I have used this forum to lookup solutions and suggestions from time to time but this is my first ever post! There are many great P and N-Channel MOSFET Questions and Answers in here, but I cannot find a solution to what I am asking. I have needed to change the low range of my Device voltage from 1.5V to 0.6V on a board I have made and have now troubles driving a P-Channel MOSFET used to turn OFF the Low Current measurements in the design by shorting the Shunt Resistor with a P-Channel MOSFET.
The Vgs for the P-Channel MOSFET is no longer enough to fully turn ON the MOSFET when the voltage for the Device is at 0.6V (Vgs = -0.6V).
Please see Schematic Named A for the working solution where the the low range of the Device Voltage is 1.5V.
I may very well have gone down a rabbit trail here with the solutions in Schematics B and C and I am asking for help to either verify my suggested solution (Schematic C) or provide a much better/different/simpler solution that I could not figure out.
Any help would be much appreciated!
Below is the somewhat long explanation of the 3 designs (I am sorry the component IDs are all different between them, copy paste does that automatically):
A – Schematic:
This is the working/current solution used on a board with a Device voltage generated in the range 1.5-5V. Q6 is shorting Shunt resistor R40 when the measured current is over 1mA to avoid a high drop voltage over R40 for higher currents. Q6 can be easily controlled here with N-Channel Q7 from a uProcessor. When Q6 Gate is grounded (via Q7) Vgs will be between -1.5V and -5V, well within the fully ON region of the P-Channel used.
B – Schematic:
This is the new design of the board where the Device voltage generated is in the range 0.6-5V, instead of as above 1.5-5V. Q8 (Same as Q6 above) can no longer be easily turned fully ON as Vgs is not enough (-0.6V). This is an attempt to lower (or increase Vgs) the Gate voltage of Q8 (Same as Q6 above) by instead of referencing Source of Q9 (Same as Q7 above) to GND, I added a negative voltage to Source. This would give a Vgs for Q8 (Same as Q6 above) between -1.6V and -6V. We can now again turn Q8 (Same as Q6 above) fully ON.
The problem is (and there is probably more that I missed) that even if I Float or High-Z the uProcessor Pin controlling Q9 (Same as Q7 above) to turn OFF Q9 (and Q8) the uProcessor pin will be at -1V. Since the uProcessor pin has protection diodes to GND and VCC this will cause the GND diode to conduct (Max is -0.5V).
C – Schematic:
Same as Schematic B but I have added Q2 as a high side switch to try to avoid the problem with the uProcessor pin explained in Schematic B. This way I should be able to control this High Side switch using the uProcessor pin which in turn would turn ON and OFF Q4 (in this schematic. Q7 and Q9 in the other ones).
Does this make sense? There are so many drivers and levels in this design and I may have just went down the wrong way and there is a much simpler solution?
Again, any help would be much appreciated! Take apart my design as much as you want.
EDIT 1 – Better details on what Powers this circuit and requirements:
The schematic is much larger as a whole. The “Device” or “VCC_Device”, which is not pictured, is a DC/DC Switcher delivering up to 2.5A that now provides 0.6 to 5V DC out, runs through the measurements shunts and deliver it to the Device Under Test (TARGET_POWER). I may not have explained the Device and the target very well, which is the risk when you try simplify or leave out parts of the schematic.
As pointed out in the comments, I should simplify to the smallest item, which has been done below by tlfong01. The only requirements are that I can turn the MOSFET fully ON using a varied Source Voltage between 0.6V to 5V. The MOSFET should be able to handle at least 2.5A running through it and the Rdson should be kept low (max 40mOhm for max 100mV drop @2.5A) to avoid heat and voltage drop. BTW, it may seem weird that I do not provide Current Measurements for the whole range 0-2.5A, but here is a reason for that (so the suggestion is good to decouple that from the problem statement!).mosfetgate-drivinglow-voltagep-channelshareedit follow flagedited 2 days agoasked Aug 24 at 0:48MWestberg2133 bronze badges New contributor
- Ah, let me see if I can (over) simply your problem. (1) Your “‘device” can output 1.5V to 0.6V to turn off a p-type MOSFET. (2) But now the (new) p-type MOSFET will not longer do the job for 0.6V as it used to. (3) Design B tries to do the following workarouond: (a) change the device’s old output range of (1.5-5V) to (0.6- 5V). But you then also need to shift down the power source by biasing (shifting down) with a negative voltage reference. / to continue, … – tlfong01 Aug 24 at 1:57
- (3) Design C basically adds a High side switch to be controlled by a uP/MCU/SBC. I confess I have not at all looked at your scary complicated schematic. Just brainstorming. – tlfong01 Aug 24 at 1:57
- (4) Please refer to the following picture: i.imgur.com/l984sdB.jpg. Do I understand correctly that only the pink dot is the problem area, stuff outside the pink dot is (almost) irrelevant? (5) Of course you can expand the pink dot by adding/changing a High side switch, to be controlled by another uP etc. – tlfong01 Aug 24 at 2:12
- 1@tlfong01: (4) Correct. The Pink is the actual problem statement. I added some more around it just to give a wider understanding. The schematic is much larger as a whole. The “Device”, which is not pictured, is a DC/DC Switcher that now provides 0.6 to 5V DC out, runs through the measurements shunts and deliver it to the Device Under Test. I may not have explained the Device and the target very well, which is the risk when you try simplify or leave out parts of the schematic. But you seem to have a very good understanding of the problem. – MWestberg 2 days ago
- #MWestberg, Thank you for your clarification. I fully agree with you, on how to present a complicated problem, especially if the complete system is hugely complex. So I suggest to first look at the big picture, then zoom into the problem area. And my usual problem solving approach is to first use the “Occam’s Razor: “Make it as simple as possible, but not simpler”. Now I would suggest to cut away all the irrelevant part of the circuit, and perhaps add the “device” part, and make a simplified question spec. / to continue, … – tlfong01 2 days ago
- Of course it is your responsibility/accountability to make sure that the pink dot can be “fully decoupled” from the rest of the world, without any undesirable “side effects” – tlfong01 2 days ago
- 1@tlfong01: You might be right. I will cut away the current design, the Current Measurements circuits and just ask the question directly. Maybe that needs to be a new post? – MWestberg 2 days ago
- I think you don’t need to write any new post. Your old post is actually very good, giving the big picture. Only thing you need to explain is the “DC/DC switch” part. So I suggest you add an “EDIT note” at the end of your question (do not change the original part, because that would confuse the old readers). The EDIT note can be what you would clarify in your earlier comment, also give more details about the DC/DC switch. Just a suggestion, of course you don’t need to agree. Comments and counter suggestions are always welcome. – tlfong01 2 days ago
If a P-Channel MOSFET’s original Vgs operation range changes from (-1.5 ~ -5V) to (-0.6 ~ -5V), what optimum changes can be made to the rest of the circuit?
After chatting, I have a better understanding the OP’s requirements, as summarized below.
User Requirements V0.1
/ to continue, …
The OP suggests two solutions, but more engineering trade offs can be made, to compare and contrast the two solutions, and other alternatives, to arrive an optimum design.
/ to continue, …
/ to continue, …
Appendix A – DMP2008UFG characteristics summary
Appendix B – The OP’s Circuit Highlighting the power MOSFET Portion
Appendix C – Spec Summary
- 1Thanks for the above. Just to make sure I follow as there is not that much text in there. You are suggesting that I stay with the original design because the Gate Threshold is -0.4? However, it may be as high as -1.0V according to the data sheet, which will not allow me to turn it ON. Even if turned ON at Vgs=-0.6V, the Rdson would probably be quite high. With the high AMP this MOSFET can handle it should not be a problem but I would prefer to have it fully ON. Not mentioned in my original post is that the DC/DC Switcher can deliver up to 2.5A. – MWestberg 2 days ago
- @MWestberg, (1) Ah, no no. I DO NOT suggest to keep the original design and do modifications (eg “biasing source”), add a high side switch (another power mosfet?) another uP etc. Patching things here and there, especially for software, are actually placing time bombs here and there. (2) The reason I asked if we can decouple the pink dot and solve the problem is to make sure if we can have “complete” freedom to redesign things. – tlfong01 2 days ago
- 1Got you, my bad. Yes, you are free to redesign all in the pink bubble. The only requirements are that I can turn the MOSFET fully ON using a varied Source Voltage between 0.6V to 5V. The MOSFET should be able to handle at least 2.5A running through it and the Rdson should be kept low (max 40mOhm for max 100mV drop @2.5A) to avoid heat and voltage drop. BTW, it may seem weird that I do not provide Current Measurements for the whole range 0-2.5A, but here is a reason for that (so your suggestion is good to decouple that from the problem statement!). Thanks for looking into this! – MWestberg 2 days ago
- Ha, I was going to point out that I would strongly disagree with any solution to only partially turn on a MOSFET, because that would damage my reputation as a ninja power mosfet guy. I am glad that our first locking down on line user requirement and functional spec meeting goes so surprisingly smooth. Ah lunch time. See you late this afternoon. Ah, one more thing before I go, I am just a friendly hobbyist doing Micky Mouse EEprojects. I have never participated in any industrial grade projects involving power MOSFET. So I am honoured to join in your project. Cheers. – tlfong01 2 days ago
- Now one thing to clarify: (1) You mentioned that “Vcc Device” is “Variable Source Voltage”, also a “DC to DC Switcher” with output range 0.6V to 5V (old range is 1.5V to 5V). (2) Let us focus on the output pattern. (a) If output is variable with a range of 0.6V to 5V, it means the output is ANALOG, can be another value between 0.6V and 5V. (b) But if output is DIGITAL, then it means output has two states: 0.6V and 5V, no value in between is allowed, or allowed bu not important, because only the values in the near 0.6V or 5V is important. . to continue, … – tlfong01 2 days ago
- Now my question is: (a) Analogue output, (b) Digital output, which one is true? – tlfong01 2 days ago
- 1The DC/DC Switcher is analog, ie the range of the DC/DC switcher output runs from 0.6V to 5V and all possible values in between. – MWestberg 2 days ago
- Let us continue this discussion in chat. – tlfong01 yesterday