MHz clock, if you use higher clock speeds , you will use more power. 2) For using this with a simple 9V battery, you need a low Quiescent current regulator , not a normal one. For Example : LM2936-5.0 (5V) , LT1521 Family (Fixed 5V, 3.3V , 3V versions or (...)
If you want LiPO power, I suggest LCD instead for low current. 8x1 or logic out are 25 Ohm drivers so no problem with 10 Ohm ESR LEDs at 2.2V FWIW... and 1 ppm TCXO is only $1
Can anybody give me a circuit that converts DC(direct current) voltage to square wave with an input voltage as low as 200mV? You can achieve this with 1. Ge-Transistors 2. depletion JFETs 3. native MOSFETs
It is also measured in mA/MHz. So the faster the clock frequency, the more current drawn. So for low current, use low speed clock.
Here is a simple chematic for controlling bipolar steppers. It has two inputs: clock: Move one step for each clock cycle The other one is direction control. In adition to this circuit, you also need: H bridge to feed current to it A oscillator (VCO) that can start with low frequensis.
What do you consider low power respectively ultra low power? What are your additional requirements for the clock signal?
Looks like the bottom has an RC characteristic (like a current sink that's run out of headroom and gone low impedance) while the top has a constant-current characteristic. But I question whether one is any better than the other if they both settle to minimum error before next clock edge.
Dear all, I want to design a 30us-delay circuit. The spec. is shown below. 1. You can only use R,C,PMOS and NMOS in IC. External components are not allowed. 2. current consumption < 1nA when VCC=4V. 3. Input is a very low clock signal with 500Hz of frequency. Output is a delayed (...)
It depends on the application. If the 220V is AC mains, the current is low and you don't mind the health risk, you can do it with two resistors. Otherwise I would suggest a 220V to 5V transformer is the best solution. Brian.
I would recommend simple relaxation oscillator. Since you are doing it for medical that is the way - all the other approaches are high current. For 1 Hz you can use few nA charging current, very low power opamp or voltage comparator. In total you should be able to sqeeze it into 1uA current max. But I would say that (...)
Are there any other specifications? Input impedance? How fast does the comparator need to transition? What temperature range does the comparator need to operate over? What is the tolerance on the 150mV? What other circuits might be available? Reference voltages? Bias currents? clock circuits? If you can handle a current pull-up (...)
There are numerous ways. Some things off the top of my head: Guard rings. Separate bond-outs for high current, low noise, or switching sections to minimize ground bounce. Careful routing to minimize ground loops. Appropriate use of bypass capacitors. Fully differential topologies Careful selection of the clock frequencies web.awrc
Hi, I want to implement a circuit to carry the clock in the chip out of the chip. The jitter spec is less than 1ps and the clock frequency is 20M~500MHz. I'm considering to use the current Mode Logic and the LVDS interface. My question is:does the LVDS interface capable of ensure such low jitter?? Or do you have any (...)
I think you've to design the registers and gates the fastest implementation you can .Fast logic families as CML (current Mode logic) ,ECL(Emitter Coupled logic) or LVDS (low Voltage Differential Signaling ) can be used .You can also try placing positive skew at long paths to increase clock speed but you have to be careful in oder not to have (...)
We design a low power d flip-flop which will consume current when clock is switching , otherwise the current is very small when the clock is a stable state. However, when the IC is startup , the clock source may rise to a meaningful voltage, say half of voltage supply(5V). Then the (...)
HI lower the cpu clock - this method might save you some current All the best Bobi


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