SCR Bridge Control - Is my Circuit Correct ?

[Okada]

That formula is mentioned in step III here.

https://www.electrical4u.com/rl-series-circuit/

 

[CataM]

Yes. Your formula and mine are the same. Keep reading the document and you will see the formula I mentioned.

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tau = L/R but without knowing L and R how can I calculate tau ?

Knowing phi you have tau. In post #57 I thought about how do you select tau.

 

[Okada]

In Series RL Circuit Analysis section strp III here

https://www.electrical4u.com/rl-series-circuit/

I got the formula but for phi = 55.83 degrees I am not getting the value you mentioned that is 4.69*10^-3

How ?

Phi = arctan(omegaL/R)

omegaL/R = tan(phi)

Thank you FvM and CataM

Yes CataM. Your calculation was correct. I got it.

Actually my calculator mode was in radians and phi was in degrees.

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@FvM and CataM

I am referring this and doing the calculations and testing in Proteus but I am not getting backemf signals on the scope.

https://www.electrical4u.com/rl-series-circuit/


1.)

VL = Ldi/dt = 15V

L = 36mH

V = 34V

Vr = V - VL = 34 - 15 = 19V

Vr = IR

I = 10mA

R = VR/I = 19V/10mA = 100


2.)

VL = Ldi/dt = 28V

L = 36mH

V = 33.9V

Vr = V - VL = 33.9 - 28 = 5.9V

Vr = IR

I = 10mA

R = VR/I = 5.9V/10mA = 59


Please help me choose the right R and L values. current can be minimum if needed but I need to get say for 34V I need to get atleast 28V backemf.

 

[CataM]

I am afraid that your calculations are all wrong. I will give you a circuit and tell me if it is good enough for you.

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Circuit:


Current through the RL load:


Voltage across the load:


Voltage across the inductor only:

 

[Okada]

Thank you for the circuit. I will check it. Can you tell how you designed it ?

Which simulator you used ? Cadence OrCAD PSpice ? I am not getting output in Proteus. Output is blank. Earlier it was working. What I did wrong ?

 

[CataM]

Can you tell how you designed it ?

...I need to get say for 34V I need to get atleast 28V backemf.

I checked if the already found inductor and resistor can achieve your spec of 28 V peak across the inductor and it turns out it does. In simulation as you can see it is a bit higher.
I have used the sinusoidal steady state to simplify things even though as you can see it is not accurate because 3*L/R is comparable to the time it takes the current to cross 0 amps.
|Z|=178.07 Ω => |I|=|V|/|Z|=0.191 A => |V_L|=|I|*|X_L|=28,14 V

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Which simulator you used ? Cadence OrCAD PSpice ?

Yes. OrCAD PSpice.

 

[Okada]

I am really sorry. I did wrong calculation. What I want is this.

V = VR + VL = IR + IXL

XL = 2 * pi * f * L

f = 50 Hz

L can be assumed as 36mH because it is a standard value and easily available.

if energy stored in inductor is 28V during positive half cycle then this voltage appears accross the load during negative half cycle but polarity reverses across the load.

I want 28V to appear across the load during negative half cycle and L = 36mH and I have to be kept minimum and R have to be calculated.

Finially V = 28V divides across R and L but voltage is measured across RL and not R and L.




This is the calculation I am doing. It is giving some back emf but not 28V. Attached is Proteus simulation. What am I doing wrong ?

V = VR + iLdi/dt

VL = iLdi/dt

During negative half cycle of AC V that is voltage across load should be 28V

28V = VR + VL

Current should be minimum

say 100mA

So, 28V = IR + IXL

XL = 2*pi*f*L

L is assumed as 36mH

XL = 2 * 3.14159265 * 50 * 36mH = 11.3097R

IXL = VL = 1.13V

So, VR = 28V - 1.13V = 26.87V

VR = IR

R = VR/I

R = 26.87/0.1 = 268.7R

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Sorry calculation was wrong. New calculation still getting only some 4V as back emf. During negative half cycle of AC I need -28V across the load which gradually decreases to 0 in say 4ms.

Just show the calculation. Current has to be kept low so that 1W or 2W resistor can be used.

V = VR + iLdi/dt

VL = iLdi/dt

During negative half cycle of AC V that is voltage across load should be -28V

28V = VR + VL

Current should be minimum

say 100mA

So, -28V = IR + IXL

XL = 2*pi*f*L

L is assumed as 36mH

XL = 2 * 3.14159265 * 50 * 36mH = 11.3097R

IXL = VL = 1.13V

So, VR = -28V - 1.13V = 29.13V

VR = IR

R = VR/I

R = 29.13/0.1 = 291.3R

 

[CataM]

Your calculations are all wrong again.

This is what you wanted. I got -27 V across the load.

Circuit:


Current:


Voltage across RL load:


Calculation process:
34*sin(180+X)=-28 V => X=55.44º => beta=180º+55.44º=4.11 rad

Alpha=90º and L=36 mH, then, introducing the full expression (transient as well) of the current through the load in a simple math tool I got the value of R.
That is a plot of the current through the load with the unknown variable beeing the resistor. I have inserted beta=4.11 rad. So , that plot tells me for what resistor the current through the load makes 0 at exactly 180º+55.44º (4.11 rad) which would give -28 V across the load before turning off.


Letting Wolfram to solve the problem, it shows this:


Solution=R=5.87 ohms

In simulation I got -27 V and not -28 V because SCRs are not ideal and the solution is a numerical one.

 

[Okada]

I am getting this. This is giving some backemf but not 28V. How can this be improved ?

phi = arctan(VL/VR)

I = -100mA (assumed)

Phi = 20 (assumed)

VL = -28V (assumed)

arctan(VL/VR) = 20

VL/VR = tan(20) = 0.36397

VR = VL/0.36397 = -28/0.36397 = -76.9294

R = VR/I = -76.9294/-0.1 = 769

VL = IXL

XL = VL/I = -28V/-0.1 = -280

XL = 2 * pi * f * L = 280

L = 280/(2 * pi * f) = 0.8912H

L = 891mH




Edit: I need to get this kind of signal but for 24V AC. The signal shown in image is for 230V AC.

 

[Okada]

I am using 24V AC for SCR circuit and also for power supply. A 220V / 24V transformer is used. A LM317T provides 5V from 34V rectified voltage. MOC3021 circuit is used to trigger the SCRs. I have connected SCR bridge output ground to PIC power supply ground. Is it ok ?

230V AC will not be used for SCRs.

 

[CataM]

Edit: I need to get this kind of signal but for 24V AC. The signal shown in image is for 230V AC.

Did you see what "kind of signal for 24 V AC" was shown in post #68 ? Why is not good enough ? You said -28 V, I gave -28 V. What do you want? -34 V ?

 

[Okada]

Yes CataM give voltage which will be like 230V simulation signal for 24V AC.

CataM, XL - omegaL = 2*pi*f*L

We are feeding pulsated dc to RL load from scr bridge. Should be take f as 100 Hz ?

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See the attached simulation result. Only L = 36mH and R = 10R is giving the required signal in Proteus for 24V AC. I want to know what will be the current through this RL load ? because it will be good if current is in milliamps otherwise 5W or 10W resistor has to be used I think.

 

[CataM]

I want to know what will be the current through this RL load ? because it will be good if current is in milliamps otherwise 5W or 10W resistor has to be used I think.

I hope you are joking. You do not know how to measure current in your simulator ?

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Should be take f as 100 Hz ?

Only if you do a Fourier Series analysis.

 

[Okada]

I am having a small problem. If I use R load then firing angle is varying but if I use RL load then firing angle is not varying. Why ?

I don't know how to place current probe in simulator.

For calculation I am using

i = V ( 1 - e^(-Rt/L) ) / R

So, for R = 10R and L = 36mH and t = 5ms (because current will be max at 90 degrees, Right ?)

i = 34 ( 1 - e^(-(10 * 0.01) / (36 * 0.01) ) / 10

i = 34 ( 1 - e^(-0.2777) ) / 10

i = 34 ( 1 - 0.758 ) / 10

i = 0.8228A

i = 822.8 mA

Is the calculation correct ?

When will be the current max ?

 

[Okada]

@FvM and CataM

Please provide a quick solution for RL Load values. See attached simulation result. Only these RL values are giving the required signal but current is high. i can't use it in real hardware.

I need exact signal and low current. L and R should be standard values.

It is urgent requirement. There is no time for experimenting with different RL values in hardware.

Right now I am only testing in Proteus. After three days I will get the PCB without RL.

Edit:

I have attached two files. Both are same but one is in color and another monochrome. In monochrome version both signals are displayed in black and it might be difficult for some people to read it and so I have attached color version also.

 

[Okada]

FvM and CataM I need to get graph as in post 75.

Please solve this equation to get the value of R using math software.

When R/L = 5t the current in the RL circuit will be maximum and I need it to be 50mA. At 5t the inductor acts as a short circuit. L I assume as 36mH because it is easily available. From the graph the voltage at 2.5ms is -25V. So, with different R and L values and load current 50mA I need to get same graph. Please solve the below equation using math software and tell me the value of R.

I = 0.05A
V @ 2.5mS = -25V
t = 2.5mS
L = 36mH
R = ?

I(t) = (V/R)(1 - e^(-Rt/L))

0.05 = (-25/R)(1 - e^((-R * 0.0025) / (36 * 10^(-3)))

 

[CataM]

Please solve the below equation using math software and tell me the value of R.

Math tools can easily solve your equations... but it would be useless because are all wrong.

You are again trying to achieve the -twenty something voltage across the load. As you have seen in post #68, the value of R was calculated.
Play around with the value of R to get your desired negative voltage or play around with the following equation of the current through the RL load to get your "low current".

 

[Okada]

Ok. I will try but why my calculation is wrong ?

In RL series circuit the current will be max at 5 * tau and if I assume it to be 50mA then current at other times will be even lesser. From the graph at 2.5ms from zero cross the voltage across RL load is 25V. L I assume as 36mH then why is my calculation wrong ? I am getting 0 Ohms for R but If I use zero ohms then it will act as a short circuit.

I used Maple to solve the equation.

https://www.electronics-tutorials.ws/inductor/lr-circuits.html

If I use your equation then theta = omega * t

what should I take for f ? I am not giving AC 50Hz but I am giving rectified DC which has f = 100 Hz. So, I have to take 100 Hz ?

Yes, my calculation is wrong because I assumed I at 5(tau) = 50mA but used t as 2.5ms. I should find out what is t at 5(tau) for that I need to know L/R but L is known and R is not known.



In your current formula what is t ? because current will be max at t = 5tau but I don't know tau and hence I don't know t. I know L but I don't know R. I have two unknowns R and t.

 

[CataM]

what should I take for f ? I am not giving AC 50Hz but I am giving rectified DC which has f = 100 Hz. So, I have to take 100 Hz ?

50 Hz.

In RL series circuit the current will be max at 5 * tau and if I assume it to be 50mA then current at other times will be even lesser.

If the source was DC, but is not the case here.

I will try but why my calculation is wrong ?

After you will have a basic Circuit Theory course, you will understand why.

 

[Okada]

Why you say source is not DC. I am giving rectified dc to RL load. The voltage doesn't have negative half wave and hence it is not alternating voltage. It is Ripple DC and Ripple frequency is 100 Hz. Please explain more clearly about the frequency. I know that properly filtered DC frequency is 0 but it is not the case here.


Ok CataM

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

Why you say I should not take voltage as -25 ? Should I take V as 0 ?, because when scr is not fired V (source voltage to RL load) is 0.