# Using forward voltage drop of diodes with linear regulator

I want to clamp the output of a adjustable three-terminal linear regulator Vout to -1.8V relative to a reference voltage Vref.

My approach is to use the forward voltage drop of three diodes but I am having trouble sizing the resistor to ground R2.

Clearly, it needs to be large enough to limit current through D1...3.

What other boundary conditions are there? I imaginge it must not be too large either to avoid pinching off the regulator's ground connection or create an voltage rise due to Iadj?

Or what are the relevant boundary conditions for sizing?

The LM338 has:

• Vref = Vout-Vref = 1.25V

simulate this circuit – Schematic created using CircuitLab

In the alternative I thought of something along the lines of:

simulate this circuit

• Can you add to the schematic how the load will be connected? Also, do you want the output to be exactly Vin - 1.8, or do you want it to be set by R1 and R2, but no more than Vin - 1.8? – The Photon May 18 '13 at 16:37
• @ThePhoton Added the load to the schematic. I want the output to be exactly Vref-1.8V. – ARF May 18 '13 at 16:41
• Remember that the regulator output is 1.2 V above the ADJ pin. So if you put 1.8 V between Vref and ADJ, you'll get an output at Vref - 0.6. – The Photon May 18 '13 at 17:13
• Is the temperature coefficient of the diodes going to be stable enough? (2mV per C i think is common...) – Spoon May 18 '13 at 18:17

Using an op-amp is probably the best way to deal with this requirement but you might be better with an emitter follower circuit: -

The op-amp will drive the NPN so that its emitter (Voltage across the load) is the same voltage as the +Vin pin on the op-amp.

I borrowed "The Photon's" drawing because although it was wrong, it was easily amended to show an emitter follower. If you did use a FET (N channel with source to load) this would work but you might find that Vin needs to be higher for it to work. BJT base drive only needs to be 0.6V above the emitter - I'd still recommend using an op-amp with rail-to-rail output capability so that Vin can be a bit lower.

I'm also assuming the op-amp is powered from Vin and GND.

• One thing with this circuit is if you really need 5 A output, like OP's choice of LM338 implies he might, the base current of the BJT could be more than most op-amps can supply. Min beta of 40 or lower is pretty common for parts able to supply that much current. Also, the op-amp output will likely have to swing very near the top rail to make this work, which puts another constraint on op-amp selection. – The Photon May 18 '13 at 19:00
• Yeah, the 5A would be a bind but, ignoring that while I think about using a darlington, I did recommend using a rail-to-rail opamp... OK if a darlington is used, the output voltage can be no closer to the power rail (Vin) than about 1.4V... Given that the OP is using a 15V reference, I would suggest that the "top" power rail is at least this value so I think it would be OK. – Andy aka May 18 '13 at 19:26
• Thanks for your answer and the link to electronics.stackexchange.com/q/69506/8534 (in comments of deleted answer). To clarify some questions on your solution I have opened the following question. (Too much to ask to be dealt with in a comment). electronics.stackexchange.com/q/69798/8534 – ARF May 19 '13 at 9:08
• To get a little closer to the power rail, you can use a Sziklai pair output stage rather than a Darlington pair, thereby avoiding the doubled-up VBE's. – Kaz Jun 7 '13 at 19:33
• @Kaz WTF - never heard of one of them - is it a darlington but the collector of the fron transistor is up to Vcc? This is a guess!! – Andy aka Jun 7 '13 at 21:18

I'm not sure I understand the reason for using the regulator in your proposal. Why not something simpler:

simulate this circuit – Schematic created using CircuitLab

(Edit: Note, diagram fixed in response to AndyAka's comment)

Edit: Again, I fixed the diagram in response to AndyAka's comment about stability. Adequate bypassing of the load at C1 is needed to make the circuit stable.

Also, while doing simulations to prove to myself that bypassing is the correct means of compensating the control loop, I realized that if the load might draw very low current, the op-amp will need to drive very close to VIN to adequately cut off M1. If VIN and the op-amp positive supply are the same, a rail-to-rail type is needed.

• That might be positive feedback but not of the sort that is desirable LOL – Andy aka May 18 '13 at 18:15
• @Andyaka, thanks. Brainf*rt on my part --- I was thinking, "I'm pretty sure I should be connecting to the +IN terminal", but when I mentally simulated the circuit my brain was taking a holiday. – The Photon May 18 '13 at 18:56
• This will be really unstable - you're adding a gain of tens or a hundred (ish) to within the feedback loop of an op-amp - it'll sing like a bird. If it were that easy, all op-amp manufacturers would put extra gain in and double or treble their GBW product. Look at this for almost definite proof: electronics.stackexchange.com/questions/69506/… - – Andy aka May 18 '13 at 19:28
• @Andyaka, you have a point. I will withdraw the answer until I have time to work out the proper compensation. – The Photon May 19 '13 at 6:41

You want to use the LM338 as a tracking regulator, to follow a control voltage ($V_c$) minus 1.8V. Maybe you like the built in current limit or thermal limit of the LM338. Or maybe you just have some LM338s available and thought why not put them to good use. Whatever the case, here you will see a simple way to implement a tracking function by adding an opamp and voltage reference to the circuit.

Most of these three terminal non-lowdrop regulators are very similar to each other. They regulate the voltage between the output ($V_o$) and the adjustment pin ($V_{\text{adj}}$) at 1.24V (or something pretty close). Typically a resistor ($R_{\text{set}}$) will be placed between $V_o$ and $V_{\text{adj}}$. $R_{\text{set}}$ defines a constant current ($I_{\text{set}}$) of $\frac{R_{\text{set}}}{\text{1.24V}}$ that will pass through a resistor ($R_{\text{adj}}$) between $V_{\text{adj}}$ and ground. The only purpose of $R_{\text{set}}$ and $I_{\text{set}}$ is to determine the voltage at $V_{\text{adj}}$. $V_o$ is then just 1.24V + $R_{\text{adj}}$ $I_{\text{set}}$, basically the regulated 1.24V is stacked on top of whatever happens to be at $V_{\text{adj}}$. If you supply $V_{\text{adj}}$ then there is no need for either $R_{\text{set}}$ or $R_{\text{adj}}$, and that's what we'll do here.

You will need an OpAmp and an adjustable reference set to 3.04V. It is 3.04V because you need to account for the 1.24V to get $V_c$ - 1.8V at $V_o$. You might use a reference like the LM4121 or you could also look at something like the LM611 which has an opamp and an adjustable reference.

Here is the OpAmp circuit:

You would insert the OpAmp into the LM338 circuit with $V_c$ hooked to what you are calling Vref, and $V_{\text{adj}}$ hooked the the Vadj pin of the LM338. And 3.04Vref would come from the reference voltage you will need to add. Resistors R1 and R2 in your circuit are not needed and would be omitted. And of course you don't need any of those diodes in your circuit either.

Be aware that tolerances add up quickly in differential amplifiers like this. For example if you want to meet 2% tolerance for the $V_{\text{adj}}$ voltage, you might need better than 1% resistors. I'll leave that as an exercise for you.