Adding voltage/current display on 3 pin-power supply design

[giasis]

Hi,

I have a 30V-2A bench top power supply with schematic I got as a gift when I finish my intern, and it has positive, negative and ground pins. Since it doesn't have any display, I am planning to add a LED display to show voltage value (and hopefully current as well after this).

I am planning to use a voltage divider (using 9kohm and 1kohm) to drop the original voltage output by 1/10, and feed it to LM324 buffer. Then, I will use ADC to detect the output of LM324 and use MCU to display on LED.


But my problem is that, I am not sure where I can add these things in the schematic since the positive and negative pins are floating. Where should I use a reference for these additional components in the schematic?

I am thinking about adding the voltage divider between positive and negative pins, and send the 1/10 voltage output to LM324 before ADC, but then should LM324 and ADC need to have negative pins as a reference point?

Thanks for your help in advance!!

[ptricks]

Connect to the ground  on the supply, measure between the same points you would use to connect a load.
edit: Thought I should clarify when I said ground, I meant the one marked negative. If you want power for the new display take that from the same supply that is used for the LM324.

[Strada916]

 Might be easier to go to eBay and buy a volt meter already made. As it sounds like you are going to need lots of help on a micro and some 7 segment displays and then some code. 

[giasis]

Connect to the ground  on the supply, measure between the same points you would use to connect a load.
edit: Thought I should clarify when I said ground, I meant the one marked negative. If you want power for the new display take that from the same supply that is used for the LM324.

Thanks for your reply. So you are saying the reference for ADC and the additional LM324 is going to be the negative pin of the power supply. Then, how about Vdd for those components? Since Vdd needs to be fixed as around 5V compared to the voltage on the negative pin, where can I get this constant supply voltage on the schematic?  (Since I am not fully understanding the schematic, could you point where would be the proper point?) I initially thought about using the Vdd of U1:A, but then U1:A is referencing the ground pin of the power supply (not negative pin). If i want to power ADC and new LM324 properly, and if I want to use negative pin as their reference, I need 5V compared to that negative pins, but not sure where I can get it...

Might be easier to go to eBay and buy a volt meter already made. As it sounds like you are going to need lots of help on a micro and some 7 segment displays and then some code. 

Thanks for your reply. I have actually already tested MCU with ADC and 7segment LED in my breadboard. So I think I just need to know where I can connect these things in my power supply.

[croberts]

Use one ADC input to measure Positive with respect to ground and one ADC input to measure Negative with respect to ground and subtract.

[vlf3]

For the MCU overall supply; you need to provide a fixed voltage regulator like the 7805 but, it will need to have a series resistor value to the input of the regulator, to drop the un-regulated DC supply, that might be at 50 volts ?... set the regulator input voltage around 7 to 10 volts, taken from the positive point at C1/11 rail, and regulator ov centre pin to ground, negative of power supply.

Decoupling electrolytic cap's would be needed on the input and output of the 7805 regulator, to keep any noise off the 5 volt supply; values around 100Mfd should do the job.

I presume your ADC input is from an I/O MCU pin, that you have already designed; with a resistor T-network to protect the input, and setting the DC reference for the desired voltage read-out consistent with your load to voltage.

[giasis]

For the MCU overall supply; you need to provide a fixed voltage regulator like the 7805 but, it will need to have a series resistor value to the input of the regulator, to drop the un-regulated DC supply, that might be at 50 volts ?... set the regulator input voltage around 7 to 10 volts, taken from the positive point at C1/11 rail, and regulator ov centre pin to ground, negative of power supply.

Decoupling electrolytic cap's would be needed on the input and output of the 7805 regulator, to keep any noise off the 5 volt supply; values around 100Mfd should do the job.

I presume your ADC input is from an I/O MCU pin, that you have already designed; with a resistor T-network to protect the input, and setting the DC reference for the desired voltage read-out consistent with your load to voltage.

Thanks for your explanation.
To avoid confusion, let me make notation clearer first. Lets call

(+): The rail coming from the rectifier
(-): The rail connected to CT of transformer.
positive pin: Red output pin from Power supply
negative pin: Black output pin from Power supply
GND: Green output pin from power supply
Vdd: 5V output rail from 7805

Based on your advice, I draw 7805 and connect the supply to the voltage divider from (+) rail, and reference to the (-) rail (= CT of the transformer) as shown in the figure attached (See yellow highlighted parts) such that 7805 will output 5V (denote Vdd) relative to the (-) rail. My next concern is that how to measure the voltage difference between two power supply's output pins (which are floating from the ground) since my ADC will use (-) rail as a reference (which is different from one of output pin's voltage) and "Vdd" as a supply.

Can I use an Instrument Amplifier to measure the differential voltage between two floating outputs here as shown in the figure? So I connected Vdd and (-) rail to the supply line for Instrument Amplifier, ADC and MCU. And instrument amplifier is measuring the differential output voltage (set gain of the amp = 1).

Does this method look ok? Or is there other conventional way to measure the floating differential voltage in commercial power supplies?

Use one ADC input to measure Positive with respect to ground and one ADC input to measure Negative with respect to ground and subtract.

I think I found Instrument Amp to measure differential voltage as explained above. Thanks

[megajocke]

I don't think it's very likely that your ground/earth terminal on the power supply is going to be connected to that ground symbol in the schematic. Check it.

Most likely the circuit is completely unconnected to chassis ground which is likely connected to the mains plug earth and the earth terminal.

The positive and negative outputs are not really floating from the node marked ground in the schematic. The current sense resistor is between the negative output and the ground node in the circuit diagram which is the most negative voltage from the raw supply consisting of transformer, rectifier and filter caps.

The center tap connected rail is a positive supply powering the control circuit and the output for low output voltages. At high output voltages the rectifier positive output, which is the most positive supply in the circuit, provides output current. This is done to reduce power dissipation in the pass transistors.

Powering the measurement circuits from between the center tap (middle) and top rails does not seem like a good solution because you have to deal with large common-mode voltages both above and below your power rails.

Best is probably to power the measurement circuit from the ground node in the schematic (which should be unrelated to the ground/earth terminal on the front) and the middle rail, or use a separate transformer if the voltage there is excessive.

The negative output will only go at most 0.4 V above circuit ground at 2 A output, and is a measurement of current. Voltage would then be the difference between positive and negative output.

[ptricks]

The outputs of the supply are not floating.  Take a look at  R19, connecting the negative terminal to ground. R19 is .2 ohms to ground , the circuit is using that for current measurement as the current flows through that resistor and the LM324 part measures the difference.

[vlf3]

The 7805 has to go to Gnd the -Neg of supply, and a potential divider is not required for the 7805 input, just a single resistor in series with the 7805 input, with at least 1/2 Watt rating... you could take the input via the transformer CT however, just in case the PSU regulating circuit outputs auto shut-down, or other events, that might disturb your meter circuit, it would be best to take the positive DC from the rectifier (+) rail.

The positive Red terminal output from Power supply, is where you need to take the ADC input, with a potential divider or T-Network as stated before; I can't see why you need to go via an Instrument Amplifier to measure the differential voltage ! unless I have missed something ?

[giasis]

I don't think it's very likely that your ground/earth terminal on the power supply is going to be connected to that ground symbol in the schematic. Check it.

Most likely the circuit is completely unconnected to chassis ground which is likely connected to the mains plug earth and the earth terminal.

The positive and negative outputs are not really floating from the node marked ground in the schematic. The current sense resistor is between the negative output and the ground node in the circuit diagram which is the most negative voltage from the raw supply consisting of transformer, rectifier and filter caps.

The center tap connected rail is a positive supply powering the control circuit and the output for low output voltages. At high output voltages the rectifier positive output, which is the most positive supply in the circuit, provides output current. This is done to reduce power dissipation in the pass transistors.

Powering the measurement circuits from between the center tap (middle) and top rails does not seem like a good solution because you have to deal with large common-mode voltages both above and below your power rails.

Best is probably to power the measurement circuit from the ground node in the schematic (which should be unrelated to the ground/earth terminal on the front) and the middle rail, or use a separate transformer if the voltage there is excessive.

The negative output will only go at most 0.4 V above circuit ground at 2 A output, and is a measurement of current. Voltage would then be the difference between positive and negative output.

megajocke. you were right!! After I check, that earth ground pin of the power supply was not connected to any of the ground symbol in the schematic at all! I didn't realize it was just a single isolated pin... Now it makes much more sense.. And I checked the voltage at the (+) and (-) rail(CT) relative to the ground symbol (which is not earth ground), and they were 60V and 30V DC. And the negative pin of the supply has almost the same voltage as the ground symbol as you said since 0.2R is so small. Then it makes sense that I can just use (-) rail's 30V and ground symbol's 0V to supply the power for ADC and MCU~! Thanks for pin pointing the problem I was confused about.

By the way, you said  "use a separate transformer if the voltage there is excessive", but do you think it's okay to use just a voltage divider or 7805 rather than transformer on the same rail(between C8 and existing LM324) supplying power to the existing LM324 in order to drop the voltage for my ADC and MCU? (Since I am not really familiar with transformer.).

The 7805 has to go to Gnd the -Neg of supply, and a potential divider is not required for the 7805 input, just a single resistor in series with the 7805 input, with at least 1/2 Watt rating... you could take the input via the transformer CT however, just in case the PSU regulating circuit outputs auto shut-down, or other events, that might disturb your meter circuit, it would be best to take the positive DC from the rectifier (+) rail.

The positive Red terminal output from Power supply, is where you need to take the ADC input, with a potential divider or T-Network as stated before; I can't see why you need to go via an Instrument Amplifier to measure the differential voltage ! unless I have missed something ?

Thanks for helping me out. vlf3! Now I understand that I can just use the positive pin as an input of ADC (powered by (-) rail's 30V and GND's 0V) through the voltage divider as you said. By the way, you meant supply 7805 from (-) and GND symbol, right? or you meant (+) supply?

So let me write how I understood.  Since (-) rail has 30V relative to GND symbol (not earth ground), I can use 7805 to drop that 30V to around 5V to supply my ADC and MCU. Is that right?

The outputs of the supply are not floating.  Take a look at  R19, connecting the negative terminal to ground. R19 is .2 ohms to ground , the circuit is using that for current measurement as the current flows through that resistor and the LM324 part measures the difference.

Thanks ptricks. I think now I understand what's going on... So R19 is the current sensing resistor. And it turns out the earth ground pin of the power supply wasn't connected to any other GND symbol in the schematic although they have the same symbol.. I think this made me confused. Since if I want to generate negative voltage, I would connect that ground pin to the red positive pin, but then it's going to almost short between the red pin and negative pin....so it didn't make sense until I realized the ground pin is totally isolated from the others...

[vlf3]

@ giasis...

There is some confusion between -Neg and Gnd here... in reality their at the same reference point, the exception is the Negative output terminal, and control input circuit where R19 0.2 ohm to Gnd provide the feed-back for current sensing, however R19 is so small in low ohms, that it makes no difference to your DC 7805 regulator... what is required is the MCU supply needs to be at Gnd, with the 7805 regulator at the same common Gnd reference point.

I have re-considered the situation, so the DC supply should now come from across the two caps C2/C12; so the Gnd end of the caps are connected to 7805 regulator centre lead/pin, it's OV common connection... then, the positive (+) from the caps, through a series resistor to further drop the voltage, onto the regulator input lead/pin, to provide the 5 volt regulated DC supply, for the MCU.

[giasis]

@ giasis...

There is some confusion between -Neg and Gnd here... in reality their at the same reference point, the exception is the Negative output terminal, and control input circuit where R19 0.2 ohm to Gnd provide the feed-back for current sensing, however R19 is so small in low ohms, that it makes no difference to your DC 7805 regulator... what is required is the MCU supply needs to be at Gnd, with the 7805 regulator at the same common Gnd reference point.

I have re-considered the situation, so the DC supply should now come from across the two caps C2/C12; so the Gnd end of the caps are connected to 7805 regulator centre lead/pin, it's OV common connection... then, the positive (+) from the caps, through a series resistor to further drop the voltage, onto the regulator input lead/pin, to provide the 5 volt regulated DC supply, for the MCU.

Thanks vlf3!  I will try that method. I start drawing schematic for my pcb while testing it, and can't wait to see the result!

[megajocke]

By the way, you said  "use a separate transformer if the voltage there is excessive", but do you think it's okay to use just a voltage divider or 7805 rather than transformer on the same rail(between C8 and existing LM324) supplying power to the existing LM324 in order to drop the voltage for my ADC and MCU? (Since I am not really familiar with transformer.).

What I mean here is that your regulator may have to dissipate lots of power if it is a linear regulator, but it depends on the display current draw. If the voltage is 30 V there you are also quite close to the 7805 maximum of 35 V. What line (utility) voltage was that measured at? The worst case for input voltage will be when the line voltage is at its upper limit and when the power supply output is unloaded. Depending on tolerances of your line voltage and what the conditions were when you measured 30 V you might be in trouble.

I remember you wrote you wanted LED displays earlier,  and those can need quite a bit of current, so power dissipation in the regulator might be substantial.

Just a resistor as a dropper on the regulator input may not be very helpful when it comes to reducing input voltage to the regulator as the current may vary a lot, but can if properly calculated take over the burden of much of the power dissipation.