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I have a 50 ohm load I am trying to drive at 7V. I currently have a 10V supply with a 10kohm potentiometer but I am expecting to run into trouble when I connect it as a voltage divider. I looked online for a 7V supply but I couldn't find much.

What is the easiest way to get 7V that can drive this load? I can purchase some equipment but would like to keep it to a minimum.

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  • \$\begingroup\$ How accurate must the 7 volts be? 7v/50ohm is 0.14 amps. A Darlington NPN in a to-220 tabbed package (to dissipate 3v * 0.14a = 0.42 watts), with the base attached to that 10K pot, may be all you need. \$\endgroup\$ – analogsystemsrf Mar 20 '19 at 2:18
  • \$\begingroup\$ Why 7 volts? Is this for a 50 ohm signal driver? \$\endgroup\$ – AnalogKid Mar 20 '19 at 3:44
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Best is hard to determine.. but here is one way for your 140mA (7V/50\$\Omega\$) load current:

You can use an old-school LM317 regulator.

enter image description here

In your case you can use 249 1% for R1 and 453 1% for R2 which should get you close to 7V out. The chip's power dissipation will be 0.15 * 3V + 0.005 * 10V = 0.5W, so a TO-220 case regulator should be okay in most circumstances without a heatsink.

The somewhat newer LM1117 could be used but you have to pay a bit more attention the the output capacitance ESR to ensure stability (and the resistor values will be different because the reference voltage is 1.25V nominal vs. 2.5V nominal for the LM317). It does give you lower dropout voltage (it's a semi-LDO regulator), but since you have 3V to play with that's not required.

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with a 50 ohm load? if you add add 21.5 ohms in series. the 21.5 ohms will get 3V and the 50 ohms will get 7V

22 ohms is a standard size, and may be close enough for your purposes. else add the 10K pot in parallel (with 100 ohms in series with te pot) and use that to trim the resistance.

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  • \$\begingroup\$ Thanks I like this idea \$\endgroup\$ – MaxwellEE Mar 20 '19 at 3:10
  • \$\begingroup\$ @MaxwellEE I started with that answer then assumed you wanted to put AC amp to drive 50 Ohms not just a DC 50 Ohm drop \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 20 '19 at 18:27
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I have a 50 ohm load I am trying to drive at 7V. I currently have a 10V supply, what is the easiest way to get 7V that can drive this load?

It depends on whether you need a regulated 7V or not:

  1. If your 10V is regulated and your 50 Ohm load does not vary then all you need is a series resistor of 21.4 Ohms.

  2. If your 10V power supply varies then you need some form of regulator to provide a regulated 7V to the load.

If you do need a regulator (load varies or Vin varies) then there are multiple solutions:

  1. A shunt regulator such as the TI TL431 along with a series resistor can allow for a small range of load (50 Ohm) and supply voltage (10V) variation.

  2. Almost any linear regulator such as the LM317 would provide for a wide variation in load resistance and power supply variations from about 5.5V to 20V or more.

  3. Any form of switching regulator that works with a relatively low Vin - Vout differential. There are endless varieties of linear regulator replacements through to modules on small PCBs. One modules I've had good success with are the ones based on the MP1484 chip such as this. Be careful that some modules have a small adjustment range (5V +/-5%) while others have 3-15V adjustment.

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If a 70% efficiency adjustable LDO is acceptable, that is the easiest.

There are many types with different current ratings and adjustable.

You can also buy variable DC-DC boards with many options incl. Voltage and current display at low cost.

You can even use a Darlington NPN pair with a pot and a cap on the base as long as you allow for at least 1W heatsink.

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  • \$\begingroup\$ Thanks. That one has an 8V max input but I am sure I can find one that will work. \$\endgroup\$ – MaxwellEE Mar 20 '19 at 2:38
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    \$\begingroup\$ @MaxwellEE Not sure why the answer said to use an LDO. With 3 V of margin, you can probably use almost any linear regulator (though you will probably need an adjustable one since 7 V is kind of unusual for a fixed-voltage regulator.) LDO's are a special, low-overhead type that are PNP-based and have some special requirements for the output (must have a minimum capacitor value and one within a specified range of ESR for stability.) Spehro covers this someone in his answer. \$\endgroup\$ – jonk Mar 20 '19 at 2:57

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