Better Way To Handle Reverse Current Flow

HighTechLab

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Recently I've acquired 2x HP/Agilent 6671A System Power Supplies. Their output is 220a @ 8v each. Let's come back to this...

Currently, I am successfully using HP 6031A 20v120a 1kw bench power supply, tasks including top-balancing and capacity testing, then recharging. It works well because the power supply can also be used to charge the batteries when they are setup in a 12v configuration as well as 3.65v thanks to 20v max limit of power supply.

The user manuals for the power supplies specifically state the following:

Quote
Battery Charging
The power supply's OVP circuit has a downprogrammer FET that discharges the power supply output whenever OVP trips. If a battery (or other external voltage source) is connected across the output and the OVP is inadvertently triggered or the output is programmed below the battery voltage, the power supply will sink current from the battery. To avoid this, insert a reverse blocking diode in series with the  output of the supply. Connect the diode cathode to the + battery terminal and the diode anode to the supply + output terminal. The diode will require a heat sink.


To accomplish this with my current setup, I have a 200a stud-type diode screwed into a heatsink with a small fan. Heat dissipation is minimal, because I'm dealing with only 120a & .7v forward voltage, so around 80 watts of heat, no big deal. I then provide voltage sense leads after the diode, connected directly to the batteries being charged and things work great.


Now that I have the two new power supplies, I can theoretically connect them in parallel, and charge 3.6v @ 440a, and due to the 8v limit if I wanted to charge at 12v, I would need to connect them in series (16v max limit). What an improvement to the speed that I can work, right?

Trouble is, the new power supplies need diodes also. If we calculate 220a of current @ .7v forward voltage of a diode per supply, I now have ~300w of heat to do something with. Where I previously had a tiny computer fan and a heat sink from a power inverter that was in the scrap bin, this new setup will require much more heat dissipation if I'm going to take the same approach of using a diode. Not that power usage of the system is a huge concern, I'm not the biggest fan of having a new 300w heater in the lab.

So I guess the question is, how can I build a more ideal diode here? I've heard that you can use PFETs (or perhaps NFETs) to provide a better solution here, but I am clueless...My area of expertise is batteries not FETs. Any guidance in the right direction is very much appreciated!
 

Zil

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Your power supply uses 120V ac to make +/- Vdc. If the battery voltage is above the set vdc output from the power supply, yes, you do need to block the reverse current.
 

timm

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I guess it's a different use case then, I use a 30v 60a version of the same power supply for charging and I don't enable OVP. The battery starts off lower (since I'm charging it) and stops at the set voltage.
 

HighTechLab

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Battery Charging
The power supply's OVP circuit has a downprogrammer FET that discharges the power supply output whenever OVP trips. If a battery (or other external voltage source) is connected across the output and the OVP is inadvertently triggered or the output is programmed below the battery voltage, the power supply will sink current from the battery. To avoid this, insert a reverse blocking diode in series with the  output of the supply. Connect the diode cathode to the + battery terminal and the diode anode to the supply + output terminal. The diode will require a heat sink.

Perhaps I'm missing something, but the manual is saying also if the output is programmed below battery voltage, this will be an issue
 

upnorthandpersonal

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You should be able to find something like this off-the-shelf:



Search for high current ideal diode, 300A ideal diode, etc.
 

Cal

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The simplest solution is to install 4 diodes in parallel. They should share current reasonable well as they have a negative temperature coefficient. Use an infrared thermometer to verify diodes operate at similar temperatures.
 

HighTechLab

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Ideal diodes sound like a perfect solution! Unfortunately, the ones on Ebay are shipped from China, and will take way too long for delivery.

By the way @timm I did a test on the power supplies - connected two in parallel, set one supply at 1v, 1a, then the second supply set at 2v and slowly dialed up the current. The first supply clamped the rail voltage down to 1v, and once supply #2 indicated 10a reverse current, I determined the test to be conclusive.

I can't risk if two supplies are in parallel, if one of them is set ever so slightly lower, that it will try to sink hundreds of amps of current. Sounds like a quick way to ruin the supply.

@smoothJoey I will be looking into the SSR-based solution, as this sounds like a good alternative. I did find a unidirectional SSR good for only 100a, but the voltage drop across it was 1v, which means even more heat than the .7v drop across the diode. Will keep digging and report back.
 

HighTechLab

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Here's a 200a SSR, unidirectional, seems like it would do the job great, only 85mv drop...but oh boy are they proud of their price.


 

Cal

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Ideal diodes sound like a perfect solution! Unfortunately, the ones on Ebay are shipped from China, and will take way too long for delivery.

By the way @timm I did a test on the power supplies - connected two in parallel, set one supply at 1v, 1a, then the second supply set at 2v and slowly dialed up the current. The first supply clamped the rail voltage down to 1v, and once supply #2 indicated 10a reverse current, I determined the test to be conclusive.

I can't risk if two supplies are in parallel, if one of them is set ever so slightly lower, that it will try to sink hundreds of amps of current. Sounds like a quick way to ruin the supply.

@smoothJoey I will be looking into the SSR-based solution, as this sounds like a good alternative. I did find a unidirectional SSR good for only 100a, but the voltage drop across it was 1v, which means even more heat than the .7v drop across the diode. Will keep digging and report back.

Not sure if the 2 supplies in parallel comment is directed to me. I suggested multiple diodes in parallel. That's no different than a solid state relay. Many FETs are in parallel in a SSR.

Still, your comment is not accurate. You program both PS's with the same current limit. The supplies will operate in current mode, not voltage mode. Both supplies will source equal current.

You also need to be careful when purchasing a SSR. Many high current devices are spec'd at instantaneous current, not continuous. Or they are spec'd at 25C (which is impossible to maintain).
 

HighTechLab

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Not sure if the 2 supplies in parallel comment is directed to me. I suggested multiple diodes in parallel. That's no different than a solid state relay. Many FETs are in parallel in a SSR.

Still, your comment is not accurate. You program both PS's with the same current limit. The supplies will operate in current mode, not voltage mode. Both supplies will source equal current.

You also need to be careful when purchasing a SSR. Many high current devices are spec'd at instantaneous current, not continuous. Or they are spec'd at 25C (which is impossible to maintain).
Sorry for any confusion, none of my comment was directed at you. As I indicated in the OP, I am trying to avoid using diodes. I don't like using diodes because they all have ~.7v drop, and no matter how many in parallel they will have the same amount of total heat dissipation. It's very different than a solid-state relay because the solid-state relay has only a .085v drop, while diode is .7v drop...which is an order of magnitude less heat dissipation when using the SSR.

The supplies need to be ran in CV mode when the batteries being charged reach 3.65 volts. Let's say one supply is set at 3.63v and the other at 3.65, due to real world tolerances (and the fact that the supplies could be slightly out of calibration), then as soon as the voltage exceeds any setpoint, the supply turns into a load. This is very undesirable.

I linked the spec sheet to an SSR I was considering, except for the price. I did make mental note the operating parameters of the SSR. It's not a good solution because the SSRs would cost more than I paid for the power supplies :lol: but it's certainly a step in the right direction toward a better solution.
 

sunshine

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Sorry for any confusion, none of my comment was directed at you. As I indicated in the OP, I am trying to avoid using diodes. I don't like using diodes because they all have ~.7v drop, and no matter how many in parallel they will have the same amount of total heat dissipation.

Use the diodes-

The simplest solution is to install 4 diodes in parallel. They should share current reasonable well as they have a negative temperature coefficient.

I use 12 x 25a diodes for a 1500w12v supply because they were the only ones I had at the time. The point you made about about temp coeff is interesting and reassuring.
I linked the spec sheet to an SSR I was considering, except for the price.

Use diodes until your order from China arrives?

Schottky Diodes in // seem to work. Not sure about ideal diodes in // ....the price does go up exponentially with the size????
 

Roswell Bob

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If you set OV trip greater than battery voltage you should not have problem. No?
The simplest solution is to install 4 diodes in parallel. They should share current reasonable well as they have a negative temperature coefficient. Use an infrared thermometer to verify diodes operate at similar temperatures.
A negative tempco will tend for the diodes not to share. Mosfets have a positive tempco that aids in sharing. Does this make any sense?
 

Cal

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A negative tempco will tend for the diodes not to share. Mosfets have a positive tempco that aids in sharing. Does this make any sense?
Yes, my bad. Paralleling diodes should still work. Just need a slight load resistance in series with each diode. Some length of cable would probably work. The diode with the smallest voltage drop will conduct more current. Cable resistance provides negative feedback. Interesting problem. If diode conducts 100A then perhaps the series resistor should drop 200 mV? That means cable resistance should be greater than 0.2 mOhm.

Another possibility is to delete the crowbar circuit. There should be service manuals/schematics around for the PS.
 

Bud Martin

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Crowbar circuit is for protection, I.E. Crowbar Diode connected across the output of the power supply incase the battery is connected to the power supply output in reverse polarity, the power supply will work without Crowbar diode but it will not have protection that is all.
 
Last edited:

Cal

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From manual:

A6 Output Filter Board​

1.
Remove the A5 Control Board as described in that procedure.
2.
On the Output Filter Board, remove the nut securing the lead from choke L900.
3.
Remove the bolt that secures the Output Filter Board + OUT out bus bar to the heat sink.
4.
Remove the screws that secure the Output Filter Board downprogrammer MOSFETs (Q901/Q902) to the heat sink.

I believe Q901/Q902 are the fets that discharge the battery when over voltage trips. You could try disabling them.

What happens when overvoltage trips and then the PS is switched off? Do the fets still discharge the battery? If not you can insert a relay in series with ac power. Relay opens when overvoltage trips. You can get the signal from Q901 or from the main board.
 

HighTechLab

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From manual:

A6 Output Filter Board​

1.
Remove the A5 Control Board as described in that procedure.
2.
On the Output Filter Board, remove the nut securing the lead from choke L900.
3.
Remove the bolt that secures the Output Filter Board + OUT out bus bar to the heat sink.
4.
Remove the screws that secure the Output Filter Board downprogrammer MOSFETs (Q901/Q902) to the heat sink.

I believe Q901/Q902 are the fets that discharge the battery when over voltage trips. You could try disabling them.

What happens when overvoltage trips and then the PS is switched off? Do the fets still discharge the battery? If not you can insert a relay in series with ac power. Relay opens when overvoltage trips. You can get the signal from Q901 or from the main board.
This is helpful, do you have the document link and page #?
 
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