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I don’t think you work with the “old” 8051, instead your derivative is more likely to be AT89S52, or something similar. They don’t have internal pull-up resistors, as resistors take up a lot of area in CMOS.
What you can find there will be a small current source made out of MOSFET(s) and limited...
Most of these circuits with IGBTs (or power MOSFETs) operate on the principle of PWM.
Somewhere in the circuit there is a comparator working as a ramp generator, and somewhere in the circuit there is a comparator generating PWM for IGBTs. This comparator needs a dc voltage reference. And that is...
Re: 433 MHz ASK RF Tx Rx using 8051 UART
Yeah, you have to invert data at both ends.
Slow down the transmission to well below rated 4800bps.
I'd go as low as 300bps, and then make sure you use at least a 2-byte sequence, don't relay on transmitting only one byte ..
:wink:
IanP
This radio-module has several GPIOs, so use one of them to drive a transistor and then a relay.
The Rx and Tx pins are designed to trasnmit serial data ..
:wink:
IanP
A follower(or a buffer) by definition has to have very low output impedance, as it has to drive any load.
If for any reason you need to increase the output impedance, just add a resistor of a desired value, but I’m afraid it will be the end of calling this thing a follower ..
:wink:
IanP
Before addressing your Qs lets make some obvious statements: in this application the transistor works as a switch ie. it replaces mechanical On-Off device. So it works in completely non-linear situation: voltage of 0.7V applied to B(ase) – switch-ON, voltage below 0.7V – switch-OFF.
In...
The correct operation of this circuit relays on propers selection of solar panel, battery and LED.
And it works like this:
When there is no voltage generated by the solar panel, or the voltage across the panel is U[battery]-0.7V, the Base voltage of the PNP transistor will be 0.7V below the...
Resistors are characterized by their resistance and max power.
That is enough to calculate max current and max voltage drop (in case you need to know it for something).
You can apply voltage-current formula, V*I, and then replace V[voltage] by R*I, and that leaves you with I*I*R - back to square...
You can call this effect – inrush current.
In all electronic devices you can find bigger or smaller storage capacitors that, as the name indicates, have to be charged first and that occurs when the power is switched on.
Once caps are charged the device’s current consumption drops to its average...
For a quick test, I’d connect output with the (-) input for a gain (1) amplifier, and try it that way –from -9 to +9 with a potentiometer ..
:wink:
IanP
All what you need for the fan is to drop 3V, from 15V to 12V, assuming the 15V source is stable enough. If so, why not to use 4 standard 1N400X diodes connected in series, and as each of diodes drops around 0.7V and can conduct current up to 1A, you will have a voltage drop of 4x0.7V=2.8V and...
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