hy,
Can anyone tell me what is the max current that can be aplied to a 7805 regulator.
I have a load of about 1.4A and i don't know which are the losses in terms of current starting from transformer itself to input pin of 7805. If i have a transformer of let's say 0.5A i get also 0.5A from output of 7805?
Is it capable to deliver 1.5 amps into a load as datasheet say?
of course with a good heatsink
Is it ok if I choose a transformer with a single secondary rated at 6V AC 18VA @ 3A? This will be i think 8.4v after rectify.
I believe I can use a higher current transformer like the one above according to output of 7805 which is 1.5A max? or transformer current can be also exactly 1.5A so 7805 can deliver max current?
Thanks,
Regards
Can anyone tell me what is the max current that can be aplied to a 7805 regulator.
I have a load of about 1.4A and i don't know which are the losses in terms of current starting from transformer itself to input pin of 7805. If i have a transformer of let's say 0.5A i get also 0.5A from output of 7805?
Is it capable to deliver 1.5 amps into a load as datasheet say?
of course with a good heatsink
Is it ok if I choose a transformer with a single secondary rated at 6V AC 18VA @ 3A? This will be i think 8.4v after rectify.
I believe I can use a higher current transformer like the one above according to output of 7805 which is 1.5A max? or transformer current can be also exactly 1.5A so 7805 can deliver max current?
Thanks,
Regards
The LM7805 is a linear regulator with output voltage regulation. It will supply to your load any current nescessary to maintain +5V on its output. If you exceed the 1.5A current limit, or the if the device overheats, it will reduce the output current and the output voltage will drop accordingly.
Being a linear regulator, the input current will be as high as the output current plus a little bit of bias (but that is microamps). The voltage difference between input and output (3.4V in your transformer example) will be dissipated as heat. Under full load that will give you 3.4 * 1.5 = 5.1W.
Using a higher current transformer will not increase the power output of the LM7805. If you need more output power, use an other device or place a pass transistor in parallel. Refer to the LM7805 typical applications mentioned in the datasheets for examples.
Kind Regards,
Bakmeel
Being a linear regulator, the input current will be as high as the output current plus a little bit of bias (but that is microamps). The voltage difference between input and output (3.4V in your transformer example) will be dissipated as heat. Under full load that will give you 3.4 * 1.5 = 5.1W.
Using a higher current transformer will not increase the power output of the LM7805. If you need more output power, use an other device or place a pass transistor in parallel. Refer to the LM7805 typical applications mentioned in the datasheets for examples.
Kind Regards,
Bakmeel
what he said, plus
there is the drop across the rectifier, anwhere from 0.5V to 1V depending on your selection and full wave or half wave rectifier topology.
6VAC is a reasonable selection for 5V DC output. Make sure to use a cap with a 1.5A or better ripple current rating if you truly need 1.5A output.
there is the drop across the rectifier, anwhere from 0.5V to 1V depending on your selection and full wave or half wave rectifier topology.
6VAC is a reasonable selection for 5V DC output. Make sure to use a cap with a 1.5A or better ripple current rating if you truly need 1.5A output.
Hi,
do you need a continuous output current of 1.5Adc?
If so, then use a 6Aac transformer to allow it to run cool.
6Vac for 5Vdc might be a little bit low if mains voltage drops significantly. However if you draw smaller currents (<=500mA) you should have no problem with 6Vac.
7.5Vac might be safer for a wider range of mains voltages.
9Vac might overheat the regulator.
do you need a continuous output current of 1.5Adc?
If so, then use a 6Aac transformer to allow it to run cool.
6Vac for 5Vdc might be a little bit low if mains voltage drops significantly. However if you draw smaller currents (<=500mA) you should have no problem with 6Vac.
7.5Vac might be safer for a wider range of mains voltages.
9Vac might overheat the regulator.
No, the circuit will not draw continuously 1.5A, but 1A surely will.
So, if my trafo has 3A, it will not affect the regulator because this can out max 1.5A, right? The regulator will draw max from the trafo only 1.5A if my load require, but also the heat dissipated will be higher
What type of capacitor value i need for this current output, how to calculate?
I guess it's not the same if the load draws a 500mA with a 1000uF 16V capacitor or it draws 1.5A with same capacitor.
Regards,
So, if my trafo has 3A, it will not affect the regulator because this can out max 1.5A, right? The regulator will draw max from the trafo only 1.5A if my load require, but also the heat dissipated will be higher
What type of capacitor value i need for this current output, how to calculate?
I guess it's not the same if the load draws a 500mA with a 1000uF 16V capacitor or it draws 1.5A with same capacitor.
Regards,
Hi,
if you need 1Adc continuous, then use a >=4Aac transformer (7.5Vac 30VA).
1000uF smoothing is very low for a 1A supply.
If the supply is 100mA continuous, then 1000uF is suitable. This will support short term 1Apk transients.
if you need 1Adc continuous, then use a >=4Aac transformer (7.5Vac 30VA).
1000uF smoothing is very low for a 1A supply.
If the supply is 100mA continuous, then 1000uF is suitable. This will support short term 1Apk transients.
So, if my circuit draws 1A or more i need a 6V AC at 4A minimum trafo?
Why trafo can't be 1.5A AC or 3A AC output
How to calculate the needed capacitor
Why trafo can't be 1.5A AC or 3A AC output
How to calculate the needed capacitor
A transformer feeding a capacitor input filter must be de-rated to 70% of rating.
The output voltage is ~=sqrt(2)*Vac.
Output power =Vdc*Idc=1.414*Vac*Idc<=0.7*VArating
Idc=VArating*0.7/1.414/Vac~= 0.5*Iac.
But if you run a transformer at full rating continuously it will run at the maximum temperature that the manufacturer has designed for.
Most recommend a maximum continuous current <50% of maximum.
This temperature limitation results in Idc<=Iac/4.
Calculate ripple voltage for the output current.
This must be added to the regulator drop out voltage to ensure continuous regulation during high current demand. small smoothing needs more transformer voltage and this gives hotter regulator chip.
The output voltage is ~=sqrt(2)*Vac.
Output power =Vdc*Idc=1.414*Vac*Idc<=0.7*VArating
Idc=VArating*0.7/1.414/Vac~= 0.5*Iac.
But if you run a transformer at full rating continuously it will run at the maximum temperature that the manufacturer has designed for.
Most recommend a maximum continuous current <50% of maximum.
This temperature limitation results in Idc<=Iac/4.
Calculate ripple voltage for the output current.
This must be added to the regulator drop out voltage to ensure continuous regulation during high current demand. small smoothing needs more transformer voltage and this gives hotter regulator chip.
Check your parts supplier for just about any 5V 3A low-dropout regulator - the choice depends more on what you can buy.
Play around with http://tangentsoft.net/elec/ps-est.html for values.
For more on ripple, see
http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/rectct.html#c4 (note they assume 60 Hz, you need to recalculate for 50 Hz or just add 20% to the cap value.)
Play around with http://tangentsoft.net/elec/ps-est.html for values.
For more on ripple, see
http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/rectct.html#c4 (note they assume 60 Hz, you need to recalculate for 50 Hz or just add 20% to the cap value.)
AndrewT said:
Recommending a 100% margin as a general rule oversimplifies things a bit, I should say - it's a conservative option to be sure, but at the current copper prices you may learn to appreciate the utility of ventilation, forced or otherwise ;-)
You always need to check the specs in question. A cheap small transformer may well run hotter than desirable at 100% specified load, but a well-designed larger one will easily handle its rated current - otherwise it would have been specified for part-time operation only!
Of course it's always a good idea to remember a transformer generates heat which needs to be removed somehow (case designers take note).
have you ever ran a larger mains transformer at full output to find out how hot it gets?wine&dine said:
Do you want YOUR transformer to run that hot?
Do you want other nearby components getting hot due to transformer dissipation?
I have looked to 78T05 rated at 3A
I think a 9VAC 3A power supply, 4x1N4004, first cap rated at 4700uF 16V and a ceramic or MKT 100nF at the input
At the output of the regulator i will insert an elko 100uF 16V cap and also a 100nF cap MKT or ceramic
To this I will add resistor and one indicator LED
My question is: At high loads meaning 2.2A or 2.5A, voltage remains the same at 5V?
I saw on some sites that voltage is dropping in loads near 3A
What is the difference between this linear supply and a switching one with a LM2576 besides the heat generated by linear device
It is more stable? I need to power a uC circuit with lots of LEDS over 400
I think a 9VAC 3A power supply, 4x1N4004, first cap rated at 4700uF 16V and a ceramic or MKT 100nF at the input
At the output of the regulator i will insert an elko 100uF 16V cap and also a 100nF cap MKT or ceramic
To this I will add resistor and one indicator LED
My question is: At high loads meaning 2.2A or 2.5A, voltage remains the same at 5V?
I saw on some sites that voltage is dropping in loads near 3A
What is the difference between this linear supply and a switching one with a LM2576 besides the heat generated by linear device
It is more stable? I need to power a uC circuit with lots of LEDS over 400
umm, what about global warming?
I feel you should stick with 6VAC. Moving to 9VAC 3A power supply will put out 12.6V peak at full load. At lighter loads you may see a much higher voltage, likely 20%. In your case running 1A DC would give on the order of 10%, rough guess. SO actual peak voltage is 13.8V.
Assume a full wave bridge with overkilled diodes, typical of a DIY application. THis gives 2 x 0.6V, so we are back around 12.6V average on the filter cap.
So, the little 'ole LM78X05 is outputing 1A, correct? If so, then it is dissapating (12.6V - 5V) * 1A = 7.6W. THis is much more than a typical 1" or 2" extruded aluminum heatsink can handle.
I really think a 6VAC 2A power supply is your best bet. Then spend the money you save on sexy Schottky rectifiers, and extra filter capacitance. Also, an LM1086 gives much better dropout performance. Its 1.5A current limit is a feature, as it protects your ciruits.
I feel you should stick with 6VAC. Moving to 9VAC 3A power supply will put out 12.6V peak at full load. At lighter loads you may see a much higher voltage, likely 20%. In your case running 1A DC would give on the order of 10%, rough guess. SO actual peak voltage is 13.8V.
Assume a full wave bridge with overkilled diodes, typical of a DIY application. THis gives 2 x 0.6V, so we are back around 12.6V average on the filter cap.
So, the little 'ole LM78X05 is outputing 1A, correct? If so, then it is dissapating (12.6V - 5V) * 1A = 7.6W. THis is much more than a typical 1" or 2" extruded aluminum heatsink can handle.
I really think a 6VAC 2A power supply is your best bet. Then spend the money you save on sexy Schottky rectifiers, and extra filter capacitance. Also, an LM1086 gives much better dropout performance. Its 1.5A current limit is a feature, as it protects your ciruits.
Yeah i know, i will use a 4A 100V bridge for this project and a 10000uF 16V elko cap for filtering
I was wondering what's the formula to calculate a fuse value for the primary assuming that trafo is 9V AC 3A
I know it needs to be slow blow cause of the caps
I was wondering what's the formula to calculate a fuse value for the primary assuming that trafo is 9V AC 3A
I know it needs to be slow blow cause of the caps
not the rated current, but the actual current
A regulator capable of 3A of current means "up to 3A" and will actually work from almost 0 to 3A. It has nothing to do with the actual current drawn.
Calculate you total load current, increase by 50% to 100%, use this as your regulator current limit. If this is 0.5A, a regulator rated for 3A provides no advantage other than increasing profit for IC manufacturer.
For example, if your circuit uses TL072 op amps, 8 (4x2) of them, the quiescent current these draw is 1.4mA each. Assume your circuit swings 10V at each of these 8 op amp outputs into 1k ohm load, then the peak current is 8 * 1.4mA + 8 * 10V/1k, or less than 100mA (92mA).
A simulation using most any op amp model should give a reasonable close estimate of total supply current.
A regulator capable of 3A of current means "up to 3A" and will actually work from almost 0 to 3A. It has nothing to do with the actual current drawn.
Calculate you total load current, increase by 50% to 100%, use this as your regulator current limit. If this is 0.5A, a regulator rated for 3A provides no advantage other than increasing profit for IC manufacturer.
For example, if your circuit uses TL072 op amps, 8 (4x2) of them, the quiescent current these draw is 1.4mA each. Assume your circuit swings 10V at each of these 8 op amp outputs into 1k ohm load, then the peak current is 8 * 1.4mA + 8 * 10V/1k, or less than 100mA (92mA).
A simulation using most any op amp model should give a reasonable close estimate of total supply current.
@thespeakerguy:
For this type of application, I prefer a "rude" power supply without voltage regolation, only a fuse, a bridge, a condenser, and series-connected LEDs to a current limiting resistor.
For very high current power supply, switching mode ICs are the solution.
adrianbodor said:
For this type of application, I prefer a "rude" power supply without voltage regolation, only a fuse, a bridge, a condenser, and series-connected LEDs to a current limiting resistor.
For very high current power supply, switching mode ICs are the solution.
- Status
- Not open for further replies.