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Why does the Vout pin in the diagram below receive 8 volts with Vcc = 0 volts, but becomes 2 volts when Vcc=5 volts. If switch SW9 is closed, circuit theory says Vout should always stay at 15 volts regardless of whether I am powering the hall-effect sensor with Vcc or not.

However, when I measure (with a multimeter) what Vout pin is getting it's never 15 volts, however if I individually measure "VpM" below (with the hall-effect sensor unplugged from breadboard) I correctly get 15 volts. As soon as I plug the hall-effect sensor back in I either get 8 volts (with Vcc=0 volts) or 2 volts (with Vcc=5).

I reason (I think) why I get 2 volts on VOUT pin when VCC=5 volts is because according to this data sheet on pg. 4 in the row marked "Pre-Programming Quiescent Voltage Output" it says it is supposed to be 2.0V.

But this still does not explain why the programming pulses I'm sending (HIGH,MID,LOW) are not at their correct values.

enter image description here

Per Brian's suggestion I have lowered impedances as shown below and ran a simulation from on circuitlab.com and here are the current readings I am getting:

enter image description here In above image R1=R6=10K Ohms, R5=R2=90K Ohms, and R3=R4=20K Ohms. The current between [R1, R2] and [R5, R6] is -300 micro amps, whereas the current between [R3,R4] is -750 micro amps.

enter image description here In above image R1=R6=1K Ohms, R5=R2=9K Ohms, and R3=R4=1K Ohms. The current between [R1, R2] and [R5, R6] is -3.0 milli amps, whereas the current between [R3,R4] is -15 milli amps.

Is there a particular range of current I need to be in? Is this specified anywhere in the data sheet? The only thing I see is on pg. 2 under the "Absolute Maximum Ratings" table for "Output Source Current" and "Output Sink Current" should be 3 and 10 milliAmps respectively.

Since all the currents I listed below are negative, this should satisfy the requirement, correct?

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    \$\begingroup\$ You may have blown an internal programming fuse, but probably not. | The programming requirements seem to be far more complex than can be a addressed with the system shown - unless the switches are processor driven and not shown here. ||| Fuse blow current is 300 mA. At a quick look though I can't see currents taken by Vout during "try" mode but they would have to be small wit R values you are using \$\endgroup\$
    – Russell McMahon ♦
    Feb 23, 2013 at 2:31
  • \$\begingroup\$ Can you explain what your attempting to achieve? The diagram is pretty confusing, why are you driving so much voltage into the output of the sensor? \$\endgroup\$
    – Mark
    Feb 23, 2013 at 6:36
  • \$\begingroup\$ Russell is correct you need way more current to blow the fuses. In another related question electronics.stackexchange.com/questions/58566/… you mentioned you'd be told to use 20-100K resistors so maybe it's a lab project and you don't want the really burn the fuses? You've gone well above those values so maybe if it is an assignment that was a hint try mode would work with those current levels, but the datasheet doesn't seem to specify if it will or not. \$\endgroup\$
    – PeterJ
    Feb 23, 2013 at 9:10
  • \$\begingroup\$ @PeterJ - I eventually want to blow fuses but my understanding is I first need to experiment with sensitivity and offset in either Hold or Try mode, then move on to Fuse Mode. My problem is in Try Mode I can't send out the correct {High, Mid, Low} pulses described in the data sheet when the hall-effect sensor is plugged in to the breadboard, but unplugging correctly sends out all 3. \$\endgroup\$
    – user1068636
    Feb 23, 2013 at 21:02
  • \$\begingroup\$ @RussellMcMahon - I don't think I have blown a fuse because the VOUT pin never exceeded 8V (at least thats what the multimeter said). To blow a fuse you need to send out high voltage pulses (at least 26V according to the data sheet). The switches are not processor driven, but why should they be? I'm not automating this (yet). I am using a power supply similar to this: ebay.com/itm/… and feeding 30 Volts to V1 in the diagram above. The current is 0A. \$\endgroup\$
    – user1068636
    Feb 23, 2013 at 21:09

1 Answer 1

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I think you are under the misapprehension that R3 and R4 define the voltage VpM when Sw9 is closed. That is not the case : R3 and R4 only define VpM as VpM = V1 * R4/(R3+R4) when Sw9 is open.

Close Sw9 and the voltage VpM is defined by R3 and the parallel combination of R4 and the output pin Vout. We don't know very much about that; the datasheet does say (on page 2) that it can sink a maximum of 10mA. Therefore you may have to assume a current source in parallel with R4 capable of sinking anything up to 10ma (i.e. sourcing up to -10ma) and designing your programming voltage sources with that in mind.

The programming instructions do refer to programmable power supplies as programming voltage sources, which suggests that low impedance sources are required.

There is also the remark about 300ma and 0.1uf being required to actually blow a fuse : the current required during the remainder of the programming operations is presumably much less than this; probably of the order of the 10ma output sink current specified.

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  • \$\begingroup\$ What you say about VpM defined by R3 and the parallel combination of R4 and VOUT makes sense when SW9 is closed. But that's a parallel combination of a resister and voltage not two resisters. So what kind of a formula am I dealing with? How do I get my custom voltage pulses to be 3V, 15V, 26V and take VOUT into consideration? I agree about the fuse blowing comment you mentioned, but before I get that far I need to perform some tests in Hold Mode or Try mode. I marked your answer correct because I think you're right, but I'm still unclear how to proceed. \$\endgroup\$
    – user1068636
    Feb 23, 2013 at 21:28
  • \$\begingroup\$ You need much lower impedance supplies. One option would be to use pairs of resistors to define the voltage, feeding an opamp wired as a voltage follower. \$\endgroup\$
    – user16324
    Feb 23, 2013 at 23:14
  • \$\begingroup\$ I used circuit lab run a simulation with lower impedances as per your suggestion and modified my question above. My original circuit with 910K Ohms was giving me currents of -29.41 microamps and -750microamps. What range of current am I trying to be in? The data sheet says no more than 10milliamps or 3 milliamps for output sink current and output source current respectively. Should I be striving to stay close to these levels? \$\endgroup\$
    – user1068636
    Feb 24, 2013 at 0:45
  • \$\begingroup\$ I wasn't clear enough : you can't do this just by fiddling with resistive dividers.You need an amplifier stage (opamp in unity gain configuration, or transistor in emitter follower configuration) after the resistors, to let the load take whatever current it needs without affecting the voltage you select. \$\endgroup\$
    – user16324
    Feb 24, 2013 at 9:16
  • \$\begingroup\$ Just one more question: If I add an amplifier stage after the resisters like you say (immediately before VOUT pin feed) is there a way I can test the amplifier is working as expected? For example, using the multimeter would kind of readings should I expect? Should I expect to get "VpM" when SW9 is closed? What should I expect current readings to look like? Should they be close to the 2nd and 3rd circuitlab figures above? \$\endgroup\$
    – user1068636
    Feb 24, 2013 at 15:52

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