I must have missed the bit about needing a controller. I only saw the part where it said a FET switch can be used instead of a mechanical switch. What does a controller consists of? Other references to a L type matching network don’t mention one.

My circuit above is based upon the Spice one shown.

 

the circuit below with a nominal 100mH inductor and the suggested P channel FET as a switch.

The inductor needs to be in series with the switch, not connected to ground (which shorts the capacitor).
And you still need the diode in series to prevent reverse flow of the current from the battery.

It would be useful if you would download the free LTspice from Analog Devices and learn to do your own simulations.

 

The controller consists of an integrated circuit such as a UC3842 which controls the amount of time that the inductor connects to the source (and a diode connects it to ground the times when it is not connected to the source).

 

The inductor needs to be in series with the switch, not connected to ground (which shorts the capacitor).
And you still need the diode in series to prevent reverse flow of the current from the battery.

It would be useful if you would download the free LTspice from Analog Devices and learn to do your own simulations.

Sure I will look into learning about using Spice.

Your simulation didn’t include a controller that apparently is required so is your simulation only part of the picture of what is going on?

The difficulty for someone in my position is that some of the facts offered appear to contradict each other - such as the 50% efficiency when the source and load are the same resistance. Hence my summaries to clarify.

 

hi T88,
What level of efficiency are aiming for in this project.?
E

 

Your simulation didn’t include a controller that apparently is required so is your simulation only part of the picture of what is going on?

Crutschow’s circuit is not an SMPS and does not need a controller. I works as is and transfers most of the energy from the capacitor to the battery. This is a simple circuit that performs wonders, why not use it as is?

 

Crutschow’s circuit is not an SMPS and does not need a controller. I works as is and transfers most of the energy from the capacitor to the battery. This is a simple circuit that performs wonders, why not use it as is?

I’m glad to hear it and that is what I was trying to show in my recent circuit but placed the inductor in the wrong place.

I couldn’t see why a controller was needed and was getting confused by an earlier post.

 

One limitation of my circuit with real parts is that for large capacitances and/or voltage, the peak inductor current can become quite high, which can saturate the inductor.
An inductor large enough to prevent that may become rather large and expensive.
That's where a buck-boost regulator may be better, as it switches rapidly, allowing the use of a much smaller inductor.

 

hi T88,
What level of efficiency are aiming for in this project.?
E

Rather than trying to obtain a specific value, I’m trying to understand the process of how a capacitor discharges to a battery when the resistances of each is quite different. One of my students has been learning about exponential charging and discharging of capacitors and RC time constants and the usual material at A level but asked me what the significance was if the two resistance were quite different. So I agreed to try and find out but it’s a lot more involved than I envisaged!

My various summaries are an attempt to bring it all together so I can then relay the facts. Some of the ‘facts’ presented don’t all agree, or appear not to, but I guess that’s to be expected with various contributors.

I have put some of these facts in post 35. I need to revise the circuit again to accurately reflect the Spice one since that is the simplest way to have the best energy transfer - or so I have been led to believe.

 

One limitation of my circuit with real parts is that for large capacitances and/or voltage, the peak inductor current can become quite high, which can saturate the inductor.
An inductor large enough to prevent that may become rather large and expensive.
That's where a buck-boost regulator may be better, as it switches rapidly, allowing the use of a much smaller inductor.

That’s a thought. Have it got it wrong or would a Buck-boost regulator still need some form of controller and could such a regulator be simulated with LTSpice?

So I can show a real world solution, is there an available Buck-Boost regulator that would work with my combo of 50mF,40V?

Tomorrow I will assemble a collection of facts to relay back to my student, which can be confirmed or contested

 

asked me what the significance was if the two resistance were quite different

Is that clear now?

A simple way to determine what happens is to look at the relative energy levels for the charged capacitor and then when it reaches the battery voltage.
The capacitor energy is 1/2 CV² and the energy transferred to the battery equals the battery voltage times the charge transferred from the capacitor (Q*V).
That charge value is the difference in the charged voltage and discharged capacitor (which is the battery) voltage times the capacitance.
(I leave those calculations as an exercise to the reader).

Since those values can be calculated with no reference to the resistance connecting the capacitor to the battery, then you can see that the resistance makes no difference, except for how fast the energy is transferred.

 

Have it got it wrong or would a Buck-boost regulator still need some form of controller and could such a regulator be simulated with LTSpice?

Yes.
The are a number of buck-boost circuits using a commercial regulator IC that should work for you

 

Rather than trying to obtain a specific value, I’m trying to understand the process of how a capacitor discharges to a battery when the resistances of each is quite different. One of my students has been learning about exponential charging and discharging of capacitors and RC time constants and the usual material at A level but asked me what the significance was if the two resistance were quite different. So I agreed to try and find out but it’s a lot more involved than I envisaged!

My various summaries are an attempt to bring it all together so I can then relay the facts. Some of the ‘facts’ presented don’t all agree, or appear not to, but I guess that’s to be expected with various contributors.

I have put some of these facts in post 35. I need to revise the circuit again to accurately reflect the Spice one since that is the simplest way to have the best energy transfer - or so I have been led to believe.

The internal resistances of the capacitor and battery simply add up and act like a single resistor between an ideal capacitor and an ideal battery. Their individual values do not matter in any way. If both are 5 or one is 0 and the other Is 10 the circuit behaves exactly the same way.

 

Good to know. I will summarise the key points tomorrow, redraw a working circuit and also hunt around for a ‘real world’ Buck-boost circuit to use with a small inductor.

 

No one has mentioned this before, but I hope you don’t expect your 1000uF capacitor charged to 30V to add any noticeable charge to your battery.

E = 1/2 CV^2 = 0.001 x 30 x 30 / 2 + 0.45J

That is how much energy you energy you give to the battery if you could get 100% of it, which you cannot.

It is enough to get 37mA for 1 second out of the battery.

 

Yes it was calculated before that 450mJ was delivered at best. (Posts 17 & 35)

 

Since I'm seeing my student soon, I thought I would collect all the suggestions and points made in a list. If anything needs correcting or refining please let me know:

1. A lower capacitor ESR will give more current but with greater I^2R losses. A lower ESR won’t change the total energy transferred, only the rate of transfer.

2: The transfer efficiency is the ratio of the Load to Source voltages. The higher the Source V the lower the efficiency.

3. In the 'Max power theorem', when the resistances or impedances are matched, it only relates to the max power in the load when the source impedance is fixed. At best efficiency here is 50% (conflicts with 2?)

4. A circuit using a Buck-Boost converter and an inductor is one efficient way to discharge the capacitor to the battery (see diagram)


5. Using a Buck-boost converter approach with a larger source capacitor (e.g. 50mF) or higher voltage (>30V) might not be possible due to the typical maximum input voltages and current capacity of these commonly available units at typically 25V, 10A.

6. Another option is to use an inductor and a diode, (inductor resonant charge circuit) although, depending on the peak instantaneous current, the inductor may be large and impractical. The primary of a >100VA transformer might be effective (see diagram).

 

You have drawn Vcc as though it is a permanent supply, but all the discussion prior to this point has been about a single charged capacitor, and, as many have pointed out, there is not enough energy is a charged capacitor to make a significant difference to the charge in the battery.
So, are you now saying (by the diagram) that you have a permanent 30V supply and what you want to do is design a battery charger?
If so, then post #1 could have said “I want to design a battery charger” and you would be a lot further forward.

 

You have drawn Vcc as though it is a permanent supply, but all the discussion prior to this point has been about a single charged capacitor, and, as many have pointed out, there is not enough energy is a charged capacitor to make a significant difference to the charge in the battery.
So, are you now saying (by the diagram) that you have a permanent 30V supply and what you want to do is design a battery charger?
If so, then post #1 could have said “I want to design a battery charger” and you would be a lot further forward.

Sorry, no it’s a one time pulse (although potentially repeatable) and I wasn’t sure how to draw the capacitor voltage to show that. How should I represent that on the circuit?

 

Sorry, no it’s a one time pulse (although potentially repeatable) and I wasn’t sure how to draw the capacitor voltage to show that. How should I represent that on the circuit?

Just leave off the Vcc connection.
I have seen a thought experiment very similar to this with two capacitors of identical value, one is charged to 100V, the other discharged.
Then they are connected together, and the charge is evenly distributed so the voltage on them both is 50V.
But what happened to the energy? (0.5C V^2)
Half of it disappeared in the process.