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I'm trying to dabble in analog circuit design, but I honestly am a complete newcomer in this area as my strengths are in CS. Sorry if this question sounds stupid, but can anyone help me with designing a voltage follower that can deliver large amounts of current to an external circuit (a purely resistive circuit). I've come up with a simple solution that doesn't quite work as the feedback load regulation is really bad (the voltage doesn't transfer over well).

Since I require a large amount of current, I can't just design an op amp voltage follower since I don't want to opt for the more expensive op amps that can provide upto 5A (if they even exist).

My attempt: My Attempt. The voltage over the resistor is close to 3V instead of the 8V that I want. The voltage over the load should stay the same regardless of the load value.

Any help in this matter would be greatly appreciated! Thank you so much!

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    \$\begingroup\$ Your circuit has no feedback from the load voltage to the op amp. \$\endgroup\$ – Ignacio Vazquez-Abrams May 22 '14 at 22:14
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    \$\begingroup\$ hook the inverting input of the op amp to the emitter of the transistor. \$\endgroup\$ – Vladimir Cravero May 22 '14 at 22:15
  • \$\begingroup\$ You also will have a Vce drop so you can never get 8V on the load from a 8V supply (load side) \$\endgroup\$ – placeholder May 22 '14 at 22:25
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    \$\begingroup\$ Please see, for example, this: electronics.stackexchange.com/a/70906/10475 \$\endgroup\$ – Alfred Centauri May 22 '14 at 22:30
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This is a good start, but it needs a few improvements. Op-amp voltage followers work because they compare the output voltage to their input voltage. In your circuit, however, the feedback is being drawn from the gate of your BJT. So, you'll see the output voltage across your load somewhere between 0.3 and 0.7 volts less than your input voltage. By using the emitter voltage as feedback, the op-amp will output a slightly higher voltage to compensate for this.

Another improvement deals with current handling. A BJT transistor usually has a current gain between 10 and 150 times. So, in a worst case scenario, the op-amp may be required to source up to 0.5 amps. We can do better. If you replace the single transistor with two transistors configured as a Darlington Pair, you get much higher current gain.

As for power dissipation, make sure you use a power transistor. Find the maximum power you expect to dissipate in the transistors, and double it, as a rule of thumb. Power transistors are designed to dissipate that much heat. Smaller BJTs are not.

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  • \$\begingroup\$ Also, be aware of the headroom your op-amp requires. Most likely, it can only output rail voltage less half a volt or so. Take off another half volt to compensate for your BJT voltage drop, and you'll probably only see up to 7 volts out, as drawn. \$\endgroup\$ – Jason_L_Bens May 22 '14 at 22:54
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What others have said, plus a few more details:

  • You could actually use a voltage regulator, if all you need is a fixed voltage. They usually have a similar circuit and an internal voltage reference.

  • In your modified circuit (Darlington pair, feedback from the emitter), you might want to add a small resistor (about 100 Ohms) in series with the base to prevent ringing (otherwise unlikely but possible, depending on parts and layout).

  • At Vce = 18 - 3 = 15 V and 5 A load the transistor will dissipate 75W, so you will need a part that's capable of doing that, and a proper heat sink. I would recommend getting at least a 100W transistor. You can also add a collector resistor, which would dissipate some of the power and provide a basic short-circuit protection. For the max load of 5A, and let's say if your battery or whatever may sag to 12 V, i.e. the voltage may be in the the range between 12 and 18 V, I would go with a 1.5 Ohm, 50 W collector resistor, and a 50 W transistor. The same approach and calculations would apply for a voltage regulator, although it already might have short-circuit protection built in, check the specs and application notes.

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  • \$\begingroup\$ What would cause the ringing without a base resistor? Also, the voltage as drawn is 8V, not 18V. I thought the same thing at first . \$\endgroup\$ – sherrellbc Jul 14 '14 at 20:24
  • \$\begingroup\$ Base-emitter capacitance + parasitic load inductance + op amps own phase shift could cause high-frequency ringing with DC offset near zero. I've seen stranger things happen :) You're right, the supply voltage is 8V, not 18, which means the circuit won't work properly for the 8V input voltage, anyway. \$\endgroup\$ – biggvsdiccvs Jul 15 '14 at 21:12

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