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Controlling the power output of a power resistor

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi,
    I am working on a temperature control system which involves using a power resistor to heat a block of aluminium and a thermocouple to measure the temperature. Both the thermocouple and the power resistor will be controlled using the NI USB-6008 and Labview.
    As my knowledge of electronics is limited, I am having trouble in designing a circuit to drive the power resistor. The resistance of the power resistor is 10Ω and I have a 12V power supply. 12/10=1.2A, so 1.2A is needed to get max power output, correct? The current o/p of the USB-6008 is 200mA so I believe I will have to amplify the current to achieve this. I have a 2N3055 power transistor to use and also some operational amplifiers.
    If anyone can point me in the right direction in identifying what the best way to achieve a regulated output voltage across the power resistor in the range of 0-12V, I would be very grateful.


    2. Relevant equations



    3. The attempt at a solution
    I am thinking of having the NI USB 6008 output input to V+ of a non-inverting amplifier, using the non inv. amp to get the required voltage gain (5V amplified to 12V) and then using the op-amp output as the input to the base of a pnp-emmiter follower circuit to get the required current gain.
     
    Last edited by a moderator: Jan 21, 2014
  2. jcsd
  3. Jan 21, 2014 #2

    rude man

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    Since you will close the loop around the thermocouple, why do you want a regulated power source?

    At half power output into the resistor the transistor will dissipate 6V x 0.6A = 3.6W which is very large. If you go that route use a power transistor like a TO-3 case and make sure it's well heat-sunk. Also make sure beta > 30 at 1.2A.

    EDIT: I didn't see the part where you plan to use an op amp to up the gain of the NI. That's OK. The op amp won't swing 0 to -12V but you should be OK. Run the load resistor from ground to the emitter, run the collector to -12V, connect the base directly to the op amp output. The op amp of course also gets powered from ground and -12V. If the op amp can output -11V you'll get (11 - 0.8)/10 = 1.02A instead of 1.2A load current. If you insist on max. current of 1.2A you need to run in the common-emitter mode which has high gain and might result in system oscillations. There is also a danger that the op amp won't put out close enough to 0V to turn ther pnp off.

    Do you have one or two 12V supplies? Things are a lot easier to design for a beginner with the op amp powered by +/- 12V supplies.

    A better idea would be a programmable switching power supply in which the output transistor is either off or saturated, both states representing a low power dissipation mode. The duty cycle represents the power to the load resistor. Since your outer loop has long time constants a switching frequency of even 60 Hz would probably be OK.

    You might need a stabilization network to stabilize your loop in any case. You'll find that out once you set the thing up.

    If you post your proposed circuit and system we can give you hints if necessary to make it work. If this is not homework you should post in the general physics or classical physics forums.
     
    Last edited: Jan 21, 2014
  4. Jan 22, 2014 #3
    Thanks for your reply.

    I have a ±12V supply.

    I don't fully understand your explanation, especially the polarity of the voltages as my knowledge is very limited.

    I have attached an attempt of a circuit that I have been working on. I am fairly sure it is incorrect but not sure why.
     

    Attached Files:

    Last edited: Jan 22, 2014
  5. Jan 22, 2014 #4

    rude man

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    OK, good that you have two power supplies.

    I see no reason why your circuit should not work, if you're OK with a max. current of 1.0A instead of 1.2A. What seems to be wrong? Give us voltages at all accessible points.
     
  6. Jan 22, 2014 #5
    Again I appreciate your help.

    Yes I think limiting the current bellow maximum to 1.0A is the right thing to do. This also helps to keep op amp from saturation due to the 12V maximum allowed.
    I am wondering if the base current, Ib is high enough to achieve my 1.0A current with my 2N3055 transistor, I have tried reading the data sheet but find it hard to understand and I'm not too sure which are the important parameters. Also is the direction of Ib correct?

    I have attached a diagram of my attempt at working out all accessible Voltage points.

    One thing that puzzles me is; what happens to the remaining 2V that is left after 10 Volts across R load?
     

    Attached Files:

  7. Jan 22, 2014 #6

    berkeman

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    You should have a base resistor in series right before the transistor's base. The voltage drop across that resistor will give you the base current value. Your current calculation of Ib and β are not correct.

    I would use something like 100 Ohm base resistor Rb to start. Note that the output current of the opamp may not be enough to provide the 50mA or so Ib to give you the 1A output current, depending on the transistor's βmin. You might end up needing to use a Darlington transistor for the output drive.

    Also, will your +/-12V power supply be able to supply the 1A+ from the +12V suppy rail?
     
  8. Jan 22, 2014 #7

    rude man

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    Base current will be around 1A/90 = 11 mA. Your computation for Ib is incorrect. You have computed the op amp output current to satisfy the two resistors only. The main drain on the op amp is the base current.

    Which brings me to suspect that your op amp (you did not specify the type) is incapable of supplying the total ~15-20 mA required. Not surprising. Should have thought of that earlier myself.

    Solution: DARLINGTON configuration! Here's how:

    Add a low-current npn transistor (e.g. 2N2222) between the op amp output and the 3055 as follows: collectors to each other, emitter of the 2222 to the base of the 3055, base of the 2222 to the op amp output directly, emitter of the 3055 to your load resistor. This unloads the op amp but will cost you another 0.7V or so across your 10 ohm load.

    The part of the 12V supply that isn't dropped across the 10 ohm load is dropped across the transistors - most of the current thru the 3055, only about 1/100 as much thru the 2222.

    PS don't let the NI put out negative voltages, it might be enough to break down one or both transistors. Keep the NI voltage range between 0 and +5Vdc. As a safety measure you could hook up a diode like a 1N4001, anode to ground and cathode to the op amp output.
     
  9. Jan 22, 2014 #8
    I am aware of the DARLINGTON configuration. I will complete the circuit and post when im finished.
    Thanks
     
  10. Jan 22, 2014 #9

    rude man

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    You might also let us see the rest of your circuit, i.e. the thermocouple and its circuit.
     
  11. Jan 23, 2014 #10
    I have only started the thermocouple circuit, reason being that the power resistor circuit has to be submitted by tomorrow and I still have a couple of days for the thermocouple circuit. I plan on using a type K thermocouple with an AD595 thermocouple amp. The AD595 will give me 10mV/°C so should be ok for my application. I

    Here is my new circuit. What do I need to consider in order to calculate Ib1? Also I am unsure where exactly to put the diode you mentioned. I attained the β values from the data sheets of the two transistors.
     

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  12. Jan 23, 2014 #11

    rude man

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    I can't open .docx files. Can you reformat to pdf?
     
  13. Jan 23, 2014 #12

    rude man

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    The ad595 is a good way to go. Just follow ad's recommended circuits.

    The AD590 is also nice; it's a stand-alone temp sensor (no thermocouple required), produces 1 uA/K and is a two-wire device. You hook up the + lead to +12Vdc and the - lead feeds an op amp summing junction (just an op amp and one feedback resistor).
     
  14. Jan 23, 2014 #13
    Sorry
     

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  15. Jan 23, 2014 #14
    Yes I have been looking at the circuits on the datasheet of the AD595 and I think I can handle them ok.
     
  16. Jan 23, 2014 #15

    rude man

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    Good. No resistor needed at base of Q2. Would just lower your output voltage even further.

    Yes, the equiv. beta is approx. the product of the betas of Q1 and Q2.
     
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