Imageshack - 85565399.png
This is a simple overload protection circuit in a power supply. Usually that configuration(the differential opamp and diode) is use to control the power supply whether function in current mode or voltage mode. However, i modified it become an overload protection circuit(the part that i drawn in the box). I have tested with multisim and it should be working theoritically.... but i didnt really try it out(in practical) because ownself make a power supply again will be costly and also busy preparing for exam. The circuit actually contain a transformer and a bridge rectifier diode but due to insufficient spaces i have deleted and just replace by a 34v supply. Here i going to explain the working principle. I constructed the overload protection mode to trigger when the load connected is less than 5ohm. The differential opamp will have an output of 2/10Vo where Vo is the load voltage which is the voltage across 2ohm resistor at here. Now consider if a large load is connected,ie 10ohm load the differential opamp will have 20v output since the initial voltage which i set to drop across the 1ohm resistor must be 2V hence a current of 2A is flowing across the 10ohm load. The outpout of U2A opamp i set it to 25v and by voltage divider formula can be seen that i can adjust the voltage to drop across the 1 ohm load to swing between 2v and 0v 2v to 0v divided by 1ohm=2A to 0A. The overload protection is activated when Vth of the 10kohm load>Vo+Vd where Vd is the forward voltage of the diode at here i suggest using germanium diode because it can pull the current to lower value during overload protection mode. When 10ohm load connected and 2A is set, the output of U3A will be 0.2x20=4V while the node voltage just above the 10kohm resistor will be 2V so the diode is now in reverse bias condition very small current can flow through the diode and hence doesnt affect the voltage reference i set which is 2V. On the other hand if we consider now we connected heavy load ie 2ohm, 4vx0.2=0.8V at the opamp output and at the node voltage just above 10khm is 2V so Vth across 10khm>Vo+Vd hence the diode is conducting and it pull down the voltage reference. This how it help to reduce the current and prevent excessive heat generated at the transistor.
The output voltage can be predicted by forming 2 equation
the node voltage at noninverting input should be same as the inverting input
Hence
IxRloadx0.2+Vd=Ix1
under overload protection mode the new current
I=Vd/(1-0.2Rload)
V=IRload
Voltage=VdRload/(1-0.2Rload)
or 2nd approach
((2xRloadx0.2+Vd1)0.2Rload+Vd2)0.2Rload+Vd3)...... and so on, i put Vd1 and Vd2 till Vdn because different output voltage at opamp will cause different forward voltage of the diode however it is obvious that the series is convergence. So until a very large term
i let the whole series will converge to a specific value and Vd also become constant and i represent the series by an unknown value X but the series is still continue to infinity
((X0.2Rload+Vd)0.2Rload+Vd)0.2Rload+vd and so on=X(0.2Rload)^N+Vd(0.2Rload)^N-1+Vd(0.2Rload)^N-2.......+Vd assume N is very big so x(0.2Rload)^N approximately 0 can be ignore. Equation become Vd(1+0.2Rload+(0.2Rload)^2+....) we can conclude this is the sum to infinity. with a=1 r=0.2Rload sum to infinity=a/(1-r)=Vd/(1-0.2Rload)
which first approach will be better since shorter=X
2nd equation will be
since the U2A i set it to 25V, we can use nodal analysis to solve it.
(25kohm-(Vo+Vd))/115kohm=(Vo+Vd)/10kohm+Ise^(Vd/nVT)
By solving the simultaneous equation we can get the predicted output voltage.
Well again the reason i post it here because even everything is working(theoritically) but i didnt practically test it(on hardware) or someone else have think before such design or come across and it is working kindly please let me know thank you. Haha, i know some people may have thought what a selfish person are you wanted people to test it on hardware for you and it is dangerous cause small explosion may occur(if overload protection circuit not working). So at here i would like to say sorry about for my selfishness=X because i dun have a own made power supply now and busy preparing for exam. I love electrical stuff very much but extremely scare explosion>.<.
I also know nowadays the overload protection are much better it can straight away cutoff the supply(which is my friend told me)rather than reduce the current to a smaller value. I have take a long time yet fail to think how to build such circuit and cannot find it on web also. If anyone have such idea or come across any web please post the link at here thank you.
This is a simple overload protection circuit in a power supply. Usually that configuration(the differential opamp and diode) is use to control the power supply whether function in current mode or voltage mode. However, i modified it become an overload protection circuit(the part that i drawn in the box). I have tested with multisim and it should be working theoritically.... but i didnt really try it out(in practical) because ownself make a power supply again will be costly and also busy preparing for exam. The circuit actually contain a transformer and a bridge rectifier diode but due to insufficient spaces i have deleted and just replace by a 34v supply. Here i going to explain the working principle. I constructed the overload protection mode to trigger when the load connected is less than 5ohm. The differential opamp will have an output of 2/10Vo where Vo is the load voltage which is the voltage across 2ohm resistor at here. Now consider if a large load is connected,ie 10ohm load the differential opamp will have 20v output since the initial voltage which i set to drop across the 1ohm resistor must be 2V hence a current of 2A is flowing across the 10ohm load. The outpout of U2A opamp i set it to 25v and by voltage divider formula can be seen that i can adjust the voltage to drop across the 1 ohm load to swing between 2v and 0v 2v to 0v divided by 1ohm=2A to 0A. The overload protection is activated when Vth of the 10kohm load>Vo+Vd where Vd is the forward voltage of the diode at here i suggest using germanium diode because it can pull the current to lower value during overload protection mode. When 10ohm load connected and 2A is set, the output of U3A will be 0.2x20=4V while the node voltage just above the 10kohm resistor will be 2V so the diode is now in reverse bias condition very small current can flow through the diode and hence doesnt affect the voltage reference i set which is 2V. On the other hand if we consider now we connected heavy load ie 2ohm, 4vx0.2=0.8V at the opamp output and at the node voltage just above 10khm is 2V so Vth across 10khm>Vo+Vd hence the diode is conducting and it pull down the voltage reference. This how it help to reduce the current and prevent excessive heat generated at the transistor.
The output voltage can be predicted by forming 2 equation
the node voltage at noninverting input should be same as the inverting input
Hence
IxRloadx0.2+Vd=Ix1
under overload protection mode the new current
I=Vd/(1-0.2Rload)
V=IRload
Voltage=VdRload/(1-0.2Rload)
or 2nd approach
((2xRloadx0.2+Vd1)0.2Rload+Vd2)0.2Rload+Vd3)...... and so on, i put Vd1 and Vd2 till Vdn because different output voltage at opamp will cause different forward voltage of the diode however it is obvious that the series is convergence. So until a very large term
i let the whole series will converge to a specific value and Vd also become constant and i represent the series by an unknown value X but the series is still continue to infinity
((X0.2Rload+Vd)0.2Rload+Vd)0.2Rload+vd and so on=X(0.2Rload)^N+Vd(0.2Rload)^N-1+Vd(0.2Rload)^N-2.......+Vd assume N is very big so x(0.2Rload)^N approximately 0 can be ignore. Equation become Vd(1+0.2Rload+(0.2Rload)^2+....) we can conclude this is the sum to infinity. with a=1 r=0.2Rload sum to infinity=a/(1-r)=Vd/(1-0.2Rload)
which first approach will be better since shorter=X
2nd equation will be
since the U2A i set it to 25V, we can use nodal analysis to solve it.
(25kohm-(Vo+Vd))/115kohm=(Vo+Vd)/10kohm+Ise^(Vd/nVT)
By solving the simultaneous equation we can get the predicted output voltage.
Well again the reason i post it here because even everything is working(theoritically) but i didnt practically test it(on hardware) or someone else have think before such design or come across and it is working kindly please let me know thank you. Haha, i know some people may have thought what a selfish person are you wanted people to test it on hardware for you and it is dangerous cause small explosion may occur(if overload protection circuit not working). So at here i would like to say sorry about for my selfishness=X because i dun have a own made power supply now and busy preparing for exam. I love electrical stuff very much but extremely scare explosion>.<.
I also know nowadays the overload protection are much better it can straight away cutoff the supply(which is my friend told me)rather than reduce the current to a smaller value. I have take a long time yet fail to think how to build such circuit and cannot find it on web also. If anyone have such idea or come across any web please post the link at here thank you.
