How do I find the resistor value to give me 15V from a DC 74V source?

Question

I need to find the value(s) for the resistor(s) needed to power an appliance that needs a 15V-19V DC input at @ 850mA, when the DC source voltage is 74V. Does @ 87 Ohms of resistance sound correct?

Answer 1

The resistor probably won't work. When the appliance is plugged in but turned off, it will use little current, the voltage drop across the resistor will be small, and the applicance will likely get smoked by the 74V source.

You either need mongo power transistors controlled by a feedback circuit to drop the voltage, or better, use a DC - DC converter. The best solution is to get a different power supply with the right voltage.

These other guys are using theoretical considerations to give you a textbook answer, but real world problems have other considerations.

Don't do the resistor nonsense unless you can afford to replace the appliance, because repairing it will be as costly as replacing it!

Answer 2

Not exactly.

A resistor of 68Ω (standard value) is what you will need.

A 68Ω resistor with 850 mA thru it will drop nearly 58 Volts, leaving 16 Volts for your appliance.

The problem is that it will drop nearly 50 Watts, so a 100 Watt resistor is recommended.  This is a BIG resistor!!

As an alternative, I would recommend a DC - DC convertor for this.  Many companies make one in this range for railroad applications (railroad locomotives in the US operate on 74VDC).

Answer 3

If you're using a resistor in series with the load to control the voltage/current ....

You're needing to drop (74-15)V. That's 59V. Plugging it (and 0.85A) into  PD=Current*Resistance and rearranging gets you a IDEAL choice of resistor to be just over 69 Ohm

You would have to combine several standard (preferred value) resistors to get that value, so it's usually easiest to try and find if the nearest single resistor (of higher/lower value) would do.

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Perhaps it would be better to seek an alternative to step the voltage down, as you'd be looking at a BIG resistor, which would dissipate a fair bit of power (Power loss= Current^2 *R) A DC to DC converter is one way of doing the voltage conversion without incurring such a large power loss.

If you can't/don't want to purchase one, it shouldn't be too difficult to make a Buck converter which covers you requirements. To MAKE one all you then need is an reference voltage (you can do that via a linear regulator, as it need not draw/dissipate that much power) a comparator, and an oscillator to drive the switching transistor, (but only when the comparator says the output is of low voltage.) Just remember to protect the transistor from voltage spikes using protective diodes!

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Not 100% sure, but you may be able to bodge what you've needing, simply by chopping  the 74V (switching it in and out, making a a square wave that's a bit like a poor quality AC), and feeding it to a SMPSU that's designed to take mains voltage, and get it to output a regulated 15V.

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If you still want to try using a resistor approach , you could protect the appliance if the resistor fails by shorting out by using a FUSE. If the voltage is high then the current will be high too.

Very good point (which I missed) by wingsdjf, that the output from the resistor will be at 74V if the resistor is connected to the 74V but not the load. Switch the load in and it will be subjected to this voltage for a short time. Could be delt with by A TVS after the appliance's switch.

Answer 4

you need 2 25v resistors and 4 2v resistors and 1 1v resestor but you have to line them in a parallel to make the resistor word

Answer 5

Try 67 Ohms.

R=E/I

R=(74-17)/.85

R=57/.85

R=67.06 Ohms

Answer 6

Input V = 74v

Output V = 15 to 19  so 17v  So you resist  74 - 17 = 57v

Out I = 850mAh  

R = V/I      = 57/0.850

    

    = 67.05 ohms