Connect with us


Discussion in 'Power Electronics' started by Kuberan, Jan 5, 2020.

Scroll to continue with content
  1. Kuberan


    Jan 5, 2020
    Hi.. I'm trying to control 2 DC motors of 120watts each (12V, 10Amps). I designed two H bridges to achieve this. I used IRF3205 N channel mosfets, BC847 Transistors and 1uf capacitors.

    One motor is controlled perfectly in both directions. But the other motor is not working in one direction. Also, while running in one direction, the high side mosfet is heating.

    Is there any problems with this circuit? [​IMG]
  2. bertus


    Nov 8, 2019

    Your image on the google drive does not show.

    I also see that you already posted your question on the AAC forum, using an hyjack, that has been split off:

    Here is the image taken from that thread:


    Are the input pulses in counter phase?
    You will run into troubles when both fets on one side are on.

    Are you trying to mimic a high side driver?
    Why not use a dedicated driver?

    Last edited: Jan 5, 2020
  3. Fish4Fun

    Fish4Fun So long, and Thanks for all the Fish!

    Aug 27, 2013

    Assuming the high-side switching device is in-fact an N-Channel mosfet; I don't know how the high-side Mosfet would ever turn on ... the data sheet shows the IRF3205 has a gate threshold Voltage (Vgst) of ~4V. This would imply you need a minimum gate voltage of 16V (with respect to ground) to turn on the High-Side Mosfet and to reach the rated Rds ON of 0.008 ohms you would need a Vgs of at least 10V (ie 22V with respect to ground).

    For low voltage (ie < ~30Vdc) H-Bridge circuits using P-Channel Mosfets for the High Sides and N-Channel Mosfets for the Low Sides usually results in the the easiest to DIY designs. As the Voltage increases (and perhaps in some cases as the Production Volume increases) using an N-Channel Mosfet for the High-Sides becomes increasingly desirable. To achieve N-Channel Mosfet High-Side switching specialized ICs are typically employed to create a floating High-Side switching voltage of Vcc + Vsw (where Vsw is typically 12V-15V).

    This Link: does an excellent job of explaining how both a High-Low Side Driver IC is used AND how a typical H-Bridge Motor controller might be designed.

    While the IRF3205 is Nominally Rated @110A Continuous, you should note this rating is @25C.... In the datasheet (Page 5 Figure 9) there is graph showing the relationship of Id (Current) to TC (Case Temperature) that clearly shows the Package Thermal Limit to be Id = 75A ... and @ 75A this implies the device is eminently close to catastrophic failure even with an infinite heat sink.

    On Page 3, Figure 4 shows the "Normalized" Rds_On with Respect to Tj (Junction Temprature) @Id = 107A is shown. While this essentially demonstrates the device in thermal runaway toward failure, it indicates a **realistic** Rds_On to Temperature relationship. Using this information can help determine heat sink requirements.

    It is also important to note that while Vgst = 4V, most of the specifications are specified @ Vgs = 10V. @ a Vgs of less than 10V you should probably de-rate the device considerably. I would expect that using a high impedance 5V logic signal to drive an IRF3205 with no heat sink would put 10A continuous Id on the wrong side of the SOA (Safe Operating Area) box.

    My abilities in thermodynamics are poor at best, but a few iterations of the conditions suggest that 10A continuous Id in an IRF3205 would certainly approach the device's 175 °C limit even with a Vgs of 10V:

    Initial conditions = 25 °C. 10A, Rds_On = .008 Ohms Vgs = 10V

    Package Heat Gain = i^2 Rds_On
    ==> 10^2 * 0.008 = 0.8W
    ===> 0.8W * 70 °C/W = 25 °C + 70 °C = 95 °C

    From Figure 4 @ 95 °C the "Normalized" Rds_On ~ 0.015 Ohms
    ==> 10^2 * 0.015 ohms = 1.5W
    ===> 1.5W * 70 °C/W = 25 °C + 105 °C = 130C

    From Figure 4 @ 105 °C the "Normalized" Rds_On ~ 0.016 Ohms
    ==> 10^2 * 0.016 ohms = 1.6W
    ===> 1.6W * 70 °C/W = 25 °C + 112 °C = 137C

    From Figure 4 @ 137 °C the "Normalized" Rds_On ~ 0.018 Ohms
    ==> 10^2 * 0.018 ohms = 1.8W
    ===> 1.8W * 70 °C/W = 25 °C + 126 °C = 151C

    From Figure 4 @ 151 °C the "Normalized" Rds_On ~ 0.019 Ohms
    ==> 10^2 * 0.019 ohms = 1.9W
    ===> 1.9W * 70 °C/W = 25 °C + 133 °C = 158C

    While an IRF3205 driven @ 5V might survive a 10A Id for some period of time w/o a heatsink, the thermal stress would almost certainly lead to premature failure. I would use rather robust heatsinks and even seriously consider using forced air (a fan) if I were going to design/build this circuit.

    Hope This Helps!

    bertus likes this.
  4. Minder


    Apr 24, 2015
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day