Replacing diode + MOSFET with BJT on High side mains

Thread Starter

tomerbr

Joined Oct 16, 2017
25
Hi everyone,

I have a working circuit that we already produced several hundreds from it.
I am preparing a new version of the board and one of the requirements was to reduce the cost as much as possible without compromising the board.

Part of the board is a LiFePO4 battery charger.
The current path is protected with a P-channel MOSFET and a diode in series.
Why both? well, we started with just a MOSFET but once it is turned on, current can flow in both directions and it burned the circuit in some situations.
Why not just a diode (schottky of course), well we have control logic that can turn on and off the charger, hence we need the MOSFET for control.

When there is no control logic (IC is offline) Q12 is ON and current flows through Q5. If the control logic sets the Q12 gate to be "0", Q12 is off and Q5 is off as well.

I was thinking on changing both of them with one PNP BJT. I have not worked much with BJTs and I know that to use them as swithces them need to be in their saturation/cut-off regions.

Below is the circuit and the relevant parts that I want to change.

1. Is my logic sound? Does it make sense to replace the MOSFET-DIODE configuration with one BJT?
2. If yes, have I placed the BJT correctly?
3. Do I need to replace R45 and R49 to enable a larger base current?
4. What other consideration do I need to take into account?


Original MOSFET-DIODE logic (Q5 - D4)

1623695466348.png

Suggested BJT logic (T1)
1623695597173.png


Full Circuit
1623695742573.png



Thanks,
Tomer
 

Irving

Joined Jan 30, 2016
2,308
You are still going to need the diode D4 for exactly the same reason, otherwise your PNP transistor will be reverse biassed and will fail fairly quickly...

Without D4 you have a very low impedance return path from the battery via RSR3/4, L1, and the source/drain diode of Q3... its not surprising you've had failures. Its not when Q5 is turned on that is the problem, its the source/drain diode of Q5 that will short the battery to whatever is connected to the 20v/5A input. Current could also flow back through the diode, through R47, turning Q12 on and also turning Q5 on.
 
Last edited:

Ian0

Joined Aug 7, 2020
3,752
I'm assuming that part of the circuit doesn't switch that oftern. If so, you could replace it by a relay. Then you wouldn't need the heatsink either.
 

Ian0

Joined Aug 7, 2020
3,752
How will that save money, which is the purpose of the redesign?
Because relays are cheaper than heatsinks. So a relay is a lot cheaper than MOSFET+Schottky+heatsink

If that heatsink labelled 573300 is a 573300D00010G, it's £2 from digikey, and a 8A relay is <50p
[EDIT] though it's 90p in Farnell, but still twice the price of a relay.
 

Irving

Joined Jan 30, 2016
2,308
I think the heatsink is for the switching MOSFETs

The control MOSFET Q5 dissipates <0.25W assuming 5A (Rds(on) is 5mOhm).
 

Thread Starter

tomerbr

Joined Oct 16, 2017
25
Thanks you everyone for your replies.
Irving, I see your point about the reverse biasing.

So, further exploring my options, would an SCR be a viable option?
I have never used an SCR and from what I saw they have a quite large VF.
Any other limitations? like the gate-emitter voltage?
 

crutschow

Joined Mar 14, 2008
28,158
So, further exploring my options, would an SCR be a viable option?
Not really.
An SCR will not turn off as long as it has DC current going through it.
You need to reduce the current to zero to reset them.
Also they have over a volt forward drop when on.
 
Last edited:

Irving

Joined Jan 30, 2016
2,308
Sorry, but I don't think you can do better without radical redesign of the overall system. Normal battery charger/discharger solutions would use a p-channel /n-channel series pair between battery and charger/load to control flow in both directions.

What's the cost saving you are trying to make? How much does replacing Q5/D4 with T1 save?
 

Irving

Joined Jan 30, 2016
2,308
Further thoughts... The BQ24650 datasheet shows a diode at your D4 position specifically for this reason, but grounds the MPPSet pin using an N-channel MOSFET to give a /CE (i.e 0 = charge enable) input to disable charging. What's the reason you've done otherwise?
 
Last edited:

Thread Starter

tomerbr

Joined Oct 16, 2017
25
Further thoughts... The BQ24650 datasheet shows a diode at your D4 position specifically for this reason, but grounds the MPPSet pin using an N-channel MOSFET to give a /CE (i.e 0 = charge enable) input to disable charging. What's the reason you've done otherwise?
You are correct but if the charger gets into a fault situation, like when trying to wake a battery from deep charge for 30 minutes without success (see 8.3.4 at page 14), the only way to restart it is by completely turning it off and back on. using the CE (MPPSet) for that option will not help
 

Thread Starter

tomerbr

Joined Oct 16, 2017
25
You are disabling the effect of the body diodes that way, but current can still flow in both directions once the MOSFETs are ON.
That was tested as well.
Once a short circuit occurs on the input the current flows back from the battery and burns the IC.
It does not happen with the protection of the diode (as the datasheet states).
 

Irving

Joined Jan 30, 2016
2,308
You are correct but if the charger gets into a fault situation, like when trying to wake a battery from deep charge for 30 minutes without success (see 8.3.4 at page 14), the only way to restart it is by completely turning it off and back on. using the CE (MPPSet) for that option will not help
Hmm, that's not how I read the datasheet. "A voltage <75mV on MPPSet resets all timers & fault conditions..." and the timing diagrams suggest that's a complete reset. So I'm still unclear why you need the MOSFet. However you still need the diode whether or not you need the MOSFET and therefore the heatsink.

@Bordodynov yes, thats what I was referring to in post #11 but it doesn't help here as it replaces D4 with the higher Vf body diode (so higher losses) in your M2 so still requires the heatsink and so no cost saving.
 

Thread Starter

tomerbr

Joined Oct 16, 2017
25
Hmm, that's not how I read the datasheet. "A voltage <75mV on MPPSet resets all timers & fault conditions..." and the timing diagrams suggest that's a complete reset. So I'm still unclear why you need the MOSFet. However you still need the diode whether or not you need the MOSFET and therefore the heatsink.

@Bordodynov yes, thats what I was referring to in post #11 but it doesn't help here as it replaces D4 with the higher Vf body diode (so higher losses) in your M2 so still requires the heatsink and so no cost saving.
I totally missed that as it was not in the original datasheet if I recall.
I will try that tomorrow and report the findings.
If it works, I can replace an expensive high power mosfet with a much cheaper one.
 

Irving

Joined Jan 30, 2016
2,308
If it works, I can replace an expensive high power mosfet with a much cheaper one.
IPD90P04P405ATAMA2 @ Mouser is £1.07ea on a reel of 2500

TJ90S04M3L @ Mouser is £0.574ea on a reel of 2000

That's 50p saving with no redesign!

Diode PDS1040-13 @ Mouser is £0.411 on a reel of 2500 (assuming you're using diode in datasheet)

Diode SDT10A45P5 @ Mouser is £0.113 on a reel of 1500
Also has lower Vf so dissipates 1W instead of 1.5W, and lower Tja so possibly can use cheaper/no heatsink if redoing PCB.

Of course YMMV if there are specific reasons for that MOSFET & diode... or if you're getting better pricing already..
 
Top