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I need to create an adjustable DC power supply that can output between 0.1 and 0.75V DC at 300 mA (possibly as little as 100 mA). My input voltage can be anything, including +/- rail.

Currently I am using an adjustable bench top power supply to supply this voltage, but I need to package this in a smaller enclosure. Once adjusted via trim pot the voltage won't change, and I do need as stable a DC source as possible (both temperature and voltage stability). Long story, but this DC power is being injected into an RF signal via a "Bias T" so an RF sin wave can be "converted" into a digital signal (-0.5V to 0.5V pk-pk becomes 0-1V and interpreted as PECL digital input. Ugly but it works!).

Anyway, all of the usual LDO / regulators / step down convertors appear to cease working at ~0.5 or 0.75V and I most likely need approximately 0.35V. Everything shares a common ground so I cannot use a floating ground for this application (i.e. 1V and 1.35V power supplies).

I am considering a unity op-amp where I set the input V+ using a trim pot. However, I can't seem to find an op-amp that provides enough mA (voltage following op-amps; I'm not familiar with current op-amps to know if that would work). Does anyone have any suggestions here? I'm somewhat at a loss.

Thanks!

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  • \$\begingroup\$ A resistor divider? (or does the voltage need to be stiff?) You can make a simple power booster for an opamp with an npn follower. \$\endgroup\$ Commented Nov 4, 2014 at 17:40

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If you have a negative supply available, you can use an LM317 or similar adjustable regulator, with the bottom of the voltage divider connected to the negative rail, rather than to ground.

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There are high current op-amps available, such as the OPA547 from Texas Instruments.

You could use that (or similar) to apply fractional gain (attenuation) to a larger voltage (say 0-8V at 10:1) from a trim-pot (8V is the minimum supply voltage, so would be convenient for 10:1 attenuation). The op-amp has a 5mV offset to take into account. That would give you a range of 5mV to 805mV.

You could also replace the trim-pot with a small precision DAC, say an MCP4821, which gives 0-2047 in 0.5mV steps. Using 2:1 attenuation, that would give 5-1028.5mV (including the 5mV offset) at 250µV resolution.

And all that at up to 500mA current.

They don't come cheap though - about 10x the price of your typical small signal op-amp.

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  • \$\begingroup\$ Thanks! This is exactly what I will do. I was going to use BUF634 but the OPA547 has a higher capacity. I'll use a resistor-divider on the input VCC to reduce the voltage and short VOUT to V- to create unity gain. Do you think it matters if I provide a negative voltage rail? \$\endgroup\$
    – bode
    Commented Nov 4, 2014 at 20:45
  • \$\begingroup\$ Using a negative voltage rail would allow you to get down to absolute zero if you wanted to, but that's about all the benefit from it. \$\endgroup\$
    – Majenko
    Commented Nov 4, 2014 at 20:50
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0-1 V does not sound like valid PECL levels. Even with 2.5 V supply, PECL levels should be something like 0.7 and 1.4 V.

If your PECL device has a VBB pin, that's there specifically for the purpose of re-biasing ac-coupled signals --- and it's tied to any Vcc variation to avoid duty cycle distortion.

Remember that PECL logic are referenced to VCC rather than ground. So you really want to generate a bias of Vcc - 1.45 V, not GND + 1.05 V.

(If your device is an FPGA, with "PECL-compatible" inputs, you can disregard this. Their levels are not true PECL and probably are ground-referenced rather than Vcc-referenced)

As other answers have pointed out, a high-current op-amp or an op-amp with a transistor buffer ought to be able to generate the bias you need --- but be sure to design the circuit to track with Vcc if your load is a true PECL part.

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  • \$\begingroup\$ Good question but I am sure it's FPGA-based. For this application we are modulating a diode laser on and off at approximately 60 MHz. So all we're trying to do is place the middle of the sin wave at the PECL "high" level, and we've found that ~.4V does this. Thanks! \$\endgroup\$
    – bode
    Commented Nov 4, 2014 at 20:48

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