CT is not measuring the comparator output.

Thread Starter

moonlystar1111

Joined Feb 1, 2024
57
I was working on a circuit where i was trying to create an equivalent circuit (of the attached .jpg file) by a comparator setup.

The comparator setup is explained as-
When the charge at capacitor c1 is increased beyond a given threshold (by voltage divider r2 and r3), it will give direct input to the power line.
I have attached a CT to the main power line which will give the value of the extra current in the power line due to the comparator output.
But the problem is the CT is not detecting any current and i want to know where I need to correct my circuit.
 

Attachments

Alec_t

Joined Sep 17, 2013
14,318
1) Since R4 is only 1 milliOhm it is effectively short-circuiting the CT secondary.
2) Why is L3 included in the CT coupling command? It is preventing the wanted CT action.
3) What is the purpose of L3?
4) Q1 and D1 and probably the comparator IC will be destroyed by being connected to V2 as shown. Note the currents being drawn.
5) If V2 is intended to be a mains supply, note that Spice uses voltage amplitude, not RMS values, to specify the V2 properties. So 120 should presumably be 120 x 1.414 = 170.
6) A CT with a smaller turns ratio might be more useful.
 

Thread Starter

moonlystar1111

Joined Feb 1, 2024
57
1) Since R4 is only 1 milliOhm it is effectively short-circuiting the CT secondary.
2) Why is L3 included in the CT coupling command? It is preventing the wanted CT action.
3) What is the purpose of L3?
4) Q1 and D1 and probably the comparator IC will be destroyed by being connected to V2 as shown. Note the currents being drawn.
5) If V2 is intended to be a mains supply, note that Spice uses voltage amplitude, not RMS values, to specify the V2 properties. So 120 should presumably be 120 x 1.414 = 170.
6) A CT with a smaller turns ratio might be more useful.
1) I have changed R4 value, but still ineffective.
The reason I have put it to be 1 mili so that all of the comparator input is measured and effect of burden is not seen on the CT.
2) L3 is included in reverse polarity so that only current differential is measured by the L2.
3) L1 L2 L3 are acting as a CT here whose primary winding is Hot and Neutral of power line. Its secondary will produce difference in the current between hot and neutral line.
4) I think you’re right. But then how to connect them?
5) Okay I have decreased the values,.
6) Done.
 

Attachments

Ian0

Joined Aug 7, 2020
9,822
So you have 12mA load and 5mA "leakage".
The ratio of the inductors is 1uH/1H (1000000:1) so you would expect 5uA in L2 (Square root of the inductance ratio), and that seems to be what you are getting.
 

Thread Starter

moonlystar1111

Joined Feb 1, 2024
57
So you have 12mA load and 5mA "leakage".
The ratio of the inductors is 1uH/1H (1000000:1) so you would expect 5uA in L2 (Square root of the inductance ratio), and that seems to be what you are getting.
Yes you're right. In the first schematic I have this setup. I have changed in the second only for better analysis
 

Thread Starter

moonlystar1111

Joined Feb 1, 2024
57
1) Since R4 is only 1 milliOhm it is effectively short-circuiting the CT secondary.
2) Why is L3 included in the CT coupling command? It is preventing the wanted CT action.
3) What is the purpose of L3?
4) Q1 and D1 and probably the comparator IC will be destroyed by being connected to V2 as shown. Note the currents being drawn.
5) If V2 is intended to be a mains supply, note that Spice uses voltage amplitude, not RMS values, to specify the V2 properties. So 120 should presumably be 120 x 1.414 = 170.
6) A CT with a smaller turns ratio might be more useful.
Can you please give an insight of how to combine transistor with the mains supply..
 

Alec_t

Joined Sep 17, 2013
14,318
You could connect R6 to ground via Q1 collector, instead of directly to ground. However, note that a 2N3904 isn't suitable for Q1 as its Vce max is only 40V.
 
Top