How to get a 6V circuit working with a 12V powersupply?
Do i:
A/ Double the size of existing resistors in the circuit(although im guessing not)
B/ Put a resistor on the plus side, between the circuit and the psu, and what resistance should it have?
Answer:
It depends on the circuit you are trying to power. The simplest way I know is to put a current limiting, low value resistor (10 ~ 100 ohms) in series with the power supply. Then put at zener diode from the end of the resistor not connected to the supply to ground. The problem here is that zeners do not come in 6V. You can get 5.6 or 6.2. You will also need to calculate the power rating of the current limiting resistor based on the current draw.
If the circuit is really simple, you might be able to do it by resizing the reisitors. If you will post a circuit diagram somewhere, you will get better answers.
I would just use a step down transformer: 2/1.
You need a 6V drop across the resistor, and you need to know how much current your circuit will be using. Then it's Ohms Law and
R = 6/I where I is the current. Note that this only works if your circuit consumes a constant amount of current. If the current demand is --not-- constant, you might be better off with a 6V Zener diode in series with the positive leg of the circuit.
Doug
try using a 6v zenner diode
Can not answer without seeing the circuit or knowing a little more about it. I assume you are talking VDC. You can use a voltage dropping resistor and a zener diode (may pull more current than you want to) or even better but a few cents more just get a fixed voltage regulator (LM7806), all done. If you do not understand, have somebody help you that does.
try a pot or a step down transformer
A = No
B = Depends. If the circuit is simple and the current the circuit uses does not change then you can use a series dropping resistor. The resistor consumes the extra potential (6V), leaving 6V for your circuit.
Measure or determine current I. Calculate resistance and power.
R_S = (12V - 6V)/I
P_S = I^2 R_S
You probably won't find a resistor that meets the exact numbers. But as long as it's close, it will work. Make sure power rating P_S of actual resistor is larger than calculations.
But a series dropping resistor will only work if the current is constant and it wastes 50% of the power to the resistor!
A buck (step-down) converter.
The best way, and far better than using resistors, is to use a linear regulator. This way the output voltage (6v) is not dependent on the current drawn. Go to National semiconductor and look for 3 terminal positive regulators. They can be purchased at Digikey.
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