NPN BJT base potentiometer?

Discussion in 'The Projects Forum' started by walk on, Jul 16, 2012.

1. walk on Thread Starter New Member

Jul 12, 2012
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Before I ask for help, I want to say thanks to all the members that post. I'm a math guy but I was able to grab ALOT of useful information that has me well on my way to designing my first "im sure i can make something just like that for much cheaper" circuit.

Here's my question. If I have the attached signal, can a place a potentiometer between it and the base of an NPN BJT in order to prevent voltages below a certain point from switching on the collecter-emitter?

The fastest "on-off-on" sequence I expect will be .1 ms so the transistor shouldn't have a problem with it. (If possible/practical I imagine that a reverse log taper pot should make tuning it much easier or should I be fine with a regular linear taper?)

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2. ErnieM AAC Fanatic!

Apr 24, 2011
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Ermmm... maybe, but it is not going to be a very exact transistion, meaning the output wave will be "sloppy" with some funny edges. A better way would be to use a comparator. Take the (linear) pot and arrange it as a voltage divider to one input, and your signal to the other input.

If that is too much then ditch the pot and put one or more diodes in series with the base: for every diode you add you move the turn on point up about .6V.

3. wayneh Expert

Sep 9, 2010
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Same two solutions I was thinking of. You could also use a zener diode on the base to control the voltage that starts to turn on the transistor.

What you're asking for is a single-transistor comparator. If you google that, I think you'll find examples.

4. walk on Thread Starter New Member

Jul 12, 2012
11
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Thanks for the replies. I attached a schematic to help.

What i'm trying to do is control the motor with PWM using the signal mentioned earlier. This signal is an IC's output. Setting the "rejection voltage" should produce a maximum duty cycle of 95-100% and minimum duty cycle of near 50%-55%. I am reading up on comparators now and have gotten to Schmitt triggers. I have a feeling i missed my exit a little while back.

Also (I should have mentioned the motor earlier, my apologies), is the concern about the neatness of the waveform still an issue? I dont think it should be. (Feel free to poke holes in my following statements.) Here's what going on in this head of mine. "On off on off, thats all i need this thing to do. So if i start to tone down the current flowing through the base, at some point the motor's going to stay off. turn the dial back a little and i'm at 50% all the time. turn it back even more and i can get it to work like a charm. As long as the motor reacts in a predictable way things should be fine." (famous last words?)

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5. wayneh Expert

Sep 9, 2010
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It matters to the transistor. Any voltage on the base (or gate, if you use a MOSFET instead, wink winkk) that doesn't fully saturate the transistor to it's "on" state will result in a voltage drop across the transistor. A voltage drop means power dissipation, potentially quite a bit. Without designing for that (big heat sinks, appropriate transistor), it'll overheat. That's the beauty of PWM - it drives the transistor fully on or fully off, so that little power is dissipated there.

6. Audioguru New Member

Dec 20, 2007
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The base-emitter turn on voltage of a transistor is different between transistors even if they have the same part number.

The turn-on voltage changes when the temperature changes.

7. walk on Thread Starter New Member

Jul 12, 2012
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the waveform holds the same shape regardless of what i set as Vmax. The highest it can be is most likely 4.5 - 5 volts. Heres the interesting part (interesting to a me at least) From my simulations it seems that i want Vreject = Vmax/2 exactly. Can I (and should I) try to get Vmax to be as small as possible so that Vreject does not saturate the base/gate? this should also help reduce heat, or am i mistaken.

In order to reduce Vmax I have to reduce Vsupply of the IC.... and in order to do that i have to make changes the voltage regulator which controls Vsupply. I am thinking that since the overall voltage drop will be the same, from start to finish I might try to spread the heat across the elements evenly. It sounds wrong but I dont have the experience to know why. (maybe because voltage regulators are expected to have a large voltage drop where as a transistor isnt?)

8. WBahn Moderator

Mar 31, 2012
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For transistors switching power, you like to either have them hard on or hard off so that you minimize the power dissipation in it. This is a nonlinear way over operating and so you can design it to achieve these goals. Something like a 7805 regulator is a linear circuit, for various reasons, and so this is not an option and you just have to deal with the power dissipation in the regulator (or in the bypass transistors, if you go that route).

I would echo the thoughts of others. Use a comparator with the an adjustable reference and use that to turn your transistor switch (be it a BJT or a FET) hard on and hard off.

If you are trying to use all discrete components and want to minimize transistor count, then that is something else and we can talk about that.

9. t_n_k AAC Fanatic!

Mar 6, 2009
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You would probably also need to consider placing a free-wheeling diode across the motor to prevent any inductive switching transients damaging the transistor in a real circuit.

10. walk on Thread Starter New Member

Jul 12, 2012
11
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transistor count is not an issue. in fact, transistors its probably the only thing i seem to have a grasp on. it may not be a good grasp but im getting there. Comparators, however, i'm still a bit fuzzy on and im still reading up on them to make sure i can get everything figured out. it seems that if i get this to work out i will get all the effects of pwm.

I should say this too. I can get the IC to produce a linear signal, but i thought this was the best way to do it without having to buy a PWM kit. #1 to save money (i will most likely need a similar setup for a future project) #2 to understand exactly what each component is doing (and to say "look what i built")

Now that I think about it, the other project will also require an output just like the one i require from the motor. Putting bragging rights aside, would it be cheaper to swallow my pride and buy a PWM kit like the one velleman offers?

11. walk on Thread Starter New Member

Jul 12, 2012
11
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you're right about the diode. i have it included but i guess i didnt save the schematic before i uploaded it.

12. ErnieM AAC Fanatic!

Apr 24, 2011
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Of course, you could use the regulator itself as the PWM element:

Using something like a LM317 the output doesn't go to zero but to the 1.25V (?) reference voltage. You can use a pair of diodes to get rid of that.

I used this scheme to control motors on a HO model train layout. I could get a unit to creep tie by tie over several minutes.

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13. walk on Thread Starter New Member

Jul 12, 2012
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Ernie, im trying to grasp what you've layed out so correct me when i say something wrong.
With your setup,when the signal is high, the voltage seen by the motor will be whatever i set the regulator to output. The motor will see the voltage from the signal alone when ...? OK never mind im lost here. My apologies, it looks simple and very similar to my circuit but im having a hard time seeing what the motor how the motor will react to the signal. Im going to run a simulation with it and see if i can have a eureka moment.

14. walk on Thread Starter New Member

Jul 12, 2012
11
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its amazing what you can learn in a few days.
OK so heres where I stand. After learning about 10% of the things op-amps can do, I realized that there has to be a way I can make this work. Heres is my plan of attack. The problem I was having with using a linear voltage signal was that 1. critical values were below the limits of sensitivity of another device. Also, 2. the current it puts out is not large enough to make it past the 10k input impedance when at lower voltages (expected outputs of 0 thru 5V with 200uA current for all output voltages).

Op amp to the rescue? I attached an simple example of what I have in mind. This should make things run smoothly. Unless i'm mistaken (which happens often) Vcc ( a regulated lantern battery, maybe some C or D) should supply enough current to get the signal through the input impedance and double the voltage. ?

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