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I am trying to create the circuit below but then the resistor is getting too hot. I think is has something to do with my power source 15V, 8A. My resistor is 51 ohms, 1/4W. Any comment please… Should I increase resistance or wattage to prevent resistor from getting very hot.
simulate this circuit – Schematic created using CircuitLab
Thanks.
They are simulating a (more or less) real 9V battery. They've modeled the battery as an ideal voltage source of 9V with a series resistance of ~1.5 ohms.
simulate this circuit – Schematic created using CircuitLab
If you work this out, the current is 9V/(11.5 ohms) = 0.783A, so the voltage across the 10 ohm resistor must be 7.83V.
You could consider using a 555 timer circuit that uses a constant current source to charge the voltage up on the charge capacitor. The constant current source requires normally 3 resistors to set the current and one of these resistors, when increased will linearly (to a point) create greater current as its value increases. A greater constant current (higher resistance) means the capacitor charges more quickly hence, F is proportional to R.
(source: next.gr)
Note that increasing R1 causes the charging current to be greater thus increasing the speed at which C1 charges and hence frequency rises. Circuit taken from here
Best Answer
For a resistor
$$p_R = \dfrac{V^2_R}{R}$$
Generally speaking, it is not a good idea to operate a resistor at its power rating.
But, assuming we do operate at 1/4W, we find:
$$V_{R_{max}} = \sqrt{0.25W \cdot 51 \Omega} = 3.57V$$
So, you really would like the voltage across the resistor to be quite a bit less than \$3.57V\$
Now, the voltage across the resistor is:
$$V_R = 15V - 4 \cdot V_D$$
where \$V_D\$ is the nominal voltage across one of the diodes. So, if
$$V_D \le \dfrac{15 - 3.57}{4} = 2.86V$$
you'll be at or exceeding the maximum power rating of the resistor which will make it very hot indeed.
If this is the case, then you must increase the resistance or power rating or both.