Vds voltage significantly high when MOSFET is ON, what could be the reason? - Page 1
Forum section: electronics / beginners

author: Harsh Chandola
author_rank: Contributor
timestamp: January 03, 2017, 04:30:11 am
content: Please refer to the diagram I have attached, the following circuit uses IRF9520 P-Type Mosfet and IRF520 N-Type Mosfet.I have designed a switching LED, i.e. when we push a button two LEDs in first branch light up when we again press the push-button LEDs in second branch light up and first branch turn off and so on, when we hold the push-button for more than 2 seconds both the branches turn ON and when we again press the push button the first branch lights up with second branch in OFF state. What I haven't shown in the circuit diagram is a push-button that toggles switching between two branches where each branch consist of two LEDs in series. The circuit's pwm is controlled via potentiometer(not shown in the circuit diagram).I am using two 3V LEDs in series with maximum current rating of 180mAmps in both the branches.I have decided the Resistance such that max current drawn be 180mAmps in each branch.Following was my calculation :at 100% duty cycle 6V in total drops across the two LEDs so rest of the voltage should drop across the series resistor. Max current rating for these LED is 180mAmp so 12-6 = 6; 6/180mAmp = ~34ohms. So I placed 34ohms resistor in series with two LEDs in each branch.Now when I connect the whole circuit up few problems arise :1> The LEDs in the branch that has P-Type MOSFET lights brighter than the LEDs in other branch initially at any PWM but at max duty cycle (100%) both shines equally but I'd want both to light up with equal brightness. why could this be happening? I measured voltages across LEDs in both the branches at same PWM by activating both branches one by one at same PWM and the LEDs in first branch with P-Mosfet one had higher voltage drop than the second branch and hence was lighting brighter than the other branch.2> Secondly what I expected was, say at 100% duty cycle, there would be 6V drop on LEDS and rest of the voltage drop should be on the resistor in series with the LEDs and a little voltage drop between drain and source but when I use multimeter to check the voltage drops at components in a branch, voltage drops at LEDs according to PWM supplied but the rest of the voltage should drop on the resistor in series with the LEDs but instead of that a very tiny voltage drops at that resistor and the remaining voltage drops between drain and source. For example if 4 volts drop across LEDs, a few volts(less than 1) drops at the series resistor and the remaining 7.x volts would drop between source and drain. I don't understand why is this happening! At ON state Vds should be less depending on the On-state resistance which from data sheet I found out it to be less than 1 ohm. And this is happening at both the branches with both P-MOSFET and N-MOSFET.I am assuming it's the problem with my understanding of transistors maybe. Any help would be greatly appreciated.

author: matseng
author_rank: Frequent Contributor
timestamp: January 03, 2017, 06:09:31 am
content: The IRF520N is not really ideal to drive from logic level. The VGS(th) can according to the datasheet be as high as 4 volts. Remember that this is the *threshold* voltage when the device just barely starts to conduct, not when it's fully open.This means that at the voltage the arduino puts out at high level might not be enough to make the fet to be at a low resistance enough to not cause a large voltage drop.For the pfet it's not a problem since the transistor pulls the gate all the way from 12 to just a few tens of a volt - making it conduct fully.

author: Seekonk
author_rank: Super Contributor
timestamp: January 03, 2017, 07:55:24 am
content: "The LEDs in the branch that has P-Type MOSFET lights brighter than the LEDs in other branch initially at any PWM but at max duty cycle (100%) both shines equally"You didn't specify resistor values. Likely gate capacitance of the N is keeping the gate voltage from reaching the maximum gate drive which is minimal to begin with. Regardless of the resistors value, the P would have a very fast turn on and slow off. The N would have a slow on never reaching the full 5V it needed during any pulsed PWM.Is this just a demonstrator circuit? It's 3am, but both channels appear to do the same thing.

author: BrianHG
author_rank: Super Contributor
timestamp: January 03, 2017, 08:07:15 am
content: Even a logic level mosfet might not be enough since R7 will lower the output voltage of the digital IO. I would recommend a logic level to mosfet gate driver IC under all circumstances since it also protects the digital IO from any signal spikes coming back through the mosfet gate.Ok, now, if you don't want the gate driver IC, for my LED lighting projects, I use the BSS806N H6327 as it is cheap and switches on completely at 5 Vgs. I would remove R7 to ensure that you get the full 5v when the output goes high, or, if you need a default off during tristate, I would make it something like 1Meg.Now, IRF520 is beefy for 2 leds and you are working with 12v, if you need more amps, I would use something like a DMN2028USS, also cheap, but more than 8 amps.

author: Harsh Chandola
author_rank: Contributor
timestamp: January 03, 2017, 09:24:07 am
content: Quote from: Seekonk on January 03, 2017, 07:55:24 amYes this is a demonstrator for now, i'll be using N-type for both the branches and might be using a gate driver. Makes sense Arduino only provides 5 volts so it could be that it's not fully turning ON but still large voltage drop on Vgs is bugging me. I have made a few changes I'm going to test it again.

author: Harsh Chandola
author_rank: Contributor
timestamp: January 03, 2017, 09:29:53 am
content: Quote from: BrianHG on January 03, 2017, 08:07:15 amMaaan I have tried it before without R7 (btw i am using 100K Ohm as R7) and it doesn't works for mosfets for some reason. I have generally tried dimming and lighting LEDs with mosfets and directly applied pwm output from arduino to gate but didn't work. and you're right about the driver, i'll be using a gate driver, i'll look into the driver you mentioned. Thanks a lot.

author: BrianHG
author_rank: Super Contributor
timestamp: January 03, 2017, 10:48:05 am
content: Quote from: Harsh Chandola on January 03, 2017, 09:29:53 amThe 2 parts I mentioned, BSS806N H6327, DMN2028US are not drivers, they are mosfets which will completely turn on at 2.5Vgs.If you still want to use the IRF520 so you may use much larger voltages than 20v, then you will need a gate driver IC something like IR4427PBF, which just so happens to cost twice as much as the mosfets I listed.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 03, 2017, 01:31:53 pm
content: As shown, L1-L2 are wired hard-on (the P channel body diode is shown conducting into the load).Tim

author: BrianHG
author_rank: Super Contributor
timestamp: January 03, 2017, 01:56:01 pm
content: Quote from: T3sl4co1l on January 03, 2017, 01:31:53 pmYes, I now see that in the schematic, the Mosfet is up-side-down. I am so used to expecting it the right way around.I wonder if Harsh Chandola just drew the schematic wrong, or, actually wired it backwards too. Those LEDs would obviously always be on full bright...

author: Seekonk
author_rank: Super Contributor
timestamp: January 04, 2017, 10:16:33 pm
content: Ask a simple question and you never get an answer. The problem is when you PWM at a mid lever, one side is brighter than the other. If it is within your capacity, what is R1. Just to give you an idea of what I am concerned about, this is a screen shot of a 1.5K resistor pulling up a gate. The opposite would be true of a resistor pulling down a gate thus extending the on time.

author: Zero999
author_rank: Super Contributor
timestamp: January 04, 2017, 10:51:08 pm
content: Why not use a constant current sink, rather than resistors?Is there any need to have one set of LEDs on the + side and the other on the 0V?And you shouldn't be driving the LEDs at their absolute current maximum rating. Knock about 10% or so off. You'll hardly notice any decrease in brightness and the LEDs will last for much longer.Using the attached schematic. R1 = 3R9 (to give about 160mA) & R2 = 1k.The transistor needs to be rated to 1W or above. The TIP32 will do.

author: BrianHG
author_rank: Super Contributor
timestamp: January 05, 2017, 04:23:19 am
content: Hero999, I like ccs, I've recently recommended them twice in 2 previous posts, but without knowing if the IO pin is only able to source 7ma to 3.3v VS sourcing 25ma or more to 5v, I might want to add another buffer switch in your circuit.(TIP32, wow, that bring back memories from the 1970s.)

author: Harsh Chandola
author_rank: Contributor
timestamp: January 05, 2017, 05:18:35 am
content: Quote from: BrianHG on January 03, 2017, 01:56:01 pmIs it? I think it's the right way round. Which part of it is wrongly wired if you could please point it out. The circuit I made on breadboard at the moment works alright. I'd really appreciate if you could point the fault out more clearly.

author: BrianHG
author_rank: Super Contributor
timestamp: January 05, 2017, 06:27:43 am
content: Quote from: Harsh Chandola on January 05, 2017, 05:18:35 amThe P-Channel mosfet Q1, you have the source connected to the LEDs instead of the +12v. Remember, when using a mosfet as a switch, we measure the voltage between the Gate and Source. In the case of a P-Channel, the Gate turn on voltage is a negative voltage. This means to conduct the source to the drain, the Source should be at 12v and the gate at 0v, hence the -12v. In your schematic, in the mosfet symbol, the pin connected to the center arrow is the Source pin, the pin on top is the drain pin.Q2 is drawn correctly...

author: Harsh Chandola
author_rank: Contributor
timestamp: January 05, 2017, 06:56:14 am
content: Quote from: BrianHG on January 03, 2017, 01:56:01 pmAlso I just checked I have put the Mosfet right way round on the hardware. I understand now I had reversed the drain and source on schematic

author: Zero999
author_rank: Super Contributor
timestamp: January 05, 2017, 08:48:43 am
content: Quote from: BrianHG on January 05, 2017, 04:23:19 amThat shouldn't be a problem. With R2 = 1k then the current drawn from the I/O pin will be 3.8mA at 5V. If you want to drive it from 3.3V, then R2 should be reduced to 560R.