Hi,
I try to design LC filter in H-bridge, where load is Peltier module. I wanted to create heat/cool options.
For values shown below, cutoff Frequency is 13kHz. I don't want to overdo. Is triangle wave sufficient or should I change freq. to get sine wave? When suppression is high, there are some issues - lower currents and voltage.

One branch of H-bridge.


I decided use 7.8kHz, then

 

Why are you using such a low frequency? 7.8kHz is very low for what is essentially a buck regulator. 100kHz would be more usual.
Choose the ripple current to be between 10% and 20% of the running current, then work out the inductor value from V/L= dI/dt

 

Increase the inductance by 10x to reduce the ripple current to 1/10.

 

Don’t know why you are making this so complicated when I gave you a resource from a manufacturer and posted by a university that said PWM at 2K or above is fine. Why do you driving it with a triangle or sine wave is better?

If you insist on a constant voltage, then build or buy a proper buck converter.

 

I don't want to overdo

But you are.

You don't need an LC filter, thus C1, C2, and L2 serve no purpose in driving the Peltier.
All you need is L1 of sufficient size to keep the ripple current at 10 to 20% of the total, as Ian0 suggested.

 

(C1 & C2 = 11uF) C1, C2 are in series across R1. The impedance of 11uF needs to be much lower than 0.7 ohms for there to be much effect. Small 22uF caps probably have more internal resistance than R1. Even if the caps were doing anything the "ESR" would cause little current flow.

I think you want 68uH to get the effect you want.

 

Hi,
I try to design LC filter in H-bridge, where load is Peltier module. I wanted to create heat/cool options.
For values shown below, cutoff Frequency is 13kHz. I don't want to overdo. Is triangle wave sufficient or should I change freq. to get sine wave? When suppression is high, there are some issues - lower currents and voltage.
View attachment 271255
One branch of H-bridge.

View attachment 271252
I decided use 7.8kHz, then
View attachment 271253

Hello,

What is it you think that C2 and L2 are doing for this?
You end up with C2 in series with the 0.7 Ohm resistor which i take it is the Peltier device. That creates a high pass filter which would pass higher frequencies and reject lower frequencies, which exactly the opposite of what you want.
At least look up what a low pass LC filter is. One L and one C, a 2nd order low pass filter which would be good enough for this project.
Most buck circuits (what this is) use one L and one C except for those that need a super smooth output then they use a second L and second C post LC filter, but the Peltier is not going to be that sensitive to frequency so one L and one C should do it.
That one C smooths the output and allows a lower value of L to be used which is smaller and cheaper.

 

Thank for all your answers. There were two capacitors and inductors, because I use H-bridge to change polarization of Peltier.

I decied to leve circuit like that:

 

I wish you'd said earlier that the 0.7Ω resistor was your load. I thought it was part of the filter.
How are you going to do the PWM? Are you going to use one side of the bridge to switch the polarity, and the other side of the bridge to produce the PWM? Or are you going to use the low side MOSFETs to determine the polarity and the high-side MOSFETs to do the PWM?

If the former, you only need one inductor and capacitor (on the PWM side).

 

I'm using two IR2104. My plan is to do the PWM on one side and set logic 0 (turn on low side MOSFET on the opposite side). To change polarization I'm going to swap that with short delay.

I wish you'd said earlier that the 0.7Ω resistor was your load. I thought it was part of the filter.

I'm really sorry. What about values of LC filter then?

 

I presume that the maximum design current is 12V/0.7Ω=17A
Allow for 1.7A to 3.4A of ripple current, that value is ΔI
Δt is half a period at your operating frequency.
Then V/L=ΔI/Δt, where V is half the supply (that works it out for the worst case, which is half the supply)
and you can work out L.
At such a low frequency, you will find that L is a very large component. Consider 50kHz or 100kHz.

 

I'm using two IR2104. My plan is to do the PWM on one side and set logic 0 (turn on low side MOSFET on the opposite side). To change polarization I'm going to swap that with short delay.

I'm really sorry. What about values of LC filter then?

Hi,

What you do with the switching pattern is to design it such that the power dissipation in equal in all transistors so that no transistor has to handle more power than the other. Think about that.

The LC values depend a lot on the frequency too. Did you decide on a frequency yet? The higher you go the better but once you get too high other problems start to creep in. A decent frequency is probably 50kHz, although 20kHz could be possible too and 100kHz with more careful design.

The equations are not too difficult i'll show some later if i can.

 

I made calculations.
f=80kHz; Vsupp = 12V; V=6V; T=12.5us; Δt = 6.25us; ΔI = 3.4A

L = V* Δt / ΔI = 11uH

What you do with the switching pattern is to design it such that the power dissipation in equal in all transistors so that no transistor has to handle more power than the other.

I use H-bridge to provide bipolar polarization. So, power dissipation won't be equal in all transistors.

I'm worried about only one inductor in circuit. When I change into cooling the voltage isn't smoothed. Also remember that I"m going to drive it sometimes with 2-20% pulse length.

What are the requirements to run it on 100kHz?

 

The inductor is in series for both directions, so why would it not filter when cooling?

 

Right, I looked on simulation.

 

Imagine if you replace the pair of transistor on the right by an electromechanical relay. It only needs to change state when you change from heating to cooling. All the variation in power is done by the pair of transistors on the left,
By the way, you can still have the capacitor.

In fact, you could replace the transistors on the right by an electromechanical relay - the overall power dissipation will be reduced. You will get really good contact life if you switch off both MOSFETs on the left side before changing the state of the relay.

 

I made calculations.
f=80kHz; Vsupp = 12V; V=6V; T=12.5us; Δt = 6.25us; ΔI = 3.4A

L = V* Δt / ΔI = 11uH


I use H-bridge to provide bipolar polarization. So, power dissipation won't be equal in all transistors.

I'm worried about only one inductor in circuit. When I change into cooling the voltage isn't smoothed. Also remember that I"m going to drive it sometimes with 2-20% pulse length.
View attachment 271461
What are the requirements to run it on 100kHz?

Hello again,

I'll quote one of your reply lines again...

"I use H-bridge to provide bipolar polarization. So, power dissipation won't be equal in all transistors."
Should that make a difference?
I think you are assuming that you must keep one side of the H bridge switch state constant while the other side switches. For example, with D3 'on' constantly while D1 and D4 switch on and off so you are effectively modulating the right side of the L and R load. What this means is that you intend to keep the right side of L and R positive while the left is negative. That makes the polarity minus on left and positive on right.
But now consider this...
Keep D1 on constantly, then switch D2 and D3. Notice that the polarity is still minus on the left and positive on the right.
This is one of the properties of an H bridge that allows you to distribute the switching losses more evenly over the four transistors.
The other property is that you get zero across L and R with either the bottom two on and the top two off, or the top two on and the bottom two off. That's another mode to consider in keeping the power distribution more evenly distributed.

Think about those two switching modes and see what you can come up with.
I apologize i havent gotten to the equations yet as i have a lot of things going on here right now.

 

I got it. Is turning on and off simultaneously both sides, ex. D1, D3 the ideal option?
However I can face some difficulties generating two PWM signals with phase shift +180 deg and different pulse width on Arduino. So, I think about not gate inverter.

 

I got it. Is turning on and off simultaneously both sides, ex. D1, D3 the ideal option?

No - you will double your switching losses.
As @MrAl says keep D1 on for heating, and PWM D2/D3 to regulate the amount.
Then for cooling, either keep D4 on and PWM D2/D3
OR
Keep D3 on and PMW D1/D4
but
Your filter is better suited to keeping D4 on and PWM D2/D3.

 

I got it. Is turning on and off simultaneously both sides, ex. D1, D3 the ideal option?
However I can face some difficulties generating two PWM signals with phase shift +180 deg and different pulse width on Arduino. So, I think about not gate inverter.

Hello again,

Im sorry i should have explained this in more detail.

If you keep D3 on and switch D1 and D4 to get the positive on the right, then keep D1 on and switch D2 and D3 other times, you get the positive on the right but you distribute the switching losses better.
So you have two choices for keeping the positive on the right side of the bridge for that drive polarity. If you alternate doing it one way some times and the the other way other times it helps to distribute the switching losses in all the transistors while keeping the drive polarity correct.
So maybe you switch D1 and D4 for 10 cycles, then switch D2 and D3 for the next 10 cycles, then repeat. Just remember you also have to switch which one stays on constantly. If this sounds too complicated then just stick to your original plan.

The equations are as follows...

Start with the equation for current in the inductor with initial current i0:
i(t)=E/R-(E*e^(-(t*R)/L))/R+i0*e^(-(t*R)/L);
and at 50 percent duty cycle we have average current:
iavg=(E/R)/2
and after some long time has past we can call the initial current as the average current so:
i0=(E/R)/2
Now if we want the peak current to be a fraction of the average current above the average current by a factor of K, the peak current is:
ip=i0+i0*K
where K is that fraction (such as 0.1 for 10 percent).
If the frequency is f then the period is:
tp=1/f
and since the peak will occur 1/4 of the time period after the wave goes through the average current, this time is:
tp4=tp/4

Put this all together and we end up with:
L=-R/(4*f*log(1-K))
for K<1.

For a quick example, with E=10 and R=7/10 and f=10000Hz and K=0.1 we get L=166uH
and again note that if K=0.1 then the peak current will be 10 percent higher than the average current.
ADDED LATER:
The peak tp peak current will be twice that which is 20 percent, so if the average current was 10 amps the peak to peak current would be 2 amps peak to peak. With K=0.05 (5 percent) the peak to peak current would be 1 amp and the value of the inductor would come out larger.