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Driving SSR ("hockey-puck") directly with Prop I/O - Page 2 — Parallax Forums

Driving SSR ("hockey-puck") directly with Prop I/O

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  • CircuitsoftCircuitsoft Posts: 1,166
    edited 2013-10-10 12:58
    ags wrote: »
    I'm totally puzzled as to why one would add a resistor in series with the MOSFET gate. The capacitance is far less than many other connections that Prop sees, so I suspect the purpose isn't to limit transient/switching current. Why do you suggest this?
    While it's more of an issue if the fet will spend any time in the switching region, the Drain-Gate capacitance along with the inductance of the leads of the mosfet can make an oscillator. The resistor inserts a zero into the control loop and damps resonance enough to prevent such oscillation.
  • agsags Posts: 386
    edited 2013-10-10 13:17
    K2 wrote: »
    Where have you seen a Prop pin directly driving a much larger capacitive load?

    Edit: A more complete answer might go something like this... There's a big difference between a circuit surviving a handful of cycles and surviving 30 trillion cycles, which is a reasonable number of cycles to expect of a MOSFET-based SMPS. A gate resistor also reduces switching noise (EMI and RFI).

    I didn't expect this was a controversial position, and admittedly my language was sloppy. I should have said that I have designs driving larger capacitive loads and have not seen any negative effects. I don't have product running 24/7 for years so my data is limited.

    A quick review shows that a single n-channel MOSFET with Vdsmax of 50V and Idsmax of 0.2A (both 10x the required capability for this use) has input capacitance of about 20pF. I have designs running that drive 6 shift registers, with all clock and latch pins driven by one Propeller pin. The input capacitance for these is about 5pF. Hardly a resounding proof of design robustness.

    Perhaps the right question is at what capacitive load do design best practices (including cost, complexity, speed) suggest use of a series resistor to limit charge/discharge current?

    Of course, my original question was why include the gate series resistor - and it was just a guess that it was for this purpose. Hopefully PhiPi will chime in and validate my assumption or correct it.
  • agsags Posts: 386
    edited 2013-10-10 13:21
    While it's more of an issue if the fet will spend any time in the switching region, the Drain-Gate capacitance along with the inductance of the leads of the mosfet can make an oscillator. The resistor inserts a zero into the control loop and damps resonance enough to prevent such oscillation.

    Got it. That makes sense, thanks for the reply... and the tip on the MUN2211.
  • Phil Pilgrim (PhiPi)Phil Pilgrim (PhiPi) Posts: 23,514
    edited 2013-10-10 13:23
    I was thinking that a MOSFET might exhibit a failure mode wherein the gate insulator is breached, allowing the drain voltage to appear on the gate pin. But that said, I've personally never used a series gate resistor with a small MOSFET, such as the 2N7000.

    -Phil
  • agsags Posts: 386
    edited 2013-10-10 15:32
    I want to thank everyone that participated in this thread. It seemed like such a simple thing when I first posted - and arguably there has been some over-engineering here (on my part, at least). But to me there was a lot of value in the thread, and hopefully others will learn from it.

    There were several solutions proposed, each with its own merit. I'm going to use the MUN2211 for simplicity (just one part) and to make sure that I'm within spec for the SSR I'm using. Overheating due to partial-on or an exposed 120VAC terminal with partial-off are things I'd like to avoid. Also, the Propeller is isolated from most failures, in high, low, reset and off conditions.

    Now I just have to figure out which of the many MUN2211 options I need to select from the alphabet soup in the datasheet.
  • agsags Posts: 386
    edited 2013-10-28 15:49
    2 reasons:
    1. The Base Emitter resistor consumes any leakage current from the Collector to Base. C to B leakage can be multiplied by the gain of the transistor resulting in some heating.
    2. Faster switching times.

    Zetex has quite a few variations. NPN, PNP, and many resister combinations.

    Duane J

    Duane, would you mind elaborating a bit more on this? A quick search of OnSemi literature (and Mouser) shows that there are many varieties available. Resistor values of 1, 2.2, 4.7, 10, 22, 47 and 100 kOhms, (with various mixing of values for Rbe and Rb) as well as some with no Rbe at all.

    I don't have any experience with digital switching BJT design. From the fundamentals, Vbe forward bias is about 0.6v, so Rbe only conducts when Vbe is below that threshold. So if there is some minor Icb leakage (the specs show Icbo (cutoff) is 100nA) that will effectively be shunted by Rb (through the Propeller output pin) and Rbe in parallel (resulting in Vbe of 0.5mV - well below forward bias). Do I have this approximately correct?

    I will be driving an SSR with 5v, using an MUN22* in open collector configuration. The SSR has (I believe - reverse engineering the specs) 2kOhm resistor in series with the 1.0v LED. The "must operate" input is 3VDC; the "must release" input is 1VDC. The question is which values for the bias resistors will work best (I'd like to minimize the on current from the Prop output) while ensuring full on/off control of the SSR?
  • Duane C. JohnsonDuane C. Johnson Posts: 955
    edited 2013-10-28 20:00
    Hi ags;
    ags wrote: »
    Duane, would you mind elaborating a bit more on this? A quick search of OnSemi literature (and Mouser) shows that there are many varieties available. Resistor values of 1, 2.2, 4.7, 10, 22, 47 and 100 kOhms, (with various mixing of values for Rbe and Rb) as well as some with no Rbe at all.
    It seems overwhelming with all the choices available.
    Let me try to break it down.
    I don't have any experience with digital switching BJT design.
    The primary difference between a switching and other types of BJTs is, (loosly), the lower saturation voltage. 0.2V in this case, (probably less but lets design to spec).
    From the fundamentals, Vbe forward bias is about 0.6v, so Rbe only conducts when Vbe is below that threshold.
    Not true.
    Rbe will conduct current anytime the voltage across Rbe is not zero even when the base is forward biased..
    R2 is primarily used, at least in this low speed application, to absorb Icb.
    Icb is specked at 100nAmax @ 50V and 25°C. They don't have a chart showing what it would be at other temperatures.
    In this case there will be little heating so lets assume its still 100nA
    R=0.5Vbe/100nA=5MΩ
    All values of R2 are less than this so any would work.
    So if there is some minor Icb leakage (the specs show Icbo (cutoff) is 100nA) that will effectively be shunted by Rb (through the Propeller output pin) and Rbe in parallel (resulting in Vbe of 0.5mV - well below forward bias). Do I have this approximately correct?
    I agree.
    We do want R2 though, (any value), in the event the control input happens to be in a high impedance state to guarantee the BJT is off. Just good practice.
    I will be driving an SSR with 5v, using an MUN22* in open collector configuration. The SSR has (I believe - reverse engineering the specs) 2kOhm resistor in series with the 1.0v LED. The "must operate" input is 3VDC; the "must release" input is 1VDC. The question is which values for the bias resistors will work best (I'd like to minimize the on current from the Prop output) while ensuring full on/off control of the SSR?
    5Vcc - 1Vdiode - 0.2Vsat = 3.8V the voltage left over to be applied to the SSR resistor.
    3.8V / 2kΩssr = 1.9mAIc this is useful for estimating the required Ib.
    There are several combinations of R1 and hFE to choose from.
    Lets see if MUN2213T1 will work as it has the highest R1 value:
    R1=47k
    R2=47k
    hFE= 80
    Ib = 1.9mAIc / 80hFE = 23uAIb minimum required base current
    (5Vcc - 0.6Vb) / (23uAIb + (0.6Vb / 47kR2 ) + 100nAIcb) = 123kR2
    Clearly 47k is less than 123k which delivers more than the required 23uAIb putting the BJT into greater saturation.

    So it looks like the MUN2213T1 should work with the minimum input current of.
    (5Vcc - 0.6Vb) / 47kR1 = 93uA
    The input current with R1 is 93uA.

    Now 93uA is pretty small.
    But I would choose one with a bit more drive current just to be safe, unless absolute minimum drive current is required.
    I would choose:
    MUN2214T1 with 440uA
    or
    MUN2233T1 with 930uA

    I hope this makes sense.

    Duane J
  • agsags Posts: 386
    edited 2013-10-28 23:35
    Not true.
    Rbe will conduct current anytime the voltage across Rbe is not zero even when the base is forward biased..
    Yes, of course. This stuck on my fingers as I typed because it's not accurate. Rbe is a fixed value resistor, and will conduct with any potential applied across it. I should have said that for Vbe<forward bias, the equivalent resistance across b and e is >> Rbe (Rbe condutivity dominates in the parallel configuration) and for Vbe>forward bias, Rbe conducts what is basically a constant current (=0.6V/Rbe).
    R2 is primarily used, at least in this low speed application, to absorb Icb.
    I'm not following. If Vin (voltage applied to the MUN22* input, the Rb terminal not connected to the BJT base terminal) is logic 1 (3.3V) then Rb limits current from the Propeller driver pin, and all but a tiny fraction of current passes through the emitter to ground (except for 0.6v/Rbe). So Rbe isn't doing much in this case.

    If Vin is logic 0, then any leakage Icb would be conducted through Rb and the Prop NMOS pulldown device to ground. With Icb=100nA, that's well within Prop I/O parameters (I presume). While having Rbe in parallel does carry some current, it still is not critical.
    We do want R2 though, (any value), in the event the control input happens to be in a high impedance state to guarantee the BJT is off. Just good practice.
    What you have clarified is that if the Prop I/O is in a hi-Z state, the Rbe does serve an important role in ensuring that Vbe is ~0V, which may be important to ensure that the connected load (my SSR in this case) is not "on" during reset or other hi-Z conditions.

    Do I have this correct so far?
    R=0.5Vbe/100nA=5MΩ
    All values of R2 are less than this so any would work.
    Lost again. What is "0.5Vbe"? Is this an approximation of the equivalent resistance from base to emitter just before the base is forward biased?
    So it looks like the MUN2213T1 should work with the minimum input current of.
    (5Vcc - 0.6Vb) / 47kR1 = 93uA
    The input current with R1 is 93uA.
    Do you mean (3.3Vprop-out - 0.6Vb)/47kR1 = 57uA?

    I hope it's clear that I'm not trying to argue, just understand the details. Thanks for taking the time to respond.
  • Duane C. JohnsonDuane C. Johnson Posts: 955
    edited 2013-10-29 07:52
    ags wrote: »
    R2 is primarily used, at least in this low speed application, to absorb Icb.
    I'm not following. If Vin (voltage applied to the MUN22* input, the Rb terminal not connected to the BJT base terminal) is logic 1 (3.3V) then Rb limits current from the Propeller driver pin, and all but a tiny fraction of current passes through the emitter to ground (except for 0.6v/Rbe). So Rbe isn't doing much in this case.

    When designing circuits one has to account for all possible weird stuff that can happen.
    If Vin was guaranteed to always be connected to a low impedance source R2 would not need to be used at all because Icb would be absorbed by R1.
    However, if the source is sometimes at high impedance, lets say when the micro is booting up or even powered off we need to make sure the BJT is off. R2 guarantees this.
    If Vin is logic 0, then any leakage Icb would be conducted through Rb and the Prop NMOS pulldown device to ground. With Icb=100nA, that's well within Prop I/O parameters (I presume). While having Rbe in parallel does carry some current, it still is not critical.
    What you have clarified is that if the Prop I/O is in a hi-Z state, the Rbe does serve an important role in ensuring that Vbe is ~0V, which may be important to ensure that the connected load (my SSR in this case) is not "on" during reset or other hi-Z conditions.

    Do I have this correct so far?
    I think you have it.
    R=0.5Vbe/100nA=5MΩ
    All values of R2 are less than this so any would work.
    Lost again. What is "0.5Vbe"? Is this an approximation of the equivalent resistance from base to emitter just before the base is forward biased?
    0.5Vbe is a bit less than 0.6Vbe so the leakage current would all go through R2 and virtually none into the base.
    0.5Vbe is in the spec for measuring Icb leakage current.
    The calculated 5MΩR2 would have been sufficient.
    However, the highest presented value of R2 is 47k.
    Do you mean (3.3Vprop-out - 0.6Vb)/47kR1 = 57uA?
    Yes it would be. I used 5V as I didn't know you were using a Prop that runs on 3.3Vdd.

    BTW, the other 2 parts would also be acceptable but would draw a bit more current from the prop pin, not necessarily a bad thing.

    Duane J
  • agsags Posts: 386
    edited 2013-10-29 08:22
    OK - I think I have it. Thanks for stepping through this.

    One more thing - if Vin (BJT base) was unconnected, and Vce of +5V (or more) applied, then any leakage current would be Ice=Ibe (KCL). What problem would this create?
  • User NameUser Name Posts: 1,451
    edited 2013-10-29 11:38
    I must be a real daredevil! I've been driving large SSRs (30kW) directly with Prop pins for nearly a decade. It has always seemed to me that Crydon et.al. have already done the engineering. :)
  • Duane C. JohnsonDuane C. Johnson Posts: 955
    edited 2013-10-29 12:08
    Hi ags;
    ags wrote: »
    OK - I think I have it. Thanks for stepping through this.

    One more thing - if Vin (BJT base) was unconnected, and Vce of +5V (or more) applied, then any leakage current would be Ice=Ibe (KCL). What problem would this create?
    I don't recognise what KCL is. But I presume its the gain factor hFE and is 80 for the suggested units.

    Assuming Vin is not connected to anything and R2 is not present:
    100nAIcb * 80 = 8000nA
    8000nA + 500nAIce = 8500nAIce total collector current

    However, this is not the case as R2 IS present and absorbs Icb and spits it out the emitter.
    100nAIcb + 500nAIce = 600nAIce total collector current.

    Now for a practical matter:
    I think I know where your going.
    Will the leakage current be to high to prevent turning off the SSR?
    600nAIce * 2Kssr = 0.0012V
    I don't know what the SSR LED voltage is passing only 600nA but it's much less than 1.1V maybe 0.1V or less.
    0.0012V + 0.1Vled = 0.1012Vssr
    This is much less than the 1V turn off voltage for the SSR.
    Furthermore, the LED will be emitting virtually no light at 600nA so the SSR will not be turned on.

    Duane J
  • User NameUser Name Posts: 1,451
    edited 2013-10-29 13:09
    Kirchhoff's Current Law
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