Linkswitch as current source

Posted by: bsala on 09/17/2008

Hello, I have built a circuit similar to that in DER172 for a battery charger. It has a 22 ohm current sense in the same position as your R4. It also has a 5.6 ohm switched across it by a FET to increase charge current. The aim was for the parallel combination to deliver 200 mA, and the 22 ohm to deliver 40 mA. Well it works beautifully (I am very impressed!) at the high setting, with about a volt across the parallel resistors and delivers 213 mA. Very nice and very stable over the full AC input range. But when I switch out the 5.6 ohm, it delivers about 100 mA. I've double checked that the 22 ohm is in fact correct. The opto has a 100 ohm as per your DER172. I've measured the voltage across the 22 ohm at about 900mV, so cannot explain how that relates to 100 mA! Ther is also a voltage clamp for when the charger is off, again following the idea in DER172. It has a 5.6 volt zener and 470 ohm in series. The output voltage when the load is removed rises to 10.3 volts. with about 8 mA flowing through the zener circuit. Can you explain how 900mV across 22 ohms results in a current of 100 mA please? Clearly the total current sense must be only 8 ohms, but I can't see how this happens. The FET which switches the 5.6 ohm resistor has its gate firmly at zero volts, no spikes as seen with a Tek TPS2024. I could try raising the 22 ohm to resolve this, but would like to know what's causing the excess current on the low charge setting. Can you assist please? Regards Brian

Comments

With 22 ohms ideally you should get about 50 mA.
Try and use 200 or 220 ohms in place of R3.

OK, I already tried increasing the 22 ohm to 56 ohm (at R4), and it dropped the current to about 82mA (with the 100 ohm as R3). After your suggestion to change R3, I made it 235 ohm (still with 56 ohms at R4) and the current is still 82mA, rising to 200mA when the 5.6 ohm is switched across the 56 ohm. If I restore R4 to 22 ohm with R3 at 235 ohm, current is 110mA, and when the 5.6 ohm is paralleled, rises to 220mA (which is fine). I still cannot see the path for the 110mA, when the 22 ohm only has 1 volt across it. Can you explain the path of the additional current please? And is it possible to reduce the output to the desired 40mA? Regards Brian

Hello bsala

Have you tried directly connecting the two resistors (removing the FET)? If not please give that a try and would you mind posting the schematic you are using.

I can think of two possibilities. 1. the voltage ripple across C4 and actual switching frequency is causing the non linearity between expected and actual output current. 2. Large ripple across C4 is saturating the opto.

The ripple voltage (Ipk x ESR) across C4 will modulate the output current. Therefore the voltage across R4 will be a function of the DC output current plus the AC content at the swtiching frequency causes by the ESR of C4. As the output current reaches the regulated value and the opto conducts the feedback current is modulated by the voltage generated across the ESR of the output capacitor and this determines if the subsequent cycle is enabled or not. We can say the converter regulates on the ESR voltage ripple of the output capacitor.

Are you using a low ESR cap for C4? If not try a low ESR type and if yes then try a larger value (reduce ESR by 50% by connecting second cap in parallel) and see if that makes a difference and post the results.

Cheers

PI-Chekov

Hello PI-Chekov, I removed the 5R6 completely and current dropped to around 46mA with R4 then at 22 ohm and R3 at 235 ohm. The capacitor at C4 is a low ESR type. I used a current probe to observe any ripple on output, plus any ripple across C4. The attached GIF shows current in yellow (20mA/div), C4 voltage in magenta, output voltage in green. When the 5R6 was replaced, even with the gate of Q5 tied to VSS, current increased, so it appears that Q5 is being partially turned on. That certainly explains why 1 volt across R4 produces more current than expected. When Q5 gate is tied to VDD, current rises to the full 220mA. Now I need to find out why Q5 is biased partially on when its gate is tied to its source. Q5 and Q4 are driven by a microprocessor. At least I now know that Ohm's Law is being obeyed. Any ideas on what is switching Q5 on would be welcome, but I can investigate properly now. Thanks, Brian

Hello PI-Chekov, I just realised the problem, which is that Q5 is conducting through its body diode and partially shunting the 22 ohm. When Q5 is fully turned on, the body diode is shorted by the FET action. I hadn't noticed that the FET had a negative voltage on its drain. Now I feel quite inadequate! Thanks for your help, the Linkswitch is brilliant, and I will be sure to use more PI products. Regards, Brian