SCR Bridge Control - Is my Circuit Correct ?

[FvM]

or if DC Motor is used with 15V reverse connected battery can I get continuous mode of operation ?

You can connect whatever you want, but I don't think it's useful.

As said, you need an "active load" (a load sourcing power to the SCR bridge) to achieve 4th quadrant operation. The lab setup is driving the load motor by a second DC motor to put it in generator mode. A negative voltage source can model an active load, but what's the purpose of using the voltage source together with a motor?

In Proteus 50mH and 12 Ohms is used. In real hardware 50mH and 10k is used.

10k is about "no load", don't know what it's good for. I don't feel responsible to fix an unsuitable test setup, however.

 

[Okada]

but what's the purpose of using the voltage source together with a motor?

Instructor od student has told to show the required inverted mode operation signag using any RL Edc load. The 120W DC Motor, Encoder and Tachogenerator are not available here. Student has to use RL Edc load or Motor with Edc. If RL Edc is used then R has to be decreased and current will increase. If current increases then higher wattage resistor is needed which is not available. So, only option left I think is to use DC Motor 12V 500mA x3 in series with a 0 to 15V Edc in reverse polarity and series to motor load.

So, will this setup work ? Only requirement is to show inverted mode signal in continuous mode operation. There is no other requirement.

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L/R = 50mH/12 = 0.004167

1H/R = 0.004167

R = 240R

Z = sqrt(240^2 + (2*pi*50*1)^2)

Z = sqrt(57600 + 98696)

Z = sqrt(156,296)

Z = 395.34 Ohms

I = V/Z = 34V/395.34 = 0.086A

P = I^2R = 0.086^2 * 395 = 2.92W

Edit

P = 0.086^2 * 240 = 1.84W

 

[CataM]

Post #140 seems to work as you want. Use values as in that simulation.

Student has to use RL Edc load or Motor with Edc

That is used in post #140. As FvM says, it has no sense, but do it if that is what the instructor wants and that's it.

will this setup work ?

If it works in simulation and you reconstruct it in real hardware, it will most likely work.

 

[Okada]

Thank you FvM and CataM

I did the calculation.

I have to see if I get these component values.

R = 0.12 1/2W
L = 100mH 300mA

L/R = 0.833

Z = sqrt(0.12^2 + (2*3.14159265*50*1)^2)

Z = sqrt(0.0144 + 986.96)

Z = 31.416 Ohms

I = V/Z = 34V/31.416 = 1.08A

P = I^2R = 1.08^2 * 0.12 = 139.9mW

Edc = 15V


Edit:

100mH 2A Inductor.

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I did some wrong calculation. it is not working.

Should I use the equations provided here to get time constant and induced emf and current ? I need to use 15V Edc source and get Inverted mode continuous conduction mode of operation.

What should be the time constant ?

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

 

[Okada]

@FvM

Please provide a solution to design a proper RLE(dc) load. How to choose R and L values to get continuous conduction mode Inverted operation of the Single Phase Fully Controlled rectifier ? E(dc) will be 15V and input to SCR Bridge is 24*1.4142 = 34V.

I am not able to get > 1mH Inductor 3A type. Please mention how to design a good RLE(dc) load.

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Here they are using big inductor and rheostat for RLE load. Where can we get such devices from ?

https://www.youtube.com/watch?v=1vkXM3ErLd4

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@FvM

Please explain how to choose proper tau = L/R for RLE load and also use 15V E. I need to get continuous mode operation in Inverter mode.

With some experimentation I have used 0.172 Ohms 1/2W and 33mH 3A successfully in simulation with 18V E. I am not able to reduce E further. If I reduce E further then I am getting discontinuous mode operation.