I'm new in electronic and electrical circuits, so maybe my question will be naive.
I have an electrical transformer that gets 220VAC as an input and gives 25VAC as an output, and can produce 20amp current.
I connect the 25VAC output to diodes bridge with capacitor(3300uF - 25v) to the positive and negative pins to eventually get 25VDC. What I want is to reduce the 25VDC to 17VDC, I thought i could build voltage divider with variable resistor to get the volt I want, but it looks like the passed 10amp damaged the resistor.
What can I do ??

Here's what I tried to do:


simulate this circuit – Schematic created using CircuitLab

This schema is pseudo, it's only to clarify the connections, circuit in real life is working as expected with the right parts.

  • 1
    \$\begingroup\$ Your rectifier will not work as it is. You have two crosses between D3 and D4. The upper one connects to D1 and D2, which will produce a shortcut between your transformer's output terminals all the time. \$\endgroup\$
    – Christoph
    Mar 12, 2013 at 10:06

2 Answers 2


25V AC means 25V effective, which is sqrt(2)*25V = 35V 0-to-top. Hence when you rectify this to DC you can get up to 35V, because the capacitor will be filled according to the top of the AC wave.

1N4148 diodes are MUCH to feeble for this application. Use at least rated for 10A plus ample margin. You might need to limit the inrush current when you first plug your power supply in.

You can indeed reduce voltage with a resistor divider, but this is only practical when the current through the resistors is much smaller than the load current, which is clearly not the case for you.

A better way is to use a voltage regulator chip, like the 780x series or an LM317. These chips behave a s series resistors, but they constantly adapt their resistance to achieve the desired output voltage. But like any resistor, they will dissipate heat according to the formula I * V = P, which is your case is 10 * (35 - 17) = 180W. That is a lot of heat, which makes this solution impractical.

The best way to reduce (or increase!) a voltage at a substantial current is to use a switching regulator. Designing such a thing for the voltages and especially the current you need is difficult, I guess it is way above your head.

Luckily, 17V adapters are available in most computer shops, because these values are often used by laptops. 10A is a bit heavy, but they might be available. I suggest you buy one :)

  • \$\begingroup\$ So there's no solution but to buy new adapter. \$\endgroup\$
    – Dabbas
    Mar 12, 2013 at 10:23
  • 1
    \$\begingroup\$ There are solutions, but designing one is quite a task, and building it will likely cost more than a laptop power brick. \$\endgroup\$ Mar 12, 2013 at 10:47
  1. A resistive / voltage divider method for dropping voltage requires a substantial amount of power to be dissipated as heat by the resistor(s).

    • For this case with 25 - 17 = 8 Volts to be dropped, the power to be dissipated at 10 Amperes is P = V x I = 80 Watts.
    • This is the equivalent of a fairly bright incandescent bulb (60 Watt bulbs run quite hot) so consider the kind of heat your resistor would be expected to dissipate.
  2. A single resistor is a very unstable way of dropping voltage: The slightest change in current drawn by the load will cause voltage to drift significantly. A resistive divider is used for better stability, but this is meaningful only if the current through such a divider is at least in the same scale as the load current.

    • In this case, a divider would have to waste around as much power as heat, as the load consumes.
    • So there are massive (and very expensive) high wattage resistors needed, plus the efficiency would be abysmal.

So, the solution:

Look for high current DC-DC buck regulators rated for the expected voltage and current ratings. This will deliver stable voltage more or less independent of the actual current drawn by the load at any given time. Buck regulators are also much more efficient than resistors, with efficiencies of 75 to 90%.

This means only about 10 to 25% of the projected power consumption will have to be accounted for in the form of heat dissipation.


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