Let the following circuit:

The diodes and SCR are supposed to be ideal, the supply source \( v(\theta) = V_m \sin(\theta)\) , let \(\phi \in (0, \pi) \) be the angular gate pulse for T1 and \( \pi + \phi \) be the angular gate pulse for T2, i want to understand why in the interval \( \theta \in (0, \phi) \) the diode D2 and SCR T2 conducts specifically they act as a freewheel diode, i understand that in that interval the polarity of B is negative, but how does that make D2 and T2 conducts?
Here is how theory works in my head:
\( V_{D_2} = v_c(\theta) - V_B , V_{T_2} = V_B - v_c(\theta) \iff V_{D_2} + V_{T_2}= 0 \) i feel like this is pure nonsense, correct me on it on the side please, but if this is valid, then this means the voltage drop around D2 and T2 is opposite, but this doesnt give us no idea if the diode or SCR is conducting!? this only give us the idea that this forms a freewheel diode right?

 

Swap D2 with T1 then D1/D2 acts as the freewheeling diode and it doesn't need controlling.
If you really want to make a controlled rectifier to that design, then put a separate diode in for freewheeling, and halve the losses (because the freewheeling current goes through one diode instead of two)

 

Swap D2 with T1 then D1/D2 acts as the freewheeling diode and it doesn't need controlling.
If you really want to make a controlled rectifier to that design, then put a separate diode in for freewheeling, and halve the losses (because the freewheeling current goes through one diode instead of two)

Thank you ian for ur answer, but it doesnt really fulfill my question though? also does swapping D2 with T1 preserve the same characteristics of the circuit? I do not want to modify my circuit i just want to know how that works in that specific interval stated.

 

does swapping D2 with T1 preserve the same characteristics of the circuit?

During the thyristor "on" time, the AC goes through one diode and one thyristor. Both circuits work the same.
During the thyristor "off" time, your circuit needs one thyristor to be triggered on to provide a path for the freewheeling current.
My circuit doesn't as the freewheeling current goes through the two diodes.
Adding a extra freewheeling diode to your circuit gives the freewheeling current an unswitched path. It now goes through only one diode instead of two so the losses are halved. (But you have ideal diodes which have no losses - I'm dealing with real ones!)
It's a phase-controlled battery charger, isn't it?

 

During the thyristor "on" time, the AC goes through one diode and one thyristor. Both circuits work the same.
During the thyristor "off" time, your circuit needs one thyristor to be triggered on to provide a path for the freewheeling current.
My circuit doesn't as the freewheeling current goes through the two diodes.
Adding a extra freewheeling diode to your circuit gives the freewheeling current an unswitched path. It now goes through only one diode instead of two so the losses are halved. (But you have ideal diodes which have no losses - I'm dealing with real ones!)
It's a phase-controlled battery charger, isn't it?

Yes i totally understand the normal settings of the circuit, but did you read my post, did u look at the equation i wrote to understand what happens in the interval \( \theta \in (0, \phi) \)?

 

Yes i totally understand the normal settings of the circuit, but did you read my post, did u look at the equation i wrote to understand what happens in the interval \( \theta \in (0, \phi) \)?

That's the time between zero crossing and the trigger point. There is no current from the AC supply, but there is freewheeling current because of the inductive load. You have to provide a path for it.

 

That's the time between zero crossing and the trigger point. There is no current from the AC supply, but there is freewheeling current because of the inductive load. You have to provide a path for it.

But why is there no current from the AC supply?(is it because there is no path for it to return from to the source?)

 

But why is there no current from the AC supply?

Because we are in the half of the cycle where voltage A is greater than voltage B. At some point (φ) T1 will be switched on and current will flow from the AC supply, but that has not yet happened.
Current is still flowing in the inductor from the previous cycle and it needs somewhere to go, after the end of the cycle.

 

Because we are in the half of the cycle where voltage A is greater than voltage B. At some point (φ) T1 will be switched on and current will flow from the AC supply, but that has not yet happened.
Current is still flowing in the inductor from the previous cycle and it needs somewhere to go, after the end of the cycle.

wym previous cycle we are talking about the initial moment the circuit getting power up.

 

wym previous cycle we are talking about the initial moment the circuit getting power up.

I'm assuming continuous operation, and that t=0 is the zero crossing point.
If zero is the point where the AC signal starts, then there will be no freewheeling current

 

I'm assuming continuous operation, and that t=0 is the zero crossing point.
If zero is the point where the AC signal starts, then there will be no freewheeling current

Yes zero is the point where the AC signal starts,did you look at my equation, does it look right?

 

did you look at my equation, does it look right?

Where are you considering to be your zero-volts reference point?

 

Where are you considering to be your zero-volts reference point?

I honestly dont know, can you tell me where to consider it?

 

I honestly dont know, can you tell me where to consider it?

It's not the easiest.
I'd put it at the negative terminal of E.
That means that B is at -0.6V when the diode D2 conducts. If you have ideal diodes it will be at zero, but I find real diodes with real voltage drops make it easier to follow.
A is at Vsin(t)-0.6V.
When T1 is triggered, The top end of R is at Vsin(t)-1.2V, and current flows according to Vsin(t)-1.2V-E across R+L.

 

It's not the easiest.
I'd put it at the negative terminal of E.
That means that B is at -0.6V when the diode D2 conducts. If you have ideal diodes it will be at zero, but I find real diodes with real voltage drops make it easier to follow.
A is at Vsin(t)-0.6V.
When T1 is triggered, The top end of R is at Vsin(t)-1.2V, and current flows according to Vsin(t)-1.2V-E across R+L.

how come you are talking about A, do you mean point B?, and how did the SCR starting suddenly conducting?

 

how did the SCR starting suddenly conducting?

Presumably something triggered it. If nothing ever triggers it, why is it there?

 

Let the following circuit:
View attachment 323587
The diodes and SCR are supposed to be ideal, the supply source \( v(\theta) = V_m \sin(\theta)\) , let \(\phi \in (0, \pi) \) be the angular gate pulse for T1 and \( \pi + \phi \) be the angular gate pulse for T2, i want to understand why in the interval \( \theta \in (0, \phi) \) the diode D2 and SCR T2 conducts specifically they act as a freewheel diode, i understand that in that interval the polarity of B is negative, but how does that make D2 and T2 conducts?
Here is how theory works in my head:
\( V_{D_2} = v_c(\theta) - V_B , V_{T_2} = V_B - v_c(\theta) \iff V_{D_2} + V_{T_2}= 0 \) i feel like this is pure nonsense, correct me on it on the side please, but if this is valid, then this means the voltage drop around D2 and T2 is opposite, but this doesnt give us no idea if the diode or SCR is conducting!? this only give us the idea that this forms a freewheel diode right?

I am not sure this answers your question but SCR's do not turn off when the gate signal is removed. They stay 'on' until the current though them goes through zero or near zero. They won't turn back on again until the gate is powered once again, and the polarity across the device is correct also.

There is another case when the SCR might turn on but it should not happen with regular line voltage AC sine waves. if the dv/dt across the cathode and anode goes too high it can trigger the gate which turns the device on. There is usually a spec for that too on the data sheet.