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I'm trying to create a circuit that will divide the input voltage in half, but still be able to supply a high current to the load.

My input voltage varies from 10-15v and I want the output voltage to variate in proportion to this. So between 5-7.5V. The load should be able to draw about 1A, so the components needs to be able to handle this.

I'm thinking it's going to be a voltage diveder and a voltage follower, but I'm not completely sure. Whats the best solution to this?

Tnx.

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  • \$\begingroup\$ Is that 1 amp load current from the mid-point generator +/- 1 amp? \$\endgroup\$
    – Andy aka
    Jun 21, 2018 at 10:20
  • \$\begingroup\$ I'm not completely sure what you mean, but load will draw up to 1A. \$\endgroup\$
    – stackman82
    Jun 21, 2018 at 10:23
  • \$\begingroup\$ So the load at the mid-point only connects to 0 volts and not up to the upper supply voltage? In other words the mid-point voltage generator only supplies current to a load that returns to 0 volts? \$\endgroup\$
    – Andy aka
    Jun 21, 2018 at 10:28
  • \$\begingroup\$ Ahh now I understand. No, the load also depend on the voltage. But the load uses the input voltage to determine different things (like what condition the system is in), so it must vary. \$\endgroup\$
    – stackman82
    Jun 21, 2018 at 11:27
  • \$\begingroup\$ FWIW, what you are asking for is an amplifier. Yeah, it's voltage gain will be less than 1, but you still can call it an "amplifier" if the power at the output potentially exceeds the power available at the input. \$\endgroup\$
    – Solomon Slow
    Jun 21, 2018 at 15:59

3 Answers 3

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If you needed better efficiency than a linear regulator, you could use a buck SMPS and use a current mirror to adjust the voltage based on the input voltage.

Here's a generalized schematic.

schematic

simulate this circuit – Schematic created using CircuitLab

The current Iadj increases linearly with the input voltage. This pulls current out of the voltage feedback net which causes the output voltage to increase.

You can pick out a buck SMPS IC that meets your needs and then design the current mirror around that.

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  • \$\begingroup\$ This is a really cool solution, tnx. There is just a couple of thing I dont understand. Would the smps not provide a fixed output voltage? Or is the purpose of R2 and R1 to "divide" the voltage, so that setting R1=R2 would give us exactly half of the input voltage? And what considerations would you need to take for selecting components? I'm guessing the transistors and the resistors need to power types to handle the high currents? \$\endgroup\$
    – stackman82
    Jun 22, 2018 at 8:41
  • \$\begingroup\$ Really cool solution! I would definitely go with this one. \$\endgroup\$
    – Stefan Wyss
    Jun 22, 2018 at 10:05
  • \$\begingroup\$ @stackman82, a VREG (whether linear or switched-mode) typically has fixed internal voltage reference and a feedback loop that samples the output voltage and compares it against its internal voltage reference. So, let's say our internal Vref=1.25 Volts. And let's say we want 5V at Vout and let R2=10k. R1 would then be 3.33 kOhms. Basically, with these values of R1 and R2 voltage divider, ADJ pin would be at 1.25V when Vout is 5V. This is how a typical adjustable voltage regulator works. Read the datasheet; typical design consideration is how much bias current goes into the ADJ pin. \$\endgroup\$
    – Vince Patron
    Jun 22, 2018 at 18:32
  • \$\begingroup\$ @stackman82, you'd have to include Iadj in your calculation. Also, you'll need some limiting circuit because, for example, if Vin shot up to 20V, Iadj current increase, which will increase your Vout up to 10V and maybe damage your load. So the circuit shown is just the basic concept; might need limiting circuit (e.g. a zener across R2) and also simulate the circuit to see if needs additional compensation is needed for stability (e.g run step load/release, power on, power off and other simulations). \$\endgroup\$
    – Vince Patron
    Jun 22, 2018 at 18:40
  • \$\begingroup\$ Tnx @VincePatron, appriciate the feedback. \$\endgroup\$
    – stackman82
    Jun 25, 2018 at 7:24
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Ok, you can do the following: Use a simple resistor divider to create 1/2 of the input voltage. Then create a simple voltage follower/buffer with a high current opamp, like e.g. the OPA548.

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  • \$\begingroup\$ 7.5 Watts of power dissipation at max load will need significant heatsinking, but yeah, it would work fine. \$\endgroup\$
    – Vince Patron
    Jun 21, 2018 at 20:45
  • \$\begingroup\$ I get the you could use a power opamp for this. Would there be any considerations to the resistors, power wise? \$\endgroup\$
    – stackman82
    Jun 22, 2018 at 9:03
  • \$\begingroup\$ No, you can choose relatively large resistors for the voltage divider, so there‘s no power issue here. \$\endgroup\$
    – Stefan Wyss
    Jun 22, 2018 at 10:02
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One solution to this is a buck regulator. It transforms the (higher potential) input voltage to a lower output voltage by means of a switch, a diode and an inductor.

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  • \$\begingroup\$ Doesn't that transform the input to a fixed output? \$\endgroup\$
    – stackman82
    Jun 21, 2018 at 14:09
  • \$\begingroup\$ Yes, but if you have e.g. a steady load current, you could set the PWM to a fixed value, defining your output to input voltage ratio. \$\endgroup\$
    – Stefan Wyss
    Jun 21, 2018 at 14:15
  • \$\begingroup\$ Okay I see. Unfortunately the load current can alter up to 1A. \$\endgroup\$
    – stackman82
    Jun 21, 2018 at 14:39

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