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enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Motor control schematic.

I am trying to replicate this circuit: Parastaltic pump driver.

The problem I have is after I lay out the circuit exactly as described, with a terminal block for the 12v power supply, the positive terminal of which goes up junctions to the Anode of the diode, then into the + of another terminal block, which runs to the pump, then comes from the - of the terminal block, junctions to the Cathode of the diode and then into the collector ([Pin 2][3]).

I have a 2.2kO resistor connected between the third and final terminal block + to the base pin, then I have the collector from the TIP120 and the - from the third terminal block going to the - from the 12v supply terminal block.

My understanding is that both power supplies need to have a common ground because obviously otherwise the + from the 3.3 PWM wont have anywhere to go and wont trigger the base of the TIP120.

Now, where it gets weird, for me...is that when I hook up the 12v power supply and I measure the voltage between the + and - on the 12v terminal block I get...12v...like the circuit is complete. The motor does nothing though. And no current is going through the base pin. If I take my pprogrammable power supply that I'm using in place of the 3.3 from the Arduino, plug it into the wall, do NOT turn it on, and touch the + lead to the + of the terminal block leading to the resistor connected to the base pin...the motor starts to slowly turn. If I touch the - from the power supply to the - terminal of the same block, it stop turning and something starts smelling funky (shut it off before I could investigate, the TIP120 is cold to the touch however).

Now, I did this in a project board and you know them, you have to do some nasty solders, so I thought, ok, I don't see any shorts or measure any shorts but lets do it in a breadboard just to be safe. Recreated the whole thing, same way, EXACT same result.

How the hell is touching the positive terminal of the powered off (if I unplug the DC PSU from the wall nothing happens when I touch the + lead to the resistor on the base) PSU causing the motor to spin? It also acts like its got plenty of current but low voltage as the motor spins up right away just very VERY slowly.

I've measured the DC motor and it takes about 375mA to get going and once its humming along with no load it draws around 80mA anything less than 375mA and it slowly ramps up, but obviously if I drop the voltage and leave the output current at 375mA it slows the RPM without impacting its ability to start.

I freely admit I'm new to this and learning but my understanding what, no base current, no circuit. But it acts like its free to carry all the voltage it wants, just no current. As to why hooking up the positive to a non powered on PSU allows it to start turning...and worse still why then also touching the negative from the powered off CPU appears to allow it to draw a LOT of current...I'm at a loss.

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  • \$\begingroup\$ It is odd behaviour. But what you're doing is a little odd too - why connect just one lead of your 3.3V PSU? Rather than getting hung up on this odd behaviour, why not connect your 3.3V PSU as you originally intended (both + and -) and see if the circuit works? If it does work then you can go back and see if you can find out why it didn't behave as you expected. \$\endgroup\$ – Steve G Aug 1 '16 at 20:51
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    \$\begingroup\$ We love schematics. There's an easy-to-use schematic editor button built into the post editor toolbar. Double-click the components to edit their properties. \$\endgroup\$ – Transistor Aug 1 '16 at 21:20
  • \$\begingroup\$ Is your 3.3V power supply ground connected to your 12V ground? It's not clear to me. You mention you know it is needed, that's all. \$\endgroup\$ – jonk Aug 1 '16 at 22:47
  • \$\begingroup\$ @transistor: Thank you so much for the tip! That is awesome! I updated the post to include a schematic. \$\endgroup\$ – alcaron Aug 2 '16 at 13:26
  • \$\begingroup\$ @jonk: Yeah I added a schematic to show that. \$\endgroup\$ – alcaron Aug 2 '16 at 13:28
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Your circuit is built and you are now in debug phase. High current faults are often a problem and to debug without burning out the components we need to limit the current to a safe value. If you have access to a lab power-supply then that's easy - just set the current limit to a low value. 100 mA might do the trick in this case. Then you can measure voltages at various points around the circuit to try and understand where all the current is going.

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Figure 1. A bench power-supply with separate voltage and current limit adjustment and readout. (The voltage setting in this case seems to consist of a coarse and fine potentiometer.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. Poor-man's current limiter.

If you can find a small car dashboard lamp, a few bicycle bulbs, Christmas-tree lamps, LEDs and resistor or even just a resistor - 120 Ω would limit you to 100 mA on 12 V - and wire them in series with your supply you can power up and test. The lamps have the advantage of instant indication of current. If you only have 1/4 W resistors the lowest value you can use can be calculated from \$ P = \frac {V^2}{R} \$. Re-arranging: \$ R = \frac {V^2}{P} = \frac {12^2}{0.25} = 576 \Omega \$. 470 Ω would be OK for a short while.

Report back.


[OP's comment:] With the power supplies, when I set the current to 100mA is it also beneficial to run them with Over Current Protection?

If the over-current protection disconnects the supply it's useless as you won't know why. Let's take an example using a lab PSU with voltage limit set at 12 V and current limit set at 100 mA.

  • \$ R = \frac {V}{I} = \frac {12}{0.1} = 120~\Omega \$. If we hook this up we will get 12 V (just) and 100 mA (just).
  • If we decrease the resistance to 100 Ω we will hit the current limit at \$ V = IR = 0.1 \times 100 = 10~V \$. The current limit LED will turn on (if there is one).
  • If we increaase the resistance to 150 Ω the voltage will rise until we hit the 12 V limit. The current will be \$ I = \frac {V}{R} = \frac {12}{150} = 80~mA \$. The constant voltage LED should light (if there is one).

Using the lamp test will allow you to measure the voltage across the diode/motor, the transistor and anything else you want. They'll all be lower than on a good circuit so you'll need to make allowances for that and interpret the readings.

It seems like it would be difficult to troubleshoot as I know its pulling more than 100 mA and its just going to kill the outputs pretty much immediately.

Whatever is giving a short-circuit will have zero-volts across it.

Or will it do a pretty good job of not letting you have any more current than this regardless.

Either the current limiter or the lamps will do this.

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  • \$\begingroup\$ Yeah I think you are right, I'm just going to have to get a second programmable power supply so I can run it without worrying. I like that idea of the light bulbs though!!! Thank you very much for all the help! With the power supplies, when I set the current to 100mA is it also beneficial to run them with Over Current Protection? It seems like it would be difficult to troubleshoot as I know its pulling more than 100mA and its just going to kill the outputs pretty much immediately. Or will it do a pretty good job of not letting you have any more current than this regardless. \$\endgroup\$ – alcaron Aug 2 '16 at 19:09
  • \$\begingroup\$ See if the updates help. \$\endgroup\$ – Transistor Aug 2 '16 at 19:43

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