0
\$\begingroup\$

I was trying out some different designs of constant current loads a.k.a dummy loads and stumbled upon this one on google images search. I do have a question or two to it's operation since on breadboard it's not working at all. I have replaced the parts in the diagram below with what I have in my parts' storage.

enter image description here

My questions:

  1. What is the use of the transistor Q1? I understand that it's an emitter follower but wouldn't I achieve the same using a voltage follower from a spare op-amp?
  2. Are R4,R8 and C1 crucial for the readout or can I just omit them out
  3. Could I use this for a DUT using AC?
  4. I built the circuit as is on breadboard but the current through R15 is not going over 100mA! Even when the reference voltage set at U1 is 5V which should give me anything close to 5A. The MOS-FET I'm using is a logic level FET according to this IRL3705N datasheet. I can confirm that the FET is original since it's from a reputable source and the op-amps are working fine in other constellations on the same breadboard. Maybe I'm overseeing something or the circuit is just baloney.

Thanks for your comments in advance

\$\endgroup\$
  • \$\begingroup\$ What is the supply voltage to the opamps? \$\endgroup\$ – Dejvid_no1 Jun 8 '17 at 16:37
  • 1
    \$\begingroup\$ Are you sure R15 is 1 ohm... looks like it is supposed to me 0.01Ohms to me. \$\endgroup\$ – Trevor_G Jun 8 '17 at 16:38
  • \$\begingroup\$ And what are the rail voltages and DUT? \$\endgroup\$ – Trevor_G Jun 8 '17 at 16:38
  • \$\begingroup\$ R4, R8, C1 feedback the output voltage from U2 and do not directly affect Vout. They cannot be omitted \$\endgroup\$ – Chu Jun 8 '17 at 16:40
  • \$\begingroup\$ @Trevor the original circuit had a 0.5mΩ shunt placed there. I don't have resistors that low and I can't measure that low either. My parts bin has a couple of 1Ω 5W, 1Ω 100W (both ±5%) so I thought the absolute value doesn't matter if I'm going with 1V = 1Ω x 1A The rail voltages are 0..15V and the DUT is at 4.2V (Li-Ion battery) \$\endgroup\$ – the_architecht Jun 8 '17 at 18:34
2
\$\begingroup\$

1.What is the use of the transistor Q1? I understand that it's an emitter follower but wouldn't I achieve the same using a voltage follower from a spare op-amp?

The circuitry around Q1 set the initial maximum current reference voltage. They presumably used a dual op-amp so had no other device to use and a transistor was cheaper than adding another one.

2.Are R4,R8 and C1 crucial for the readout or can I just omit them out

U1 provides a differential amplifier subtracting the current feedback from the required value, so yes R4,R8 and C1 are required.

3.Could I use this for a DUT using AC?

No it will not work on AC.

  1. I built the circuit as is on breadboard but the current through R15 is not going over 100mA! Even when the reference voltage set at U1 is 5V which should give me anything close to 5A. The MOS-FET I'm using is a logic level FET according to this IRL3705N datasheet. I can confirm that the FET is original since it's from a reputable source and the op-amps are working fine in other constellations on the same breadboard. Maybe I'm overseeing something or the circuit is just baloney.

Are you sure R15 is 1 ohm... looks like it is supposed to me 0.01Ohms to me

as for baloney... the circuit could do with a little work.

Here is a simpler block diagram of your circuit.

enter image description here

\$\endgroup\$
1
\$\begingroup\$

Let's briefly analyze how this works:

We are trying to put transistor Q2 into the linear region, effectively behaving as a variable resistor such that there's always a given current passing through it.

To do this we need: 1. To measure the current. 2. To control the current, i.e. the transistor.

To achieve (1) we use R15. Since it is 1 ohm, for every amp it will have 1V. R11/C4 provide lowpass filtering to smooth the signal.

U2 amplifies the signal since it is a non-inverting op amp.The output gain will be (1+(R12+R13)/R14). This is a value between 100 and 150. This is quite ridiculous to be honest. Doesn't look good to be honest.

But to answer your questions:

(3). No. Don't use it on AC. As you can see due to constructions reasons the transistor has a diode in anti-parallel. If you reverse the voltage (as happens in AC) you will just have a diode conducting which will effectively short your power supply.

(2). R4 and R8 are crucial, mostly in their relative values since they determine the gain of the noninverting op-amp. Biggest difference is that in this case is not referenced to ground as in U2 (with R13 grounded) but referenced to the output of the opamp. C1 is used so as to make a lowpass filter with R4 and R8. Here relative values are more important than absolute values.

(4). TBH I don't like the circuit. I would use something much simpler. Check this out: https://www.youtube.com/watch?v=8xX2SVcItOA

(1). The emmitter follower, or common collector acts as a buffer since its gain is roughly 1. I don't think you need it at all. I'd refer to answer 4 for a simpler circuit. If that doesn't satisfy your needs then move to something more sophisticated.

Cheers

\$\endgroup\$
  • \$\begingroup\$ Thanks, I had already had a look at that video and Dave's version works fine, with some errors on the breadboard due to the resistor tolerances but its okay. I was just trying some other constellations and this one sparked my interest because of the Transistor used as a buffer. \$\endgroup\$ – the_architecht Jun 8 '17 at 18:40
1
\$\begingroup\$

There are several errors in this circuit, some are pretty gross. First, the single-supply voltage requires the op-amps to work voltage levels a little bit higher than zeroV, not most op-amps can do it, 358 and 324 will not!

Then, R15 should produce something higher than 50mV per Ampere, or the op-amp will not answer by working pretty close to rail. Suppose R15 is 1Ω, then 1A will deliver 1V on the non-inverting input of U2, its output will raise until the inverting input would be also 1V. For that to happen, the resistors R12+R13 and R14 form a voltage divider, what makes U2 to amplify by the factor 1+((R12+R13)/R14).

Lets see, suppose R13 is set to zero, so U2 gain will be 1+(100k/1k), or, 101. Then, to have 1V on the inverting input, U2 output should be 101V, what is ridiculous, will never happens for two reasons, first, VCC is only 15V, and second, U2 would never support such high voltage. So, that statement 1V/A at the output is completely wrong. For that statement to be true, R15 should be 0.01Ω, not 1Ω.

But then, you go into another problem. Most old op-amps can not work very well with voltages close to a rail, in this case, single supply, it limit low voltages on the input and output. With 0.01Ω at R15, it means 10mV at U2 non-inverting input, it will not work well, some op-amps have more than that just as offset error between the inputs.

I would recommend to use 0.1Ω, 0.5Ω or even 1Ω at R15 for those reasons. If using 0.5Ω, the U2 gain could be set to 2 in order to have 1V/A at the output. So, R11 and R14 could be 10k, R12=8k2 and R13=5k (4k7) potentiometer, in order to have a fine adjustment of U2 gain, with R13 centered for U2 gain of 2, having 1V/A at output.

Then comes the circuit of Q1. It seems the designer had a full bag of 50kΩ potentiometers, since he used them all over without any calculation. Pot R1 is totally unnecessary, since you can control the voltage at the emitter of Q1 by R5 adjustment. Also, serving as a emitter follower bias base resistor, R1 should have a lower value, a maximum of 10k will be in order. R6 makes no sense in series with the base, it is a protection for the base if R5 cursor touches the 15V on the supply, so, R6 should be between R5 and VCC.

Also, a 12V (15V) zener diode must be inserted between gate and source of the mosfet, when you power on the circuit, if DUT is higher than 20V, a pulse of current through the internal capacitor Drain to Gate (CDG) will deliver a voltage on gate that could be higher than the gate could support, damaging the mosfet.

Also, the left side of R4 could be connected directly to the top of R15 (mosfet source), making both things independent; the display of V/A of U2 and the feedback control of the mosfet via U1, so, in case of any problem (short circuit or low impedance) at the monitor, will not interfere with the current regulation. I am pretty sure that this circuit as shown was never tested, it will not work.

\$\endgroup\$
0
\$\begingroup\$

If you're powering the opamps from ground and the +15v supply make sure you use one that includes ground as an acceptable input common mode range.

The LM358 shown on the schematic is OK but the RC4558 is NOT, its input common mode range will prevent it working. Other types such as the old 741 will also not work.

\$\endgroup\$
  • \$\begingroup\$ Got it. I guess I'll try it with ±15V from my PSU and report back \$\endgroup\$ – the_architecht Jun 8 '17 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.