In this post I have explained the main electrical features, pinout specifications, datasheet, and application circuit of the IC 723.
The IC 723 is a general purpose, extremely versatile voltage regulator IC, which can be used for making various types of regulated power supplies such as:
- Positive Voltage Regulator
- Negative Voltage Regulator
- Switching Regulator
- Foldback Current Limiter
Main Features
- The minimum voltage that can be achieved from IC 723 Regulator Circuit is 2 V, and the maximum is around 37 V.
- The peak voltage that can handled by the IC is 50 V in pulsed form, and 40 V is the maximum continuous voltage limit.
- The maximum output current from this IC is 150 mA which can be upgraded to as high as10 amps through an external series pass transistor integration.
- The maximum tolerable dissipation of this IC 500 mW, therefore it should be mounted on a suitable heatsink in order to allow optimal performance from the device.
- Being a linear regulator, the IC 723 needs an input supply that should be at least 3 V higher than the desired output voltage, and the maximum difference between the input and the output voltage should never be allowed to exceed 37 V.
ABSOLUTE MAXIMUM RATINGS
- Pulse Voltage from V+ to V- (50 ms) = 50V
- Continuous Voltage from V+ to V- = 40V
- Input-Output Voltage Differential = 40V
- Maximum Amplifier Input Voltage (Either Input) = 8.5V
- Maximum Amplifier Input Voltage (Differential) = 5V
- Current from Vz 25 mA Current from VREF = 15 mA
- Internal Power Dissipation Metal Can = 800 mW
- CDIP = 900 mW
- PDIP = 660 mW
- Operating Temperature Range LM723 = -55°C to +150°C
- Storage Temperature Range Metal Can = -65°C to +150°C P DI P -55°C to +150°C
- Lead Temperature (Soldering, 4 sec. max.) Hermetic Package = 300°C Plastic
- Package 260°C ESD Tolerance = 1200V (Human body model, 1.5 k0 in series with 100 pF)
Block Diagram
Referring to the above block diagram of the internal circuitry of the IC 723, we can see that the device is internally configured with a highly stable reference voltage at 7 V, created through advanced circuitry using op amp, buffer amplifier, and transistor current limiting stages.
We can also visualize that instead of creating a feedback stabilization by directly connecting the inverting input pin of the op amp with the output pinout of the IC, the inverting pin is rather terminated with a separate individual pinout of the IC.
This inverting pin facilitates integration with the center pin of an external potentiometer, while the other outer pins of the pot is linked with output pinout of the device and ground respectively.
How Potentiometer Adjusts the Output Voltage
The potentiometer can then be used for accurately setting or adjusting the internal reference level of the IC 723, and therefore a stabilized output from the IC in the following manner:
- Gradually lowering the slider center arm of the pot towards ground interacts with the inverting pin of the opamp to raise the output voltage
- If the slider of the potentiometer is lowered down its track, instead of causing a stabilization of the output at a potential identical to the reference voltage, the feedback regulates the inverting input of the op amp at the potential developed by the potentiometer.
- Due to a decreased potential across the potentiometer pins, the output is prompted to increase to a greater potential so that it allows the inverting input to adjust at the correct suitable voltage level.
- If the pot center wiper arm is moved further down, causes a proportionately higher voltage drop, which prompts the output to climb even higher, causing the output voltage from the IC to become higher.
- To understand the working better, let's imagine, the center wiper of the pot is moved 2/3rd section in the lower direction. This may cause a feedback voltage to the inverting pin of the internal op amp to be just 1/3rd of the output voltage.
- This enables the output to become stabilized and constant at a potential that's 3 times higher than the reference voltage and allows an appropriate voltage level to be established on the inverting input of the internal op amp.
- Therefore this feedback control through a potentiometer facilitates the user to get the intended adjustable output voltage, along with a very high and efficient level of output stabilization.
Calculating the Output Voltage using Formula
In case the output needs to be a fixed constant stabilized voltage, the pot could be replaced with a potential divider network using R1 and R2 resistors as shown below:
The formula 7 (R1 + R2)/R2 volts determines the desired constant output voltages, where the resistor R1 is connected between the output and the inverting input of the operational amplifier, while the resistor R2 is wired between the inverting input and the negative supply line of the device.
This implies that the reference voltage is directly associated with the non-inverting input of the IC 723 internal op amp.
The number 7 in the formula indicates the reference value, and also minimum output voltage the IC can deliver. For getting fixed output voltages lower than 7 V, this number in the formula could be replaced by the desired minimum voltage value.
However, this minimum output voltage value for IC 723 cannot be less than 2 V, therefore the formula for fixing 2 V at the output will be: 2 (R1 + R2)/R2
Understanding Current Limit Feature in IC 723
The IC 723 enables the user to get a precisely adjustable current control at the output depending on the load requirement.
An array of discretely calculated resistors are employed for sensing and limiting current to the desired levels.
The formula for calculating the current limiting resistor is simple, and as given below:
Rsc = 0.66 / Maximum Current
IC 723 Application Circuit
The above application circuit using IC 723 demonstrates a practical example of a useful bench power supply which can deliver an output voltage range from 3.5 V to 20 volts, and a optimum output current of 1.5 amps. 3 step switchable ranges of current limiting, accessible through 15 mA., 150 mA., and 1.5A current ranges (approximately).
How it Works
The mains AC input supply is stepped down by the transformer T1 to 20 volts with a maximum current of 2 amps. A full wave rectifier built using by D1 to D4, and a filter capacitor C1 converts the 20 V RMS AC to 28 V DC .
As discussed earlier, to be able to achieve the minimum 3.5 volts range at the output it is necessary to associate the reference source of the IC at pin 6 with the non-inverting pin 5 of the IC through a calculated potential divider stage.
This is implemented through the network created by R1 and R2 which are selected with identical values. Due to the identical values of the R1/R2 divider, the 7 V reference at pin 6 gets divided by 2 to produce a minimum effective output range of 3.5 volts.
The positive supply line from the bridge rectifier is attached to the pin 12, Vcc of the IC, and also with the pin12 buffer amplifier input of the ICI through fuse FS1.
Since the power handling specification of the IC alone is quite low, it is not suitable for making a bench power supply directly. Due to this reason output terminal pin10 of the IC 723 is upgraded with an external emitter follower transistor Tr1.
This allows the IC output to get upgraded to much higher current depending on the rating of the transistor. However, to ensure this high current is now controlled as per the needs of the output load specs, it is passed through a selectable current limiter stage having 3 switchable current sensing resistors.
ME1 is actually a mV meter which is used like an ammeter. It measures the voltage drop across the current sensing resistors and translates it to the amount of current drawn by the load. R4 can be used for calibrating the full scale range in the order of 20 mA., 200 mA., and 2A, as determined by the limiting R5, R6, R7 resistors .
This allows a more accurate and efficient reading of the current compared to having a single full scale range of 0 to 2A.
VR1 and R3 is used for achieving the desired output voltage, which could be continuously varied from aapproximately 3.5 volts to 23 volts.
It is advised to use 1% resistors for R1, R2 and R3 to ensure higher accuracy of the output regulation with minimum errors and deviations.
C2 works like a compensation capacitor for the in-built compensation op amp stage of the IC, for complementing enhanced stability to the output.
ME2 is configured like a voltmeter for reading the output volts. The associated resistor R8 is used for fine tuning and setting the full scale voltage range of the meter to about 25 volts. A 100 micro amp meter works great for this through a calibration of one division per volt.
Parts List
Resistors
R1 = 2.7k 1/4 watt 2% or better
R2 = 2.7k 1/4 watt 2% or better
R3 lk 1/4 watt 2% or better
R4 = 10k 0.25 watt preset
R5 = 0.47 ohms 2 watt 5%
R6 = 4.7 ohms 1/4 watt 5%
R7 = 47 ohms 1/4 watt 5%
R8 = 470k 0.25 watt preset
VR1 = 4.7k or 5k lin. carbon
Capacitors
C1 = 4700 AF 50V
C2 = 120 pF ceramic Disc
Semiconductors
IC1 = 723C (14 pin DIL)
Tr1 = TIP33A
D1 to D4 = 1N5402 (4 off)
Transformer
T1 Standard mains primary, 20 volt 2 amp secondary
Switches
S1 = D.P.S.T. rotary mains or toggle type
S2 = 3 way single pole rotary type capable of switching
FS1 = 1.5A 20mm quick blow type
Lamp
Neon Lamp indicator neon having integral series resistor
for use on 240V mains
Meters
MEI, ME2 100 µA. moving coil panel meters (2 off)
Miscellaneous
Cabinet, output sockets, veroboard, mains cord, wire, 20mm
chassis mounting fuseholder, solder etc.
Automatic Ambient Light Illumination Adjust
This circuit will automatically adjust the illumination of an incandescent lamp with respect to the available ambient or reference light conditions. This can be ideal for instrument panel lights, bedroom clock lighting and related purposes.
The circuit was created for 6-24 V lamps; the overall current should never go beyond 1 amp. The ambient light adjuster works as explained in the following points.
LDR 1 scans and detects the ambient light. LDR 2 is connected optically to an incandescent lamp. The circuit tries to balance as soon as the two LDR 1 and LDR 2 detect the identical level of illumination.
The circuit should, nevertheless, induce the external lamp(s) to be higher in brightness than the intensity of the ambient light. Due to this specific reason L1 needs to have rated with lower current than L2, L3 etc; or, if this is not followed, a small screen (small page of paper) could be positioned between the lamp (L1) and the LDR inside the opto-coupler.
The 0.68 ohm resistor restricts the lamp current; the 1 nF capacitor inhibits the circuit from going into oscillating mode. The circuit should be powered by a minimum of 8.5 volts; lower voltages that this might affect the operation of the IC LM723.
We advise to employ a supply that is higher than the lamp voltage specs by at least 3 volts. The zener (Z1) is selected to complement the lamp voltage; for 6 V lamps the in-built zener of the IC can be exploited by connecting terminal 9 of the IC to ground.
Reducing Dissipation in IC 723 Power Supply Circuit
The IC 723 is a quite commonly used IC regulator. For this reason the below circuit, that is designed to minimize power dissipation while the chip is applied through an external transistor, should be really popular.
Based on the company datasheets the supply voltage to the IC 723 must strictly be a minimum of 8.5 V to guarantee proper functioning of the chip's in-built 7.5 V reference and also the IC's internal differential amplifier.
While using chip 723 in a low-voltage high-current mode, through an outer series transistor working through the existing power supply lines used by the IC 723, usually leads to abnormal heat dissipation on the series external transistor.
As an illustration, in a 5 V, 2 A supply for TTL approximately 3.5 V could well be dropped over the external transistor and a staggering 7 watt power would be wasted through heat at full load current conditions.
Additionally, the filter capacitor has to be bigger than required to stop the 723 voltage supply from dropping under 8.5 V within the ripple troughs. Actually the supply voltage to the external transistor is required to be hardly 0.5 V higher than regulated output voltage, to enable its saturation.
The answer is to make use of another 8.5 V supply for your device 723 and a lower voltage supply to the external transistor. Instead of working with individual transformer windings for the a pair of supplies, the supply source to the IC 723 is basically extracted through a peak rectifier network comprising of D1/C1.
Due to the fact the 723 requires just a tiny current C1 can quickly charge to essentially the peak voltage through the bridge rectifier, 1.414X the transformer RMS voltage minus the drop in voltage across the bridge rectifier.
The transformer voltage specification as a result has to be a minimum of 7 V to allow an 8.5 V source to the IC 723. On the other hand, through appropriate selection of the filter capacitor C2 the ripple around the mains unregulated supply could be implemented in a way that the voltage drops to around 0.5 V higher than the regulated output voltage within the ripple troughs.
The average voltage given to the external pass transistor may consequently be lower than 8.5 V and the heat dissipation shall be tremendously minimized.
The C1 value is dependent upon the highest base current that this 723 has to source to the series output transistor. As a general guideline allow around 10 uF per mA. The base current could be determined by dividing the highest output current by the transistor gain or the hFE. An appropriate number for the mains filter capacitor C2 may be between 1500 uF and 2200 uF per amp of output current.
Another Cool 723 Power Supply Design
The following great IC 723 based power design was submitted by Mr. Roya, who is one of the avid readers of this blog. The details are furnished below, as compiled by Mr. Roya
For the past 30 years, I've been using a very nice professional adjustable 0 to 35V, 4A power supply. Three potentiometers and an LM723 IC are included in the circuit.
The IC and two of the potentiometers burned out a few days ago. I wanted to design its circuit while fixing it. Now I'd want to send you an email with the circuit I created, so you may post it on your website and share it with your visitors.
As you are probably aware, if the output voltage is adjusted to a significant proportion, such as 30 volts, we may draw nearly 4 amps of current from the power supply without any problems.
Unfortunately, when we wish to draw 3 or 4 amps from the device when the output voltage is adjusted to low voltages like 2 or 4 volts, a serious scenario emerges. Transistors are required to convert more than 100 watts of electrical power "within them" into heat in this condition, a power that would be deadly to power-transistors and end up causing them to smoke out.
Not a good example for using the 723. No current limit circuit, no overvoltage detection, no reverse voltage protection. A better option is to use the LM338 – higher current capability, better circuit protection, and much simpler.
723 has a current limit, overload, over voltage protections. But LM338 is much simpler and has better in-built protections.
ben!!! before anything happens ,LM338 packs off during a short circuit….very difficult to choose a genuine IC…national semi symbol will be there…even though Texas instruments took it… i think LM 723 is a better choice with current limit option…swagatam sir, have you made a negative reg using LM723 for -12V,4A with current limit? Thanks
Hi Ramana,
Currently I do not have a negative 723 power supply circuit, if I get one will surely update it in the above article.
Design a power supply with following specifications. Input: 230V,50Hz supply. Output: 45V(+/-1V),1A
Hello Sir Swagatam
I am going to assemble the circuit ( Another cool 723 power s… ). I have a 3A transformer. Please tell me does it matter if I add a 3A fuse at 220V side or at the 15.0.15V side of it. I would be very thankful to have your reply.
sincerely yours
Veronika
Thank you Veronika,
You will have to divide the wattage of the load with the DC output voltage of the power supply or the 220V, this will provide you the values of the fuses which can be used at the DC side or at the 220V side.
Hello Sir Swagatam
Thank you for your so soon response. please assure me if I am right
DC: W = 30v x 3A >> W = 90 watt
AC: 90 = 220 x I >> I = 400 mA
I should therefore use a 500mA fuse at the 220V side for my 30V, 3A power supply
Wish you all the best
sincerely yours
Veronika
That’s perfectly correct Veronika!
Hello dear Swagatam
Thank you so much Sir. With best wishes for you who are so kind and a good man.
Sincerely yours
Veronika
Thank you Veronika, I am glad to help!
Sir engineer Swagatam
Hello. good evening dear Sir
Two weeks ago, I assembled your newest circuit under title “Another Cool 723 Power Supply Design”. The circuit is very good, exact, and practical. It was working well and everything was right until yesterday that I faced a problem after connecting a 12V DC fan to the power supply. At first, whenever I rotated the current and voltage potentiometers, the speed of the fan obeyed these two components. I do not know what happened after a while that these two components no longer increased the voltage and current more than 14v and 900mA (voltage varies from 1v to 14V yet ). Regarding the fact that I have physically prevented the current potentiometer from decreasing the current not less than 30 percents of it’s rotating angle, would you please do a favor and tell me which components have been damaged and need to be replaced?
Thank you so much for your kind reply, I remain
Best wishes
Mynah
Hello Mynah,
Thank you for building this circuit and glad it worked for you. There are probably only two things that can get damaged in the circuit and those are the transistors or the IC itself.
Try changing the transistors, and if you still find the problem not solved then you may have to replace the IC. I don’t think any of those resistors will burn at 12V, but if possible please check the resistors also. It is difficult to pin point a specific part for troubleshooting. According to me most probably it could be the IC itself.
Sir Engineer Swagatm
Hello. I hope you are healthy. First of all please accept my excuse for replying so late to your letter. I acted as you had told. After replacing LM723 and BD138 which did not affect the working of the power supply, I accidentally noticed that the 3A fuse of the circuit is burned. After replacing it, circuit began working normally but new instances occurred which I was not able to troubleshoot it and you would certainly find the response.
I should tell you that I have mounted the two transistors of the circuit on a graphic card’s heat sink and have mounted a small 12v fan on it for cooling . Feeding of the fan is supplied by IC7809 which is fed through 15V AC of the 15.0.15 transformer. moreover, I have added a digital V.A.meter to the circuit.
For more investigation, I drew the wiring of the circuit that I had done, on a paper and sent it on your Email.
Instances:
1. Maximum output voltage is now 30V while it was 35v during the two weeks ago.
2- I took the fuse out of the circuit (please see the attached photo of wiring) and noticed that the utmost voltage of it was 30v again !!! which astonished me a lot.
3. I disconnected both 15V wires of the transformer from C and D terminals of the SW, soldered those to A and B terminals while simultaneously disconnecting the wires from A and B and soldered them to C and D terminals. I then disconnected the four wires ( I did not took apart the one that consists of 2 wires from each other) that go to the 4 terminals of SW and soldered those again in various positions. I connected my mini 12V drill which pulls about 1A and the strange thing is that the circuit works strongly with and without Fuse and the utmost voltage is 30V. I assure you that when I take off the fuse there is no connection between fuse terminals.
With best wishes
He who never forgets your kindness
Mynah
Thanks Mynah,
I saw your email diagrams they looked OK to me, all connections are correct.
That is strange! Did you check the voltage at the output of the bridge rectifier? Is it 35V? Remember the transistors will drop around 1.5 V, and the pot settings may also drop a few voltage, due to inefficiencies. If the output from the bridge is 35V then the circuit parts may be dropping 5V on its way. If the output of the bridge is showing 30 V then you will have to check the voltage from the transformer wires and calculate the peak value by multiplying the transformer voltage with 1.41 to confirm the bridge rectifier peak output.
Sir engineer Swagatam
Hello. Thank you very much for your so valuable tips. Here are the result of what you had ordered me to be checked:
1- Input AC = 210V >> output voltage of transformer across 2200uf capacitor = 40V
2- Input AC = 210V >> voltage of transformer wires = 30V
3- Input AC = 210V >> output voltage of bridge ( without 2200uf Capacitor connected) = 28V.
And 28 x 1.41 = 39.48 while the output of the power supply is still 30.5 volts ; Should we conclude that 1N5402 diodes are defective?
Dear Sir Swagatam my second question is that to Please do a favor, take another look at my diagram and tell me what is the reason of not ceasing the output of the power supply while I remove the fuse out of the circuit?! in other words, when I remove the fuse, the power supply is still working well and I have the output voltage and current
Wish you all the best
Mynah
Thank you Mynah,
You must connect the 2200uF also to check the peak voltage, without it the bridge output will continue to show the RMS value.
I checked your diagram again,and found no issues. The 7809 IC bridge rectifier is getting 0-15V from the transformer, while the 5402 bridge is getting getting the 15-0-15V from the transformer…so I am also confused how the current is passing even after you remove the fuse??
You can remove the switch, and connect the transformer wires directly across the 1N4007 bridge and the 1N5402 bridge and see if the problem still persists.
Or use two separate SPDT switches for the two connections
Dear Sir Swagatam
Hello
I did all you dear had instructed: RMS voltage is 40V with 2200uf capacitor connected. I replaced the bridge and capacitor with new ones but no change occurred. I also removed SW and connected wires of transformer directly to bridge, then removed the fuse but the current still passes and and makes me astonished. I don’t know what else I should do.
Thank you so much for you endeavors and kindness.
Wish you health and joy
Mynah
Dear Mynah,
Please check the voltage at the collector/ground of the T2, and then check the voltage at the emitter/ground of the T2, if the collector/ground shows 40V and the emitter/ground shows 30V then its the transistors that are dropping 10V, we cannot do anything about it.
Did you connect the bridge rectifier in the following manner? I cannot see anything wrong with the diagram. If you cut any one input wires to the bridge then that particular bridge will not get the supply that is natural.
Dear Sir Swagatam
Thank you a lot for your kind response. I think how a man on the earth could be so good who spends his precious time with no expectation to help others. It is hard to imagine but I see that it is happening now. I wish you and your dears; specially your family all the best and continuous health.
Well, these are the results of what I experienced
1- Where the input voltage is 210V, the collector/ground voltage is 41v and emitter/ground voltage is only 23v. Does it mean that 2N3055 should be replaced?
2- Yes I have exactly wired as you have drawn, meaning the above bridge in your diagram goes to feed 7809 IC ( and 12v fan ) and V.A. meter; and the other one has been used to feed the circuit. I will send a few related pictures on your Email.
3. I added a 500mA non fixed fuse before input of the transformer for security of the circuit. It means that I am completely disappointed with the functioning of the 3A fuse.
It is so hard to thank you but I know no alternative word for it. So I say I thank you and admire you.
Sincerely yours
Mynah
Thank you Mynah, No problem at all, I am always glad to help!
1) you can also measure the base/ground voltage of 2N3055, if it is 1V higher than its emitter/ground voltage then the transistor is good, the problem could be happening from the IC or due to the specific design of the circuit. We cannot do much about it since troubleshooting the whole working of the circuit can be quite difficult.
2) In the transformer bridge wiring you can see that if we cut any one of the green wires then it affects the specific bridge rectifiers and current supply is completely cut to that specific bridge. for example if you cut the green wire which is connected to the lower bridge then the lower bridge cannot get any current from the transformer. So adding or removing the fuse should definitely cause the specific bridge to switch ON/OFf.
Yes I have seen the images, they look very good, but it can be difficult to understand the wiring details from the images.
Dear Sir Swagatam
Thank you a lot for all your endeavor and help.
The day after I installed the digital voltmeter/ammeter, I noticed that whenever I turned the voltage potentiometer from maximum to minimum it took about a minute for the display to count down to zero volts.
Unfortunately, the day I sent you the wiring diagram and the tests that I did about the fuse, I had forgotten this point, which confused both of us; I am so sorry and apologize.
So now, when the potentiometer is on 30 volts, after removing the fuse, the display voltage drops to about 14 volts, but if I turn the potentiometer to zero volts, the voltage display becomes zero.
In the video that I will send on your Email, you will see that when I remove the fuse , the current does not pass in the upper bridge ( your diagram ), but the voltage remains constant at about 14 volts.
The base/ground voltage of 2N3055 is 2.4 volts and the base/emitter voltage of it is 2.8 volts.
Thank you again and wish you health and joy
Truly
Mynah
Thank you Mynah,
I saw the video it looks very strange.
I think the problem could be with the meter itself. If you remove the fuse and the if the current cannot entering the bridge then how will the output get any voltage??
Once the fuse is removed there should be absolutely 0V across all the points of the circuit.
As you can see once the fuse is removed only one wire from the transformer is connected to the bridge. With only one wire the circuit cannot complete. If the circuit cannot complete the output will have zero voltage.
You can do one thing. Remove and isolate the fan bridge rectifier completely from the transformer, and after this remove the fuse and check the circuit output. Check the output with analogue meter, not with the digital meter. This will prove the actual situation.
Once the fuse is removed there should no voltage output across any point of the circuit not even across the transistor base or emitter.
Dear Sir Swagatam
Hello. Hope you are fine. Here are the results of new tests
1. When I disconnected the zero wire (blue one on your diagram) of the transformer from the upper bridge, the value of voltage on digital Voltmeter/Ammeter remained 30 volts. then; when I removed the fuse, the current passing through the whole circuit stopped at once and the digital V/A went out too. I soldered this wire again and did the 2nd test:
2. When I disconnected the 15v wire of the transformer (green one) from the upper bridge, nothing happened but when I removed the fuse, the voltage began counting down while simultaneously the digits of V/A began blinking and the voltage stopped reducing at about 14 volts. I hope the blinking is visible on the video that I will send on your Email.
3. I substituted the 2N3055 with two new ones but the maximum voltage of the power supply was still 30V.
Thanks a lot for you precious help dear Swagatam
Wish you health and joy
Truly
Mynah
Dear Mynah,
If you remove any one of the input wires from the bridge then circuit through the bridge cannot complete so the output of the bridge should become zero volts. I am not how and why the meter is behaving in this way?
I truly wish I could solve your problem,
But without checking your circuit practically it is very difficult for me to troubleshoot the fault.
If you want I can you can give you a fully transistorized power supply circuit without any IC and that should give you around 38V output
Dear Sir Swagatam
Hello. Hope you are fine.I thank you so much for your kind response.
1. The reason ” why and how the meter …. “, may be because it is made by an unauthorized Chinese Co. I can add another SPDT switch for connecting and disconnecting the zero wire of the transformer and fasten the levers/handles of both switches to each other in a way that work simultaneously. Thus the problem would be solved.
2. Thank you very much for your favor; also your propose to give a power supply but I am not living in India and it is not possible for me to pay the price. please remember that It is my utmost desire that I could see you a day face to face, and press the warm hands of a man who is so good and kind.
I will never forget your endeavor and times you spent to help me.
wish you health and joy
Mynah
No problem Mynah,
I appreciate your dedicated efforts in the field of electronics.
Yes you can add another switch to completely disable the power from entering the output and the meter.
Surely, with the grace of God we can meet someday face to face. I truly appreciate your kind thoughts.
Dear Sir Swagatam
Your kind words make me proud and I am very happy and grateful to God for creating such a good and generous man. Be sure that there are so very many people like me who love you sincerely though many of them do not write that to you.
I will never forget your kindness
Best wishes
Mynah
Thank you so much Mynah,
I truly appreciate your kind thoughts. Please keep up the good work!
Why don’t you guys leave schematics that we can download and play with in circuit simulators? Change our components and test the results?
It’s silly to have to manually recreate these when you clearly drew them only to save as an image file and then slap a watermark on it..