SCR Bridge Control - Is my Circuit Correct ?

[CataM]

I am giving rectified dc to RL load. The voltage doesn't have negative half wave and hence it is not alternating voltage.

That is why you need a basic Circuit Theory course.

In RL load according to my source at what time the current will be maximum ?

I gave you the formula. Can you figure that out with simple derivative of the current with respect to time ?

Why you say I should not take voltage as -25 ?

I have never said that. I am saying that who cares if it is -25 or -28. The advantage is that you already know the resistor value for the -28 V. Play around a bit with the resistor if you want -25 V.

 

[Okada]

That is why you need a basic Circuit Theory course.

Long back I studied electronics but did not study analog electronics much because it was full of equations.

I gave you the formula. Can you figure that out with simple derivative of the current with respect to time ?

You cave i(theta) or i(t) formula. I know calculus but don't know how to apply it in this case but if you provide the exact differential equation then I will solve it.

you already know the resistor value for the -28 V

Where is the resistor and inductor value for -28V. The ones I am using in circuit (36mH and 10R) will make high current to flow through it.

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I did some calculations usin gthe equation I = V/Z. I kept I as 100mA, V as 34V because I is directly proportinol to V and hence if Z is constant and V decreases then i decreases but as V will not exceed 34V so the current will be max at this voltage for a given impedance.

For 100mA, 34V and 36mH I found R to be 340 Ohms but these value doesn't provide the required waveform.

I think that dc motor has to be used to get this kind of signal. In dc motor there will be no heat generated.

So, I will go with dc motor for the load.



When we studied electronics we were give problems for series RL, RC, and RLC circuits which used either AC voltage source or DC (steady state). We never was taught how to solve RL, RC, RLC circuits for rectified but not filtered ripple dc source.

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Edit:

Thank you FvM and CataM

My problem is solved. I got the right signal. I used I = V/Z formula.

The simulation result is shown in the attached file. In the PDFs it is the top most graph. It is mentioned for R load but it is actually for RL Load. I see a large spike which is making the actual graph small but in oscilloscope it is showing the signal clearly.

I used 1k resistor and 36mH Inductor which limited the current to 3.4mA. I used voltage as 34V. So, Power dissipated in resistor came to be 1.15W and so I can safely use 2W resistor.


One last question. See the top most graph and also the oscilloscope signal and tell me what is causing the 540V spike.



Please discard all previous files and consider only Revision E file and answer my questions.

1.) For L = 36mH, R = 1k, V = 34V, I is 3.4mA but when in the bottom most graph it is showing 26.2A ?

I see that Voltage is leading the current in the graphs for RL load (middle and bottom graphs) but I have a doubt. is the current graph correct ? because I guess there should be no current when scr is off.

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Sorry, forgot to attach Revision E files.

 

[Okada]

Everything seems to be correct except the RL load current. The Ammeter and Current Graph are showing 23A and 26A(peak). R is 10k and L is 36mH. Please check the attached file and reply. Use the color pdf for checking which signal is which and use monochrome pdf for meter readings.

Discard all previous files and consider only Revision G file.

 

[Okada]

See attached image. I am using I = V/Z equation. My calculations are correct. I am getting I = 3.39mA for R = 10k and L = 36uH. Why Proteus is showing 23A in Ammeter and around 26A in Current Graph for RL Load current. In series RL circuit current through R and L are same.

V = Vr + Vl

I = V/Z

Vr = I*R
Vl = I*XL

Pr = I^2*R
Pl = Vl*I

 

[CataM]

Why Proteus is showing 23A in Ammeter and around 26A in Current Graph for RL Load current

V4 across an ON state SCR gives a large current.

 

[Okada]

@CataM

But in my hardware I am using 12-0-12V 500mA transformer for testing. I will get the board in 3 days. I have used 10k for R and 39mH for L in RL Load. Using the formula I = V/Z, I got I = 3.4mA and power dissipated in R of RL load is 115.6mW and so I have used 250mW resistor.

Is it safe to do the testing or should I change something to limit the 26A current ?

 

[CataM]

Option A: Do not use the center tapped terminal i.e. do not wire it. Use only the extreme terminals.
Option B: do not add the ground to the controlled full bridge rectifier.

 

[Okada]

I am using same 12-0-12V 500mA transformer to power the PIC power supply and SCRs but center tap of the transformer is not used.

The SCR output ground is connected to PIC power supply ground. I can't change this. Already sent the layout for PCB manufacturing.

 

[Okada]

See the attached file. See the top most graph on the right side. The BC337 output is taking 2.3ms to go low after Zero Cross and after BC337 goes low that is INT pin triggers then even though there are no delays for firing SCRs it is taking approx 400us to trigger SCR. Why is this ? I know there will be propogation delays but how can I reduce transistor switching time ? Should I use high speed switching transistor ? If yes, what suitable transistor can I use ? This 2.3ms to 2.7ms delay is affecting the conduction angle.

2N3866 is reducing the delay very much. I think I will have to use it to get better conduction angle.

 

[FvM]

The BC337 output is taking 2.3ms to go low after Zero Cross.

This is just expectable according to R1/R2 dimensioning. Far off from reasonable values.

 

[Okada]

According to BC337 datasheet Vbe(max) is 1.2V. So, I took Vbe as 0.7V. Can I reduce this.

My Pulsated DC varies between 0 and 34V and so I have to divide 34V into (34 - 0.7V) and (0.7V).

So, I assumed Ib as 1mA

So, R1 = (34 - 0.7) / 1mA = 33.3K

R2 = 0.7 / 1mA = 700R

How to choose Proper R1/R2. My DC to base-emitter circuit is pulsated DC whose max value is 34V. Vbe should not exceed 1.2V at any time otherwise tranistor will get damaged. I am using Voltage divider bias formula for calculation.

Please provide suitable formula for calculating R1 and R2 for Pulsated DC input.

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Edit:

It is very difficult to calculate the resistor values for pulsated dc. Vb should not exceet 1.2V for BC337. I tried using

Vb = ((R2/(R1 + R2))V

V == 34V

R2 I assumed as 1k

Vb I assumed as 0.7V

but it is not working.

So, I removed R2 and used just R1 and so the whole V - 0.7 drop across R1. I used 10k for R1. It is working fine.

 

[Okada]

Please check this simulation and tell me why RL load is not working properly ? You can set the rotary switch to R load and then simulate and vary the pot. The signals on oscilloscope and also graph will be fine but if load is changed to RL and simulated then at approx 90 degrees a spike is coming and I guess it is due to gate 1 and due to this the conduction angle doesn't vary after 90 degrees.

Do I need a RC snubber to eliminate the spike ? If yes please mention which application note I can refer to for snubber designing.

 

[FvM]

Don't know how many Edaboard members are using Proteus and who is willing to review your simulation files. You'll probably raise the chances to get help by posting a generally understood problem description, e.g. simulation waveforms.

 

[Okada]

@FvM

Ok. Here I port the graphs.

See the two graphs for RL Load. If firing angle is less than 90 degrees then it ts working fine but if firing angle is greater than 90 degrees then you can see in the 2nd graph for RL Load that at 4.9ms (Simulation starts at 1 sec and spike comes at 1.0049 sec) a spike comes and this is making the conduction angle not to change. Actually this spike is making the conduction angle to be 90 degrees all the time if firing angle is greater than 90 degrees. If firing angle is greater than 90 degrees then it is neglected and this spike is making SCRS to conduct for 90 degrees fixed conduction angle.

Inductow was changed to 50mH as 36mH and 39mH was not available.

 

[Okada]

PCB has been fabricated. Components has been purchased. I will get the soldered PCB in 3 days. I will then do the testing using a 12-0-12V 500mA transformer.

https://www.mediafire.com/?du2ie49wpf9qrcb

 

[KlausST]

Hi,

in your simulation:
V3 as well as V4 are related to GND.
In reality this is not the case, I assume.

Because of this:
In simulation you could omit U7 and U10. They are useless.
But in reality you need them.

For sure - especially with L1 - you will see much difference between simulation and reality.

Klaus

 

[Okada]

@Klaus

Yes, V3 and V4 are related to ground only in simulation. In real hardware it is 12-0-12V 500mA transformer without using center tap. The same 24V goes to power supply bridge, ZC detection circuit bridge and SCRs.

Ok. I will try omitting U7 and U10 in simulation. It will be used in real hardware.

 

[CataM]

I will try omitting U7 and U10 in simulation. It will be used in real hardware.

Why are you simulating a circuit that you will not use in real life?
Why don't you get rid of the ground at U7 and U10 ?

 

[Okada]

@CataM

I will use the same simulation circuit in hardware only difference is V3 and V4 will not be grounded. it will be a 12-0-12V 500mA transformer without using center tap.

Ok. In simulation I will get rid of the ground at U7 and U10 and test.

 

[CataM]

I will use the same simulation circuit in hardware only difference is V3 and V4 will not be grounded. it will be a 12-0-12V 500mA transformer without using center tap.

If so, why are you not simulating as such? I see no reason for using 2 voltage sources.