How do I interface a low power controller with a higher power load?

Question

I am not an electrical engineer (rather a photonics engineer), so I am looking for some direction on how to attack an issue I have.

Basically, I need to drive a 150 Ω load with a computerized controller. The controller I have (a National Instruments I/O device) produces a series of voltage step functions over time, but it's not rated up to the current and power that I need at my load.

What do I need between the controller and the load to fully power the load, without damaging the controller? And of what flavor, like common emitter, voltage follower, etc.?

Note: Precise multiplication of the input signal is not necessary, as the controller can be easily calibrated to account for any incidental offsets. (In fact, Vin does not need to equal Vout.) And fast switching is also not necessary.

Thanks!

Answer 1

The obvious solution (to an engineer in this field) is an emitter follower:

^{simulate this circuit – Schematic created using CircuitLab}

For reasons I explain below, that solution has some failings. You could have found this answer on a thousand different web sites, so I presume it's not good enough. I propose a simple enhancement, with the addition of an op-amp:

^{simulate this circuit}

Here are the things you should know about it:

1. The voltage at OUT equals the voltage at IN, to within a few millivolts. There's no voltage attenuation or gain. There is however a maximum output voltage: Here you can see that maximum output potential is 7V. It has two causes; firstly the op-amp U1 (a cheap LM358) is unable to get its output all the way to the positive supply of 8V. Op-amps exist that can do much better (like the TLC2272), and this application could use either. The second cause is the voltage \$V_{BE}\$ dropped from base to emitter of Q1, which will be about 0.7V.

2. Increasing the power supply above 8V will raise that upper limit imposed on the output, so you may use any power supply, up to the op-amp's limit of 32V. However, Q1 will dissipate more power. See note 5.

3. Q1 is a common collector "emitter follower". Its purpose is to relieve the op-amp of any appreciable load, and take on the task of current provision itself, sourced directly from the power supply instead of the signal source.

4. An emitter follower might be all that's needed, but that drop from base to emitter of \$V_{BE} \approx 0.7V\$ is appreciable, and will vary slightly as load current and temperature changes. It will also differ from transistor to transistor. The addition of U1 linearises and equalises the input-to-output relationship. Note the negative feedback from overall output to inverting input. All I've done here is to make a regular "voltage follower", but with an extra current gain stage inside the feedback loop, compensating for \$V_{BE}\$ (and any variation in it), and rendering it moot.

5. Transistor Q1 is going to dissipate power, because its role is to "waste" whatever voltage the load doesn't need. This is true whether it's on its own (as in my first circuit above) or part of a closed loop (used with an op-amp). Since current through Q1 and the load is about equal, the product of wasted voltage and current is the power it will have to consume. Here's a graph of the power in Q1 for the same sinusoidal input shown above: It peaks at just over 100mW, which means Q1 will get warm, but that's well within its capabilities. Here's a graph of power dissipated by Q1 using a power supply of 15V: This peaks at just under 400mW, which is getting close to the limits of what a small transistor like this can tolerate without a heatsink. If you do decide to use a power supply over 15V, you should use a beefier transistor, like a TIP31.

6. You should take care not to take IN beyond the power supply voltage. That is, don't allow your signal source to exceed 8V, or whatever the power supply voltage (to the op-amp ) is. The behaviour of the op-amp is not defined for that condition. There are some measures you can take to ensure this never happens, such as attenuating the input signal with a resistor divider:

^{simulate this circuit}

Resistors R2 and R3 attentuate the potential at IN: $$ V_X = V_{IN} \times \frac{R_3}{R_2 + R_3} = V_{IN} \times \frac{10k\Omega}{10k\Omega + 8.2k\Omega} = 0.55 V_{IN} $$ This has the effect of bringing the full 10V input down to 5.5V, other values being similarly scaled. This is the new input and output relationship:

Answer 2

With a 150 Ohm resistance, that is pretty easy to buffer with a low impedance voltage driver that can drive 40 mA with low saturation limits on supply. some Op Amps can do this, but if you are flixible on inputs any 1A NPN emitter Follower will work. But heat dissipation in device is not the load power but the difference between supply and output V times current.

Essentially you need to lower the output impedance of your voltage driver from 10V/10mA= 1k to <<10 Ohms. This is easily done with an emitter follower with a large case size to keep only warm to touch with high Beta current gain and collector to 6V limiting input to 6V for a 5.3V output or use 6.2V to get 5.5 output. A Beta of 100 is minimum needed to reduce 1k to 10 Ohms, higher the Beta, the better tracking of input voltage changes with a 0.7V approx offset.

There are many more precise ways with current sensing and error feedback but this is the simplest. the NPN must be rated for >1A and not short circuited.

Assuming you have some steps between 0 & 10V with a 1k source resistance, here is an attenuation and Beta = 100 voltage follower simulation to produce 0 to 5.5V output into 150 ohms BUT using 1 to 10V input.