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It's not clear what you are really trying to do, but if you're trying to replace the left circuit with the right one, that's a bad idea: The left circuit switches to appear to pass partial volta...
Answer
#4: Post edited
by
Lorenzo Donati
·
2023-07-30T18:28:01Z (over 1 year ago)
Typos.
- It's not clear what you are really trying to do, but if you're trying to replace the left circuit with the right one, that's a bad idea:
- <img src="https://electrical.codidact.com/uploads/dLUNkUNrMeuh98yxAKSCXxqR">
- The left circuit <i>switches</i> to appear to pass partial voltage. That means it doesn't dissipate much power. Power into something is the voltage across it times the current thru it. A fully off switch doesn't dissipate power because the current thru it is 0. A fully on switch doesn't dissipate power because the voltage across it is 0.
- The second circuit always "eats up" a fixed voltage. When in series with a load, it will dissipate the 32 V or so it drops times the load current. That could be quite significant power.
- Step back and show the original circuit, and explain what you are trying to accomplish.
- <h2>Response to schematic</h2>
- Now that you have posted a schematic, we can see that you want to replace the part that triggers the pass element, not the pass element itself.
Two back to back Zener diodes are not the same as the diac. Both can be arranged to have a threshold below which they won't conduct. However, the diac has the special property that once conduction is triggered, it the voltage drop becomes much lower and the device stays on until the current falls below some threshold.There are some other devices that exhibit this kind of foldback behavior, but probably not appropriate for your case. Neon bulbs have this property, but usually require about 80 V to trigger for the common type. Spark gaps have this property too, but are much higher voltage devices. It would be difficult to make a spark gap reliably trigger at only 32 V.You could try another triac with the gate being driven from a resistor divider between the anode and cathode. Or use a Zener diode between anode and gate. This will be more finnicky than a diac, but could possibly be made to work when that's all you've got. You can think of a diac as a triac that self-triggers at a pre-determined voltage threshold.
- It's not clear what you are really trying to do, but if you're trying to replace the left circuit with the right one, that's a bad idea:
- <img src="https://electrical.codidact.com/uploads/dLUNkUNrMeuh98yxAKSCXxqR">
- The left circuit <i>switches</i> to appear to pass partial voltage. That means it doesn't dissipate much power. Power into something is the voltage across it times the current thru it. A fully off switch doesn't dissipate power because the current thru it is 0. A fully on switch doesn't dissipate power because the voltage across it is 0.
- The second circuit always "eats up" a fixed voltage. When in series with a load, it will dissipate the 32 V or so it drops times the load current. That could be quite significant power.
- Step back and show the original circuit, and explain what you are trying to accomplish.
- <h2>Response to schematic</h2>
- Now that you have posted a schematic, we can see that you want to replace the part that triggers the pass element, not the pass element itself.
- Two back to back Zener diodes are not the same as the diac. Both can be arranged to have a threshold below which they won't conduct. However, the diac has the special property that once conduction is triggered, the voltage drop becomes much lower and the device stays on until the current falls below some threshold.
- There are some other devices that exhibit this kind of foldback behavior, but probably not appropriate for your case. Neon bulbs have this property, but usually require about 80 V to trigger for the common type. Spark gaps have this property too, but are much higher voltage devices. It would be difficult to make a spark gap reliably trigger at only 32 V.
- You could try another triac with the gate being driven from a resistor divider between the anode and cathode. Or use a Zener diode between anode and gate. This will be more finicky than a diac, but could possibly be made to work when that's all you've got. You can think of a diac as a triac that self-triggers at a pre-determined voltage threshold.
#3: Post edited
by
Olin Lathrop
·
2020-07-22T19:54:10Z (over 4 years ago)
- It's not clear what you are really trying to do, but if you're trying to replace the left circuit with the right one, that's a bad idea:
- <img src="https://electrical.codidact.com/uploads/dLUNkUNrMeuh98yxAKSCXxqR">
- The left circuit <i>switches</i> to appear to pass partial voltage. That means it doesn't dissipate much power. Power into something is the voltage across it times the current thru it. A fully off switch doesn't dissipate power because the current thru it is 0. A fully on switch doesn't dissipate power because the voltage across it is 0.
- The second circuit always "eats up" a fixed voltage. When in series with a load, it will dissipate the 32 V or so it drops times the load current. That could be quite significant power.
Step back and show the original circuit, and explain what you are trying to accomplish.
- It's not clear what you are really trying to do, but if you're trying to replace the left circuit with the right one, that's a bad idea:
- <img src="https://electrical.codidact.com/uploads/dLUNkUNrMeuh98yxAKSCXxqR">
- The left circuit <i>switches</i> to appear to pass partial voltage. That means it doesn't dissipate much power. Power into something is the voltage across it times the current thru it. A fully off switch doesn't dissipate power because the current thru it is 0. A fully on switch doesn't dissipate power because the voltage across it is 0.
- The second circuit always "eats up" a fixed voltage. When in series with a load, it will dissipate the 32 V or so it drops times the load current. That could be quite significant power.
- Step back and show the original circuit, and explain what you are trying to accomplish.
- <h2>Response to schematic</h2>
- Now that you have posted a schematic, we can see that you want to replace the part that triggers the pass element, not the pass element itself.
- Two back to back Zener diodes are not the same as the diac. Both can be arranged to have a threshold below which they won't conduct. However, the diac has the special property that once conduction is triggered, it the voltage drop becomes much lower and the device stays on until the current falls below some threshold.
- There are some other devices that exhibit this kind of foldback behavior, but probably not appropriate for your case. Neon bulbs have this property, but usually require about 80 V to trigger for the common type. Spark gaps have this property too, but are much higher voltage devices. It would be difficult to make a spark gap reliably trigger at only 32 V.
- You could try another triac with the gate being driven from a resistor divider between the anode and cathode. Or use a Zener diode between anode and gate. This will be more finnicky than a diac, but could possibly be made to work when that's all you've got. You can think of a diac as a triac that self-triggers at a pre-determined voltage threshold.
#2: Post edited
by
Olin Lathrop
·
2020-07-22T13:11:30Z (over 4 years ago)
It's not clear what you are really trying to do, but if you're trying to replace the left circuit with the right one, that's a bad ide:- <img src="https://electrical.codidact.com/uploads/dLUNkUNrMeuh98yxAKSCXxqR">
- The left circuit <i>switches</i> to appear to pass partial voltage. That means it doesn't dissipate much power. Power into something is the voltage across it times the current thru it. A fully off switch doesn't dissipate power because the current thru it is 0. A fully on switch doesn't dissipate power because the voltage across it is 0.
- The second circuit always "eats up" a fixed voltage. When in series with a load, it will dissipate the 32 V or so it drops times the load current. That could be quite significant power.
- Step back and show the original circuit, and explain what you are trying to accomplish.
- It's not clear what you are really trying to do, but if you're trying to replace the left circuit with the right one, that's a bad idea:
- <img src="https://electrical.codidact.com/uploads/dLUNkUNrMeuh98yxAKSCXxqR">
- The left circuit <i>switches</i> to appear to pass partial voltage. That means it doesn't dissipate much power. Power into something is the voltage across it times the current thru it. A fully off switch doesn't dissipate power because the current thru it is 0. A fully on switch doesn't dissipate power because the voltage across it is 0.
- The second circuit always "eats up" a fixed voltage. When in series with a load, it will dissipate the 32 V or so it drops times the load current. That could be quite significant power.
- Step back and show the original circuit, and explain what you are trying to accomplish.
#1: Initial revision
by
Olin Lathrop
·
2020-07-22T13:10:05Z (over 4 years ago)
It's not clear what you are really trying to do, but if you're trying to replace the left circuit with the right one, that's a bad ide: <img src="https://electrical.codidact.com/uploads/dLUNkUNrMeuh98yxAKSCXxqR"> The left circuit <i>switches</i> to appear to pass partial voltage. That means it doesn't dissipate much power. Power into something is the voltage across it times the current thru it. A fully off switch doesn't dissipate power because the current thru it is 0. A fully on switch doesn't dissipate power because the voltage across it is 0. The second circuit always "eats up" a fixed voltage. When in series with a load, it will dissipate the 32 V or so it drops times the load current. That could be quite significant power. Step back and show the original circuit, and explain what you are trying to accomplish.