Showing posts with label current. Show all posts
Showing posts with label current. Show all posts
Saturday, November 8, 2014
Solar Panel Current Meter
This circuit is used to measure the current from a solar panel. It has very low power loss for currents in the 0-10A range. It also works as a general purpose DC current meter. The circuit can be used on either the positive or negative side of a DC circuit.
Solar Panel Current Meter
Specifications
Measured Current: 0-10 Amps DC
Circuit Voltage: Will work with DC circuits at any practical voltage.
Accuracy: approximately 2%, depending on the meter movement.
Theory
The current to be measured flows through the 0.01 ohm resistor which causes a small voltage drop across the resistor. The 100 microamp meter is set up with the series 50 ohm and 500 ohm variable resistor in a voltage measurement configuration to measure this voltage drop. The 500 ohm variable resistor is used to adjust the meters full scale reading. The 50 ohm resistor limits the maximum current to the meter no matter what setting is on the 500 ohm resistor, this protects the meter from passing too much current and burning up. The series resistance of the meter, 500 ohm (or less) variable resistor and 50 ohm resistor should total 1000 ohms. Different meters may require a different variable resistor to achieve the 1000 ohm value.
Construction
Build the meter into a metal box with the meter and two connectors mounted on the outside of the box.
Alignment
Put the meter circuit in series with a known current meter such as a digital VOM meter set to measure current. Run a known current through both meters. Adjust the 500 ohm resistor until both meters read the same current. A good way to get a known current is to put a 12V lead acid battery in series with a 2 ohm 100 watt current limiting resistor. This will produce approximately 6 Amps of current. Put the two meters in series with this loop and adjust for the same reading. Beware, the resistor will get fairly hot in a short time.
Use
Connect this circuit in series with a nominal 12V or 24V solar panel array. The meter can go in either the positive or negative side of the solar panel circuit. The current flowing through the solar panel to the load will be shown on the meter.
Parts
1x 100 microamp DC meter
1x 0.01 ohm 5 W resistor
1x 50 ohm 1/4 W resistor
1x 500 ohm 10 turn variable resistor
2x banana plugs or a 2 pin screw type terminal block.
1x metal box
Sunday, October 5, 2014
Power Current Limiter 13v 20A Circuit
This PSU has long been particularly created for current-hungry ham radio transceivers. It delivers safely around 20Amps at thirteen.8V. For reduced currents, a separate existing limiting output, able to 15ma as much as a total of 20A has long been extra. The energy transformer should be able to provide at least 25A at 17.5 to 20V. The decrease the voltage, the reduce energy dissipation.
The rectified existing will be "ironed" by C1, whose capacity will need to not be less than 40.000uF, (a golden rule of close to 2000uF/A), but we suggest fifty.000uF. This capacity could be constructed up by many smaller capacitors in parallel. The base of this design is a basic 12V regulator (7812). The output voltage might be introduced to ideal value (here thirteen.8V) by two exterior resistors (R5 and R6) using this formula: U= 12(1+R5/R6). The very low currents (here 15mA) will maintain the 7812 in its normal perform. As quickly because the present rises over 15ma, the voltage drop on R4 will "open" the Q3, basically handling the high output present. This is actually a PNP transistor (Ic > 25) and current amplification factor of a minimum of twenty. The one which has been tested and confirmed right here is the 2N5683. The existing limiting resistance RL, for that maximum output of twenty Amps must be 0.03 Ohms, rated a minimum of 15W. You possibly can make use of the resistance wire or swap numerous resistors in parallel, totaling the resistance/power values. Values for other currents could be calculated by the rule: RL=0.7/Imax
The RL and Q2 (3A PNP which include BD330) form a brief circuit automated fuse. As quickly as the maximum current reaches 20Amps, the voltage drop more than the resistor RL will open Q2, and thus restrict the B-E Existing of Q3. Parallel to Q2 is Q1, which lights the LED one when the present limiting circuit is active. Once the fuse is active, the Q2 bridges the R3, so the full present would circulation by way of the IC1, and harm it. As a result the R4 is inserted, as to limit the IC1 existing to 15mA. This may make it feasible to run the IC1 without having any cooling aid. The LED two will light up every single time the PSU is switched on.
There is an adjustable current limiter in parallel for the fixed output, thus supplying adjustable present source for scaled-down currents. This circuit is incredibly straightforward also. You might discover that there is no current sensing resistor. But it is definitely there, inside a type of the Rds-on resistance of the N-channel FET, which basically handles the load cutoff from the supply. The perform of the FET is proven in the diagram 2. When the existing Id is rising, the pressure Uds over the resistance Rds rises especially gradually in the starting, but especially quick immediately after the knick. This means, that just before the knick the FET behaves as being a resistor but right after it, works as continuous existing source. The D2, R3 and B-E connection of the Q4 will feeling the Uds voltage with the FET1. When the voltage rises sufficient, the Q4 will shortcut the FET1 gate to mass, and reduce the current circulation via the FET 1 off.
Nonetheless, to allow the FET1 to open, there is particular gate voltage required, which with this case is brought up through the voltage divider consisting of R8, Z1, P1 and R9. So the optimum Gate voltage will probably be the one with the Z1, along with the minimal is going to be around 3V6. The Z1 voltage (Uz1) will therefore establish the max current flowing via the FET one. The diagram two will indicate that for 5 Amps the Uz1 really should be 5V6, and for 20Amps close to 9V6.The Capacitor C4 will determine the “velocity” or the reaction time with the limiter. a hundred uF will make the response time to be about 100ms, and 1n will allow it to be 1us. Inside the created limits, the P1 will restrict the existing output within the array of 15mA to 20A.
You possibly can use each output simultaneously, but the complete output current will be restricted by the value of the RL. This PSU can be built also for greater outputs, as long as the transformer will deal with the present needs, and also you provide adequate cooling for your Q3.
The rectified existing will be "ironed" by C1, whose capacity will need to not be less than 40.000uF, (a golden rule of close to 2000uF/A), but we suggest fifty.000uF. This capacity could be constructed up by many smaller capacitors in parallel. The base of this design is a basic 12V regulator (7812). The output voltage might be introduced to ideal value (here thirteen.8V) by two exterior resistors (R5 and R6) using this formula: U= 12(1+R5/R6). The very low currents (here 15mA) will maintain the 7812 in its normal perform. As quickly because the present rises over 15ma, the voltage drop on R4 will "open" the Q3, basically handling the high output present. This is actually a PNP transistor (Ic > 25) and current amplification factor of a minimum of twenty. The one which has been tested and confirmed right here is the 2N5683. The existing limiting resistance RL, for that maximum output of twenty Amps must be 0.03 Ohms, rated a minimum of 15W. You possibly can make use of the resistance wire or swap numerous resistors in parallel, totaling the resistance/power values. Values for other currents could be calculated by the rule: RL=0.7/Imax
The RL and Q2 (3A PNP which include BD330) form a brief circuit automated fuse. As quickly as the maximum current reaches 20Amps, the voltage drop more than the resistor RL will open Q2, and thus restrict the B-E Existing of Q3. Parallel to Q2 is Q1, which lights the LED one when the present limiting circuit is active. Once the fuse is active, the Q2 bridges the R3, so the full present would circulation by way of the IC1, and harm it. As a result the R4 is inserted, as to limit the IC1 existing to 15mA. This may make it feasible to run the IC1 without having any cooling aid. The LED two will light up every single time the PSU is switched on.
There is an adjustable current limiter in parallel for the fixed output, thus supplying adjustable present source for scaled-down currents. This circuit is incredibly straightforward also. You might discover that there is no current sensing resistor. But it is definitely there, inside a type of the Rds-on resistance of the N-channel FET, which basically handles the load cutoff from the supply. The perform of the FET is proven in the diagram 2. When the existing Id is rising, the pressure Uds over the resistance Rds rises especially gradually in the starting, but especially quick immediately after the knick. This means, that just before the knick the FET behaves as being a resistor but right after it, works as continuous existing source. The D2, R3 and B-E connection of the Q4 will feeling the Uds voltage with the FET1. When the voltage rises sufficient, the Q4 will shortcut the FET1 gate to mass, and reduce the current circulation via the FET 1 off.
Nonetheless, to allow the FET1 to open, there is particular gate voltage required, which with this case is brought up through the voltage divider consisting of R8, Z1, P1 and R9. So the optimum Gate voltage will probably be the one with the Z1, along with the minimal is going to be around 3V6. The Z1 voltage (Uz1) will therefore establish the max current flowing via the FET one. The diagram two will indicate that for 5 Amps the Uz1 really should be 5V6, and for 20Amps close to 9V6.The Capacitor C4 will determine the “velocity” or the reaction time with the limiter. a hundred uF will make the response time to be about 100ms, and 1n will allow it to be 1us. Inside the created limits, the P1 will restrict the existing output within the array of 15mA to 20A.
You possibly can use each output simultaneously, but the complete output current will be restricted by the value of the RL. This PSU can be built also for greater outputs, as long as the transformer will deal with the present needs, and also you provide adequate cooling for your Q3.
Thursday, September 25, 2014
IC 723 Power Supply Circuit with Current Indicator
- This can be . increased readily by fitting a resistor in parallel with Ra, until the output voltage is5.0 V precisely Switches S1 and Sz are preferably SPDT types with a centre position, but three-way rotary switches should also do if in both cases the centre contact is not used.
- A difficulty arises if it is intended to provide an overcurrent indication for the shutdown circuit with the aid of an external transistor fitted in parallel onto pins 2 and 3, since the external and internal transistor are highly unlikely to have precisely the same characteristics.
- A further transistor, T3, has been added to keep the base current for T2 as low as possible.
- The output voltages of the supply may not be as accurate as required, and this is mainly due to the use of resistors from the E12 series. Close tolerance is especially important in the 5 V range, since the value shown for R3 gives a theoretical output of 4.9 V
- The three output voltages from this supply are probably the , most commonly used for testing asymmetrically fed designs: 5 volt for many TTL and CMOS circuits, 9 volt for battery operated equipment or logic circuits equipped with a 7805 regulator (this requires, an input of at least 8.5 V), and 12 volt for RS232 drivers, and miscellaneous opamp or tran- sistor based circuits.
- Since the base-emitter junction then has a diode characteristic, the associated voltage drop is always lower than that of the transistor internal to the 723.
- The current limiter can be set to l0 mA, 100 mA, or l A for safely power- ing- experimental circuits. Power regulator T1 should be fitted with a heat sink sized at least l0 ><‘10 cm. leds dv (green) and ds (red) are the power on and current overload indicator, respectively.
- When the internal transistor has the low B—E voltage of the two, the indication will not work, while in the other case the external transistor takes away the base current for the internal transistor, so that the current limiter is rendered ineffective. In this design of a power supply, a current overload indication was realized by fitting the external transistor with a high value base resistor, Rv, to ensure that the current limiter in the 723 is not disabled. IC 723 Power Supply Circuit with Current Indicator
Wednesday, September 3, 2014
Power Current Limiter 13v 20A Circuit
This PSU has long been particularly created for current-hungry ham radio transceivers. It delivers safely around 20Amps at thirteen.8V. For reduced currents, a separate existing limiting output, able to 15ma as much as a total of 20A has long been extra. The energy transformer should be able to provide at least 25A at 17.5 to 20V. The decrease the voltage, the reduce energy dissipation.
The rectified existing will be "ironed" by C1, whose capacity will need to not be less than 40.000uF, (a golden rule of close to 2000uF/A), but we suggest fifty.000uF. This capacity could be constructed up by many smaller capacitors in parallel. The base of this design is a basic 12V regulator (7812). The output voltage might be introduced to ideal value (here thirteen.8V) by two exterior resistors (R5 and R6) using this formula: U= 12(1+R5/R6). The very low currents (here 15mA) will maintain the 7812 in its normal perform. As quickly because the present rises over 15ma, the voltage drop on R4 will "open" the Q3, basically handling the high output present. This is actually a PNP transistor (Ic > 25) and current amplification factor of a minimum of twenty. The one which has been tested and confirmed right here is the 2N5683. The existing limiting resistance RL, for that maximum output of twenty Amps must be 0.03 Ohms, rated a minimum of 15W. You possibly can make use of the resistance wire or swap numerous resistors in parallel, totaling the resistance/power values. Values for other currents could be calculated by the rule: RL=0.7/Imax
The RL and Q2 (3A PNP which include BD330) form a brief schema automated fuse. As quickly as the maximum current reaches 20Amps, the voltage drop more than the resistor RL will open Q2, and thus restrict the B-E Existing of Q3. Parallel to Q2 is Q1, which lights the LED one when the present limiting schema is active. Once the fuse is active, the Q2 bridges the R3, so the full present would circulation by way of the IC1, and harm it. As a result the R4 is inserted, as to limit the IC1 existing to 15mA. This may make it feasible to run the IC1 without having any cooling aid. The LED two will light up every single time the PSU is switched on.
There is an adjustable current limiter in parallel for the fixed output, thus supplying adjustable present source for scaled-down currents. This schema is incredibly straightforward also. You might discover that there is no current sensing resistor. But it is definitely there, inside a type of the Rds-on resistance of the N-channel FET, which basically handles the load cutoff from the supply. The perform of the FET is proven in the diagram 2. When the existing Id is rising, the pressure Uds over the resistance Rds rises especially gradually in the starting, but especially quick immediately after the knick. This means, that just before the knick the FET behaves as being a resistor but right after it, works as continuous existing source. The D2, R3 and B-E connection of the Q4 will feeling the Uds voltage with the FET1. When the voltage rises sufficient, the Q4 will shortcut the FET1 gate to mass, and reduce the current circulation via the FET 1 off.
Nonetheless, to allow the FET1 to open, there is particular gate voltage required, which with this case is brought up through the voltage divider consisting of R8, Z1, P1 and R9. So the optimum Gate voltage will probably be the one with the Z1, along with the minimal is going to be around 3V6. The Z1 voltage (Uz1) will therefore establish the max current flowing via the FET one. The diagram two will indicate that for 5 Amps the Uz1 really should be 5V6, and for 20Amps close to 9V6.The Capacitor C4 will determine the “velocity” or the reaction time with the limiter. a hundred uF will make the response time to be about 100ms, and 1n will allow it to be 1us. Inside the created limits, the P1 will restrict the existing output within the array of 15mA to 20A.
You possibly can use each output simultaneously, but the complete output current will be restricted by the value of the RL. This PSU can be built also for greater outputs, as long as the transformer will deal with the present needs, and also you provide adequate cooling for your Q3.
The rectified existing will be "ironed" by C1, whose capacity will need to not be less than 40.000uF, (a golden rule of close to 2000uF/A), but we suggest fifty.000uF. This capacity could be constructed up by many smaller capacitors in parallel. The base of this design is a basic 12V regulator (7812). The output voltage might be introduced to ideal value (here thirteen.8V) by two exterior resistors (R5 and R6) using this formula: U= 12(1+R5/R6). The very low currents (here 15mA) will maintain the 7812 in its normal perform. As quickly because the present rises over 15ma, the voltage drop on R4 will "open" the Q3, basically handling the high output present. This is actually a PNP transistor (Ic > 25) and current amplification factor of a minimum of twenty. The one which has been tested and confirmed right here is the 2N5683. The existing limiting resistance RL, for that maximum output of twenty Amps must be 0.03 Ohms, rated a minimum of 15W. You possibly can make use of the resistance wire or swap numerous resistors in parallel, totaling the resistance/power values. Values for other currents could be calculated by the rule: RL=0.7/Imax
The RL and Q2 (3A PNP which include BD330) form a brief schema automated fuse. As quickly as the maximum current reaches 20Amps, the voltage drop more than the resistor RL will open Q2, and thus restrict the B-E Existing of Q3. Parallel to Q2 is Q1, which lights the LED one when the present limiting schema is active. Once the fuse is active, the Q2 bridges the R3, so the full present would circulation by way of the IC1, and harm it. As a result the R4 is inserted, as to limit the IC1 existing to 15mA. This may make it feasible to run the IC1 without having any cooling aid. The LED two will light up every single time the PSU is switched on.
There is an adjustable current limiter in parallel for the fixed output, thus supplying adjustable present source for scaled-down currents. This schema is incredibly straightforward also. You might discover that there is no current sensing resistor. But it is definitely there, inside a type of the Rds-on resistance of the N-channel FET, which basically handles the load cutoff from the supply. The perform of the FET is proven in the diagram 2. When the existing Id is rising, the pressure Uds over the resistance Rds rises especially gradually in the starting, but especially quick immediately after the knick. This means, that just before the knick the FET behaves as being a resistor but right after it, works as continuous existing source. The D2, R3 and B-E connection of the Q4 will feeling the Uds voltage with the FET1. When the voltage rises sufficient, the Q4 will shortcut the FET1 gate to mass, and reduce the current circulation via the FET 1 off.
Nonetheless, to allow the FET1 to open, there is particular gate voltage required, which with this case is brought up through the voltage divider consisting of R8, Z1, P1 and R9. So the optimum Gate voltage will probably be the one with the Z1, along with the minimal is going to be around 3V6. The Z1 voltage (Uz1) will therefore establish the max current flowing via the FET one. The diagram two will indicate that for 5 Amps the Uz1 really should be 5V6, and for 20Amps close to 9V6.The Capacitor C4 will determine the “velocity” or the reaction time with the limiter. a hundred uF will make the response time to be about 100ms, and 1n will allow it to be 1us. Inside the created limits, the P1 will restrict the existing output within the array of 15mA to 20A.
You possibly can use each output simultaneously, but the complete output current will be restricted by the value of the RL. This PSU can be built also for greater outputs, as long as the transformer will deal with the present needs, and also you provide adequate cooling for your Q3.
Subscribe to:
Posts (Atom)