Charge pump switch runs current in opposite direction – is that OK?

Divide by 2, H-Bridge, step-down, switched-capacitor, charge-pump - Operation principle of is shown the attached jpg.
Suppose we look at node “bottom”.
In steady-state it toggles between voltages (Vin+ε)/2 and 0V, while Vout is at ~(Vin-ε)/2.
i.e. the bottom right transistor experiences OFF and ON states in opposite polarities !

The involved voltages are >>5V so NMOS and PMOS are asymmetrical, due to the multiple wells that enable the high VDS operation.

For example for PMOS:
VSD>0 and VSG>Vth => Isd>0 (ON state)
VSD>0 and VSG<Vth => Isd~0 (OFF state)
But when VSD<0 regardless of VSG, the diodes between Drain and Bulk will conduct.
I.e. reverse SD polarity enable current through diodes.

Should the Bottom Right switch be realized by a PMOS?
And if yes, do you agree it will run negative direction current through the its diodes?
Is it OK?
Should it work like that?

MOSFET can well run current in both directions, it's the case in many switching topologies. In the present circuit, it doesn't depend on P or N MOS.

So when an asymmetrical MOS is OFF
AND
Source/Drain swap polarity:
Channel is pinched, bulk diodes are forward running the current -
and that is normal?
Power PMOS are designed to operate that way?

Surely not. You run reverse current in on-state. Source and drain connection are chosen so that the bulk diode is reverse biased in off state.

My simulation shows current flowing up from 0V ground to the bottom left switch, through the charge pump capacitor.

For the switch I have an NPN which does not need to get a clock signal. Instead it receives bias from collector leg attached to 0V ground. Bias resistor 500 ohms. Such unconventional biasing works because the emitter goes into the negative, pulling current through the transistor even though its upper section is at 0V gnd.

Likewise an NPN can be at bottom right (with similar biasing).

One clock is needed. It drives both uppermost transistors. The one at upper left is a PNP.

- - - Updated - - -

At upper right is NPN.

Re: Charge pump switch runs current in opposite direction – is that OK?

Realization with bipolar is indeed an option. But bipolars, clocked or un-clocked, have a Vbe drop in forward state, therefore the Vout<Vin/2 and efficiency suffers.

The question is if it is possible to realize (suppose we focus on bottom right switch) with MOSFET?

(Keeping in mind that high voltage MOSFETs are asymmetrical, and have body diode in reverse polarity)

Suppose that switch is clocked control into OFF state in normal polarity, and ON in reverse polarity.
Is it normal to have "two competing current paths" in the device (1. The channel 2. the forward diodes).
What type of device and what polarity would you put for the bottom right switch?

Hi,

Not sure fully get the gist of this issue, so sorry if that is so. Could you use back-to-back MOSFETS here?

Your schematic is a useful concept which steps down supply voltage 50 percent. As drawn with switching devices we think it needs a full H-bridge. However my simulation has a half-bridge and 2 diodes. It appears to perform the same.



Furthermore the diodes can be omitted. Connect the right-hand end of C1 to the top of C2.

This may reduce efficiency in the circuit, as resistance is in the transistors. Adjusting duty cycle can be a way to change voltage at the load.

Hi,

Have you seen this h-bridge version that switches top left + bottom left then top right + bottom right? 1/2 VIN Charge Pump

The schematic you posted in #1 reminds me of charge pump doublers but I am not a charge pump expert.

What fascinates me is how very little charge pump educational material actually shows the polarity of switches used, I remember from researching doublers ages ago and a quick look now for a suitable schematic for this thread, all you ever get (and in in-depth documents as well) is that useless picture of four switches...

d123 said:

Hi,

Not sure fully get the gist of this issue, so sorry if that is so. Could you use back-to-back MOSFETS here?

Click to expand...

If a specific switch (say - bottom right) is composed of back-to-back devices, for ON state, both need to be ON, with one of them in reverse polarity, hence in that device a current "may" run in channel AND in body diode.
In OFF state, one will be in reversed polarity, and body diode will be open.

BradtheRad said:

Your schematic is a useful concept which steps down supply voltage 50 percent. As drawn with switching devices we think it needs a full H-bridge. However my simulation has a half-bridge and 2 diodes. It appears to perform the same.

View attachment 148961

Furthermore the diodes can be omitted. Connect the right-hand end of C1 to the top of C2.

This may reduce efficiency in the circuit, as resistance is in the transistors. Adjusting duty cycle can be a way to change voltage at the load.

Click to expand...

Using such bipolars/diodes topology certainly bypasses the raised issue in#1, but the the Vout will always be Vin/2-(Vbe+Vdiode), i.e. less than Vin/2 (loss of efficiency)

d123 said:

Hi,

Have you seen this h-bridge version that switches top left + bottom left then top right + bottom right? 1/2 VIN Charge Pump

The schematic you posted in #1 reminds me of charge pump doublers but I am not a charge pump expert.

What fascinates me is how very little charge pump educational material actually shows the polarity of switches used, I remember from researching doublers ages ago and a quick look now for a suitable schematic for this thread, all you ever get (and in in-depth documents as well) is that useless picture of four switches...

Click to expand...

H-bridge for doubler is possible (Vout=2*Vin) with Bottom-Left tied to Vin, and Bottom-Right tied to ground.
The raised issue in #1 is more critical there.

mmwicant said:

Divide by 2, H-Bridge, step-down, switched-capacitor, charge-pump - Operation principle of is shown the attached jpg.
Suppose we look at node “bottom”.
In steady-state it toggles between voltages (Vin+ε)/2 and 0V, while Vout is at ~(Vin-ε)/2.
i.e. the bottom right transistor experiences OFF and ON states in opposite polarities !

The involved voltages are >>5V so NMOS and PMOS are asymmetrical, due to the multiple wells that enable the high VDS operation.

For example for PMOS:
VSD>0 and VSG>Vth => Isd>0 (ON state)
VSD>0 and VSG<Vth => Isd~0 (OFF state)
But when VSD<0 regardless of VSG, the diodes between Drain and Bulk will conduct.
I.e. reverse SD polarity enable current through diodes.

Should the Bottom Right switch be realized by a PMOS?
And if yes, do you agree it will run negative direction current through the its diodes?
Is it OK?
Should it work like that?

Click to expand...

-------------------------------------------------------------
Following FvM replies @ #2 & #4 (please confirm/reject) my summary is:
///////
in circuits like switched-capacitors charge-pumps, which are based on
high-voltage,asymmetrical-MOSFETs,with reverse polarity body-diodes:

*Devices OFTEN experience straight and reverse polarity (Vsd>0 and Vsd<0) - depend on topology.

*For OFF state - they MUST be ONLY in straight polarity (otherwise the body diodes will open).

*For ON state - they are allowed to experience reverse polarity. In this condition, the channel will run current in the reverse direction, and if Ion*Ron>Vdiode, then, in addition, the diode will open and run forward current in the reverse polarity. For Ion*Ron>>Vdiode, most of the current will run through the body diode.////

- Was it an accurate statement?
- Is such current through a body diode typically safe for asymmetrical MOSFET?
- Is it typically modeled in MOSFETs? By discrete devices vendors and BCD IC process fabs?
Thanks

Following replies from FvM @ #2 & #4 (please confirm/reject) my summary is:
///////
in circuits like switched-capacitors charge-pumps, which are based on
high-voltage,asymmetrical-MOSFETs,with reverse polarity body-diodes:

*Devices OFTEN experience straight and reverse polarity (Vsd>0 and Vsd<0) - depend on topology.

*For OFF state - they MUST be ONLY in straight polarity (otherwise the body diodes will open).

*For ON state - they are allowed to experience reverse polarity. In this condition, the channel will run current in the reverse direction, and if Ion*Ron>Vdiode, then, in addition, the diode will open and run forward current in the reverse polarity. For Ion*Ron>>Vdiode, most of the current will run through the body diode.////

- Was it an accurate statement?
- Is such current through a body diode typically safe for asymmetrical MOSFET?
- Is it typically modeled in MOSFETs? By discrete devices vendors and BCD IC process fabs?
Thanks