Switched capacitive boost converter analysis

[Hina Gohar]

Plz guide how I can solve below circuit i.e. its transient analysis..final expression for load voltage..it is actually mode 3 of my capacitor based boost converter ..in mode 3 T2 transistor on while T1 off..so output would be double..I need expression for instantenous

 

[Baluncore]

You can redraw your circuit to give a common ground reference. See attached.
First change C1 to connect with the positive side of the load.
Then move the series diode D2 to the positive side of the load.
You now have a common ground reference with a charge pump that is easier to analyse.

Can you now analyse the charge pump circuit ?
If you still have problems, ask further questions.

 

[Hina Gohar]

You can redraw your circuit to give a common ground reference. See attached.
First change C1 to connect with the positive side of the load.
Then move the series diode D2 to the positive side of the load.
You now have a common ground reference with a charge pump that is easier to analyse.

Can you now analyse the charge pump circuit ?
If you still have problems, ask further questions.

I need help in its voltage,current and power expression of whole circuit so that I would find cpacitors value and which transistor to be used..

 

[Baluncore]

Component selection for a voltage doubler charge pump needs more information.
Maximum permissible output ripple, Vr, or minimum output voltage, Vm, must be specified.

The supply voltage is Vs. The maximum possible output voltage is Vs*2.
The minimum output voltage is Vs*2 – Vr.
Vr will include two diode voltage drops of Vpn and droop in the capacitors during charge transfer.
We specify the capacitor droop voltage as Vd, then Vr = 2 * (Vpn + Vd)
We can model Vpn, (expect about 0.8V), so compute the maximum value of Vd = Vr/2 – Vpn.

Output voltage, Vo, across the load resistance, Ro, sinks a maximum current of Io = 2*Vs / Ro.
The half bridge is driven at a frequency, f, and so has a period of T.
Capacitance, C = Q / V; and charge, Q = I * T.
Therefore C1 = Io * T / Vd. C2 can use the same value.

Some form of current limiting must be modeled or introduced in the charge pump path through C1.
This could be current limiters in the half bridge, inductance or resistance in the C1 lead.
That will set the maximum transistor and diode current.

 

[Hina Gohar]

Component selection for a voltage doubler charge pump needs more information.
Maximum permissible output ripple, Vr, or minimum output voltage, Vm, must be specified.

The supply voltage is Vs. The maximum possible output voltage is Vs*2.
The minimum output voltage is Vs*2 – Vr.
Vr will include two diode voltage drops of Vpn and droop in the capacitors during charge transfer.
We specify the capacitor droop voltage as Vd, then Vr = 2 * (Vpn + Vd)
We can model Vpn, (expect about 0.8V), so compute the maximum value of Vd = Vr/2 – Vpn.

Output voltage, Vo, across the load resistance, Ro, sinks a maximum current of Io = 2*Vs / Ro.
The half bridge is driven at a frequency, f, and so has a period of T.
Capacitance, C = Q / V; and charge, Q = I * T.
Therefore C1 = Io * T / Vd. C2 can use the same value.

Some form of current limiting must be modeled or introduced in the charge pump path through C1.
This could be current limiters in the half bridge, inductance or resistance in the C1 lead.
That will set the maximum transistor and diode current.

I have attached main paper of this circuit in my previous thread "Integration of below expression" if you kindly look there.Some expression for load average and instantaneous voltage is given while for current and capacitors value I need equations like for boost dc-dc converter circuits whole analysis. Plus to this experimentally and in simulation also I checked this circuit although voltage is doubling but current is very low i.e. in mA which is not correct because load which will be attached to this circuit is basically 12V lead acid battery of 100AH which means this converter should provide 10A current for proper charging.So how I can increase it's current or some modification in circuit.

 

[Baluncore]

Your other thread with link to paper … Integration of below expression

Plus to this experimentally and in simulation also I checked this circuit although voltage is doubling but current is very low i.e. in mA which is not correct because load which will be attached to this circuit is basically 12V lead acid battery of 100AH which means this converter should provide 10A current for proper charging. So how I can increase it's current or some modification in circuit.

Charging a battery will be different to driving a resistive load.
A 12V battery will probably need 13.5 to 14 volts before it takes a charge current of 10A.

I can answer your question better if I have the following information.
What is the Vs used in your experiment and simulation ?
What half-bridge clock frequency did you use ?

 

[Hina Gohar]

Your other thread with link to paper … Integration of below expression


Charging a battery will be different to driving a resistive load.
A 12V battery will probably need 13.5 to 14 volts before it takes a charge current of 10A.

I can answer your question better if I have the following information.
What is the Vs used in your experiment and simulation ?
What half-bridge clock frequency did you use ?

well Vs I used different 5,10,15 etc from dc power supply..while Vmax of available solar panel as mentioned in datasheet is 16.1V at standard condition.Right now I am using dc source as input while latter it will be replaced by two solar panel in parallel of total 40W..components selection should be such that there will be power matching between source and load side i.e. around 40W..switching frequency which I used as mentioned in paper 1kHz..these two transistors will actual act as a switch on or off ..duty cycle is 50% for both..in half time one will be on and in other half time 2nd transistor will be on..

 

[Hina Gohar]

well Vs I used different 5,10,15 etc from dc power supply..while Vmax of available solar panel as mentioned in datasheet is 16.1V at standard condition.Right now I am using dc source as input while latter it will be replaced by two solar panel in parallel of total 40W..components selection should be such that there will be power matching between source and load side i.e. around 40W..switching frequency which I used as mentioned in paper 1kHz..these two transistors will actual act as a switch on or off ..duty cycle is 50% for both..in half time one will be on and in other half time 2nd transistor will be on..

yes in place of resistive load I have to use 12V battery which will be charged through this charge controller from solar panel during day time..and after battery there will be dc load in parallel..so solar panel then voltage doubler then some stabilizing circuit which will provide constant 12V to battery then load..

 

[Hina Gohar]

Your other thread with link to paper … Integration of below expression


Charging a battery will be different to driving a resistive load.
A 12V battery will probably need 13.5 to 14 volts before it takes a charge current of 10A.

I can answer your question better if I have the following information.
What is the Vs used in your experiment and simulation ?
What half-bridge clock frequency did you use ?

yes in place of resistive load I have to use 12V battery which will be charged through this charge controller from solar panel during day time..and after battery there will be dc load in parallel..so solar panel then voltage doubler then some stabilizing circuit which will provide constant 12V to battery then load..

 

[Baluncore]

Sorry about the delay in my reply, but since this thread has significantly changed, I have needed to collect my thoughts.

To put it simply, the referenced switched capacitor circuit is controlled by a sinewave so as to prevent step edges with infinite switch and capacitor currents. This requires that the switch transistors are turned partially ON which will lead to significant power dissipation in the transistors. It does explain why 2N3055 transistors were specified. You are discussing here a 25 year old technology that has been superseded by very fast IGBT or MOSFET transistors, with very small high frequency inductors.

The efficiency analysis in the paper shows higher efficiency at lower frequencies, that requires bigger capacitors. A switching converter has a higher switching frequency and so can use smaller inductors and capacitors with lower output voltage ripple. The reason for this difference is that with the switched capacitor filter, the significant losses occur while switching, because the transistor is not operating with either zero current or zero voltage, but instead is partially ON for a significant time and so has a V*I power loss.

yes in place of resistive load I have to use 12V battery which will be charged through this charge controller from solar panel during day time..and after battery there will be dc load in parallel..so solar panel then voltage doubler then some stabilizing circuit which will provide constant 12V to battery then load..

The 16.1Vmax from the cell will be sufficient to charge a lead acid battery without a voltage doubler. The stabilising circuit will need to be a switching converter so why also use the doubler. Running two PV modules in parallel to feed a doubler is unnecessary when you can simply run the two modules in series.

I would never consider a capacitive voltage doubler to charge a lead acid battery. The reason being that it does not find the optimum MPP of the PV panel. Also, it can only double the voltage, not multiply efficiently by say 1.35 or 0.75.

If you have two stages, one with an efficiency of 80%, the other with 90%, then the resulting efficiency will be about 72%. So I cannot see why you do not throw out the 80% efficient voltage doubler and settle for a 90% efficient switching converter alone. I would use a switching converter to do the regulation in one step. Indeed I use a cheap switching buck regulator, (with a power diode for polarity protection), to charge batteries.

There are several requirements on your PV battery charger.

1. The current must be limited to about 10 amps. This is satisfied by selecting a PV panel with a short circuit current specification that is close to, but below 10 amp.

2. The PV must be loaded at close to it's MPP voltage. That can be achieved by using a 16 volt panel to charge a 12V battery to about 14 volts.

3. To prevent gassing, the battery voltage must not exceed a battery temperature dependent maximum. When voltage exceeds the specified maximum, the unused current can be shunted through a resistive load.

 

[Hina Gohar]

Sorry about the delay in my reply, but since this thread has significantly changed, I have needed to collect my thoughts.

To put it simply, the referenced switched capacitor circuit is controlled by a sinewave so as to prevent step edges with infinite switch and capacitor currents. This requires that the switch transistors are turned partially ON which will lead to significant power dissipation in the transistors. It does explain why 2N3055 transistors were specified. You are discussing here a 25 year old technology that has been superseded by very fast IGBT or MOSFET transistors, with very small high frequency inductors.

The efficiency analysis in the paper shows higher efficiency at lower frequencies, that requires bigger capacitors. A switching converter has a higher switching frequency and so can use smaller inductors and capacitors with lower output voltage ripple. The reason for this difference is that with the switched capacitor filter, the significant losses occur while switching, because the transistor is not operating with either zero current or zero voltage, but instead is partially ON for a significant time and so has a V*I power loss.



The 16.1Vmax from the cell will be sufficient to charge a lead acid battery without a voltage doubler. The stabilising circuit will need to be a switching converter so why also use the doubler. Running two PV modules in parallel to feed a doubler is unnecessary when you can simply run the two modules in series.

I would never consider a capacitive voltage doubler to charge a lead acid battery. The reason being that it does not find the optimum MPP of the PV panel. Also, it can only double the voltage, not multiply efficiently by say 1.35 or 0.75.

If you have two stages, one with an efficiency of 80%, the other with 90%, then the resulting efficiency will be about 72%. So I cannot see why you do not throw out the 80% efficient voltage doubler and settle for a 90% efficient switching converter alone. I would use a switching converter to do the regulation in one step. Indeed I use a cheap switching buck regulator, (with a power diode for polarity protection), to charge batteries.

There are several requirements on your PV battery charger.

1. The current must be limited to about 10 amps. This is satisfied by selecting a PV panel with a short circuit current specification that is close to, but below 10 amp.

2. The PV must be loaded at close to it's MPP voltage. That can be achieved by using a 16 volt panel to charge a 12V battery to about 14 volts.

3. To prevent gassing, the battery voltage must not exceed a battery temperature dependent maximum. When voltage exceeds the specified maximum, the unused current can be shunted through a resistive load.

Solar panel Vmax =16V but it is in standard condition as mentioned in its datasheet while in normal condition it is varying 8-13V and would be minimum as compared to mentioned one..that's why put doubler circuit.
Plus MPPT I am not doing right now as I studied so I found that for bigger system or higher power MPP need while for low power there is no need of MPPT..in my case
panel total power is 40W which will simply operate dc load of 40W..
http://www.solarsystem.pk/chargecontroller.html..there you can find about charge "For bigger systems like tubewell or water pumping they are used high voltage charge controllers, called MPPT(Maximum Power Point Tracker) charge controllers. MPPT charge controller can convert dc-dc voltage. Example used in 24, 38, 48 volt systems. MPPT charge controllers are used with higher power panels like 135 watt and higher panels."